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NCERT Solutions for Class 12 Maths Chapter 4 - Determinants

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NCERT Solutions for Class 12 Maths Chapter 4 - Determinants

The fundamental concepts and applications of determinants are explained in NCERT Solutions for Class 12 Maths Chapter 4 Determinants. Determinants are an important part of maths studies because they have many applications. A determinant is a function that can be used to associate a real or complex number with any square matrix. This concept is used to determine whether or not a system of equations has a unique solution. Aside from that, determinants have a wide range of applications in engineering, science, economics, and social science. As a result, gaining a thorough understanding of this topic and its applications is critical. NCERT Solutions Class 12 Maths Chapter 4 teaches students the fundamentals of determinants.


Class:

NCERT Solutions for Class 12

Subject:

Class 12 Maths

Chapter Name:

Chapter 4 - Determinants

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2023-24

Medium:

English and Hindi

Available Materials:

  • Chapter Wise

  • Exercise Wise

Other Materials

  • Important Questions

  • Revision Notes



NCERT Solutions for Class Maths Chapter-4 PDF is available to download now. The PDF file can be downloaded from the official website of Vedantu. Determinants Class 12 NCERT PDF relating to all the solutions have been provided for determinants. All the answers related to Class 12 Maths Chapter 4 can be found here. The Determinants Class 12 NCERT Solutions PDF Download are created by the subject matter experts who have years of teaching experience. They have the proper expertise to explain the questions thoroughly and efficiently while providing adequate justifications and solutions to the problem. If a student has any sort of queries, they can always go to the website and drop their questions online. They can also download NCERT Solutions for Class 12 Maths Chapter 4 PDF file for better understanding. Also, the online helping staff will get in touch with the student if a detailed look is required into the matter. 


Overview of Deleted Syllabus for CBSE Class 12 Determinants 2023-24

Exercise 4.3

Properties of Determinants

Example Problem

Miscellaneous Examples 30, 31, 32 and 34

Miscellaneous Exercise

Ques. 2, 4, 5, 6, 11, 12, 13, 14, 15 and 17 

Summary Points

4, 5, 6, 7, 8, 9, 10, 11



Determinants Chapter at a Glance - Class 12 NCERT Solutions

  • Determinant of a matrix $A=\left[a_{11}\right]_{11}$ is given by $\left|a_{11}\right|=a_{11}$

  • Determinant of a matrix $A=\left[\begin{array}{ll}a_{11} & a_2 \\ a_{21} & a_{22}\end{array}\right]$ is given by $|A|=\left|\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{21}\end{array}\right|=a_{11} a_{22}-a_0 a_{21}$

  • Determinant of a matrix $A=\left[\begin{array}{ccc}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right]$ is given by (expanding along $k_1$ )

$|A|=\left|\begin{array}{lll} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array}\right|=a_1\left|\begin{array}{ll} b_2 & c_2 \\ c_3 & c_3 \end{array}\right|-b_3\left|\begin{array}{cc} a_2 & c_2 \\ a_3 & c_3 \end{array}\right|+c_1\left|\begin{array}{cc} a_2 & b_2 \\ a_3 & b_3 \end{array}\right|$

For any square matrix $A$, the $|A|$ satisfy following properties.

  • $\left|A^{\prime}\right|=|1|$, where $A^{\prime}=$ transpose of $A$.

  • If we interchange any two rows(or columns), then the sign of the determinant changes.

  • If any two rows or any two columns are identical or proportional, then value of determinant is zero

  • If we multiply each element of a row or column of determinant by constant, $k$, then value of determinant is multiplied by $k$

  • Multiplying a determinant by $k$ means multiplying elements of only one row (or one column) by $k$.

  • If $A=\left[a_i\right]$, then $|f-1|=k^3|A|$

  • If elements of a row or a column in a determinant can be expressed as sum or two or more elements, then the given determinant can be expressed as sum of two or more determinants.

  • If to each element of a row or a column of a determinant, the equimultiples of corresponding elements of other rows or columns are added, then the value of the determinant remains same.

  • Minor of an element $a_{i y}$ of the determinant of matrix $A$ is the determinant obtained by deleting $j^{\text {th }}$ row and $j^{\text {th }}$ column and denoted by $M_{i q}$ -

  • Soffector of $a_{i y}$ is given by $A_y=(-1)^{2+i} \mathrm{M}_4$ -

  • Value of the determinant of a matrix $\mathrm{A}$ is obtained by the sum of the product of elements of a row (or a column) with corresponding cofactors.

  • For example- $|1|=a_{11} A_{11}+a_{12} A_2+a_{13} A_{12}$

  • If elements of one row (or column) are multiplied with cofactors of elements of any other row (or column), then their sum is zero for example, $a_{11} A_{11}+a_1 A_2+a_{12} A_2=0$.

  • $A($ ad $A)=($ and $A) A=|A| f$, where $A$ is a square matrix of order $w$.

  • A square matrix $A$ is said to be singular or non-singular according as $|A|=0$ or $|A|=0$.

  • If $A B=B A=8$, where $B$ is square matrix, then $B$ is called the inverse of $A$. $A$ koo $A^{-1}=B$ or $B^{-1}=A$ and hence $\left(A^{-1}\right)^{-1}=A$

  • A square matrix $A$ has an inverse if and only if $A$ is non-singular.

  • $A^{-1}=\frac{1}{4}\left(\operatorname{arc}{ }^A\right)$

  • If $a_1 x+b_2 y+c_1 z=d_1, a_2 x+b_1 y+c_2 z=d_1, a_2 x+b_2 y+c_3 z=d_3$, Then these equations can be written as $A X=B$, whene $A=\left[\begin{array}{lll}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{l}d_1 \\ d_2 \\ d_3\end{array}\right]$

  • Unique solution of equation $A X=B$ is given by $X=A^{-1} B$, where $|A|=0$.

  • A system of equations is consistent or inconsistent according to whether its solution exists or not.

  • For a square matrix $A$ in the matrix equation $A X=B$.

(i) ||$=0$, there exists a unique solution.

(ii) $|1|=0$ and $($ ad $A) B-0$, then there exists no solution.

(ii) $|A|=0$ and $($ and $A) B=0$, then system may or may not be consistent


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Exercises under NCERT Class 12 Maths Chapter 4 Determinants

Exercise 4.1: This exercise introduces the concept of determinants and their notation. Students will learn how to find the determinants of 2x2 and 3x3 matrices using various methods, including the rule of Sarrus and the cofactor expansion method.

Exercise 4.2: In this exercise, students will learn about the properties of determinants and practice using these properties to solve problems. They will learn about the properties of determinants under row and column operations, the property of determinants of triangular matrices, and the property of determinants of transpose matrices. (Not in the current syllabus)

Exercise 4.2: This exercise focuses on the application of determinants in solving systems of linear equations. Students will learn about Cramer's rule, which allows them to solve systems of linear equations using determinants.

Exercise 4.3: In this exercise, students will learn about the inverse of a matrix and how to find it using determinants. They will also practice using determinants to find the area of a triangle and the volume of a parallelepiped.

Exercise 4.4: This exercise covers the concept of minors and cofactors of a matrix. Students will learn how to find the minors and cofactors of a matrix and use them to find the inverse of a matrix.

Exercise 4.5: In this exercise, students will learn about the adjoint of a matrix and its properties. They will also practice finding the adjoint of a matrix and using it to find the inverse of a matrix.

Miscellaneous Exercise: This exercise includes a mix of questions covering all the concepts taught in the chapter. Students will have to apply their knowledge of determinants to solve various problems and answer questions. They will also practice using determinants to find the inverse of a matrix, solve systems of linear equations, and find the area/volume of geometric figures.


Overall, this chapter is an important topic in linear algebra and covers the fundamental concepts of determinants, including determinant notation, determinant properties, the application of determinants in solving systems of linear equations, the inverse of a matrix using determinants, and the adjoint of a matrix.


NCERT Solutions for Class 12 Maths Chapter 4 - Determinants

Exercise (4.1)

1. Evaluate the determinants in Exercise $\mathbf{1}$ and $\mathbf{2}$. 

\[\mathbf{\left| \begin{matrix}   2 & 4  \\   -5 & -1  \\ \end{matrix} \right|}\]

Ans: Solving the determinant 

\[\left| \begin{matrix}  \text{2} & \text{4}  \\ \text{-5} & \text{-1}  \\\end{matrix} \right|\], 

we have:

\[\Rightarrow \left| \begin{matrix} \text{2} & \text{4}  \\ \text{-5} & \text{-1}  \\ \end{matrix} \right|\text{= -2+20}\]

\[\therefore \left| \begin{matrix} \text{2} & \text{4}  \\ \text{-5} & \text{-1}  \\ \end{matrix} \right|\text{=18}\]


2. Evaluate the determinants in Exercise $1$ and $2$.

  1. \[\mathbf{\left| \begin{matrix}\cos \theta  & -\sin \theta   \\\sin \theta  & \cos \theta   \\ \end{matrix} \right|}\].

Ans: Solving the determinant 

\[\left| \begin{matrix} \cos \theta  & -\sin \theta   \\\sin \theta  & \cos \theta   \\ \end{matrix} \right|\], 

we have:

\[\Rightarrow \left| \begin{matrix}\cos \theta  & -\sin \theta   \\\sin \theta  & \cos \theta   \\\end{matrix} \right|=\left( \cos \theta  \right)\left( \cos \theta  \right)-\left( -\sin \theta  \right)\left( \sin \theta  \right)\]

\[\Rightarrow \left| \begin{matrix}\cos \theta  & -\sin \theta   \\\sin \theta  & \cos \theta   \\\end{matrix} \right|=\text{ }{{\cos }^{2}}\theta +{{\sin }^{2}}\theta \]

We know, \[{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\]

\[\therefore \left| \begin{matrix}\cos \theta  & -\sin \theta   \\\sin \theta  & \cos \theta   \\\end{matrix} \right|=1\]

  1. \[\mathbf{\left| \begin{matrix}{{\text{x}}^{\text{2}}}\text{-x+1} & \text{x-1}  \\\text{x+1} & \text{x+1}  \\\end{matrix} \right|}\].

Ans: Solving the determinant 

\[\left| \begin{matrix}{{\text{x}}^{\text{2}}}\text{-x+1} & \text{x-1}  \\\text{x+1} &\text{x+1}  \\\end{matrix} \right|\], 

we have:

\[\Rightarrow \left| \begin{matrix}{{\text{x}}^{\text{2}}}\text{-x+1} & \text{x-1}  \\\text{x+1} & \text{x+1}  \\\end{matrix} \right|\text{=}\left( {{\text{x}}^{\text{2}}}\text{-x+1} \right)\left( \text{x+1} \right)\text{-}\left( \text{x-1} \right)\left( \text{x+1} \right)\]

\[\Rightarrow \left| \begin{matrix}{{\text{x}}^{\text{2}}}\text{-x+1} & \text{x-1}  \\\text{x+1} & \text{x+1}  \\\end{matrix} \right|\text{= }{{\text{x}}^{\text{3}}}\text{-}{{\text{x}}^{\text{2}}}\text{+x+}{{\text{x}}^{\text{2}}}\text{-x+1-}\left( {{\text{x}}^{\text{2}}}\text{-1} \right)\]

So, 

\[\left| \begin{matrix}{{\text{x}}^{\text{2}}}\text{-x+1} & \text{x-1}  \\\text{x+1} & \text{x+1}  \\\end{matrix} \right|\text{= }{{\text{x}}^{\text{3}}}\text{+1-}{{\text{x}}^{\text{2}}}\text{+1}\]

\[\therefore \left| \begin{matrix} {{\text{x}}^{\text{2}}}\text{-x+1} & \text{x-1}  \\ \text{x+1} & \text{x+1}  \\ \end{matrix} \right|\text{= }{{\text{x}}^{\text{3}}}\text{-}{{\text{x}}^{\text{2}}}\text{+2}\].


3. If \[\mathbf{\text{A=}\left[ \begin{matrix} \text{1} & \text{2}  \\ \text{4} & \text{2}  \\ \end{matrix} \right]}\]. , then show that \[\mathbf{\left| \text{2A} \right|\text{=4}\left| \text{A} \right|}\].

Ans: Given that, 

\[\text{A=}\left[ \begin{matrix} \text{1} & \text{2}  \\ \text{4} & \text{2}  \\ \end{matrix} \right]\]

Multiplying $\text{A}$ by $2$, we have:

\[\Rightarrow \text{2A= 2}\left[ \begin{matrix} \text{1} & \text{2}  \\ \text{4} & \text{2}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{2} & \text{4}  \\ \text{8} & \text{4}  \\ \end{matrix} \right]\]

\[\Rightarrow \text{2A=}\left[ \begin{matrix} \text{2} & \text{4}  \\ \text{8} & \text{4}  \\ \end{matrix} \right]\]

\[\therefore L.H.S \text{=}\left| \text{2A} \right|\text{=}\left| \begin{matrix}\text{2} & \text{4}  \\\text{8} & \text{4}  \\\end{matrix} \right|\]

\[\Rightarrow \left| \text{2A} \right|\text{=}2\times 4-4\times 8\]

\[\Rightarrow \left| \text{2A} \right|\text{=}8-32\]

\[\therefore \left| \text{2A} \right|\text{=}-24\]

The value of determinant $\text{A}$ is

\[\Rightarrow \left| \text{A} \right|\text{=}\left| \begin{matrix} \text{1} & \text{2}  \\ \text{4} & \text{2}  \\ \end{matrix} \right|\]

\[\Rightarrow \left| \text{A} \right|\text{=}2-8\]

\[\therefore \left| \text{A} \right|\text{=}-6\]

R.H.S is given as $\text{4}\left| \text{A} \right|$.

\[\therefore \text{4}\left| \text{A} \right|\text{=4 }\!\!\times\!\!\text{ }\left( \text{-6} \right)\text{=-24}\]

Hence, we have L.H.S $=$ R.H.S

\[\therefore \left| \text{2A} \right|\text{=4}\left| \text{A} \right|\].


4. If \[\mathbf{\text{A=}\left[ \begin{matrix} \text{1} & \text{0} & \text{1}  \\ \text{0} & \text{1} & \text{2}  \\ \text{0} & \text{0} & \text{4}  \\ \end{matrix} \right]}\]. , 

then show that $\mathbf{\left | 3A \right |=27\left | A \right |}$.

Ans: Given, 

\[\text{A=}\left[ \begin{matrix} \text{1} & \text{0} & \text{1}  \\ \text{0} & \text{1} & \text{2}  \\ \text{0} & \text{0} & \text{4}  \\ \end{matrix} \right]\]

Determining the value of determinant $\text{A}$, by expanding along the first column, i.e., \[C_1\], we get:

\[\Rightarrow \left| \text{A} \right|\text{=1}\left| \begin{matrix} \text{1} & \text{2}  \\ \text{0} & \text{4}  \\ \end{matrix} \right|\text{-0}\left| \begin{matrix} \text{0} & \text{1}  \\ \text{0} & \text{4}  \\ \end{matrix} \right|\text{+0}\left| \begin{matrix} \text{0} & \text{1}  \\ \text{1} & \text{2}  \\ \end{matrix} \right|\]

\[\Rightarrow \left| \text{A} \right|\text{=}1\left( 4-0 \right)-0+0\]

\[\therefore \left| \text{A} \right|\text{=4}\]

Hence, $27\left| A \right|=27\times 4$

$\Rightarrow 27\left| A \right|=108$   ……(1)

The value of $\left| 3\text{A} \right|$ is obtained as:

\[\Rightarrow \text{3A=3}\left[ \begin{matrix} \text{1} & \text{0} & \text{1}  \\ \text{0} & \text{1} & \text{2}  \\ \text{0} & \text{0} & \text{4}  \\ \end{matrix} \right]\] 

\[\Rightarrow \text{3A=}\left[ \begin{matrix} \text{3} & \text{0} & \text{3}  \\ \text{0} & \text{3} & \text{6}  \\ \text{0} & \text{0} & \text{12}  \\ \end{matrix} \right] \]

\[\therefore \left| \text{3A} \right|\text{=3}\left| \begin{matrix} \text{3} & \text{6}  \\ \text{0} & \text{12}  \\ \end{matrix} \right|\text{-0}\left| \begin{matrix} \text{0} & \text{3}  \\ \text{0} & \text{12}  \\ \end{matrix} \right|\text{+0}\left| \begin{matrix} \text{0} & \text{3}  \\ \text{3} & \text{6}  \\ \end{matrix} \right|\]

\[\Rightarrow \text{3}\left( \text{36-0} \right)\text{+0+0}\]

\[\Rightarrow \left| \text{3A} \right|\text{=3}\times \text{36}\]

Thus, \[\left| \text{3A} \right|\text{=}108\]   ……(2)  

From equations (1) and (2), we have:

\[\left| \text{3A} \right|\text{=27}\left| \text{A} \right|\]

Hence proved.

5. Evaluate the determinants

  1. \[\mathbf{\left| \begin{matrix} \text{3} & \text{-1} & \text{-2}  \\ \text{0} & \text{0} & \text{-1}  \\ \text{3} & \text{-5} & \text{0}  \\ \end{matrix} \right|}\]

Ans: Let \[\text{A=}\left| \begin{matrix} \text{3} & \text{-1} & \text{-2}  \\ \text{0} & \text{0} & \text{-1}  \\ \text{3} & \text{-5} & \text{0}  \\ \end{matrix} \right|\]

Determining the value of $\text{A}$ by expanding along the third row, we have:

\[\Rightarrow \left| \text{A} \right|\text{=3}\left| \begin{matrix}  -1 & -2  \\  0 & -1  \\ \end{matrix} \right|-\left( -5 \right)\left| \begin{matrix}  3 & -2  \\ 0 & -1  \\ \end{matrix} \right|+0\left| \begin{matrix} 3 & -1  \\ 0 & 0  \\ \end{matrix} \right|\]

\[\Rightarrow \left| \text{A} \right|\text{=}\left( \text{3-15} \right)\]

\[\therefore \left| \text{A} \right|\text{=-12}\]

  1. \[\mathbf{\left| \begin{matrix} \text{3} & \text{-4} & \text{5}  \\ \text{1} & \text{1} & \text{-2}  \\ \text{2} & \text{3} & \text{1}  \\ \end{matrix} \right|}\]

Ans: Let \[\text{A=}\left| \begin{matrix} \text{3} & \text{-4} & \text{5}  \\ \text{1} & \text{1} & \text{-2}  \\ \text{2} & \text{3} & \text{1}  \\ \end{matrix} \right|\]

Determining the value of $\text{A}$ by expanding along the first row, we have:

\[\Rightarrow \left| \text{A} \right|\text{=3}\left| \begin{matrix} \text{1} & \text{-2}  \\ \text{3} & \text{1}  \\ \end{matrix} \right|\text{+4}\left| \begin{matrix} \text{1} & \text{-2}  \\ \text{2} & \text{1}  \\ \end{matrix} \right|\text{+5}\left| \begin{matrix} \text{1} & \text{1}  \\ \text{2} & \text{3}  \\ \end{matrix} \right|\]

\[\Rightarrow \left| \text{A} \right|\text{=3}\left( \text{1+6} \right)\text{+4}\left( \text{1+4} \right)\text{+5}\left( \text{3-2} \right)\]

\[\Rightarrow \left| \text{A} \right|\text{=21+20+5}\]

\[\therefore \left| \text{A} \right|\text{=}46\]

  1. \[\mathbf{\left| \begin{matrix} \text{0} & \text{1} & \text{2}  \\ \text{-1} & \text{0} & \text{-3}  \\ \text{-2} & \text{3} & \text{0}  \\ \end{matrix} \right|}\].

Ans: Let \[\text{A=}\left| \begin{matrix} \text{0} & \text{1} & \text{2}  \\ \text{-1} & \text{0} & \text{-3}  \\ \text{-2} & \text{3} & \text{0}  \\ \end{matrix} \right|\]

Determining the value of $\text{A}$ by expanding along the first row, we have:

\[\Rightarrow \left| \text{A} \right|\text{=0}\left| \begin{matrix} \text{0} & \text{-3}  \\ \text{3} & \text{0}  \\ \end{matrix} \right|\text{-1}\left| \begin{matrix} \text{-1} & \text{-3}  \\ \text{-2} & \text{0}  \\ \end{matrix} \right|\text{+2}\left| \begin{matrix} \text{-1} & \text{0}  \\ \text{-2} & \text{3}  \\ \end{matrix} \right|\]

\[\Rightarrow \left| \text{A} \right|\text{=0}\left( 9 \right)-\left( -6 \right)+2\left( -3 \right)\]

\[\therefore \left| \text{A} \right|\text{=0}\]

  1. \[\mathbf{\left| \begin{matrix} \text{2} & \text{-1} & \text{-2}  \\ \text{0} & \text{2} & \text{-1}  \\ \text{3} & \text{-5} & \text{0}  \\ \end{matrix} \right|}\]

Ans: Let $\text{A=}\left| \begin{matrix} \text{2} & \text{-1} & \text{-2}  \\ \text{0} & \text{2} & \text{-1}  \\ \text{3} & \text{-5} & \text{0}  \\ \end{matrix} \right|$

Determining the value of $\text{A}$ by expanding along the first column, we have:

\[\Rightarrow \left| \text{A} \right|\text{=2}\left| \begin{matrix} \text{2} & \text{-1}  \\ \text{-5} & \text{0}  \\ \end{matrix} \right|\text{-0}\left| \begin{matrix} \text{-1} & \text{-2}  \\ \text{-5} & \text{0}  \\ \end{matrix} \right|\text{+3}\left| \begin{matrix} \text{-1} & \text{-2}  \\ \text{2} & \text{-1}  \\ \end{matrix} \right|\]

\[\Rightarrow \left| \text{A} \right|\text{=}2\left( -5 \right)-0+3\left( 5 \right)\]

\[\Rightarrow \left| \text{A} \right|\text{=}-\text{10+15}\]

\[\therefore \left| \text{A} \right|\text{=5}\]


6.  If \[\mathbf{\text{A=}\left[ \begin{matrix} \text{1} & \text{1} & \text{-2}  \\ \text{2} & \text{1} & \text{-3}  \\ \text{5} & \text{4} & \text{-9}  \\ \end{matrix} \right]}\] , 

find \[\mathbf{\left| \text{A} \right|}\].

Ans: Given, \[\text{A=}\left[ \begin{matrix} \text{1} & \text{1} & \text{-2}  \\ \text{2} & \text{1} & \text{-3}  \\ \text{5} & \text{4} & \text{-9}  \\ \end{matrix} \right]\]

Determining the value of $\text{A}$ by expanding along the first row, we have:

\[\Rightarrow \left| \text{A} \right|\text{=1}\left| \begin{matrix} \text{1} & \text{-3}  \\ \text{4} & \text{-9}  \\ \end{matrix} \right|\text{-1}\left| \begin{matrix} \text{2} & \text{-3}  \\ \text{5} & \text{-9}  \\ \end{matrix} \right|\text{-2}\left| \begin{matrix} \text{2} & \text{1}  \\ \text{5} & \text{4}  \\ \end{matrix} \right|\]

\[\Rightarrow \left| \text{A} \right|\text{=}1\left( -9+12 \right)-1\left( -18+15 \right)-2\left( 8-5 \right)\]

\[\Rightarrow \left| \text{A} \right|\text{=}3+3-6\]

\[\therefore \left| \text{A} \right|\text{=}0\]


7. Find values of \[x\] , if

  1. \[\mathbf{\left| \begin{matrix} \text{2} & \text{4}  \\ \text{5} & \text{1}  \\ \end{matrix} \right|\text{=}\left| \begin{matrix} \text{2x} & \text{4}  \\ \text{6} & \text{x}  \\ \end{matrix} \right|}\]

Ans: Given, 

\[\left| \begin{matrix} \text{2} & \text{4}  \\ \text{5} & \text{1}  \\ \end{matrix} \right|\text{=}\left| \begin{matrix} \text{2x} & \text{4}  \\ \text{6} & \text{x}  \\ \end{matrix} \right|\]

 Solving it, we have:

\[\Rightarrow \left( 2\times 1 \right)-\left( 5\times 4 \right)=\left( 2x\times x \right)-\left( 6\times 4 \right)\]

\[\Rightarrow 2-20=2{{x}^{2}}-24\]

\[\Rightarrow -18+24=2{{x}^{2}}\]

\[\Rightarrow 3={{x}^{2}}\]

Applying square root on both the sides, we obtain:

\[\Rightarrow \text{x =  }\!\!\pm\!\!\text{ }\sqrt{\text{3}}\]

  1. \[\mathbf{\left| \begin{matrix} \text{2} & \text{3}  \\ \text{4} & \text{5}  \\ \end{matrix} \right|\text{=}\left| \begin{matrix} \text{x} & \text{3}  \\ \text{2x} & \text{5}  \\ \end{matrix} \right|}\]

Ans: Given, 

\[\left| \begin{matrix} \text{2} & \text{3}  \\ \text{4} & \text{5}  \\n\end{matrix} \right|\text{=}\left| \begin{matrix} \text{x} & \text{3}  \\ \text{2x} & \text{5}  \\ \end{matrix} \right|\]

Solving it, we have:

\[\Rightarrow \left( \text{2 }\!\!\times\!\!\text{ 5} \right)-\left( \text{3 }\!\!\times\!\!\text{ 4} \right)\text{=}\left( \text{x }\!\!\times\!\!\text{ 5} \right)-\left( \text{3 }\!\!\times\!\!\text{ 2x} \right)\]

\[\Rightarrow \text{10}-\text{12=5x}-\text{6x}\]9

\[\Rightarrow -\text{2=}-\text{x}\]

Multiplying by $\left( -1 \right)$ on both the sides, we obtain:

\[\Rightarrow \text{x = 2}\]


8. If \[\mathbf{\left| \begin{matrix} \text{x} & \text{2}  \\ \text{18} & \text{x}  \\ \end{matrix} \right|\text{=}\left| \begin{matrix} \text{6} & \text{2}  \\ \text{18} & \text{6}  \\\end{matrix} \right|}\] , 

then \[\mathbf{x}\] is equal to

  1. \[\mathbf{\text{6}}\]

  2. \[\mathbf{\text{ }\!\!\pm\!\!\text{ 6}}\]

  3. \[\mathbf{\text{-6}}\]

  4. \[\mathbf{\text{0}}\]

Ans: Given, 

$\begin{vmatrix}x &2 \\ 18 &x \end{vmatrix}=\begin{vmatrix}6 &2 \\ 18 &6 \end{vmatrix}$

Solving it, we have:

\[\Rightarrow {{\text{x}}^{\text{2}}}-\text{36 = 36}-\text{36}\]

\[\Rightarrow {{\text{x}}^{\text{2}}}-\text{36=0}\]

\[\Rightarrow {{\text{x}}^{\text{2}}}\text{=36}\]

Applying square root on both the sides, we obtain:

\[\Rightarrow \text{x= }\!\!\pm\!\!\text{ 6}\]

Hence, B. \[\text{ }\!\!\pm\!\!\text{ 6}\] is the correct answer.

Exercise (4.2)

1. Using the property of determinants and without expanding, prove that:

\[\mathbf{\left| \begin{matrix} \text{x} & \text{a} & \text{x+a}  \\ \text{y} & \text{b} & \text{y+b}  \\ \text{z} & \text{c} & \text{z+c}  \\ \end{matrix} \right|\text{=0}}\].

Ans: Given matrix 

\[\left| \begin{matrix} \text{x} & \text{a} & \text{x+a}  \\ \text{y} & \text{b} & \text{y+b}  \\ \text{z} & \text{c} & \text{z+c}  \\ \end{matrix} \right|\].

Applying the Sum Property of determinants, we have

\[\left| \begin{matrix} \text{x} & \text{a} & \text{x+a}  \\ \text{y} & \text{b} & \text{y+b}  \\ \text{z} & \text{c} & \text{z+c}  \\ \end{matrix} \right|\text{=}\left| \begin{matrix} \text{x} & \text{a} & \text{x}  \\ \text{y} & \text{b} & \text{y}  \\ \text{z} & \text{c} & \text{z}  \\ \end{matrix} \right|\text{+}\left| \begin{matrix} \text{x} & \text{a} & \text{a}  \\ \text{y} & \text{b} & \text{b}  \\ \text{z} & \text{c} & \text{c}  \\ \end{matrix} \right|\]

We know, if two rows or columns of a determinant are identical, then the value of the determinant is zero. 

Since, the two columns in both the determinants are identical, thus its determinant would be zero.

\[\Rightarrow \left| \begin{matrix} \text{x} & \text{a} & \text{x+a}  \\ \text{y} & \text{b} & \text{y+b}  \\ \text{z} & \text{c} & \text{z+c}  \\ \end{matrix} \right|=0+0\]

\[\therefore \left| \begin{matrix} \text{x} & \text{a} & \text{x+a}  \\ \text{y} & \text{b} & \text{y+b}  \\ \text{z} & \text{c} & \text{z+c}  \\ \end{matrix} \right|=0\]

2. Using the property of determinants and without expanding, prove that:

\[\mathbf{\left| \begin{matrix} \text{a-b} & \text{b-c} & \text{c-a}  \\ \text{b-c} & \text{c-a} & \text{a-b}  \\ \text{c-a} & \text{a-b} & \text{b-c}  \\ \end{matrix} \right|\text{=0}}\]

Ans: Let \[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{a}-\text{b} & \text{b}-\text{c} & \text{c}-\text{a}  \\ \text{b}-\text{c} & \text{c}-\text{a} & \text{a}-\text{b}  \\ \text{c}-\text{a} & \text{a}-\text{b} & \text{b}-\text{c}  \\ \end{matrix} \right|\]

Applying row operation, 

\[{{\text{R}}_{1}}\to \text{ }{{\text{R}}_{1}}\text{+}{{\text{R}}_{2}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{a}-\text{b+b}-\text{c} & \text{b}-\text{c+c}-a & \text{c}-\text{a+a}-b  \\ \text{b}-\text{c} & \text{c}-\text{a} & \text{a}-\text{b}  \\ \text{c}-\text{a} & \text{a}-\text{b} & \text{b}-\text{c}  \\ \end{matrix} \right|\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{a}-c & \text{b}-a & \text{c}-b  \\ \text{b}-\text{c} & \text{c}-\text{a} & \text{a}-\text{b}  \\ c-a & a-b & b-c  \\ \end{matrix} \right|\]

\[\Rightarrow \Delta =\left| \begin{matrix} a-c & b-a & c-b  \\ b-c & c-a & a-b  \\ -\left( a-c \right) & -\left( b-a \right) & -\left( c-b \right)  \\ \end{matrix} \right|\]

Multiplying the third row by $\left( -1 \right)$, we get:

\[\Rightarrow \Delta \text{=}\left| \begin{matrix} \text{a}-c & \text{b}-a & \text{c}-b  \\ \text{b}-\text{c} & \text{c}-\text{a} & \text{a}-\text{b}  \\ \text{a}-\text{c} & \text{b}-\text{a} & \text{c}-\text{b}  \\ \end{matrix} \right|\]

We know, if two rows or columns of a determinant are identical, then the value of the determinant is zero. 

Since, the two rows \[{{\text{R}}_{1}}\] and \[{{\text{R}}_{3}}\] are identical.

\[\therefore \text{ }\!\!\Delta\!\!\text{ =0}\]

Hence, 

\[\left| \begin{matrix} \text{a}-\text{b} & \text{b}-\text{c} & \text{c}-\text{a}  \\ \text{b}-\text{c} & \text{c}-\text{a} & \text{a}-\text{b}  \\ \text{c}-\text{a} & \text{a}-\text{b} & \text{b}-\text{c}  \\ \end{matrix} \right|=0\].

3. Using the property of determinants and without expanding, prove that:

\[\mathbf{\left| \begin{matrix} \text{2} & \text{7} & \text{65}  \\ \text{3} & \text{8} & \text{75}  \\ \text{5} & \text{9} & \text{86}  \\ \end{matrix} \right|\text{=0}}\]

Ans: Let 

$\Delta =\left| \begin{matrix} \text{2} & \text{7} & \text{65}  \\ \text{3} & \text{8} & \text{75}  \\ \text{5} & \text{9} & \text{86}  \\ \end{matrix} \right|$

\[\therefore \Delta \text{=}\left| \begin{matrix} \text{2} & \text{7} & \text{63+2}  \\ \text{3} & \text{8} & \text{72+3}  \\ \text{5} & \text{9} & \text{81+5}  \\ \end{matrix} \right|\]

Applying the Sum Property of determinants, we get

\[\Rightarrow \Delta \text{=}\left| \begin{matrix} \text{2} & \text{7} & \text{63}  \\ \text{3} & \text{8} & \text{72}  \\ \text{5} & \text{9} & \text{81}  \\ \end{matrix} \right|\text{+}\left| \begin{matrix} \text{2} & \text{7} & \text{2}  \\ \text{3} & \text{8} & \text{3}  \\ \text{5} & \text{9} & \text{5}  \\ \end{matrix} \right|\]

The two columns of the second determinant are identical, thus it’s value becomes zero.

Hence,

\[\Rightarrow \Delta \text{=}\left| \begin{matrix} \text{2} & \text{7} & 63  \\ \text{3} & \text{8} & 72  \\ \text{5} & \text{9} & 81  \\ \end{matrix} \right|\text{+0}\]            

\[\Rightarrow \Delta \text{=}\left| \begin{matrix} \text{2} & \text{7} & 9\left( 7 \right)  \\ \text{3} & \text{8} & 9\left( 8 \right)  \\ \text{5} & \text{9} & 9\left( 9 \right)  \\ \end{matrix} \right|\]

Taking $9$ common from the third column, we have

\[\Rightarrow \Delta =9\left| \begin{matrix}  2 & 7 & 7  \\ 3 & 8 & 8  \\ 5 & 9 & 9  \\ \end{matrix} \right|\]     

Since, the two columns \[{{C}_{2}}\] and \[{{C}_{3}}\]

 are identical.

$\therefore \Delta =0$

Hence, 

$\left| \begin{matrix} \text{2} & \text{7} & \text{65}  \\ \text{3} & \text{8} & \text{75}  \\ \text{5} & \text{9} & \text{86}  \\ \end{matrix} \right|=0$

4. Using the property of determinants and without expanding, prove that:\[\mathbf{\left| \begin{matrix} \text{1} & \text{bc} & \text{a}\left( \text{b+c} \right)  \\ \text{1} & \text{ca} & \text{b}\left( \text{c+a} \right)  \\ \text{1} & \text{ab} & \text{c}\left( \text{a+b} \right)  \\ \end{matrix} \right|\text{=0}}\].

Ans: Let 

\[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{1} & \text{bc} & \text{a}\left( \text{b+c} \right)  \\ \text{1} & \text{ca} & \text{b}\left( \text{c+a} \right)  \\ \text{1} & \text{ab} & \text{c}\left( \text{a+b} \right)  \\ \end{matrix} \right|\]

Applying the column operation, \[{{\text{C}}_{\text{3}}}\to {{\text{C}}_{\text{3}}}\text{+}{{\text{C}}_{\text{2}}}\].

\[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{1} & \text{bc} & \text{ab+bc+ca}  \\ \text{1} & \text{ca} & \text{ab+bc+ca}  \\ \text{1} & \text{ab} & \text{ab+bc+ca}  \\ \end{matrix} \right|\]

Taking $\left( ab+bc+ca \right)$ common from the third column, we get:

\[\text{ }\!\!\Delta\!\!\text{ =}\left( ab+bc+ca \right)\left| \begin{matrix} \text{1} & \text{bc} & 1  \\ \text{1} & \text{ca} & 1  \\ \text{1} & \text{ab} & 1  \\ \end{matrix} \right|\]

Since, the two columns \[{{C}_{1}}\] and \[{{C}_{3}}\] are identical.

\[\therefore \text{ }\!\!\Delta\!\!\text{ =0}\]

Hence, 

\[\left| \begin{matrix} \text{1} & \text{bc} & \text{a}\left( \text{b+c} \right)  \\ \text{1} & \text{ca} & \text{b}\left( \text{c+a} \right)  \\ \text{1} & \text{ab} & \text{c}\left( \text{a+b} \right)  \\ \end{matrix} \right|=0\].

5. Using the property of determinants and without expanding, prove that:

\[\mathbf{\left| \begin{matrix} \text{b+c} & \text{q+r} & \text{y+z}  \\ \text{c+a} & \text{r+p} & \text{z+x}  \\ \text{a+b} & \text{p+q} & \text{x+y}  \\ \end{matrix} \right|\text{=2}\left| \begin{matrix} \text{a} & \text{p} & \text{x}  \\ \text{b} & \text{q} & \text{y}  \\ \text{c} & \text{r} & \text{z}  \\ \end{matrix} \right|}\].

Ans: Let 

\[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{b+c} & \text{q+r} & \text{y+z}  \\ \text{c+a} & \text{r+p} & \text{z+x}  \\ \text{a+b} & \text{p+q} & \text{x+y}  \\ \end{matrix} \right|\]

Applying the Sum Property, we get

\[\Rightarrow \Delta \text{=}\left| \begin{matrix} \text{b+c} & \text{q+r} & \text{y+z}  \\ \text{c+a} & \text{r+p} & \text{z+x}  \\ \text{a} & \text{p} & \text{x}  \\ \end{matrix} \right|\text{+}\left| \begin{matrix} \text{b+c} & \text{q+r} & \text{y+z}  \\ \text{c+a} & \text{r+p} & \text{z+x}  \\ \text{b} & \text{q} & \text{y}  \\ \end{matrix} \right|\]

Suppose \[\Delta \text{=}\Delta \text{1+ }\!\!\Delta\!\!\text{ 2}\]   ……(1)

Now, 

\[{{\text{ }\!\!\Delta\!\!\text{ }}_{1}}\text{= }\left| \begin{matrix} \text{b+c} & \text{q+r} & \text{y+z}  \\ \text{c+a} & \text{r+p} & \text{z+x}  \\ \text{a} & \text{p} & \text{x}  \\ \end{matrix} \right|\]

Applying the row operation, \[{{\text{R}}_{2}}\to {{\text{R}}_{2}}-{{\text{R}}_{3}}\]

\[\Rightarrow {{\text{ }\!\!\Delta\!\!\text{ }}_{1}}\text{= }\left| \begin{matrix} \text{b+c} & \text{q+r} & \text{y+z}  \\ \text{c} & \text{r} & \text{z}  \\ \text{a} & \text{p} & \text{x}  \\ \end{matrix} \right|\]

Again, applying the row operation, \[{{\text{R}}_{1}}\to {{\text{R}}_{1}}-{{\text{R}}_{2}}\]

\[\Rightarrow \Delta_1=\left|\begin{matrix} \text{b}&\text{q}&\text{y}\\ \text{c}&\text{r}&\text{z}\\ \text{a}&\text{p}&\text{x}\\ \end{matrix}  \right|\]

We know that  if any two rows or columns of a determinant are interchanged, the value of the determinant is multiplied by $\left( -1 \right)$.

Hence, interchanging the rows, 

\[{{\text{R}}_{\text{1}}}\leftrightarrow {{\text{R}}_{\text{2}}}\] and \[{{\text{R}}_{\text{2}}}\leftrightarrow {{\text{R}}_{\text{3}}}\], we have

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ 1=}{{\left( \text{-1} \right)}^{\text{2}}}\left| \begin{matrix} \text{a} & \text{p} & \text{x}  \\ \text{b} & \text{q} & \text{y}  \\ \text{c} & \text{r} & \text{z}  \\ \end{matrix} \right|\]

\[\therefore {{\Delta }_{1}}\text{=}\left| \begin{matrix} \text{a} & \text{p} & \text{x}  \\ \text{b} & \text{q} & \text{y}  \\ \text{c} & \text{r} & \text{z}  \\ \end{matrix} \right|\]  ……(2)

We have, 

\[{{\text{ }\!\!\Delta\!\!\text{ }}_{1}}\text{=}\left| \begin{matrix} \text{b+c} & \text{q+r} & \text{y+z}  \\ \text{c+a} & \text{r+p} & \text{z+x}  \\ \text{b} & \text{q} & \text{y}  \\ \end{matrix} \right|\]

Applying the row operation, \[{{\text{R}}_{1}}\to {{\text{R}}_{1}}-{{\text{R}}_{3}}\]

\[\Rightarrow {{\text{ }\!\!\Delta\!\!\text{ }}_{2}}=\left| \begin{matrix} \text{c} & \text{r} & \text{z}  \\ \text{c+a} & \text{r+p} & \text{z+x}  \\ \text{b} & \text{q} & \text{y}  \\ \end{matrix} \right|\]

Applying the row operation, \[{{\text{R}}_{2}}\to {{\text{R}}_{2}}-{{\text{R}}_{1}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ 2=}\left| \begin{matrix} \text{c} & \text{r} & \text{z}  \\ \text{a} & \text{p} & \text{x}  \\ \text{b} & \text{q} & \text{y}  \\ \end{matrix} \right|\]

Interchanging the rows, \[{{\text{R}}_{\text{1}}}\leftrightarrow {{\text{R}}_{\text{2}}}\] and \[{{\text{R}}_{\text{2}}}\leftrightarrow {{\text{R}}_{\text{3}}}\]

\[\Rightarrow {{\text{ }\!\!\Delta\!\!\text{ }}_{2}}\text{=}{{\left( \text{-1} \right)}^{\text{2}}}\left| \begin{matrix} \text{a} & \text{p} & \text{x}  \\ \text{b} & \text{q} & \text{y}  \\ \text{c} & \text{r} & \text{z}  \\ \end{matrix} \right|\]

\[\therefore {{\text{ }\!\!\Delta\!\!\text{ }}_{2}}\text{=}\left| \begin{matrix} \text{a} & \text{p} & \text{x}  \\ \text{b} & \text{q} & \text{y}  \\ \text{c} & \text{r} & \text{z}  \\ \end{matrix} \right|\]   …… (3)

From (2) and (3), we get:

\[\Rightarrow {{\text{ }\!\!\Delta\!\!\text{ }}_{1}}={{\text{ }\!\!\Delta\!\!\text{ }}_{2}}=\left| \begin{matrix} \text{a} & \text{p} & \text{x}  \\ \text{b} & \text{q} & \text{y}  \\ \text{c} & \text{r} & \text{z}  \\ \end{matrix} \right|\]

From (1), we have:

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =2}{{\Delta }_{1}}\] \[\therefore \Delta \text{=2}\left| \begin{matrix} \text{a} & \text{p} & \text{x}  \\ \text{b} & \text{q} & \text{y}  \\ \text{c} & \text{r} & \text{z}  \\ \end{matrix} \right|\].

Hence, \[\left| \begin{matrix} \text{b+c} & \text{q+r} & \text{y+z}  \\ \text{c+a} & \text{r+p} & \text{z+x}  \\ \text{a+b} & \text{p+q} & \text{x+y}  \\ \end{matrix} \right|\text{=2}\left| \begin{matrix} \text{a} & \text{p} & \text{x}  \\ \text{b} & \text{q} & \text{y}  \\ \text{c} & \text{r} & \text{z}  \\ \end{matrix} \right|\].

6. By using properties of determinants, show that: (Not in the current syllabus)

\[\mathbf{\left| \begin{matrix} \text{0} & \text{a} & \text{-b}  \\ \text{-a} & \text{0} & \text{-c}  \\ \text{b} & \text{c} & \text{0}  \\ \end{matrix} \right|\text{=0}}\].

Ans: Given, 

\[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{0} & \text{a} & -\text{b}  \\ -\text{a} & \text{0} & -\text{c}  \\ \text{b} & \text{c} & \text{0}  \\ \end{matrix} \right|\]

We know that if we multiply the elements of a matrix by a scalar $c$, then we will multiply the matrix by the scalar, $\dfrac{1}{c}$.

Applying \[{{\text{R}}_{1}}\to \text{c}{{\text{R}}_{1}}\]:

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{c}}\left| \begin{matrix} \text{0} & \text{ac} & -\text{bc}  \\ -\text{a} & \text{0} & -\text{c}  \\ \text{b} & \text{c} & \text{0}  \\ \end{matrix} \right|\]

Applying \[{{\text{R}}_{1}}\to {{\text{R}}_{1}}-\text{b}{{\text{R}}_{2}}\]

\[\text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{c}}\left| \begin{matrix} \text{ab} & \text{ac} & \text{0}  \\ -\text{a} & \text{0} & -\text{c}  \\ \text{b} & \text{c} & \text{0}  \\ \end{matrix} \right|\]

Taking $a$ common from the first row, we have:

\[\Rightarrow \Delta \text{=}\dfrac{\text{a}}{\text{c}}\left| \begin{matrix} \text{b} & \text{c} & \text{0}  \\ -\text{a} & \text{0} & -\text{c}  \\ \text{b} & \text{c} & \text{0}  \\ \end{matrix} \right|\]

Since, the two rows \[{{\text{R}}_{1}}\] and \[{{\text{R}}_{3}}\] are identical.

\[\therefore \text{ }\!\!\Delta\!\!\text{ =0}\]

Hence, \[\left| \begin{matrix} \text{0} & \text{a} & -\text{b}  \\ -\text{a} & \text{0} & -\text{c}  \\ \text{b} & \text{c} & \text{0}  \\ \end{matrix} \right|=0\].

7. By using properties of determinants, show that:

\[\mathbf{\left| \begin{matrix} \text{-}{{\text{a}}^{\text{2}}} & \text{ab} & \text{ac}  \\ \text{ba} & \text{-}{{\text{b}}^{\text{2}}} & \text{bc}  \\ \text{ca} & \text{cb} & \text{-}{{\text{c}}^{\text{2}}}  \\ \end{matrix} \right|\text{=4}{{\text{a}}^{\text{2}}}{{\text{b}}^{\text{2}}}{{\text{c}}^{\text{2}}}}\].

Ans: Let \[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{-}{{\text{a}}^{\text{2}}} & \text{ab} & \text{ac}  \\ \text{ba} & \text{-}{{\text{b}}^{\text{2}}} & \text{bc}  \\ \text{ca} & \text{cb} & \text{-}{{\text{c}}^{\text{2}}}  \\ \end{matrix} \right|\]

Taking out $a,b,c$ from \[{{\text{R}}_{1}}\],  \[{{\text{R}}_{2}}\] and \[{{\text{R}}_{3}}\] respectively, we have:

\[\Rightarrow \Delta \text{=abc}\left| \begin{matrix} \text{-a} & \text{b} & \text{c}  \\ \text{a} & \text{-b} & \text{c}  \\ \text{a} & \text{b} & \text{-c}  \\ \end{matrix} \right|\]    

Similarly, taking out $a,b,c$ from \[{{C}_{1}}\],  \[{{C}_{2}}\] and \[{{C}_{3}}\] respectively, we have:

\[\Rightarrow \Delta \text{=}{{\text{a}}^{\text{2}}}{{\text{b}}^{\text{2}}}{{\text{c}}^{\text{2}}}\left| \begin{matrix} \text{-1} & \text{1} & \text{1}  \\ \text{1} & \text{-1} & \text{1}  \\ \text{1} & \text{1} & \text{-1}  \\ \end{matrix} \right|\]

Applying the row operations \[{{\text{R}}_{2}}\to {{\text{R}}_{2}}\text{+}{{\text{R}}_{1}}\] and \[{{\text{R}}_{3}}\to {{\text{R}}_{3}}\text{+}{{\text{R}}_{1}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}{{\text{a}}^{\text{2}}}{{\text{b}}^{\text{2}}}{{\text{c}}^{\text{2}}}\left| \begin{matrix} \text{-1} & \text{1} & \text{1}  \\ \text{0} & \text{0} & \text{2}  \\ \text{0} & \text{2} & \text{0}  \\ \end{matrix} \right|\]

Solving it along the first column, ${{C}_{1}}$ we get:

\[\Rightarrow \Delta \text{=}{{\text{a}}^{\text{2}}}{{\text{b}}^{\text{2}}}{{\text{c}}^{\text{2}}}\left( \text{-1} \right)\left| \begin{matrix} \text{0} & \text{2}  \\ \text{2} & \text{0}  \\ \end{matrix} \right|\]

\[\Rightarrow \Delta \text{=}{{\text{a}}^{\text{2}}}{{\text{b}}^{\text{2}}}{{\text{c}}^{\text{2}}}\left( \text{0-4} \right)\]

\[\therefore \Delta \text{=4}{{\text{a}}^{\text{2}}}{{\text{b}}^{\text{2}}}{{\text{c}}^{\text{2}}}\]

Hence, \[\left| \begin{matrix} \text{-}{{\text{a}}^{\text{2}}} & \text{ab} & \text{ac}  \\ \text{ba} & \text{-}{{\text{b}}^{\text{2}}} & \text{bc}  \\ \text{ca} & \text{cb} & \text{-}{{\text{c}}^{\text{2}}}  \\ \end{matrix} \right|\text{=4}{{\text{a}}^{\text{2}}}{{\text{b}}^{\text{2}}}{{\text{c}}^{\text{2}}}\].

8. By using properties of determinants, show that:

  1. \[\mathbf{\left| \begin{matrix} \text{1} & \text{a} & {{\text{a}}^{\text{2}}}  \\ \text{1} & \text{b} & {{\text{b}}^{\text{2}}}  \\ \text{1} & \text{c} & {{\text{c}}^{\text{2}}}  \\ \end{matrix} \right|=\left( \text{a}-\text{b} \right)\left( \text{b}-\text{c} \right)\left( \text{c}-\text{a} \right)}\]

Ans: Let \[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{1} & \text{a} & {{\text{a}}^{\text{2}}}  \\ \text{1} & \text{b} & {{\text{b}}^{\text{2}}}  \\ \text{1} & \text{c} & {{\text{c}}^{\text{2}}}  \\ \end{matrix} \right|\]

Applying the row operations \[{{\text{R}}_{1}}\to {{\text{R}}_{1}}-{{\text{R}}_{3}}\] and \[{{\text{R}}_{2}}\to {{\text{R}}_{2}}-{{\text{R}}_{3}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{0} & \text{a-c} & {{\text{a}}^{\text{2}}}\text{-}{{\text{c}}^{\text{2}}}  \\ \text{0} & \text{b-c} & {{\text{b}}^{\text{2}}}\text{-}{{\text{c}}^{\text{2}}}  \\ \text{1} & \text{c} & {{\text{c}}^{\text{2}}}  \\ \end{matrix} \right|\]

We know, ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$

Thus,

\[\Rightarrow \Delta \text{=}\left( \text{c}-\text{a} \right)\left( \text{b}-c \right)\left| \begin{matrix} \text{0} & -\text{1} & -\left( a+c \right)  \\ \text{0} & \text{1} & \text{b+c}  \\ \text{1} & \text{c} & {{\text{c}}^{\text{2}}}  \\ \end{matrix} \right|\]

Applying \[{{R}_{1}}\to {{R}_{1}}+{{R}_{2}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left( \text{b}-\text{c} \right)\left( \text{c}-\text{a} \right)\left| \begin{matrix} \text{0} & \text{0} & -a+b  \\ \text{0} & \text{1} & \text{b+c}  \\ \text{1} & \text{c} & {{\text{c}}^{\text{2}}}  \\ \end{matrix} \right|\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left( \text{b}-\text{c} \right)\left( \text{c}-\text{a} \right)\left| \begin{matrix} \text{0} & \text{0} & -\left( a-b \right)  \\ \text{0} & \text{1} & \text{b+c}  \\ \text{1} & \text{c} & {{\text{c}}^{\text{2}}}  \\ \end{matrix} \right|\]

Taking out \[\left( \text{a}-\text{b} \right)\] common from \[{{R}_{1}}\]

\[\Rightarrow \Delta \text{=}\left( \text{a}-\text{b} \right)\left( \text{b}-\text{c} \right)\left( \text{c}-\text{a} \right)\left| \begin{matrix} \text{0} & \text{0} & -\text{1}  \\ \text{0} & \text{1} & \text{b+c}  \\ \text{1} & \text{c} & {{\text{c}}^{\text{2}}}  \\ \end{matrix} \right|\]

Expanding along \[{{\text{C}}_{1}}\],

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left( \text{a-b} \right)\left( \text{b-c} \right)\left( \text{c-a} \right)\left| \begin{matrix} \text{0} & \text{1}  \\ \text{1} & \text{b+c}  \\ \end{matrix} \right|\]

\[\therefore \Delta \text{=}\left( \text{a-b} \right)\left( \text{b-c} \right)\left( \text{c-a} \right)\]

Hence, \[\left| \begin{matrix} \text{1} & \text{a} & {{\text{a}}^{\text{2}}}  \\ \text{1} & \text{b} & {{\text{b}}^{\text{2}}}  \\ \text{1} & \text{c} & {{\text{c}}^{\text{2}}}  \\ \end{matrix} \right|=\left( \text{a}-\text{b} \right)\left( \text{b}-\text{c} \right)\left( \text{c}-\text{a} \right)\]

  1. \[\mathbf{\left| \begin{matrix} \text{1} & \text{1} & \text{1}  \\ \text{a} & \text{b} & \text{c}  \\ {{\text{a}}^{\text{3}}} & {{\text{b}}^{\text{3}}} & {{\text{c}}^{\text{3}}}  \\ \end{matrix} \right|\text{=}\left( \text{a-b} \right)\left( \text{b-c} \right)\left( \text{c-a} \right)\left( \text{a+b+c} \right)}\]

Ans: Let \[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{1} & \text{1} & \text{1}  \\ \text{a} & \text{b} & \text{c}  \\ {{\text{a}}^{\text{3}}} & {{\text{b}}^{\text{3}}} & {{\text{c}}^{\text{3}}}  \\ \end{matrix} \right|\]

Applying the column operations, \[{{C}_{1}}\to {{C}_{1}}-{{C}_{3}}\] and \[{{C}_{2}}\to {{C}_{2}}-{{C}_{3}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{0} & \text{0} & \text{1}  \\ \text{a-c} & \text{b-c} & \text{c}  \\ {{\text{a}}^{\text{3}}}\text{-}{{\text{c}}^{\text{3}}} & {{\text{b}}^{\text{3}}}\text{-}{{\text{c}}^{\text{3}}} & {{\text{c}}^{\text{3}}}  \\ \end{matrix} \right|\]

We know, ${{x}^{3}}-{{y}^{3}}=\left( x-y \right)\left( {{x}^{2}}+{{y}^{2}}+xy \right)$.

\[\Rightarrow \Delta \text{=}\left| \begin{matrix} \text{0} & \text{0} & \text{1}  \\ \text{a-c} & \text{b-c} & \text{c}  \\ \left( \text{a-c} \right)\left( {{\text{a}}^{\text{2}}}\text{+ac+}{{\text{c}}^{\text{2}}} \right) & \left( \text{b-c} \right)\left( {{\text{b}}^{\text{2}}}\text{+bc+}{{\text{c}}^{\text{2}}} \right) & {{\text{c}}^{\text{3}}}  \\ \end{matrix} \right|\]

Taking out \[\left( c-a \right)\] and \[\left( b-c \right)\] common from \[{{C}_{1}}\] and \[{{C}_{2}}\] respectively,

\[\Rightarrow \Delta \text{=}\left( c-a \right)\left( b-c \right)\left| \begin{matrix} \text{0} & \text{0} & \text{1}  \\ -1 & \text{1} & \text{c}  \\ -\left( {{\text{a}}^{\text{2}}}\text{+ac+}{{\text{c}}^{\text{2}}} \right) & \left( {{\text{b}}^{\text{2}}}\text{+bc+}{{\text{c}}^{\text{2}}} \right) & {{\text{c}}^{\text{3}}}  \\ \end{matrix} \right|\]

Applying \[{{C}_{1}}\to {{C}_{1}}+{{C}_{2}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left( c-a \right)\left( b-c \right)\left| \begin{matrix} \text{0} & \text{0} & \text{1}  \\ 0 & \text{1} & \text{c}  \\ \left( {{b}^{2}}-{{a}^{2}} \right)+c\left( b-a \right) & \left( {{\text{b}}^{\text{2}}}\text{+bc+}{{\text{c}}^{\text{2}}} \right) & {{\text{c}}^{\text{3}}}  \\ \end{matrix} \right|\]

Taking out $\left( a-c \right)$ common from \[{{C}_{1}}\], we get:

\[\Rightarrow \Delta \text{=}\left( a-b \right)\left( c-a \right)\left( b-c \right)\left| \begin{matrix} \text{0} & \text{0} & \text{1}  \\ \text{0} & \text{0} & \text{c}  \\ \text{-}\left( \text{a+b+c} \right) & \left( {{\text{b}}^{\text{2}}}\text{+bc+}{{\text{c}}^{\text{2}}} \right) & {{\text{c}}^{\text{3}}}  \\ \end{matrix} \right|\]

Again taking out $\left( a+b+c \right)$ common from \[{{C}_{1}}\], we get:

\[\Rightarrow \Delta \text{=}\left( a-b \right)\left( b-c \right)\left( c-a \right)\left( a+b+c \right)\left| \begin{matrix} \text{0} & \text{0} & \text{1}  \\ \text{0} & \text{0} & \text{c}  \\ -1 & \left( {{\text{b}}^{\text{2}}}\text{+bc+}{{\text{c}}^{\text{2}}} \right) & {{\text{c}}^{\text{3}}}  \\ \end{matrix} \right|\]

Expanding along \[{{C}_{1}}\], we get:

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left( a-b \right)\left( b-c \right)\left( c-a \right)\left( \text{a+b+c} \right)\left( \text{-1} \right)\left| \begin{matrix} \text{0} & \text{1}  \\ \text{1} & \text{c}  \\ \end{matrix} \right|\]

\[\therefore \Delta \text{=}\left( \text{a-b} \right)\left( \text{b-c} \right)\left( \text{c-a} \right)\left( \text{a+b+c} \right)\]

Hence, \[\left| \begin{matrix} \text{1} & \text{1} & \text{1}  \\ \text{a} & \text{b} & \text{c}  \\ {{\text{a}}^{\text{3}}} & {{\text{b}}^{\text{3}}} & {{\text{c}}^{\text{3}}}  \\ \end{matrix} \right|\text{=}\left( \text{a-b} \right)\left( \text{b-c} \right)\left( \text{c-a} \right)\left( \text{a+b+c} \right)\]

9. By using properties of determinants, show that:

\[\mathbf{\left| \begin{matrix} \text{x} & {{\text{x}}^{\text{2}}} & \text{yz}  \\ \text{y} & {{\text{y}}^{\text{2}}} & \text{zx}  \\ \text{z} & {{\text{z}}^{\text{2}}} & \text{xy}  \\ \end{matrix} \right|\text{=}\left( \text{x-y} \right)\left( \text{y-z} \right)\left( \text{z-x} \right)\left( \text{xy+yz+zx} \right)}\].

Ans: Let \[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{x} & {{\text{x}}^{\text{2}}} & \text{yz}  \\ \text{y} & {{\text{y}}^{\text{2}}} & \text{zx}  \\ \text{z} & {{\text{z}}^{\text{2}}} & \text{xy}  \\ \end{matrix} \right|\]

Applying the row operations \[{{\text{R}}_{2}}\to {{\text{R}}_{2}}-{{\text{R}}_{1}}\] and \[{{\text{R}}_{3}}\to {{\text{R}}_{3}}-{{\text{R}}_{1}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{x} & {{\text{x}}^{\text{2}}} & \text{yz}  \\ \text{y-x} & {{\text{y}}^{\text{2}}}\text{-}{{\text{x}}^{\text{2}}} & \text{zx-yz}  \\ \text{z-x} & {{\text{z}}^{\text{2}}}\text{-}{{\text{x}}^{\text{2}}} & \text{xy-yz}  \\ \end{matrix} \right|\]

\[\Rightarrow \Delta \text{=}\left| \begin{matrix} \text{x} & {{\text{x}}^{\text{2}}} & \text{yz}  \\ \text{-}\left( \text{x-y} \right) & \text{-}\left( \text{x-y} \right)\left( \text{x+y} \right) & \text{z}\left( \text{x-y} \right)  \\ \left( \text{z-x} \right) & \left( \text{z-x} \right)\left( \text{z+x} \right) & \text{-y}\left( \text{z-x} \right)  \\ \end{matrix} \right|\]

Taking out \[\left( x-y \right)\] and \[\left( z-x \right)\] common from \[{{R}_{2}}\] and \[{{R}_{3}}\] respectively

\[\Rightarrow \Delta \text{=}\left( x-y \right)\left( z-x \right)\left| \begin{matrix} \text{x} & {{\text{x}}^{\text{2}}} & \text{yz}  \\ -1 & -x-y & \text{z}  \\ \text{1} & z+x & -y  \\ \end{matrix} \right|\]

Applying \[{{R}_{3}}\to {{R}_{3}}+{{R}_{2}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left( x-y \right)\left( z-x \right)\left| \begin{matrix} \text{x} & {{\text{x}}^{\text{2}}} & \text{yz}  \\ \text{-1} & \text{-x-y} & \text{z}  \\ 0 & z-y & z-y  \\ \end{matrix} \right|\]

Taking out \[\left( z-y \right)\] common from \[{{R}_{3}}\], we get:

\[\Rightarrow \Delta \text{=}\left( x-y \right)\left( z-x \right)\left( z-y \right)\left| \begin{matrix} \text{x} & {{\text{x}}^{\text{2}}} & \text{yz}  \\ -1 & -x-y & \text{z}  \\ \text{0} & \text{1} & \text{1}  \\ \end{matrix} \right|\]

Expanding along \[{{R}_{3}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left[ \left( x-y \right)\left( z-x \right)\left( z-y \right) \right]\left[ \left( -1 \right)\left| \begin{matrix} \text{x} & \text{yz}  \\ -1 & \text{z}  \\ \end{matrix} \right|\text{+1}\left| \begin{matrix} \text{x} & {{\text{x}}^{\text{2}}}  \\ -1 & -x-y  \\ \end{matrix} \right| \right]\]

\[\Rightarrow \Delta \text{=}\left( \text{x-y} \right)\left( \text{z-x} \right)\left( \text{z-y} \right)\left[ \left( \text{-xz-yz} \right)\text{+}\left( \text{-}{{\text{x}}^{\text{2}}}\text{-xy+}{{\text{x}}^{\text{2}}} \right) \right]\]

\[\Rightarrow \Delta \text{=}-\left( x-y \right)\left( z-x \right)\left( z-y \right)\left( \text{xy+yz+zx} \right)\]

\[\therefore \Delta \text{=}\left( x-y \right)\left( y-z \right)\left( z-x \right)\left( \text{xy+yz+zx} \right)\]

Hence,\[\left| \begin{matrix} \text{x} & {{\text{x}}^{\text{2}}} & \text{yz}  \\ \text{y} & {{\text{y}}^{\text{2}}} & \text{zx}  \\ \text{z} & {{\text{z}}^{\text{2}}} & \text{xy}  \\ \end{matrix} \right|\text{=}\left( x-y \right)\left( y-z \right)\left( z-x \right)\left( \text{xy+yz+zx} \right)\]

10. By using properties of determinants, show that:

  1. \[\mathbf{\left| \begin{matrix} \text{x+4} & \text{2x} & \text{2x}  \\ \text{2x} & \text{x+4} & \text{2x}  \\ \text{2x} & \text{2x} & \text{x+4}  \\ \end{matrix} \right|\text{=}\left( \text{5x+4} \right){{\left( \text{4-x} \right)}^{\text{2}}}}\]

Ans: Let \[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{x+4} & \text{2x} & \text{2x}  \\ \text{2x} & \text{x+4} & \text{2x}  \\ \text{2x} & \text{2x} & \text{x+4}  \\ \end{matrix} \right|\]

Applying the row operation, \[{{R}_{1}}\to {{R}_{1}}\text{+}{{\text{R}}_{2}}\text{+}{{\text{R}}_{3}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{5x+4} & \text{5x+4} & \text{5x+4}  \\ \text{2x} & \text{x+4} & \text{2x}  \\ \text{2x} & \text{2x} & \text{x+4}  \\ \end{matrix} \right|\]

Taking out $\left( 5x+4 \right)$ common from \[{{R}_{1}}\]

\[\Rightarrow \Delta \text{=}\left( \text{5x+4} \right)\left| \begin{matrix} \text{1} & 1 & 1  \\ \text{2x} & \text{x+4} & 2x  \\ \text{2x} & \text{0} & \text{x+4}  \\ \end{matrix} \right|\]

Applying the column operations, \[{{C}_{2}}\to {{C}_{2}}-{{C}_{1}}\] and \[{{C}_{3}}\to {{C}_{3}}-{{C}_{1}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left( \text{5x+4} \right)\left| \begin{matrix} \text{1} & \text{0} & \text{0}  \\ \text{2x} & 4-x & \text{0}  \\ \text{2x} & \text{0} & 4-x  \\ \end{matrix} \right|\]

Taking out $\left( 4-x \right)$ common from \[{{C}_{2}}\] and \[{{C}_{3}}\] respectively,

\[\Rightarrow \Delta \text{=}\left( \text{5x+4} \right)\left( \text{4-x} \right)\left( \text{4-x} \right)\left| \begin{matrix} \text{1} & \text{0} & \text{0}  \\ \text{2x} & \text{1} & \text{0}  \\ \text{2x} & \text{0} & \text{1}  \\ \end{matrix} \right|\]

Expanding along \[{{C}_{3}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left( \text{5x+4} \right){{\left( 4-x \right)}^{\text{2}}}\left| \begin{matrix} \text{1} & \text{0}  \\ \text{2x} & \text{1}  \\ \end{matrix} \right|\]

\[\therefore \Delta \text{=}\left( \text{5x+4} \right){{\left( 4-x \right)}^{\text{2}}}\]

Hence,\[\left| \begin{matrix} \text{x+4} & \text{2x} & \text{2x}  \\ \text{2x} & \text{x+4} & \text{2x}  \\ \text{2x} & \text{2x} & \text{x+4}  \\ \end{matrix} \right|\text{=}\left( \text{5x+4} \right){{\left( \text{4-x} \right)}^{\text{2}}}\]

  1. \[\mathbf{\left| \begin{matrix} \text{y+k} & \text{y} & \text{y}  \\ \text{y} & \text{y+k} & \text{y}  \\ \text{y} & \text{y} & \text{y+k}  \\ \end{matrix} \right|\text{=}{{\text{k}}^{\text{2}}}\left( \text{3x+k} \right)}\]

Ans: Let \[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{y+k} & \text{y} & \text{y}  \\ \text{y} & \text{y+k} & \text{y}  \\ \text{y} & \text{y} & \text{y+k}  \\ \end{matrix} \right|\]

Applying the row operation, \[{{R}_{1}}\to {{R}_{1}}\text{+}{{\text{R}}_{2}}\text{+}{{\text{R}}_{3}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{3y+k} & \text{3y+k} & \text{3y+k}  \\ \text{y} & \text{y+k} & \text{y}  \\ \text{y} & \text{y} & \text{y+k}  \\ \end{matrix} \right|\]

Taking out $\left( 3y+k \right)$ common from \[{{R}_{1}}\]

\[\Rightarrow \Delta \text{=}\left( \text{3y+k} \right)\left| \begin{matrix} \text{1} & \text{1} & \text{1}  \\ \text{y} & \text{y+k} & \text{y}  \\ \text{y} & \text{y} & \text{y+k}  \\ \end{matrix} \right|\]

Applying the column operations, \[{{C}_{2}}\to {{C}_{2}}-{{C}_{1}}\] and \[{{C}_{3}}\to {{C}_{3}}-{{C}_{1}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left( \text{3y+k} \right)\left| \begin{matrix} \text{1} & \text{0} & \text{0}  \\ \text{y} & \text{k} & \text{0}  \\ \text{y} & \text{0} & \text{k}  \\ \end{matrix} \right|\]

Taking out $\left( k \right)$ common from \[{{C}_{2}}\] and \[{{C}_{3}}\] respectively,

\[\Rightarrow \Delta \text{=}{{\text{k}}^{\text{2}}}\left( \text{3x+k} \right)\left| \begin{matrix} \text{1} & \text{0} & \text{0}  \\ \text{y} & \text{1} & \text{0}  \\ \text{y} & \text{0} & \text{1}  \\ \end{matrix} \right|\]

Expanding along \[{{C}_{3}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}{{\text{k}}^{\text{2}}}\left( \text{3x+k} \right)\left| \begin{matrix} \text{1} & \text{0}  \\ \text{y} & \text{1}  \\ \end{matrix} \right|\]

\[\therefore \Delta \text{=}{{\text{k}}^{\text{2}}}\left( \text{3x+k} \right)\]

Hence,\[\left| \begin{matrix} \text{y+k} & \text{y} & \text{y}  \\ \text{y} & \text{y+k} & \text{y}  \\ \text{y} & \text{y} & \text{y+k}  \\ \end{matrix} \right|\text{=}{{\text{k}}^{\text{2}}}\left( \text{3x+k} \right)\].

11. By using properties of determinants, show that:

  1. \[\mathbf{\left| \begin{matrix} \text{a-b-c} & \text{2a} & \text{2a}  \\ \text{2b} & \text{b-c-a} & \text{2b}  \\ \text{2c} & \text{2c} & \text{c-a-b}  \\ \end{matrix} \right|\text{=}{{\left( \text{a+b+c} \right)}^{\text{3}}}}\]

Ans: Let \[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{a+b+c} & \text{a+b+c} & \text{a+b+c}  \\ \text{2b} & \text{b-c-a} & \text{2b}  \\ \text{2c} & \text{2c} & \text{c-a-b}  \\ \end{matrix} \right|\]

Applying the row operation, \[{{R}_{1}}\to {{R}_{1}}\text{+}{{\text{R}}_{2}}\text{+}{{\text{R}}_{3}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{a+b+c} & \text{a+b+c} & \text{a+b+c}  \\ \text{2b} & \text{b-c-a} & \text{2b}  \\ \text{2c} & \text{2c} & \text{c-a-b}  \\ \end{matrix} \right|\]

Taking out $\left( a+b+c \right)$ common from \[{{R}_{1}}\]

\[\Rightarrow \Delta \text{=}\left( \text{a+b+c} \right)\left| \begin{matrix} \text{1} & \text{1} & \text{1}  \\ \text{2b} & \text{b-c-a} & \text{2b}  \\ \text{2c} & \text{2c} & \text{c-a-b}  \\ \end{matrix} \right|\]

Applying the column operations, \[{{C}_{2}}\to {{C}_{2}}-{{C}_{1}}\] and \[{{C}_{3}}\to {{C}_{3}}-{{C}_{1}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left( \text{a+b+c} \right)\left| \begin{matrix} \text{1} & \text{0} & \text{0}  \\ \text{2b} & \text{-}\left( \text{a+b+c} \right) & \text{0}  \\ \text{2c} & \text{0} & \text{-}\left( \text{a+b+c} \right)  \\ \end{matrix} \right|\]

Taking out $\left( a+b+c \right)$ common from \[{{C}_{2}}\] and \[{{C}_{3}}\] respectively,

\[\Rightarrow \Delta \text{=}{{\left( \text{a+b+c} \right)}^{\text{3}}}\left| \begin{matrix} \text{1} & \text{0} & \text{0}  \\ \text{2b} & \text{-1} & \text{0}  \\ \text{2c} & \text{0} & \text{-1}  \\ \end{matrix} \right|\]

Expanding along \[{{C}_{3}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}{{\left( \text{a+b+c} \right)}^{\text{3}}}\left( \text{-1} \right)\left( \text{-1} \right)\]

\[\therefore \Delta \text{=}{{\left( \text{a+b+c} \right)}^{\text{3}}}\]

Hence,\[\left| \begin{matrix} \text{a-b-c} & \text{2a} & \text{2a}  \\ \text{2b} & \text{b-c-a} & \text{2b}  \\ \text{2c} & \text{2c} & \text{c-a-b}  \\ \end{matrix} \right|\text{=}{{\left( \text{a+b+c} \right)}^{\text{3}}}\].

  1. \[\mathbf{\left| \begin{matrix} \text{x+y+2z} & \text{x} & \text{y}  \\ \text{z} & \text{y+z+2z} & \text{y}  \\ \text{z} & \text{x} & \text{z+x+2y}  \\ \end{matrix} \right|\text{=2}{{\left( \text{x+y+z} \right)}^{\text{3}}}}\]

Ans: Let \[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{x+y+2z} & \text{x} & \text{y}  \\ \text{z} & \text{y+z+2z} & \text{y}  \\ \text{z} & \text{x} & \text{z+x+2y}  \\ \end{matrix} \right|\]

Applying the column operation, \[{{C}_{1}}\to {{C}_{1}}\text{+}{{\text{C}}_{2}}\text{+}{{\text{C}}_{3}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{2}\left( \text{x+y+z} \right) & \text{x} & \text{y}  \\ \text{2}\left( \text{x+y+z} \right) & \text{y+z+2z} & \text{y}  \\ \text{2}\left( \text{x+y+z} \right) & \text{x} & \text{z+x+2y}  \\ \end{matrix} \right|\]

Taking out $2\left( x+y+z \right)$ common from \[{{C}_{1}}\]

\[\Rightarrow \Delta \text{=2}\left( \text{x+y+z} \right)\left| \begin{matrix} \text{1} & \text{x} & \text{y}  \\ \text{1} & \text{y+z+2z} & \text{y}  \\ \text{1} & \text{x} & \text{z+x+2y}  \\ \end{matrix} \right|\]

Applying the row operations \[{{\text{R}}_{2}}\to {{\text{R}}_{2}}-{{\text{R}}_{1}}\] and \[{{\text{R}}_{3}}\to {{\text{R}}_{3}}-{{\text{R}}_{1}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =2}\left( \text{x+y+z} \right)\left| \begin{matrix} \text{1} & \text{x} & \text{y}  \\ \text{0} & \text{x+y+z} & \text{0}  \\ \text{0} & \text{0} & \text{x+y+z}  \\ \end{matrix} \right|\]

Taking out $\left( x+y+z \right)$ common from \[{{R}_{2}}\] and \[{{R}_{3}}\] respectively,

\[\Rightarrow \Delta \text{=2}\left( \text{x+y+z} \right)\left| \begin{matrix} \text{1} & \text{x} & \text{y}  \\ \text{0} & \text{1} & \text{0}  \\ \text{0} & \text{0} & \text{1}  \\ \end{matrix} \right|\]

Expanding along \[{{R}_{3}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =2}{{\left( \text{x+y+z} \right)}^{\text{3}}}\left( \text{1} \right)\left( \text{1-0} \right)\]

\[\therefore \Delta \text{=2}{{\left( \text{x+y+z} \right)}^{\text{3}}}\]

Hence,\[\left| \begin{matrix} \text{x+y+2z} & \text{x} & \text{y}  \\ \text{z} & \text{y+z+2z} & \text{y}  \\ \text{z} & \text{x} & \text{z+x+2y}  \\ \end{matrix} \right|\text{=2}{{\left( \text{x+y+z} \right)}^{\text{3}}}\].

12. By using properties of determinants, show that:\[\mathbf{\left| \begin{matrix} \text{1} & \text{x} & {{\text{x}}^{\text{2}}}  \\ {{\text{x}}^{\text{2}}} & \text{1} & \text{x}  \\ \text{x} & {{\text{x}}^{\text{2}}} & \text{1}  \\ \end{matrix} \right|\text{=}{{\left( \text{1-}{{\text{x}}^{\text{3}}} \right)}^{\text{2}}}}\].

Ans:Let \[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{1} & \text{x} & {{\text{x}}^{\text{2}}}  \\ {{\text{x}}^{\text{2}}} & \text{1} & \text{x}  \\ \text{x} & {{\text{x}}^{\text{2}}} & \text{1}  \\ \end{matrix} \right|\]

Applying \[{{R}_{1}}\to {{R}_{1}}\text{+}{{\text{R}}_{2}}\text{+}{{\text{R}}_{3}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{1+x+}{{\text{x}}^{\text{2}}} & \text{1+x+}{{\text{x}}^{\text{2}}} & \text{1+x+}{{\text{x}}^{\text{2}}}  \\ {{\text{x}}^{\text{2}}} & \text{1} & \text{x}  \\ \text{x} & {{\text{x}}^{\text{2}}} & \text{1}  \\ \end{matrix} \right|\]

Taking out $\left( \text{1+x+}{{\text{x}}^{\text{2}}} \right)$ common from \[{{R}_{1}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left( \text{1+x+}{{\text{x}}^{\text{2}}} \right)\left| \begin{matrix} 1 & 1 & 1  \\ {{\text{x}}^{\text{2}}} & \text{1} & \text{x}  \\ \text{x} & {{\text{x}}^{\text{2}}} & \text{1}  \\ \end{matrix} \right|\]

Applying \[{{\text{C}}_{2}}\to {{C}_{2}}-{{\text{C}}_{1}}\] and \[{{\text{C}}_{3}}\to {{C}_{3}}-{{C}_{1}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left( \text{1+x+}{{\text{x}}^{\text{2}}} \right)\left| \begin{matrix} \text{1} & \text{0} & \text{0}  \\ {{\text{x}}^{\text{2}}} & \text{1-}{{\text{x}}^{\text{2}}} & \text{x-}{{\text{x}}^{\text{2}}}  \\ \text{x} & {{\text{x}}^{\text{2}}}\text{-x} & \text{1-x}  \\ \end{matrix} \right|\]

Taking out $\left( 1-x \right)$ common from \[{{C}_{2}}\] and \[{{C}_{3}}\] respectively,

\[\Rightarrow \Delta \text{=}\left( \text{1+x+}{{\text{x}}^{\text{2}}} \right)\left( \text{1-x} \right)\left( \text{1-x} \right)\left| \begin{matrix} \text{1} & \text{0} & \text{0}  \\ {{\text{x}}^{\text{2}}} & \text{1+x} & \text{x}  \\ \text{x} & \text{-x} & \text{1}  \\ \end{matrix} \right|\]

\[\Rightarrow \Delta \text{=}\left( \text{1-}{{\text{x}}^{\text{3}}} \right)\left( \text{1-x} \right)\left| \begin{matrix} \text{1} & \text{0} & \text{0}  \\ {{\text{x}}^{\text{2}}} & \text{1+x} & \text{x}  \\ \text{x} & \text{-x} & \text{1}  \\ \end{matrix} \right|\]

Expanding along \[{{R}_{1}}\].

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left( \text{1-}{{\text{x}}^{\text{3}}} \right)\left( 1-x \right)\left( \text{1} \right)\left| \begin{matrix} \text{1+x} & \text{x}  \\ -x & \text{1}  \\ \end{matrix} \right|\]

\[\Rightarrow \Delta \text{=}\left( \text{1-}{{\text{x}}^{\text{3}}} \right)\left( \text{1-x} \right)\left( \text{1+x+}{{\text{x}}^{\text{2}}} \right)\]

\[\Rightarrow \Delta \text{=}\left( \text{1-}{{\text{x}}^{\text{3}}} \right)\left( \text{1-}{{\text{x}}^{\text{3}}} \right)\]

\[\therefore \Delta \text{=}{{\left( \text{1-}{{\text{x}}^{\text{3}}} \right)}^{\text{2}}}\]

Hence,\[\left| \begin{matrix} \text{1} & \text{x} & {{\text{x}}^{\text{2}}}  \\ {{\text{x}}^{\text{2}}} & \text{1} & \text{x}  \\ \text{x} & {{\text{x}}^{\text{2}}} & \text{1}  \\ \end{matrix} \right|\text{=}{{\left( \text{1-}{{\text{x}}^{\text{3}}} \right)}^{\text{2}}}\].

13. By using properties of determinants, show that: \[\mathbf{\left| \begin{matrix} \text{1+}{{\text{a}}^{\text{2}}}\text{-}{{\text{b}}^{\text{2}}} & \text{2ab} & \text{-2b}  \\ \text{2ab} & \text{1-}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}} & \text{2a}  \\ \text{2b} & \text{-2a} & \text{1-}{{\text{a}}^{\text{2}}}\text{-}{{\text{b}}^{\text{2}}}  \\ \end{matrix} \right|\text{=}{{\left( \text{1+}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}} \right)}^{\text{3}}}}\].

Ans: Let \[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{1+}{{\text{a}}^{\text{2}}}\text{-}{{\text{b}}^{\text{2}}} & \text{2ab} & \text{-2b}  \\ \text{2ab} & \text{1-}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}} & \text{2a}  \\ \text{2b} & \text{-2a} & \text{1-}{{\text{a}}^{\text{2}}}\text{-}{{\text{b}}^{\text{2}}}  \\ \end{matrix} \right|\]

Applying the row operations, \[{{R}_{1}}\to {{R}_{1}}\text{+b}{{R}_{3}}\] and \[{{R}_{2}}\to {{R}_{2}}-{{R}_{3}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{1+}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}} & \text{0} & \text{-b}\left( \text{1+}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}} \right)  \\ \text{0} & \text{1+}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}} & \text{a}\left( \text{1+}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}} \right)  \\ \text{2b} & \text{-2a} & \text{1-}{{\text{a}}^{\text{2}}}\text{-}{{\text{b}}^{\text{2}}}  \\ \end{matrix} \right|\]

Taking out $\left( \text{1+}{{\text{a}}^{2}}\text{+}{{\text{b}}^{\text{2}}} \right)$ common from \[{{R}_{1}}\] and \[{{R}_{2}}\] respectively,

\[\Rightarrow \Delta \text{=}{{\left( \text{1+}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}} \right)}^{2}}\left| \begin{matrix} \text{1} & \text{0} & \text{-b}  \\ \text{0} & \text{1} & \text{a}  \\ \text{2b} & \text{-2a} & \text{1-}{{\text{a}}^{\text{2}}}\text{-}{{\text{b}}^{\text{2}}}  \\ \end{matrix} \right|\]

Expanding along \[{{R}_{1}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left( \text{1+}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}} \right)\left[ \left( \text{1} \right)\left| \begin{matrix} \text{1} & \text{a}  \\ -\text{2a} & \text{1}-{{\text{a}}^{\text{2}}}-{{\text{b}}^{\text{2}}}  \\ \end{matrix} \right|-b\left| \begin{matrix} \text{0} & \text{1}  \\ \text{2b} & -2a  \\ \end{matrix} \right| \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}{{\left( \text{1+}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}} \right)}^{2}}\left[ \text{1-1-}{{\text{a}}^{\text{2}}}\text{-}{{\text{b}}^{\text{2}}}\text{+2}{{\text{a}}^{\text{2}}}\text{-b}\left( \text{-2b} \right) \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}{{\left( \text{1+}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}} \right)}^{2}}\left( \text{1+}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}} \right)\]

\[\therefore \text{ }\!\!\Delta\!\!\text{ =}{{\left( \text{1+}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}} \right)}^{\text{3}}}\]

Hence,\[\left| \begin{matrix} \text{1+}{{\text{a}}^{\text{2}}}\text{-}{{\text{b}}^{\text{2}}} & \text{2ab} & \text{-2b}  \\ \text{2ab} & \text{1-}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}} & \text{2a}  \\ \text{2b} & \text{-2a} & \text{1-}{{\text{a}}^{\text{2}}}\text{-}{{\text{b}}^{\text{2}}}  \\ \end{matrix} \right|\text{=}{{\left( \text{1+}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}} \right)}^{\text{3}}}\].



14. By using properties of determinants, show that:

\[\mathbf{\left| \begin{matrix} {{\text{a}}^{\text{2}}}\text{+1} & \text{ab} & \text{ac}  \\ \text{ab} & {{\text{b}}^{\text{2}}}\text{+1} & \text{bc}  \\ \text{ca} & \text{cb} & {{\text{c}}^{\text{2}}}\text{+1}  \\ \end{matrix} \right|\text{=1+}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}}\text{+}{{\text{c}}^{\text{2}}}}\].

Ans: Let \[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} {{\text{a}}^{\text{2}}}\text{+1} & \text{ab} & \text{ac}  \\ \text{ab} & {{\text{b}}^{\text{2}}}\text{+1} & \text{bc}  \\ \text{ca} & \text{cb} & {{\text{c}}^{\text{2}}}\text{+1}  \\ \end{matrix} \right|\]

Taking out \[\text{a,b}\] and \[\text{c}\] from \[{{R}_{1}}\] , \[{{R}_{2}}\] and \[{{R}_{3}}\] respectively

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =abc}\left| \begin{matrix} \text{a+}\dfrac{\text{1}}{\text{a}} & \text{b} & \text{c}  \\ \text{a} & \text{b+}\dfrac{\text{1}}{\text{b}} & \text{c}  \\ \text{a} & \text{b} & \text{c+}\dfrac{\text{1}}{\text{c}}  \\ \end{matrix} \right|\]

Applying the row operations \[{{\text{R}}_{2}}\to {{\text{R}}_{2}}-{{\text{R}}_{1}}\] and \[{{\text{R}}_{3}}\to {{\text{R}}_{3}}-{{\text{R}}_{1}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =abc}\left| \begin{matrix} \text{a+}\dfrac{\text{1}}{\text{a}} & \text{b} & \text{c}  \\ \text{-}\dfrac{\text{1}}{\text{a}} & \dfrac{\text{1}}{\text{b}} & \text{0}  \\ \text{-}\dfrac{\text{1}}{\text{a}} & \text{0} & \dfrac{\text{1}}{\text{c}}  \\ \end{matrix} \right|\]

Applying \[{{\text{C}}_{1}}\to \text{a}{{\text{C}}_{1}}\] , \[{{\text{C}}_{2}}\to \text{b}{{\text{C}}_{2}}\] and \[{{\text{C}}_{3}}\to \text{c}{{\text{C}}_{3}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =abc }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{abc}}\left| \begin{matrix} {{\text{a}}^{\text{2}}}\text{+1} & {{\text{b}}^{\text{2}}} & {{\text{c}}^{\text{2}}}  \\ \text{-1} & \text{1} & \text{0}  \\ \text{-1} & \text{0} & \text{1}  \\ \end{matrix} \right|\]

Expanding along \[{{\text{C}}_{3}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =-1}\left| \begin{matrix} {{\text{b}}^{\text{2}}} & {{\text{c}}^{\text{2}}}  \\ \text{1} & \text{0}  \\ \end{matrix} \right|\text{+1}\left| \begin{matrix} {{\text{a}}^{\text{2}}}\text{+1} & {{\text{b}}^{\text{2}}}  \\ \text{-1} & \text{1}  \\ \end{matrix} \right|\]

\[\Rightarrow \Delta \text{=1}\left( \text{-}{{\text{c}}^{\text{2}}} \right)\text{+}\left( {{\text{a}}^{\text{2}}}\text{+1+}{{\text{b}}^{\text{2}}} \right)\]

\[\therefore \Delta \text{=1+}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}}\text{+}{{\text{c}}^{\text{2}}}\]

Hence, \[\left| \begin{matrix} {{\text{a}}^{\text{2}}}\text{+1} & \text{ab} & \text{ac}  \\ \text{ab} & {{\text{b}}^{\text{2}}}\text{+1} & \text{bc}  \\ \text{ca} & \text{cb} & {{\text{c}}^{\text{2}}}\text{+1}  \\ \end{matrix} \right|\text{=1+}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}}\text{+}{{\text{c}}^{\text{2}}}\].


15. Choose the correct answer. Let \[\text{A}\] be a square matrix of order \[\text{3 }\!\!\times\!\!\text{ 3}\] , then 

  1. \[\left| \text{kA} \right|\] is equal to \[\text{k}\left| \text{A} \right|\]

  2. \[{{\text{k}}^{\text{2}}}\left| \text{A} \right|\]

  3. \[{{\text{k}}^{\text{3}}}\left| \text{A} \right|\]

  4. \[\text{3k}\left| \text{A} \right|\]

Ans: Since, \[\text{A}\] is a square matrix of order \[\text{3 }\!\!\times\!\!\text{ 3}\].

Let us suppose\[\text{A=}\left[ \begin{matrix} {{\text{a}}_{1}} & {{\text{b}}_{1}} & {{\text{c}}_{1}}  \\ {{\text{a}}_{2}} & {{\text{b}}_{2}} & {{\text{c}}_{2}}  \\ {{\text{a}}_{3}} & {{\text{b}}_{3}} & {{\text{c}}_{3}}  \\ \end{matrix} \right]\]

Thus, \[\text{kA=}\left[ \begin{matrix} \text{k}{{\text{a}}_{1}} & \text{k}{{\text{b}}_{1}} & \text{k}{{\text{c}}_{1}}  \\ \text{k}{{\text{a}}_{2}} & \text{k}{{\text{b}}_{2}} & \text{k}{{\text{c}}_{2}}  \\ \text{k}{{\text{a}}_{3}} & \text{k}{{\text{b}}_{3}} & \text{k}{{\text{c}}_{3}}  \\ \end{matrix} \right]\]

\[\therefore \left| \text{kA} \right|\text{=}\left| \begin{matrix} \text{k}{{\text{a}}_{1}} & \text{k}{{\text{b}}_{1}} & \text{k}{{\text{c}}_{1}}  \\ \text{k}{{\text{a}}_{2}} & \text{k}{{\text{b}}_{2}} & \text{k}{{\text{c}}_{2}}  \\ \text{k}{{\text{a}}_{3}} & \text{k}{{\text{b}}_{3}} & \text{k}{{\text{c}}_{3}}  \\ \end{matrix} \right|\]

Taking out $\left( k \right)$ common from each row, we have:

 \[\Rightarrow \left| kA \right|\text{=}{{\text{k}}^{\text{3}}}\left| \begin{matrix} {{\text{a}}_{1}} & {{\text{b}}_{1}} & {{\text{c}}_{1}}  \\ {{\text{a}}_{2}} & {{\text{b}}_{2}} & {{\text{c}}_{2}}  \\ {{\text{a}}_{3}} & {{\text{b}}_{3}} & {{\text{c}}_{3}}  \\ \end{matrix} \right|\]       

\[\therefore \left| \text{kA} \right|\text{=}{{\text{k}}^{\text{3}}}\left| \text{A} \right|\]

Hence, B. \[{{\text{k}}^{\text{3}}}\left| \text{A} \right|\] is the correct option.


16. Which of the following is correct?

  1. Determinant is a square matrix.

  2. Determinant is a number associated to a matrix.

  3. Determinant is a number associated to a square matrix.

  4. None of these.

Ans: For every square matrix, \[\text{A=}\left[ {{\text{a}}_{ij}} \right]\] of order \[\text{n}\], we can determine or associate a value which is termed as determinant of square matrix $A$, where \[{{\text{a}}_{ij}}\text{=}{{\left( \text{i,j} \right)}^{\text{th}}}\] element of \[\text{A}\] .

Thus, the determinant is a number associated to a square matrix.

Hence, C. Determinant is a number associated to a square matrix is the correct option.


Exercise (4.3)

1. Find area of the triangle with vertices at the point given in each of the following:

  1. \[\mathbf{\left( \text{1,0} \right)\text{,}\left( \text{6,0} \right)\text{,}\left( \text{4,3} \right)}\]

Ans: Given vertices, \[\left( \text{1,0} \right)\text{,}\left( \text{6,0} \right)\text{,}\left( \text{4,3} \right)\]

We know, if we have three points $\left( {{x}_{1}},{{y}_{1}} \right)$, $\left( {{x}_{2}},{{y}_{2}} \right)$ and $\left( {{x}_{3}},{{y}_{3}} \right)$, then the area of the triangle is given by,

$\Delta =\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1  \\ {{x}_{2}} & {{y}_{2}} & 1  \\ {{x}_{3}} & {{y}_{3}} & 1  \\ \end{matrix} \right|$

Thus, the area of the triangle is given by,

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left| \begin{matrix} \text{1} & \text{0} & \text{1}  \\ \text{6} & \text{0} & \text{1}  \\ \text{4} & \text{3} & \text{1}  \\ \end{matrix} \right|\]

\[\Rightarrow \Delta =\dfrac{\text{1}}{\text{2}}\left[ \text{1}\left( \text{0-3} \right)\text{-0}\left( \text{6-4} \right)\text{+1}\left( \text{18-0} \right) \right]\]

\[\Rightarrow \Delta \text{=}\dfrac{\text{1}}{\text{2}}\left[ \text{-3+18} \right]\]

\[\Rightarrow \Delta \text{=}\dfrac{\text{15}}{\text{2}}\] square units

$\therefore $Area of the triangle with vertices \[\left( \text{1,0} \right)\text{,}\left( \text{6,0} \right)\text{,}\left( \text{4,3} \right)\] is \[\dfrac{\text{15}}{\text{2}}\] square units.

  1. \[\mathbf{\left( \text{2,7} \right)\text{,}\left( \text{1,1} \right)\text{,}\left( \text{10,8} \right)}\]

Ans: Given vertices, \[\left( \text{2,7} \right)\text{,}\left( \text{1,1} \right)\text{,}\left( \text{10,8} \right)\]

We know, if we have three points $\left( {{x}_{1}},{{y}_{1}} \right)$, $\left( {{x}_{2}},{{y}_{2}} \right)$ and $\left( {{x}_{3}},{{y}_{3}} \right)$, then the area of the triangle is given by,

$\Delta =\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1  \\ {{x}_{2}} & {{y}_{2}} & 1  \\ {{x}_{3}} & {{y}_{3}} & 1  \\ \end{matrix} \right|$

Thus, the area of the triangle is given by,

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left| \begin{matrix} \text{2} & \text{7} & \text{1}  \\ \text{1} & \text{1} & \text{1}  \\ \text{10} & \text{8} & \text{1}  \\ \end{matrix} \right|\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left[ \text{2}\left( \text{1-8} \right)\text{-7}\left( \text{1-10} \right)\text{+1}\left( \text{8-10} \right) \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left[ \text{2}\left( \text{-7} \right)\text{-7}\left( \text{-9} \right)\text{+1}\left( \text{-2} \right) \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left[ \text{-16+63} \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{47}}{\text{2}}\] square units

$\therefore $Area of the triangle with vertices \[\left( \text{2,7} \right)\text{,}\left( \text{1,1} \right)\text{,}\left( \text{10,8} \right)\] is \[\dfrac{47}{\text{2}}\] square units.

  1. \[\mathbf{\text{(-2,-3),}\left( \text{3,2} \right)\text{,(-1,-8)}}\]

Ans: Given vertices, \[\text{(-2,-3),}\left( \text{3,2} \right)\text{,(-1,-8)}\]

We know, if we have three points $\left( {{x}_{1}},{{y}_{1}} \right)$, $\left( {{x}_{2}},{{y}_{2}} \right)$ and $\left( {{x}_{3}},{{y}_{3}} \right)$, then the area of the triangle is given by,

$\Delta =\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1  \\ {{x}_{2}} & {{y}_{2}} & 1  \\ {{x}_{3}} & {{y}_{3}} & 1  \\ \end{matrix} \right|$

Thus, the area of the triangle with vertices \[\text{(-2,-3),}\left( \text{3,2} \right)\text{,(-1,-8)}\] is given by,

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left| \begin{matrix} \text{-2} & \text{-3} & \text{1}  \\ \text{3} & \text{2} & \text{1}  \\ \text{-1} & \text{-8} & \text{1}  \\ \end{matrix} \right|\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left[ \text{-2}\left( \text{2+8} \right)\text{+3}\left( \text{3+1} \right)\text{+1}\left( \text{-24+2} \right) \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left[ \text{-20+12-22} \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ }=-\dfrac{\text{30}}{\text{2}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =-15}\]

$\therefore $The area of the triangle with vertices \[\left( \text{2,7} \right)\text{,}\left( \text{1,1} \right)\text{,}\left( \text{10,8} \right)\]is \[\left| \text{-15} \right|\text{=15}\] square units.


2. Show that points \[\mathbf{\text{A}\left( \text{a,b+c} \right)\text{,B}\left( \text{b,c +a} \right)\text{,C}\left( \text{c,a +b} \right)}\] are collinear.

Ans: To show that the points \[\text{A}\left( \text{a,b+c} \right)\text{,B}\left( \text{b,c +a} \right)\text{,C}\left( \text{c,a +b} \right)\] are collinear, the area of the triangle formed by these points as vertices should be zero.

$\therefore $ Area of \[\text{ }\!\!\Delta\!\!\text{ ABC}\] is given by,

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left| \begin{matrix} \text{a} & \text{b+a} & \text{1}  \\ \text{b} & \text{c+a} & \text{1}  \\ \text{c} & \text{a+b} & \text{1}  \\ \end{matrix} \right|\]

Applying the row operations \[{{\text{R}}_{2}}\to {{\text{R}}_{2}}-{{\text{R}}_{1}}\] and \[{{\text{R}}_{3}}\to {{\text{R}}_{3}}-{{\text{R}}_{1}}\]

\[\Rightarrow \Delta \text{=}\dfrac{\text{1}}{\text{2}}\left| \begin{matrix} \text{a} & \text{b+c} & \text{1}  \\ \text{b-a} & \text{a-b} & \text{0}  \\ \text{c-a} & \text{a-c} & \text{0}  \\ \end{matrix} \right|\]

Taking out $\left( a-b \right)$ and $\left( c-a \right)$ common from \[R_2\] and \[R_3\] respectively,

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left( \text{a-b} \right)\left( \text{c-a} \right)\left| \begin{matrix} \text{a} & \text{b+c} & \text{1}  \\ \text{-1} & \text{1} & \text{0}  \\ \text{1} & \text{-1} & \text{0}  \\ \end{matrix} \right|\]

Applying the row operation \[{{\text{R}}_{3}}\to {{\text{R}}_{3}}\text{+}{{\text{R}}_{2}}\]

\[\Rightarrow \Delta \text{=}\dfrac{\text{1}}{\text{2}}\left( \text{a-b} \right)\left( \text{c-a} \right)\left| \begin{matrix} \text{a} & \text{b+c} & \text{1}  \\ \text{-1} & \text{1} & \text{0}  \\ \text{0} & \text{0} & \text{0}  \\ \end{matrix} \right|\]

Since all the elements of the last row of the matrix are zero then the value of the determinant will be $0$.

\[\therefore \Delta \text{=0}\]                                           

Thus, the area of the triangle formed by points \[\text{A}\] , \[\text{B}\] and \[\text{C}\] is zero.

Hence, the points \[\text{A}\left( \text{a,b+c} \right)\text{,B}\left( \text{b,c +a} \right)\text{,C}\left( \text{c,a +b} \right)\] are collinear.


3. Find values of \[\mathbf{\text{k}}\] if area of triangle is \[\mathbf{\text{4}}\] square units and vertices are

  1. \[\mathbf{\left( \text{k,0} \right)\text{,}\left( \text{4,0} \right)\text{,}\left( \text{0,2} \right)}\]

Ans: Given vertices are \[\left( \text{k,0} \right)\text{,}\left( \text{4,0} \right)\text{,}\left( \text{0,2} \right)\].

We know, if we have three points $\left( {{x}_{1}},{{y}_{1}} \right)$, $\left( {{x}_{2}},{{y}_{2}} \right)$ and $\left( {{x}_{3}},{{y}_{3}} \right)$, then the area of the triangle is given by,

$\Delta =\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1  \\ {{x}_{2}} & {{y}_{2}} & 1  \\ {{x}_{3}} & {{y}_{3}} & 1  \\ \end{matrix} \right|$

Thus, the area of the triangle is given by,

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left| \begin{matrix} \text{k} & \text{0} & \text{1}  \\ \text{4} & \text{0} & \text{1}  \\ \text{0} & \text{2} & \text{1}  \\ \end{matrix} \right|\]

\[\Rightarrow \Delta \text{=}\dfrac{\text{1}}{\text{2}}\left[ \text{k}\left( \text{0-2} \right)\text{-0}\left( \text{4-0} \right)\text{+1}\left( \text{8-0} \right) \right]\]

\[\Rightarrow \Delta \text{=}\dfrac{\text{1}}{\text{2}}\left[ \text{-2k+8} \right]\]

\[\therefore \Delta =-k+4\]

Since the area is given to be \[\text{4}\] square units, thus

$-k+4=\pm 4$

When \[-k+4=-4\]

\[\therefore k=8\].

When \[-k+4=4\]

\[\therefore k=0\].

Hence, \[\text{k=0,8}\].

  1. \[\mathbf{\text{(-2,0),}\left( \text{0,4} \right)\text{,}\left( \text{0,k} \right)}\]

Ans: Given vertices are \[\text{(-2,0),}\left( \text{0,4} \right)\text{,}\left( \text{0,k} \right)\].

We know, if we have three points $\left( {{x}_{1}},{{y}_{1}} \right)$, $\left( {{x}_{2}},{{y}_{2}} \right)$ and $\left( {{x}_{3}},{{y}_{3}} \right)$, then the area of the triangle is given by,

$\Delta =\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1  \\ {{x}_{2}} & {{y}_{2}} & 1  \\ {{x}_{3}} & {{y}_{3}} & 1  \\ \end{matrix} \right|$

The area of the triangle is given by,

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left| \begin{matrix} -2 & \text{0} & \text{1}  \\ \text{0} & \text{4} & \text{1}  \\ \text{0} & \text{k} & \text{1}  \\ \end{matrix} \right|\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left[ -2\left( 4-k \right) \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ }=k-4\]

Since the area is given to be \[\text{4}\] square units, thus

\[k-4=\pm 4\]

When \[k-4=-4\]

\[\therefore k=0\].

When \[k-4=4\]

\[\therefore k=8\].

Hence, \[k=0,8\].


3. Determine the following:

  1. Find equation of line joining \[\mathbf{\left( \text{1,2} \right)\] and \[\left( \text{3,6} \right)}\] using determinants.

Ans: Let us assume a point, \[\text{P}\left( \text{x, y} \right)\] on the line joining points \[\text{A}\left( \text{1,2} \right)\] and \[\text{B}\left( \text{3,6} \right)\] .

Then, the point \[\text{A}\],$B$ and $P$ are collinear.

Thus, the area of triangle \[\text{ABP}\] will be zero.

\[\therefore \dfrac{\text{1}}{\text{2}}\left| \begin{matrix} \text{1} & \text{2} & \text{1}  \\ \text{3} & \text{6} & \text{1}  \\ \text{x} & \text{y} & \text{1}  \\ \end{matrix} \right|\text{=0}\]

\[\Rightarrow \dfrac{\text{1}}{\text{2}}\left[ \text{1}\left( \text{6-y} \right)\text{-2}\left( \text{3-x} \right)\text{+1}\left( \text{3y-6x} \right) \right]\text{=0}\]

\[\Rightarrow \text{6-y-6+2x+3y-6x=0}\]

\[\Rightarrow \text{2y-4x=0}\]

\[\Rightarrow \text{y=2x}\]

$\therefore $ The equation of the line joining the given points is \[y=2x\] .

  1. Find equation of line joining \[\mathbf{\left( \text{3,1} \right)\] and \[\left( \text{9,3} \right)}\] using determinants.

Ans: Let us assume a point, \[\text{P}\left( \text{x, y} \right)\] on the line joining points \[\text{A}\left( \text{3,1} \right)\] and \[\text{B}\left( \text{9,3} \right)\]. 

Then, the point \[\text{A}\],$B$ and $P$ are collinear.

Thus, the area of the triangle \[\text{ABP}\] will be zero.

\[\therefore \dfrac{1}{2}\left|\begin{matrix} \text{3}&\text{1}&\text{1}\\ \text{9}&\text{3}&\text{1}\\ \text{x}&\text{y}&\text{1}\\ \end{matrix}  \right|=0\]

\[\Rightarrow \dfrac{1}{2}\left[ 3(3-\text{y})-1(9-\text{x})+1(9\text{y}-3\text{x})\right]=0\]

\[\Rightarrow 9-3\text{y}-9+\text{x}+9\text{y}-3\text{x}=0\]

\[\Rightarrow 6\text{y}-2\text{x}=0\]

\[\Rightarrow \text{x}-3\text{y}=0\]

\[\therefore\] The equation of the line joining the given points is \[\text{x}-3\text{y}=0\].


4. If the area of triangle is \[\mathbf{\text{35}}\] square units with vertices \[\mathbf{\text{(2,-6)}}\] , \[\mathbf{\left( \text{5,4} \right)}\] and \[\mathbf{\left( \text{k,4} \right)}\] . Then \[\mathbf{\text{k}}\] is

  1. \[\mathbf{\text{12}}\]

  2. \[\mathbf{\text{-2}}\]

  3. \[\mathbf{\text{-12,-2}}\]

  4. \[\mathbf{\text{12,-2}}\]

Ans: Given vertices, \[\text{(2,-6)}\],\[\left( \text{5,4} \right)\] and \[\left( \text{k,4} \right)\]

We know, if we have three points $\left( {{x}_{1}},{{y}_{1}} \right)$, $\left( {{x}_{2}},{{y}_{2}} \right)$ and $\left( {{x}_{3}},{{y}_{3}} \right)$, then the area of the triangle is given by,

$\Delta =\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1  \\ {{x}_{2}} & {{y}_{2}} & 1  \\ {{x}_{3}} & {{y}_{3}} & 1  \\ \end{matrix} \right|$

The area of the triangle is given by,

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left| \begin{matrix} \text{2} & \text{-6} & \text{1}  \\ \text{5} & \text{4} & \text{1}  \\ \text{k} & \text{4} & \text{1}  \\ \end{matrix} \right|\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left[ \text{2}\left( \text{4-4} \right)\text{+6}\left( \text{5-k} \right)\text{+1}\left( \text{20-4k} \right) \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left[ \text{50-10k} \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =25-5k}\]

Given, the area of the triangle is \[\text{35}\] square units .

Thus, we have:

\[\Rightarrow 25-5k=\pm 35\]

\[\Rightarrow 5\left( 5-k \right)=\pm 35\]

\[\Rightarrow 5-k=\pm 7\].

When \[5-k=7\]

\[\therefore k=-2\].

When \[5-k=-7\]

\[\therefore k=12\].

Hence, \[k=12,-2\] .

Thus, D. \[12,-2\] is the correct option.


Exercise (4.4)

1. Write Minors and Cofactors of the elements of following determinants:

  1. \[\mathbf{\left| \begin{matrix} \text{2} & \text{-4}  \\ \text{0} & \text{3}  \\ \end{matrix} \right|}\]

Ans: Given, \[\left| \begin{matrix} \text{2} & \text{-4}  \\ \text{0} & \text{3}  \\ \end{matrix} \right|\]

Minor of an element is termed as the determinant obtained by removing the row and the column in which that element is present.

Minor of element \[{{\text{a}}_{ij}}\] is denoted by \[{{M}_{ij}}\],where

$i$ and $j$ denotes the row and the column of the determinant respectively.

\[\therefore {{\text{M}}_{11}}\text{=3}\]

\[{{\text{M}}_{12}}\text{=0}\]

\[{{\text{M}}_{21}}\text{=-4}\]

\[{{\text{M}}_{22}}\text{=2}\]

Cofactor of an element is termed as the determinant obtained by removing the row and the column in which that element is present preceded by a negative or a positive sign based on the position of the element.

Thus,

Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\]

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{\text{1+1}}}{{\text{M}}_{11}}\]

\[\Rightarrow {{\text{A}}_{11}}={{\left( \text{-1} \right)}^{\text{2}}}\left( \text{3} \right)\]

\[\therefore {{\text{A}}_{11}}\text{=3}\]

Similarly,

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{\text{1+2}}}{{\text{M}}_{12}}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{\text{3}}}\left( \text{0} \right)\]

\[\therefore {{\text{A}}_{12}}\text{=0}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{\text{2+1}}}{{M}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{\text{3}}}\left( \text{-4} \right)\]

\[\therefore {{\text{A}}_{21}}\text{=4}\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{\text{2+2}}}{{M}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{\text{4}}}\left( \text{2} \right)\]

\[\therefore {{\text{A}}_{22}}\text{=2}\]

  1. \[\mathbf{\left| \begin{matrix} \text{a} & \text{c}  \\ \text{b} & \text{d}  \\ \end{matrix} \right|}\]

Ans: Given, \[\left| \begin{matrix} \text{a} & \text{c}  \\ \text{b} & \text{d}  \\ \end{matrix} \right|\]

Minor of element \[{{\text{a}}_{ij}}\] is denoted by \[{{M}_{ij}}\].

\[\therefore {{\text{M}}_{11}}\text{=d}\]

\[{{\text{M}}_{12}}\text{=b}\]

\[{{\text{M}}_{21}}\text{=c}\]

\[{{\text{M}}_{22}}\text{=a}\]

Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\]

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{\text{1+1}}}{{\text{M}}_{11}}\]

\[\Rightarrow {{\text{A}}_{11}}={{\left( \text{-1} \right)}^{\text{2}}}\left( d \right)\]

\[\therefore {{\text{A}}_{11}}\text{=d}\]

Similarly,

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{\text{1+2}}}{{\text{M}}_{12}}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{\text{3}}}\left( b \right)\]

\[\therefore {{\text{A}}_{12}}\text{=}-b\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{\text{2+1}}}{{M}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{\text{3}}}\left( c \right)\]

\[\therefore {{\text{A}}_{21}}\text{=}-\text{c}\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{\text{2+2}}}{{M}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{\text{4}}}\left( a \right)\]

\[\therefore {{\text{A}}_{22}}\text{=a}\]


2. Write Minors and Cofactors of the elements of following determinants:

  1. \[\mathbf{\left| \begin{matrix} \text{1} & \text{0} & \text{0}  \\ \text{0} & \text{1} & \text{0}  \\ \text{0} & \text{0} & \text{1}  \\ \end{matrix} \right|}\]

Ans: Given determinant, \[\left| \begin{matrix} \text{1} & \text{0} & \text{0}  \\ \text{0} & \text{1} & \text{0}  \\ \text{0} & \text{0} & \text{1}  \\ \end{matrix} \right|\].

Minor of element \[{{\text{a}}_{ij}}\] is denoted by \[{{M}_{ij}}\].

\[\therefore {{\text{M}}_{11}}\text{=}\left| \begin{matrix} \text{1} & \text{0}  \\ \text{0} & \text{1}  \\ \end{matrix} \right|\text{=1}\]

\[\Rightarrow {{\text{M}}_{12}}\text{=}\left| \begin{matrix} \text{0} & \text{0}  \\ \text{0} & \text{1}  \\ \end{matrix} \right|\text{=0}\]

\[\Rightarrow {{\text{M}}_{13}}\text{=}\left| \begin{matrix} \text{0} & \text{1}  \\ \text{0} & \text{0}  \\ \end{matrix} \right|\text{=0}\]

\[\Rightarrow {{\text{M}}_{21}}\text{=}\left| \begin{matrix} \text{0} & \text{0}  \\ \text{0} & \text{1}  \\ \end{matrix} \right|\text{=0}\]

\[\Rightarrow {{\text{M}}_{22}}\text{=}\left| \begin{matrix} \text{1} & \text{0}  \\ \text{0} & \text{1}  \\ \end{matrix} \right|\text{=1}\]

\[\Rightarrow {{\text{M}}_{23}}\text{=}\left| \begin{matrix} \text{1} & \text{0}  \\ \text{0} & \text{0}  \\ \end{matrix} \right|\text{=0}\]

\[\Rightarrow {{\text{M}}_{31}}\text{=}\left| \begin{matrix} \text{0} & \text{0}  \\ \text{1} & \text{0}  \\ \end{matrix} \right|\text{=0}\]

\[\Rightarrow {{\text{M}}_{32}}\text{=}\left| \begin{matrix} \text{1} & \text{0}  \\ \text{0} & \text{0}  \\ \end{matrix} \right|\text{=0}\]

\[\Rightarrow {{\text{M}}_{33}}\text{=}\left| \begin{matrix} \text{1} & \text{0}  \\ \text{0} & \text{1}  \\ \end{matrix} \right|\text{=1}\]

Cofactor of \[{{\text{a}}_{ij}}\] is \[{{A}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{M}_{ij}}\]. 

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}=1\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}=0\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}=0\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}=0\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}=1\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}=0\]

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}=0\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}=0\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}=1\]


  1. \[\mathbf{\left| \begin{matrix} \text{1} & \text{0} & \text{4}  \\ \text{3} & \text{5} & \text{-1}  \\ \text{0} & \text{1} & \text{2}  \\ \end{matrix} \right|}\]

Ans: Given determinant, \[\left| \begin{matrix} \text{1} & \text{0} & \text{4}  \\ \text{3} & \text{5} & \text{-1}  \\ \text{0} & \text{1} & \text{2}  \\ \end{matrix} \right|\]

Minor of element \[{{\text{a}}_{ij}}\] is denoted by \[{{M}_{ij}}\].

\[\Rightarrow {{\text{M}}_{11}}\text{=}\left| \begin{matrix} \text{5} & \text{-1}  \\ \text{1} & \text{2}  \\ \end{matrix} \right|\text{=10+1=11}\]

\[\Rightarrow {{\text{M}}_{12}}\text{=}\left| \begin{matrix} \text{3} & \text{-1}  \\ \text{0} & \text{2}  \\ \end{matrix} \right|\text{=6-0=6}\]

\[\Rightarrow {{\text{M}}_{13}}\text{=}\left| \begin{matrix} \text{3} & \text{5}  \\ \text{0} & \text{1}  \\ \end{matrix} \right|\text{=3-0=3}\]

\[\Rightarrow {{\text{M}}_{21}}\text{=}\left| \begin{matrix} \text{0} & \text{4}  \\ \text{1} & \text{2}  \\ \end{matrix} \right|\text{=0-4=-4}\]

\[\Rightarrow {{\text{M}}_{22}}\text{=}\left| \begin{matrix} \text{1} & \text{4}  \\ \text{0} & \text{2}  \\ \end{matrix} \right|\text{=2-0=2}\]

\[\Rightarrow {{\text{M}}_{23}}\text{=}\left| \begin{matrix} \text{1} & \text{0}  \\ \text{0} & \text{1}  \\ \end{matrix} \right|\text{=1-0=1}\]

\[\Rightarrow {{\text{M}}_{31}}\text{=}\left| \begin{matrix} \text{0} & \text{4}  \\ \text{5} & \text{-1}  \\ \end{matrix} \right|\text{=0-20=-20}\]

\[\Rightarrow {{\text{M}}_{32}}\text{=}\left| \begin{matrix} \text{1} & \text{4}  \\ \text{3} & \text{-1}  \\ \end{matrix} \right|\text{=-1-12=-13}\]

\[\Rightarrow {{\text{M}}_{33}}\text{=}\left| \begin{matrix} \text{1} & \text{0}  \\ \text{3} & \text{5}  \\ \end{matrix} \right|\text{=5-0=5}\]

Cofactor of \[{{\text{a}}_{ij}}\] is \[{{A}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{M}_{ij}}\]. 

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}=11\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}=6\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}=3\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}=-4\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}=2\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}=1\]

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}=-20\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}=-13\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}=5\]


3. Using Cofactors of elements of second row, evaluate \[\mathbf{\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{5} & \text{3} & \text{8}  \\ \text{2} & \text{0} & \text{1}  \\ \text{1} & \text{2} & \text{3}  \\ \end{matrix} \right|}\].

Ans: Given determinant, \[\left| \begin{matrix} \text{5} & \text{3} & \text{8}  \\ \text{2} & \text{0} & \text{1}  \\ \text{1} & \text{2} & \text{3}  \\ \end{matrix} \right|\]

Determining the minors and cofactors, we get:

\[\Rightarrow {{\text{M}}_{21}}\text{=}\left| \begin{matrix} \text{3} & \text{8}  \\ \text{2} & \text{3}  \\ \end{matrix} \right|=9-16=-7\]

\[\therefore {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{\text{2+1}}}{{M}_{21}}\text{=7}\]

\[\Rightarrow {{\text{M}}_{22}}\text{=}\left| \begin{matrix} \text{5} & \text{8}  \\ \text{1} & \text{3}  \\ \end{matrix} \right|=15-8=7\]

\[\therefore {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{\text{2+2}}}{{M}_{22}}\text{=7}\]

\[\Rightarrow {{\text{M}}_{23}}\text{=}\left| \begin{matrix} \text{5} & \text{3}  \\ \text{1} & \text{2}  \\ \end{matrix} \right|=10-3=7\]

\[\therefore {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{\text{2+1}}}{{M}_{23}}=-7\]

Since, \[\text{ }\!\!\Delta\!\!\text{ }\] is equal to the sum of the product of the elements of the second row with their corresponding cofactors.

\[\therefore \Delta ={{a}_{21}}{{A}_{21}}+{{a}_{22}}{{A}_{22}}+{{a}_{23}}{{A}_{23}}\]

\[\Rightarrow \Delta \text{=2}\left( \text{7} \right)\text{+0}\left( \text{7} \right)\text{+1}\left( \text{7} \right)\]

Hence, \[\Delta \text{=21}\].


4. Using Cofactors of elements of third column, evaluate  \[\mathbf{\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{1} & \text{x} & \text{yz}  \\ \text{1} & \text{y} & \text{zx}  \\ \text{1} & \text{z} & \text{xy}  \\ \end{matrix} \right|}\].

Ans: Given determinant, \[\left| \begin{matrix} \text{1} & \text{x} & \text{yz}  \\ \text{1} & \text{y} & \text{zx}  \\ \text{1} & \text{z} & \text{xy}  \\ \end{matrix} \right|\]

Determining the minors and cofactors, we get:

\[\Rightarrow {{\text{M}}_{13}}\text{=}\left| \begin{matrix} 1 & y  \\ 1 & z  \\ \end{matrix} \right|=z-y\]

\[\therefore {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{\text{1+3}}}{{M}_{13}}\text{=}\left( z-y \right)\]

\[\Rightarrow {{\text{M}}_{23}}\text{=}\left| \begin{matrix} 1 & x  \\ \text{1} & z  \\ \end{matrix} \right|=z-x\]

\[\therefore {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{\text{2+3}}}{{M}_{23}}\text{=}\left( x-z \right)\]

\[\Rightarrow {{\text{M}}_{33}}\text{=}\left| \begin{matrix} 1 & x  \\ \text{1} & y  \\ \end{matrix} \right|=y-x\]

\[\therefore {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{\text{3+3}}}{{M}_{33}}=\left( y-x \right)\]

Since, \[\text{ }\!\!\Delta\!\!\text{ }\] is equal to the sum of the product of the elements of the first row with their corresponding cofactors. 

\[\therefore \Delta ={{a}_{13}}{{A}_{13}}+{{a}_{23}}{{A}_{23}}+{{a}_{33}}{{A}_{33}}\]

\[\Rightarrow \Delta =yz\left( z-y \right)+zx\left( x-z \right)+xy\left( y-x \right)\]

\[\Rightarrow \Delta =y{{z}^{2}}-{{y}^{2}}z+{{x}^{2}}z-x{{z}^{2}}+x{{y}^{2}}-{{x}^{2}}y\]

\[\Rightarrow \Delta =\left( {{x}^{2}}z-{{y}^{2}}z \right)+\left( y{{z}^{2}}-x{{z}^{2}} \right)+\left( x{{y}^{2}}-{{x}^{2}}y \right)\]

\[\Rightarrow \Delta =\left( x-y \right)\left[ zx+zy-{{z}^{2}}-xy \right]\]

\[\Rightarrow \Delta =\left( x-y \right)\left[ z\left( x-z \right)+y\left( z-x \right) \right]\]

Thus, \[\text{ }\!\!\Delta\!\!\text{ =}\left( \text{x-y} \right)\left( \text{y-z} \right)\left( \text{z-x} \right)\].


5. For the matrices \[\mathbf{\text{A}}\] and \[\mathbf{\text{B}}\] , verify that \[\mathbf{\left( \text{AB} \right)\text{ }\!\!'\!\!\text{ =B }\!\!'\!\!\text{ A }\!\!'\!\!\text{ }}\] where

If $\mathbf{\Delta =\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}}  \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}}  \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}}  \\ \end{matrix} \right|}$ and ${{A}_{ij}}$ is Cofactors of ${{a}_{ij}}$ , then value of $\Delta $ is given by

  1. \[\mathbf{{{a}_{11}}{{A}_{31}}+{{a}_{12}}{{A}_{32}}+{{a}_{13}}{{A}_{33}}}\]

  2. \[\mathbf{{{a}_{11}}{{A}_{11}}+{{a}_{12}}{{A}_{21}}+{{a}_{13}}{{A}_{31}}}\]

  3. \[\mathbf{{{a}_{21}}{{A}_{11}}+{{a}_{22}}{{A}_{12}}+{{a}_{23}}{{A}_{13}}}\]

  4. \[\mathbf{{{a}_{11}}{{A}_{11}}+{{a}_{21}}{{A}_{21}}+{{a}_{31}}{{A}_{31}}}\]

Ans: It is given that $\Delta =\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}}  \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}}  \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}}  \\ \end{matrix} \right|$.

The value of $\Delta $ by expanding along first column is obtained as,

\[{{a}_{11}}\left( {{a}_{22}}.{{a}_{33}}-{{a}_{23}}.{{a}_{32}} \right)-{{a}_{21}}\left( {{a}_{12}}.{{a}_{33}}-{{a}_{13}}.{{a}_{32}} \right)+{{a}_{31}}\left( {{a}_{12}}.{{a}_{23}}-{{a}_{13}}.{{a}_{22}} \right)\] ….. 1

Now, the cofactor ${{A}_{ij}}$ of element ${{a}_{ij}}$ is given by ${{\left( -1 \right)}^{i+j}}{{M}_{ij}}$, where ${{M}_{ij}}$ is the minor. Minor is the determinant obtained by cancelling the ith row and jth column of the original matrix.

Now for element ${{a}_{11}}$, the minor is ${{M}_{11}}=\left| \begin{matrix} {{a}_{22}} & {{a}_{23}}  \\ {{a}_{32}} & {{a}_{33}}  \\ \end{matrix}\right|={{a}_{22}}.{{a}_{33}}-{{a}_{23}}.{{a}_{32}}$and the cofactor is ${{A}_{11}}={{\left( -1 \right)}^{1+1}}\left( {{a}_{22}}.{{a}_{33}}-{{a}_{23}}.{{a}_{32}} \right)\Rightarrow {{A}_{11}}=\left( {{a}_{22}}.{{a}_{33}}-{{a}_{23}}.{{a}_{32}} \right)$.

Next for element ${{a}_{21}}$, the minor is ${{M}_{21}}=\left| \begin{matrix} {{a}_{12}} & {{a}_{13}}  \\ {{a}_{32}} & {{a}_{33}}  \\ \end{matrix}\right|={{a}_{12}}.{{a}_{33}}-{{a}_{13}}.{{a}_{32}}$ and the cofactor is ${{A}_{21}}={{\left( -1 \right)}^{2+1}}\left( {{a}_{12}}.{{a}_{33}}-{{a}_{13}}.{{a}_{32}} \right)\Rightarrow {{A}_{21}}=-\left( {{a}_{12}}.{{a}_{33}}-{{a}_{13}}.{{a}_{32}} \right)$.

Next for element ${{a}_{31}}$, the minor is ${{M}_{31}}=\left| \begin{matrix} {{a}_{12}} & {{a}_{13}}  \\ {{a}_{22}} & {{a}_{23}}  \\ \end{matrix}\right|={{a}_{12}}.{{a}_{23}}-{{a}_{13}}.{{a}_{22}}$ and the cofactor is ${{A}_{31}}={{\left( -1 \right)}^{3+1}}\left( {{a}_{12}}.{{a}_{23}}-{{a}_{13}}.{{a}_{22}} \right)\Rightarrow {{A}_{31}}=\left( {{a}_{12}}.{{a}_{23}}-{{a}_{13}}.{{a}_{22}} \right)$.

Now substituting the terms as obtained from above computation in equation 1,

\[{{a}_{11}}{{A}_{11}}-{{a}_{21}}\left( -{{A}_{21}} \right)+{{a}_{31}}{{A}_{31}}\]

\[{{a}_{11}}{{A}_{11}}+{{a}_{21}}{{A}_{21}}+{{a}_{31}}{{A}_{31}}\]

This matches with option d.


Exercise (4.5)

1. Find the adjoint of each of the matrices. \[\mathbf{\left[ \begin{matrix} \text{1} & \text{2}  \\ \text{3} & \text{4}  \\ \end{matrix} \right]}\]

Ans: Let \[\text{A=}\left[ \begin{matrix} \text{1} & \text{2}  \\ \text{3} & \text{4}  \\ \end{matrix} \right]\]

Since, Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\].

Thus,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{A}_{11}}\text{=4}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{A}_{12}}=-3\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}-2\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=1\]

We know that adjoint of a matrix is the transpose of its cofactor matrix.

Thus,\[\text{adjA=}{{\left[ \begin{matrix} {{\text{A}}_{11}} & {{\text{A}}_{12}}  \\ {{\text{A}}_{21}} & {{\text{A}}_{22}}  \\ \end{matrix} \right]}^{T}}\]

\[\therefore adjA\text{=}\left[ \begin{matrix} \text{4} & \text{-2}  \\ \text{-3} & \text{1}  \\ \end{matrix} \right]\].


2. Find adjoint of each of the matrices \[\mathbf{\left[ \begin{matrix} \text{1} & \text{-1} & \text{2}  \\ \text{2} & \text{3} & \text{5}  \\ \text{-2} & \text{0} & \text{1}  \\ \end{matrix} \right]}\].

Ans: Let \[\text{A=}\left[ \begin{matrix} \text{1} & \text{-1} & \text{2}  \\ \text{2} & \text{3} & \text{5}  \\ \text{-2} & \text{0} & \text{1}  \\ \end{matrix} \right]\]

Since, Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\].

Thus,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{A}_{11}}\text{=}\left| \begin{matrix} \text{3} & \text{5}  \\ \text{0} & \text{1}  \\ \end{matrix} \right|=3-0=3\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{A}_{12}}=\left| \begin{matrix} \text{2} & \text{5}  \\ \text{-2} & \text{1}  \\ \end{matrix} \right|=-\left( \text{2+10} \right)=-12\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]

\[\Rightarrow {{A}_{13}}=\left| \begin{matrix} 2 & 3  \\ -2 & 0  \\ \end{matrix} \right|\text{=0+6=6}\]

Similarly,

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}\left| \begin{matrix} \text{-1} & \text{2}  \\ \text{0} & \text{1}  \\ \end{matrix} \right|=-\left( -1-0 \right)\text{=1}\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=\left| \begin{matrix} \text{1} & \text{2}  \\ -2 & \text{1}  \\ \end{matrix} \right|\text{=1+4=5}\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}\left| \begin{matrix} \text{1} & \text{-1}  \\ \text{-2} & 0  \\ \end{matrix} \right|\text{=}\left( 0-2 \right)\text{=2}\]

and

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]

\[\Rightarrow {{\text{A}}_{31}}=\left| \begin{matrix} -1 & \text{2}  \\ \text{2} & \text{5}  \\ \end{matrix} \right|=-5-4=-9\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]

\[\Rightarrow {{\text{A}}_{32}}=\left| \begin{matrix} \text{1} & \text{2}  \\ \text{2} & \text{5}  \\ \end{matrix} \right|=-\left( 5-4 \right)=-1\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]

\[\Rightarrow {{\text{A}}_{33}}=\left| \begin{matrix} \text{1} & \text{-1}  \\ \text{2} & \text{3}  \\ \end{matrix} \right|\text{=3+2=5}\].

We know that adjoint of a matrix is the transpose of its cofactor matrix.

Thus, \[\text{adjA=}{{\left[ \begin{matrix} A11 & A\text{12} & A\text{13}  \\ A\text{21} & {{\text{A}}_{\text{22}}} & A23  \\ A\text{31} & {{\text{A}}_{\text{32}}} & A\text{33}  \\ \end{matrix} \right]}^{T}}\text{=}\left[ \begin{matrix} \text{3} & -12 & 6  \\ 1 & \text{5} & 2  \\ -9 & -1 & \text{5}  \\ \end{matrix} \right]\]


3. Verify \[\mathbf{\text{A}\left( \text{adjA} \right)\text{=}\left( \text{adjA} \right)\text{A=}\left| \text{A} \right|I}\]. \[\mathbf{\left[ \begin{matrix} \text{2} & \text{3}  \\ \text{-4} & \text{-6}  \\ \end{matrix} \right]}\]

Ans: Given,\[\text{A=}\,\left[ \begin{matrix} \text{2} & \text{3}  \\ \text{-4} & \text{-6}  \\ \end{matrix} \right]\]

\[\therefore \left| \text{A} \right|=-12-\left( -12 \right)\]

\[\Rightarrow \left| \text{A} \right|=0\]

Hence, \[\left| A \right|I=0\left[ \begin{matrix} 1 & 0  \\ 0 & 1  \\ \end{matrix} \right]\]

\[\Rightarrow \left| A \right|I=\left[ \begin{matrix} 0 & 0  \\ 0 & 0  \\ \end{matrix} \right]\]

Since, Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\].

Then,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{A}_{11}}\text{=}-6\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{A}_{12}}=4\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}-3\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=2\]

Cofactor matrix is $\left[ \begin{matrix} -6 & 4  \\ -3 & 2  \\ \end{matrix} \right]$.

We know that adjoint of a matrix is the transpose of its cofactor matrix.

Thus, \[\text{adj}\,\text{A=}\,\left[ \begin{matrix} -6 & -3  \\ 4 & \text{2}  \\ \end{matrix} \right]\]

Now, multiplying $A$ with its adjoint, we have:

\[\Rightarrow \text{A}\left( \text{adjA} \right)\text{=}\left[ \begin{matrix} \text{2} & \text{3}  \\ \text{-4} & \text{-6}  \\ \end{matrix} \right]\left[ \begin{matrix} -6 & -3  \\ 4 & \text{2}  \\ \end{matrix} \right]\]

\[\Rightarrow \text{A}\left( \text{adjA} \right)\text{=}\left[ \begin{matrix} -12+12 & -6+6  \\ 24-24 & 12-12  \\ \end{matrix} \right]\]

\[\therefore \text{A}\left( \text{adjA} \right)\text{=}\left[ \begin{matrix} \text{0} & \text{0}  \\ \text{0} & \text{0}  \\ \end{matrix} \right]\]

Similarly, multiplying \[\left( adjA \right)\] with \[A\], we get:

\[\Rightarrow \left( \text{adjA} \right)\text{A=}\left[ \begin{matrix} -6 & -3  \\ 4 & \text{2}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{2} & \text{3}  \\ -4 & -6  \\ \end{matrix} \right]\]

\[\Rightarrow \left( \text{adjA} \right)\text{A=}\left[ \begin{matrix} -12+12 & -18+18  \\ 8-8 & 12-12  \\ \end{matrix} \right]\]

\[\therefore \left( \text{adjA} \right)\text{A=}\left[ \begin{matrix} \text{0} & \text{0}  \\ \text{0} & \text{0}  \\ \end{matrix} \right]\]

Thus, \[\text{A}\left( \text{adjA} \right)\text{=}\left( \text{adjA} \right)\text{A=}\left| \text{A} \right|I\]

Hence verified.


4. Verify \[\mathbf{\text{A}\left( \text{adjA} \right)\text{=}\left( \text{adjA} \right)\text{A=}\left| \text{A} \right|I}\]. \[\mathbf{\left[ \begin{matrix} \text{1} & -1 & \text{2}  \\ \text{3} & \text{0} & -2  \\ \text{1} & \text{0} & \text{3}  \\ \end{matrix} \right]}\].

Ans: Let \[\text{A=}\left[ \begin{matrix} \text{1} & \text{-1} & \text{2}  \\ \text{3} & \text{0} & \text{-2}  \\ \text{1} & \text{0} & \text{3}  \\ \end{matrix} \right]\]

\[\Rightarrow \left| \text{A} \right|\text{=1}\left( \text{0-0} \right)\text{+1}\left( \text{9+2} \right)\text{+2}\left( \text{0-0} \right)\]

\[\therefore \left| \text{A} \right|\text{=11}\]

\[\left| \text{A} \right|\text{I=11}\left[ \begin{matrix} \text{1} & \text{0} & \text{0}  \\ \text{0} & \text{1} & \text{0}  \\ \text{0} & \text{0} & \text{1}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{11} & \text{0} & \text{0}  \\ \text{0} & \text{11} & \text{0}  \\ \text{0} & \text{0} & \text{11}  \\ \end{matrix} \right]\]

Since, Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\].

Thus,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{A}_{11}}\text{=0}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{A}_{12}}=-\left( 9+2 \right)=-11\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]

\[\Rightarrow {{A}_{13}}=0\]

Similarly,

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}-\left( -3+0 \right)=3\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=3-2=1\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}-\left( 0+1 \right)=-1\]

and

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]

\[\Rightarrow {{\text{A}}_{31}}=2-0=2\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]

\[\Rightarrow {{\text{A}}_{32}}=-\left( -2-6 \right)=8\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]

\[\Rightarrow {{\text{A}}_{33}}=0+3=3\].

Cofactor matrix is \[\left[ \begin{matrix} 0 & -11 & 0  \\ 3 & 1 & -1  \\ 2 & 8 & 3  \\ \end{matrix} \right]\].

We know that adjoint of a matrix is the transpose of its cofactor matrix.

\[\Rightarrow adjA={{\left[ \begin{matrix} 0 & -11 & 0  \\ 3 & 1 & -1  \\ 2 & 8 & 3  \\ \end{matrix} \right]}^{T}}\]

\[\therefore \text{adj}\,\text{A=}\left[ \begin{matrix} \text{0} & 3 & 2  \\ -11 & \text{1} & \text{8}  \\ \text{0} & -1 & \text{3}  \\ \end{matrix} \right]\]

Now, multiplying $A$ with its adjoint, we have:

\[\Rightarrow \text{A}\left( \text{adj}\,\text{A} \right)\text{=}\left[ \begin{matrix} \text{1} & \text{-1} & \text{2}  \\ \text{3} & \text{0} & \text{-2}  \\ \text{1} & \text{0} & \text{3}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{0} & \text{3} & \text{2}  \\ \text{-11} & \text{1} & \text{8}  \\ \text{0} & \text{-1} & \text{3}  \\ \end{matrix} \right]\]

\[\Rightarrow \text{A}\left( \text{adj}\,\text{A} \right)\text{=}\left[ \begin{matrix} \text{0+11+0} & \text{3-1-2} & \text{2-8+6}  \\ \text{0+0+0} & \text{9+0+2} & \text{6+0-6}  \\ \text{0+0+0} & \text{3+0-3} & \text{2+0+9}  \\ \end{matrix} \right]\]

\[\therefore \text{A}\left( \text{adj}\,\text{A} \right)\text{=}\left[ \begin{matrix} \text{11} & \text{0} & \text{0}  \\ \text{0} & \text{11} & \text{0}  \\ \text{0} & \text{0} & \text{11}  \\ \end{matrix} \right]\]

Similarly, multiplying \[\left( adjA \right)\] with \[A\], we get:

\[\Rightarrow \left( \text{adj}\,\text{A} \right)\text{A=}\left[ \begin{matrix} \text{0} & \text{3} & \text{2}  \\ \text{-11} & \text{1} & \text{8}  \\ \text{0} & \text{-1} & \text{3}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{1} & \text{-1} & \text{2}  \\ \text{3} & \text{0} & \text{-2}  \\ \text{1} & \text{0} & \text{3}  \\ \end{matrix} \right]\]

\[\Rightarrow \left( \text{adj}\,\text{A} \right)\text{A=}\left[ \begin{matrix} \text{0+9+2} & \text{0+0+0} & \text{0-6+6}  \\ \text{-11+3+8} & \text{11+0+0} & \text{-22-2+24}  \\ \text{0-3+3} & \text{0+0+0} & \text{0+2+9}  \\ \end{matrix} \right]\]

\[\therefore \left( \text{adj}\,\text{A} \right)\text{A=}\left[ \begin{matrix} \text{11} & \text{0} & \text{0}  \\ \text{0} & \text{11} & \text{0}  \\ \text{0} & \text{0} & \text{11}  \\ \end{matrix} \right]\]

Thus, \[\text{A}\left( \text{adjA} \right)\text{=}\left( \text{adjA} \right)\text{A=}\left| \text{A} \right|I\]

Hence verified.

5. Find the inverse of each of the matrices (if it exists). $\mathbf{\left[ \begin{matrix} 2 & -2  \\ 4 & 3  \\ \end{matrix} \right]}$

Ans: Let \[\text{A=}\left[ \begin{matrix} 2 & -2  \\ 4 & 3  \\ \end{matrix} \right]\]

\[\Rightarrow \left| \text{A} \right|=6+8\]

\[\therefore \left| \text{A} \right|=14\]

Since, Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\].

Then,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{\text{A}}_{11}}\text{=3}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=-4}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}=2\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=2\]

Cofactor matrix is $\left[ \begin{matrix} 3 & -4  \\ 2 & 2  \\ \end{matrix} \right]$.

We know that adjoint of a matrix is the transpose of its cofactor matrix.

\[\therefore \text{adjA=}\left[ \begin{matrix} 3 & 2  \\ -4 & 2  \\ \end{matrix} \right]\]

Hence, the inverse of the matrix $A$ is given by,

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\text{adjA}\]

\[\therefore {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{14}}\left[ \begin{matrix} 3 & 2  \\ \text{-4} & 2  \\ \end{matrix} \right]\].


6. Find the inverse of each of the matrices (if it exists). \[\mathbf{\left[ \begin{matrix} \text{-1} & \text{5}  \\ \text{-3} & \text{2}  \\ \end{matrix} \right]}\]

Ans: Let \[\text{A=}\left[ \begin{matrix} \text{-1} & \text{5}  \\ \text{-3} & \text{2}  \\ \end{matrix} \right]\]

\[\Rightarrow \left| \text{A} \right|=-2+15\]

\[\therefore \left| \text{A} \right|=13\]

Since, Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\].

Then,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{\text{A}}_{11}}\text{=2}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=3}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}=-5\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=-1\]

Cofactor matrix is $\left[ \begin{matrix} 2 & 3  \\ -5 & -1  \\ \end{matrix} \right]$.

We know that adjoint of a matrix is the transpose of its cofactor matrix.

\[\therefore \text{adjA=}\left[ \begin{matrix} \text{2} & \text{-5}  \\ \text{3} & \text{-1}  \\ \end{matrix} \right]\]

Hence, the inverse of the matrix $A$ is given by,

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\text{adjA}\]

\[\therefore {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{13}}\left[ \begin{matrix} \text{2} & \text{-5}  \\ \text{3} & \text{-1}  \\ \end{matrix} \right]\].


7. Find the inverse of each of the matrices (if it exists). \[\mathbf{\left[ \begin{matrix} \text{1} & \text{2} & \text{3}  \\ \text{0} & \text{2} & \text{4}  \\ \text{0} & \text{0} & \text{5}  \\ \end{matrix} \right]}\]

Ans: Let \[\text{A=}\left[ \begin{matrix} \text{1} & \text{2} & \text{3}  \\ \text{0} & \text{2} & \text{4}  \\ \text{0} & \text{0} & \text{5}  \\ \end{matrix} \right]\]

Then,

\[\Rightarrow \left| \text{A} \right|\text{=1}\left( 10-0 \right)-2\left( 0-0 \right)\text{+3}\left( 0-0 \right)\]

\[\therefore \left| \text{A} \right|\text{=10}\]

Since, Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\].

Thus,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{A}_{11}}\text{=}10-0=10\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{A}_{12}}=-\left( 0+0 \right)=0\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]

\[\Rightarrow {{A}_{13}}\text{=0}\]

Similarly,

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}-\left( 10-0 \right)\text{=}-10\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=5-0=5\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}-\left( 0-0 \right)=0\]

And

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]

\[\Rightarrow {{\text{A}}_{31}}=8-6=2\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]

\[\Rightarrow {{\text{A}}_{32}}=-\left( 4-0 \right)=-4\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]

\[\Rightarrow {{\text{A}}_{33}}=2-0=2\].

Cofactor matrix is $\left[ \begin{matrix} 10 & 0 & 0  \\ -10 & 5 & 0  \\ 2 & -4 & 2  \\ \end{matrix} \right]$.

We know that adjoint of a matrix is the transpose of its cofactor matrix.

\[\therefore \text{adjA=}\left[ \begin{matrix} \text{10} & -10 & \text{2}  \\ \text{0} & \text{5} & -4  \\ \text{0} & \text{0} & \text{2}  \\ \end{matrix} \right]\]

Hence, the inverse of the matrix $A$ is given by,

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\text{adjA}\]

\[\therefore {{\text{A}}^{\text{-1}}}=\dfrac{\text{1}}{\text{10}}\left[ \begin{matrix} \text{10} & \text{-10} & \text{2}  \\ \text{0} & \text{5} & \text{-4}  \\ \text{0} & \text{0} & \text{2}  \\ \end{matrix} \right]\]


8. Find the inverse of each of the matrices (if it exists). \[\mathbf{\left[ \begin{matrix} \text{1} & \text{0} & \text{0}  \\ \text{3} & \text{3} & \text{0}  \\ \text{5} & \text{2} & \text{-1}  \\ \end{matrix} \right]}\]

Ans: Let \[\text{A=}\left[ \begin{matrix} \text{1} & \text{0} & \text{0}  \\ \text{3} & \text{3} & \text{0}  \\ \text{5} & \text{2} & \text{-1}  \\ \end{matrix} \right]\]

Then,

\[\Rightarrow \left| \text{A} \right|\text{=1}\left( -3-0 \right)\text{-0+0}\]

\[\therefore \left| \text{A} \right|=-3\]

Since, Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\].

Thus,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{A}_{11}}\text{=}-3-0=-3\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{A}_{12}}=-\left( -3-0 \right)=3\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]

\[\Rightarrow {{A}_{13}}=6-15=-9\]

Similarly,

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}-\left( 0+0 \right)=0\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=-1-0=-1\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}-\left( 2-0 \right)=-2\]

and

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]

\[\Rightarrow {{\text{A}}_{31}}=0-0=0\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]

\[\Rightarrow {{\text{A}}_{32}}=-\left( 0-0 \right)=0\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]

\[\Rightarrow {{\text{A}}_{33}}=3-0=3\].

Cofactor matrix is $\left[ \begin{matrix} -3 & 3 & -9  \\ 0 & -1 & -2  \\ 0 & 0 & 3  \\ \end{matrix} \right]$.

We know that adjoint of a matrix is the transpose of its cofactor matrix.

\[\Rightarrow adjA={{\left[ \begin{matrix} -3 & 3 & -9  \\ 0 & -1 & -2  \\ 0 & 0 & 3  \\ \end{matrix} \right]}^{T}}\]

\[\therefore \text{adjA=}\left[ \begin{matrix} \text{-3} & \text{0} & \text{0}  \\ \text{3} & \text{-1} & \text{0}  \\ \text{-9} & \text{-2} & \text{3}  \\ \end{matrix} \right]\]

Hence, the inverse of the matrix $A$ is given by,

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\text{adjA}\]

\[\therefore {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{10}}\left[ \begin{matrix} \text{-3} & \text{0} & \text{0}  \\ \text{3} & \text{-1} & \text{0}  \\ \text{-9} & \text{-2} & \text{3}  \\ \end{matrix} \right]\]


9. Find the inverse of each of the matrices (if it exists). \[\mathbf{\left[ \begin{matrix} \text{2} & \text{1} & \text{3}  \\ \text{4} & \text{-1} & \text{0}  \\ \text{-7} & \text{2} & \text{1}  \\ \end{matrix} \right]}\]

Ans: Let \[\text{A=}\left[ \begin{matrix} \text{2} & \text{1} & \text{3}  \\ \text{4} & \text{-1} & \text{0}  \\ \text{-7} & \text{2} & \text{1}  \\ \end{matrix} \right]\]

Thus,

\[\Rightarrow \left| \text{A} \right|\text{=2}\left( \text{-1-0} \right)\text{-1}\left( \text{4-0} \right)\text{+3}\left( \text{8-7} \right)\]

\[\Rightarrow \left| A \right|=2\left( -1 \right)-1\left( 4 \right)+3\left( 1 \right)\]

\[\therefore \left| \text{A} \right|=-3\]

Since, Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\].

Thus,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{A}_{11}}\text{=}-1-0=-1\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{A}_{12}}=-\left( 4-0 \right)=-4\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]

\[\Rightarrow {{A}_{13}}=8-7=1\]

Similarly,

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}-\left( 1-6 \right)=5\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=2+21=23\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}-\left( 4+7 \right)=-11\]

and

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]

\[\Rightarrow {{\text{A}}_{31}}=0+3=3\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]

\[\Rightarrow {{\text{A}}_{32}}=-\left( 0-12 \right)=12\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]

\[\Rightarrow {{\text{A}}_{33}}=-2-4=-6\].

Cofactor matrix is $\left[ \begin{matrix} -1 & -4 & 1  \\ 5 & 23 & -11  \\ 3 & 12 & -6  \\ \end{matrix} \right]$.

We know that adjoint of a matrix is the transpose of its cofactor matrix.

$\Rightarrow adjA={{\left[ \begin{matrix} -1 & -4 & 1  \\ 5 & 23 & -11  \\ 3 & 12 & -6  \\ \end{matrix} \right]}^{T}}$

\[\therefore \text{adjA=}\left[ \begin{matrix} \text{-1} & \text{5} & \text{3}  \\ \text{-4} & \text{23} & \text{12}  \\ \text{1} & \text{-11} & \text{-6}  \\ \end{matrix} \right]\]

Hence, the inverse of the matrix $A$ is given by,

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\text{adjA}\]

\[\therefore {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{10}}\left[ \begin{matrix} \text{-1} & \text{5} & \text{3}  \\ \text{-4} & \text{23} & \text{12}  \\ \text{1} & \text{-11} & \text{-6}  \\ \end{matrix} \right]\]


10. Find the inverse of each of the matrices (if it exists). \[\mathbf{\left[ \begin{matrix} \text{1} & \text{-1} & \text{2}  \\ \text{0} & \text{2} & \text{-3}  \\ \text{3} & \text{-2} & \text{4}  \\ \end{matrix} \right]}\]

Ans: Let \[\text{A=}\left[ \begin{matrix} \text{1} & \text{-1} & \text{2}  \\ \text{0} & \text{2} & \text{-3}  \\ \text{3} & \text{-2} & \text{4}  \\ \end{matrix} \right]\]

Expanding along column \[{{\text{C}}_{1}}\],

\[\Rightarrow \left| \text{A} \right|\text{=1}\left( 8-6 \right)\text{-0+3}\left( 3-4 \right)\]

\[\therefore \left| \text{A} \right|=-1\]

Since, Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\].

Thus,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{A}_{11}}\text{=8-6=2}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{A}_{12}}=-\left( 0+9 \right)=-9\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]

\[\Rightarrow {{A}_{13}}=0-6=-6\]

Similarly,

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}-\left( -4+4 \right)=0\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=4-6=-2\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}-\left( -2+3 \right)=-1\]

and

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]

\[\Rightarrow {{\text{A}}_{31}}=3-4=-1\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]

\[\Rightarrow {{\text{A}}_{32}}=-\left( -3-0 \right)=3\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]

\[\Rightarrow {{\text{A}}_{33}}=2-0=2\].

Cofactor matrix is \[\left[ \begin{matrix} 2 & -9 & -6  \\ 0 & -2 & -1  \\ -1 & 3 & 2  \\ \end{matrix} \right]\].

We know that adjoint of a matrix is the transpose of its cofactor matrix.

\[\Rightarrow adjA={{\left[ \begin{matrix} 2 & -9 & -6  \\ 0 & -2 & -1  \\ -1 & 3 & 2  \\ \end{matrix} \right]}^{T}}\]

\[\Rightarrow adjA=\left[ \begin{matrix} 2 & 0 & -1  \\ -9 & -2 & 3  \\ -6 & -1 & 2  \\ \end{matrix} \right]\]

The inverse of the matrix $A$ is given by,

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\text{adjA}\]

$\therefore A^{-1}=-1\begin{bmatrix}2 &0  &-1 \\ -9 &-2  &3 \\ -6 &-1  &2 \end{bmatrix}$

$\text{Hence},A^{-1}=\begin{bmatrix}-2 &0  &1 \\ 9 &2  &-3 \\ 6 &1  &-2 \end{bmatrix}$


11. Find the inverse of each of the matrices (if it exists). \[\mathbf{\left[ \begin{matrix} \text{1} & \text{0} & \text{0}  \\ \text{0} & \text{cos}\,\text{a} & \text{sin}\,\text{a}  \\ \text{0} & \text{sin}\,\text{a} & \text{-cos}\,\text{a}  \\ \end{matrix} \right]}\]

Ans: Let \[\text{A=}\left[ \begin{matrix} \text{1} & \text{0} & \text{0}  \\ \text{0} & \text{cos a} & \text{sin a}  \\ \text{0} & \text{sin a} & \text{-cos a}  \\ \end{matrix} \right]\]

Expanding along column, \[{{C}_{1}}\]

\[\Rightarrow \left| \text{A} \right|\text{=1}\left( \text{-co}{{\text{s}}^{\text{2}}}\text{a-si}{{\text{n}}^{\text{2}}}\text{a} \right)\]

\[\Rightarrow \left| \text{A} \right|=-\left( \text{co}{{\text{s}}^{\text{2}}}\text{a+si}{{\text{n}}^{\text{2}}}\text{a} \right)\]

\[\therefore \left| \text{A} \right|=-1\]

Since, Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\].

Thus,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{A}_{11}}\text{=}-\text{co}{{\text{s}}^{\text{2}}}\text{a}-\text{si}{{\text{n}}^{\text{2}}}\text{a=}-1\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{A}_{12}}=0\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]

\[\Rightarrow {{A}_{13}}=0\]

Similarly,

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=0}\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=-\cos a\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}-\sin a\]

and

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]

\[\Rightarrow {{\text{A}}_{31}}=0\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]

\[\Rightarrow {{\text{A}}_{32}}=-\sin a\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]

\[\Rightarrow {{\text{A}}_{33}}=\cos a\].

Cofactor matrix is \[\left[ \begin{matrix} -1 & 0 & 0  \\ 0 & -\cos a & -\sin a  \\ 0 & -\sin a & \cos a  \\ \end{matrix} \right]\].

We know that adjoint of a matrix is the transpose of its cofactor matrix.

\[\Rightarrow adjA={{\left[ \begin{matrix} -1 & 0 & 0  \\ 0 & -\cos a & -\sin a  \\ 0 & -\sin a & \cos a  \\ \end{matrix} \right]}^{T}}\]

\[\therefore \text{adjA=}\left[ \begin{matrix} \text{-1} & \text{0} & \text{0}  \\ \text{0} & \text{-cos}\,\text{a} & \text{-sin}\,\text{a}  \\ \text{0} & \text{-sin}\,\text{a} & \text{cos}\,\text{a}  \\ \end{matrix} \right]\]

The inverse of the matrix $A$ is given by,

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\text{adjA}\]

\[\therefore {{\text{A}}^{\text{-1}}}\text{=}-\text{1}\left[ \begin{matrix} \text{-1} & \text{0} & \text{0}  \\ \text{0} & \text{-cos}\,\text{a} & \text{-sin}\,\text{a}  \\ \text{0} & \text{-sin}\,\text{a} & \text{cos}\,\text{a}  \\ \end{matrix} \right]\]

Hence, \[{{\text{A}}^{-1}}\text{=}\left[ \begin{matrix} \text{1} & \text{0} & \text{0}  \\ \text{0} & \text{cos}\,\text{a} & \text{sin}\,\text{a}  \\ \text{0} & \text{sin}\,\text{a} & \text{-cos}\,\text{a}  \\ \end{matrix} \right]\].

12. Let \[\mathbf{\text{A=}\left[ \begin{matrix} \text{3} & \text{7}  \\ \text{2} & \text{5}  \\ \end{matrix} \right]}\] and \[\mathbf{\text{B=}\left[ \begin{matrix} \text{6} & \text{8}  \\ \text{7} & \text{9}  \\ \end{matrix} \right]}\] . Verify that \[\mathbf{{{\left( \text{AB} \right)}^{\text{-1}}}\text{=}{{\text{B}}^{\text{-1}}}{{\text{A}}^{\text{-1}}}}\].

Ans: Let \[\text{A=}\left[ \begin{matrix} \text{3} & \text{7}  \\ \text{2} & \text{5}  \\ \end{matrix} \right]\]

Thus, determining the value of \[\left| \text{A} \right|\],

\[\Rightarrow \left| \text{A} \right|\text{=}15-14\]

\[\therefore \left| \text{A} \right|=1\]

Since, Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\].

Thus,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{\text{A}}_{11}}\text{=5}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}-2\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}=-7\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=3\]

We know that adjoint of a matrix is the transpose of its cofactor matrix.

\[\therefore \text{adjA=}{{\left[ \begin{matrix} \text{5} & -2  \\ -7 & \text{3}  \\ \end{matrix} \right]}^{T}}\]

\[\Rightarrow \text{adjA=}\left[ \begin{matrix} \text{5} & -7  \\ -2 & \text{3}  \\ \end{matrix} \right]\]

The inverse of a matrix is given by, \[{{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\text{adjA}\]

Hence, \[{{\text{A}}^{\text{-1}}}\text{=}\left[ \begin{matrix} \text{5} & \text{-7}  \\ \text{-2} & \text{3}  \\ \end{matrix} \right]\]

For \[\text{B=}\left[ \begin{matrix} \text{6} & \text{8}  \\ \text{7} & \text{9}  \\ \end{matrix} \right]\]

\[\Rightarrow \left| \text{B} \right|\text{=54}-\text{56}\]

\[\therefore \left| \text{B} \right|\text{=}-\text{2}\]

Thus,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{\text{A}}_{11}}\text{=9}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}-7\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}=-8\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=6\]

We know that adjoint of a matrix is the transpose of its cofactor matrix.

\[\therefore \text{adjA=}{{\left[ \begin{matrix} 9 & -7  \\ -8 & 6  \\ \end{matrix} \right]}^{T}}\]

\[\Rightarrow \text{adjA=}\left[ \begin{matrix} 9 & -8  \\ -7 & 6  \\ \end{matrix} \right]\]

Hence, \[\text{adjB=}\left[ \begin{matrix} \text{9} & \text{-8}  \\ \text{-7} & \text{6}  \\ \end{matrix} \right]\]

\[\therefore {{\text{B}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{B} \right|}\text{adjB=}-\dfrac{\text{1}}{\text{2}}\left[ \begin{matrix} \text{9} & -\text{8}  \\ -\text{7} & \text{6}  \\ \end{matrix} \right]\]

Thus, \[{{\text{B}}^{\text{-1}}}=\left[ \begin{matrix} -\dfrac{\text{9}}{\text{2}} & \text{4}  \\ \dfrac{\text{7}}{\text{2}} & -\text{3}  \\ \end{matrix} \right]\].

Now, multiplying ${{B}^{-1}}$ and ${{A}^{-1}}$, we get:

\[\Rightarrow {{\text{B}}^{\text{-1}}}{{\text{A}}^{\text{-1}}}\text{=}\left[ \begin{matrix} \text{-}\dfrac{\text{9}}{\text{2}} & \text{4}  \\ \dfrac{\text{7}}{\text{2}} & \text{-3}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{5} & \text{-7}  \\ \text{-2} & \text{3}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{B}}^{\text{-1}}}{{\text{A}}^{\text{-1}}}\text{=}\left[ \begin{matrix} \text{-}\dfrac{\text{45}}{\text{2}}\text{-8} & \dfrac{\text{63}}{\text{2}}\text{+12}  \\ \dfrac{\text{35}}{\text{2}}\text{+6} & \text{-}\dfrac{\text{49}}{\text{2}}\text{-9}  \\ \end{matrix} \right]\]

\[\therefore {{\text{B}}^{\text{-1}}}{{\text{A}}^{\text{-1}}}\text{=}\left[ \begin{matrix} \text{-}\dfrac{\text{61}}{\text{2}} & \dfrac{\text{87}}{\text{2}}  \\ \dfrac{\text{47}}{\text{2}} & \text{-}\dfrac{\text{67}}{\text{2}}  \\ \end{matrix} \right]\]   ……(1)

Similarly, multiplying the matrices $A$ and $B$, we get:

\[\Rightarrow \text{AB=}\left[ \begin{matrix} \text{3} & \text{7}  \\ \text{2} & \text{5}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{6} & \text{8}  \\ \text{7} & \text{9}  \\ \end{matrix} \right]\]

\[\Rightarrow \text{AB=}\left[ \begin{matrix} \text{18+49} & \text{24+63}  \\ \text{12+35} & \text{16+45}  \\ \end{matrix} \right]\]

\[\therefore \text{AB=}\left[ \begin{matrix} \text{67} & \text{87}  \\ \text{47} & \text{61}  \\ \end{matrix} \right]\]

The value of \[\left| \text{AB} \right|\] is 

\[\Rightarrow \left| \text{AB} \right|\text{=67 }\!\!\times\!\!\text{ 61-87 }\!\!\times\!\!\text{ 47}\]

\[\Rightarrow \left| \text{AB} \right|\text{=4087-4089}\]

\[\therefore \left| \text{AB} \right|=-2\]

The adjoint of $\left( AB \right)$ is given by,

\[\Rightarrow \text{adj}\left( \text{AB} \right)\text{=}\left[ \begin{matrix} \text{61} & \text{-87} \\\text{-47} & \text{67} \\\end{matrix} \right]\]

Thus, the inverse is,

\[\Rightarrow {{\left( \text{AB} \right)}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{AB} \right|}\text{adj}\left( \text{AB} \right)\]

\[\Rightarrow {{\left( \text{AB} \right)}^{\text{-1}}}-\dfrac{\text{1}}{\text{2}}\left[ \begin{matrix} \text{61} & \text{-87}  \\ \text{-47} & \text{67}  \\ \end{matrix} \right]\]

\[\therefore {{\left( \text{AB} \right)}^{\text{-1}}}\text{=}\left[ \begin{matrix} \text{-}\dfrac{\text{61}}{\text{2}} & \dfrac{\text{87}}{\text{2}}  \\ \dfrac{\text{47}}{\text{2}} & \text{-}\dfrac{\text{67}}{\text{2}}  \\ \end{matrix} \right]\]   ……. (2)

From (1) and (2), we have:

\[{{\left( \text{AB} \right)}^{\text{-1}}}\text{=}{{\text{B}}^{\text{-1}}}{{\text{A}}^{\text{-1}}}\]

Hence proved.

13. If $A = \begin{bmatrix} 3&1 \\ -1 &2 \end{bmatrix} , \text{show that}\;\; A^2-5A+7I=0. \text{Hence find} A^{-1}$.

Ans: Given, \[\text{A=}\left[ \begin{matrix} \text{3} & \text{1}  \\ \text{-1} & \text{2}  \\ \end{matrix} \right]\]

We can write, \[{{\text{A}}^{\text{2}}}\text{=A}\text{.A}\]

\[\Rightarrow {{\text{A}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{3} & \text{1}  \\ \text{-1} & \text{2}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{3} & \text{1}  \\ \text{-1} & \text{2}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{9-1} & \text{3+2}  \\ \text{-3-2} & \text{-1+4}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{8} & \text{5}  \\ \text{-5} & \text{3}  \\ \end{matrix} \right]\]

\[\therefore \] The value of \[{{\text{A}}^{\text{2}}}\text{-5A+7I}\] is:

\[\Rightarrow {{\text{A}}^{\text{2}}}\text{-5A+7I=}\left[ \begin{matrix} \text{8} & \text{5}  \\ \text{-5} & \text{3}  \\ \end{matrix} \right]\text{-5}\left[ \begin{matrix} \text{3} & \text{1}  \\ \text{-1} & \text{2}  \\ \end{matrix} \right]\text{+7}\left[ \begin{matrix} \text{1} & \text{0}  \\ \text{0} & \text{1}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{2}}}\text{-5A+7I=}\left[ \begin{matrix} \text{-7} & \text{0}  \\ \text{0} & \text{-7}  \\ \end{matrix} \right]\text{+}\left[ \begin{matrix} \text{7} & \text{0}  \\ \text{0} & \text{7}  \\ \end{matrix} \right]\]

\[\therefore {{\text{A}}^{\text{2}}}\text{-5A+7I=}\left[ \begin{matrix} \text{0} & \text{0}  \\ \text{0} & \text{0}  \\ \end{matrix} \right]\]

Hence, \[{{\text{A}}^{\text{2}}}\text{-5A+7I=0}\]

\[\Rightarrow {{\text{A}}^{\text{2}}}\text{-5A=-7I}\]

Multiplying by \[{{\text{A}}^{\text{-1}}}\] on both the sides, we have:

\[\Rightarrow \text{AA}\left( {{\text{A}}^{\text{-1}}} \right)-\text{5A}{{\text{A}}^{\text{-1}}}\text{=}-\text{7I}{{\text{A}}^{\text{-1}}}\]    

\[\Rightarrow \text{A}\left( \text{A}{{\text{A}}^{\text{-1}}} \right)-\text{5I=}-\text{7I}{{\text{A}}^{\text{-1}}}\]

\[\Rightarrow \text{AI}-\text{5I=}-\text{7I}{{\text{A}}^{\text{-1}}}\]

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}-\dfrac{\text{1}}{\text{7}}\left( \text{A}-\text{5I} \right)\]

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{7}}\left( \text{5I}-\text{A} \right)\]

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{7}}\left( \left[ \begin{matrix} \text{5} & \text{0}  \\ \text{0} & \text{5}  \\ \end{matrix} \right]-\left[ \begin{matrix} \text{3} & \text{1}  \\ \text{-1} & \text{2}  \\ \end{matrix} \right] \right)\]

\[\therefore {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{7}}\left[ \begin{matrix} \text{2} & \text{-1}  \\ \text{1} & \text{3}  \\ \end{matrix} \right]\]

14. For the matrix\[\mathbf{\text{A=}\left[ \begin{matrix} \text{3} & \text{2}  \\ \text{1} & \text{1}  \\ \end{matrix} \right]}\] .find the number \[\mathbf{\text{a}}\] and \[\mathbf{\text{b}}\] such that. \[\mathbf{{{\text{A}}^{\text{2}}}\text{+aA+bI=0}}\].

Ans: Given \[\text{A=}\left[ \begin{matrix} \text{3} & \text{2}  \\ \text{1} & \text{1}  \\ \end{matrix} \right]\]

We can write, \[{{\text{A}}^{\text{2}}}\text{=A}\text{.A}\]

\[\therefore {{\text{A}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{3} & \text{2}  \\ \text{1} & \text{1}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{3} & \text{2}  \\ \text{1} & \text{1}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{9+2} & \text{6+2}  \\ \text{3+1} & \text{2+1}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{11} & \text{8}  \\ \text{4} & \text{3}  \\ \end{matrix} \right]\]

Solving \[{{\text{A}}^{\text{2}}}\text{+aA+bI=0}\] by multiplying the whole equation by \[{{A}^{-1}}\].

\[\Rightarrow \left( \text{AA} \right){{\text{A}}^{\text{-1}}}\text{+aA}{{\text{A}}^{\text{-1}}}\text{+bI}{{\text{A}}^{\text{-1}}}\text{=0}\]

\[\Rightarrow \text{A}\left( \text{A}{{\text{A}}^{\text{-1}}} \right)\text{+aI+b}\left( \text{I}{{\text{A}}^{\text{-1}}} \right)\text{=0}\]

\[\Rightarrow \text{AI+aI+b}{{\text{A}}^{\text{-1}}}\text{=0}\]

\[\Rightarrow \text{A+aI=-b}{{\text{A}}^{\text{-1}}}\]

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{b}}\left( \text{A+aI} \right)\]

Now, determining the value of  \[{{\text{A}}^{\text{-1}}}\].

We know that the adjoint of a square matrix is the transpose of its cofactor matrix.

Hence, the adjoint of matrix $A$ is:

$\therefore adjA=\left[ \begin{matrix} 1 & -2  \\ -1 & 3  \\ \end{matrix} \right]$

The inverse is given by, \[{{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}adjA\].

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{1}}\left[ \begin{matrix} \text{1} & \text{-2}  \\ \text{-1} & \text{3}  \\ \end{matrix} \right]\]

\[\therefore {{\text{A}}^{\text{-1}}}\text{=}\left[ \begin{matrix} \text{1} & \text{-2}  \\ \text{-1} & \text{3}  \\ \end{matrix} \right]\]

Thus,

\[\Rightarrow \left[ \begin{matrix} \text{1} & \text{-2}  \\ \text{-1} & \text{3}  \\ \end{matrix} \right]\text{=-}\dfrac{\text{1}}{\text{b}}\left( \left[ \begin{matrix} \text{3} & \text{2}  \\ \text{1} & \text{1}  \\ \end{matrix} \right]\text{+}\left[ \begin{matrix} \text{a} & \text{0}  \\ \text{0} & \text{a}  \\ \end{matrix} \right] \right)\]

\[\Rightarrow \left[ \begin{matrix} \text{1} & \text{-2}  \\ \text{-1} & \text{3}  \\ \end{matrix} \right]\text{=-}\dfrac{\text{1}}{\text{b}}\left[ \begin{matrix} \text{3+a} & \text{2}  \\ \text{1} & \text{1+a}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{1} & \text{-2}  \\ \text{-1} & \text{3}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \dfrac{\text{-3-a}}{\text{b}} & \text{-}\dfrac{\text{2}}{\text{b}}  \\ \text{-}\dfrac{\text{1}}{\text{b}} & \dfrac{\text{-1-a}}{\text{b}}  \\ \end{matrix} \right]\]

Equating the corresponding elements of the two matrices, we get:

\[\Rightarrow \text{-}\dfrac{\text{1}}{\text{b}}\text{=-1}\]

\[\therefore \text{b=1}\]

\[\Rightarrow \dfrac{\text{-3-a}}{\text{b}}=1\]

\[\therefore \text{a=}-4\]

Thus, \[-4\] and \[1\] are the required values of \[\text{a}\] and \[\text{b}\] respectively.

15. For the matrix \[\mathbf{\text{A=}\left[ \begin{matrix} \text{1} & \text{1} & \text{1}  \\ \text{1} & \text{2} & \text{-3}  \\ \text{2} & \text{-1} & \text{3}  \\ \end{matrix} \right]}\] show that \[\mathbf{{{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+5A+11I=0}}\] . Hence, \[\mathbf{{{\text{A}}^{\text{-1}}}}\].

Ans: Given, \[\text{A=}\left[ \begin{matrix} \text{1} & \text{1} & \text{1}  \\ \text{1} & \text{2} & \text{-3}  \\ \text{2} & \text{-1} & \text{3}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{1} & \text{1} & \text{1}  \\ \text{1} & \text{2} & \text{-3}  \\ \text{2} & \text{-1} & \text{3}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{1} & \text{1} & \text{1}  \\ \text{1} & \text{2} & \text{-3}  \\ \text{2} & \text{-1} & \text{3}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{1+1+1} & \text{1+2-1} & \text{1-3+3}  \\ \text{1+2-6} & \text{1+4+3} & \text{1-6-9}  \\ \text{2-1+6} & \text{2-2-3} & \text{2+3+9}  \\ \end{matrix} \right]\]

\[\therefore {{\text{A}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{4} & \text{2} & \text{1}  \\ \text{-3} & \text{8} & \text{-14}  \\ \text{7} & \text{-3} & \text{14}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{3}}}\text{=}{{\text{A}}^{\text{2}}}\text{.A=}\left[ \begin{matrix} \text{4} & \text{2} & \text{1}  \\ \text{-3} & \text{8} & \text{-14}  \\ \text{7} & \text{-3} & \text{14}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{1} & \text{1} & \text{1}  \\ \text{1} & \text{2} & \text{-3}  \\ \text{2} & \text{-1} & \text{3}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{3}}}\text{=}\left[ \begin{matrix} \text{4+2+2} & \text{4+4-1} & \text{4-6+3}  \\ \text{-3+8-28} & \text{-3+16+14} & \text{-3-24-42}  \\ \text{7-3+28} & \text{7-6-14} & \text{7+9+42}  \\ \end{matrix} \right]\]

\[\therefore {{\text{A}}^{\text{3}}}\text{=}\left[ \begin{matrix} \text{8} & \text{7} & \text{1}  \\ \text{-23} & \text{27} & \text{-69}  \\ \text{32} & \text{-13} & \text{58}  \\ \end{matrix} \right]\]

Substituting the values for \[{{\text{A}}^{\text{3}}}\], \[{{\text{A}}^{2}}\] and \[\text{A}\] in \[{{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+5A+11I}\], we have:

\[\Rightarrow {{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+5A+11I=}\left[ \begin{matrix} \text{8} & \text{7} & \text{1}  \\ \text{-23} & \text{27} & \text{-69}  \\ \text{32} & \text{-13} & \text{58}  \\ \end{matrix} \right]\text{-6}\left[ \begin{matrix} \text{4} & \text{2} & \text{1}  \\ \text{-3} & \text{8} & \text{-14}  \\ \text{7} & \text{-3} & \text{14}  \\ \end{matrix} \right]\text{+5}\left[ \begin{matrix} \text{1} & \text{1} & \text{1}  \\ \text{1} & \text{2} & \text{-3}  \\ \text{2} & \text{-1} & \text{3}  \\ \end{matrix} \right]\text{+11}\left[ \begin{matrix} \text{1} & \text{0} & \text{0}  \\ \text{0} & \text{1} & \text{0}  \\ \text{0} & \text{0} & \text{1}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+5A+11I=}\left[ \begin{matrix} \text{8} & \text{7} & \text{1}  \\ \text{-23} & \text{27} & \text{-69}  \\ \text{32} & \text{-13} & \text{58}  \\ \end{matrix} \right]\text{-}\left[ \begin{matrix} \text{24} & \text{12} & \text{6}  \\ \text{-18} & \text{48} & \text{-84}  \\ \text{42} & \text{-18} & \text{84}  \\ \end{matrix} \right]\text{+}\left[ \begin{matrix} \text{5} & \text{5} & \text{5}  \\ \text{5} & \text{10} & \text{-15}  \\ \text{2} & \text{-5} & \text{15}  \\ \end{matrix} \right]\text{+11}\left[ \begin{matrix} \text{11} & \text{0} & \text{0}  \\ \text{0} & \text{11} & \text{0}  \\ \text{0} & \text{0} & \text{11}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+5A+11I=}\left[ \begin{matrix} \text{24} & \text{12} & \text{6}  \\ \text{-18} & \text{48} & \text{-84}  \\ \text{42} & \text{-18} & \text{84}  \\ \end{matrix} \right]\text{-}\left[ \begin{matrix} \text{24} & \text{12} & \text{6}  \\ \text{-18} & \text{48} & \text{-84}  \\ \text{42} & \text{-18} & \text{84}  \\ \end{matrix} \right]\]

\[\therefore {{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+5A+11I=}\left[ \begin{matrix} \text{0} & \text{0} & \text{0}  \\ \text{0} & \text{0} & \text{0}  \\ \text{0} & \text{0} & \text{0}  \\ \end{matrix} \right]\text{=0}\]

Thus, \[{{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+5A+11I=0}\]

Since, \[{{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+5A+11I=0}\]. 

Multiplying the whole equation by \[{{A}^{-1}}\], we have:

\[\Rightarrow \left( \text{AAA} \right){{\text{A}}^{\text{-1}}}\text{-6}\left( \text{AA} \right){{\text{A}}^{\text{-1}}}\text{+5A}{{\text{A}}^{\text{-1}}}\text{+11I}{{\text{A}}^{\text{-1}}}\text{=0}\] 

\[\Rightarrow \text{AA}\left( \text{A}{{\text{A}}^{\text{-1}}} \right)\text{-6A}\left( \text{A}{{\text{A}}^{\text{-1}}} \right)\text{+5}\left( \text{A}{{\text{A}}^{\text{-1}}} \right)\text{=11}\left( \text{I}{{\text{A}}^{\text{-1}}} \right)\]

\[\Rightarrow {{\text{A}}^{\text{2}}}\text{-6A+5I=-11}{{\text{A}}^{\text{-1}}}\]

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=-}\dfrac{\text{1}}{\text{11}}\left( {{\text{A}}^{\text{2}}}\text{-6A+5I} \right)\]   …. (1)

Now, \[{{\text{A}}^{\text{2}}}\text{-6A+5I}\] is given by:

\[\Rightarrow {{\text{A}}^{\text{2}}}\text{-6A+5I=}\left[ \begin{matrix} \text{4} & \text{2} & \text{1}  \\ \text{-3} & \text{8} & \text{-14}  \\ \text{7} & \text{-3} & \text{14}  \\ \end{matrix} \right]\text{-6}\left[ \begin{matrix} \text{1} & \text{1} & \text{1}  \\ \text{1} & \text{2} & \text{-3}  \\ \text{2} & \text{-1} & \text{3}  \\ \end{matrix} \right]\text{+5}\left[ \begin{matrix} \text{1} & \text{0} & \text{0}  \\ \text{0} & \text{1} & \text{0}  \\ \text{0} & \text{0} & \text{1}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{2}}}\text{-6A+5I=}\left[ \begin{matrix} \text{4} & \text{2} & \text{1}  \\ \text{-3} & \text{8} & \text{-14}  \\ \text{7} & \text{-3} & \text{14}  \\ \end{matrix} \right]\text{-}\left[ \begin{matrix} \text{6} & \text{6} & \text{6}  \\ \text{6} & \text{12} & \text{-18}  \\ \text{12} & \text{6} & \text{18}  \\ \end{matrix} \right]\text{+}\left[ \begin{matrix} \text{5} & \text{0} & \text{0}  \\ \text{0} & \text{5} & \text{0}  \\ \text{0} & \text{0} & \text{5}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{2}}}\text{-6A+5I=}\left[ \begin{matrix} \text{4} & \text{2} & \text{1}  \\ \text{-3} & \text{13} & \text{-14}  \\ \text{7} & \text{-3} & \text{19}  \\ \end{matrix} \right]\text{-}\left[ \begin{matrix} \text{6} & \text{6} & \text{6}  \\ \text{6} & \text{12} & \text{-18}  \\ \text{12} & \text{-6} & \text{18}  \\ \end{matrix} \right]\]

\[\therefore {{\text{A}}^{\text{2}}}\text{-6A+5I=}\left[ \begin{matrix} \text{3} & \text{-4} & \text{-5}  \\ \text{-9} & \text{1} & \text{4}  \\ \text{-5} & \text{3} & \text{1}  \\ \end{matrix} \right]\]

Substituting for \[{{\text{A}}^{\text{2}}}\text{-6A+5I}\] equation (1), we get

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=-}\dfrac{\text{1}}{\text{11}}\left[ \begin{matrix} \text{3} & \text{-4} & \text{-5}  \\ \text{-9} & \text{1} & \text{4}  \\ \text{-5} & \text{3} & \text{1}  \\ \end{matrix} \right]\]

\[\therefore {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{11}}\left[ \begin{matrix} \text{-3} & \text{4} & \text{5}  \\ \text{9} & \text{-1} & \text{-4}  \\ \text{5} & \text{-3} & \text{-1}  \\ \end{matrix} \right]\]

16. If \[\mathbf{\text{A=}\left[ \begin{matrix} \text{2} & \text{-1} & \text{1}  \\ \text{-1} & \text{2} & \text{-1}  \\ \text{1} & \text{-1} & \text{2}  \\ \end{matrix} \right]}\] verify that \[\mathbf{{{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+9A+4I=0}}\] and hence find \[\mathbf{{{\text{A}}^{\text{-1}}}}\].

Ans: Given,\[\text{A=}\left[ \begin{matrix} \text{2} & \text{-1} & \text{1}  \\ \text{-1} & \text{2} & \text{-1}  \\ \text{1} & \text{-1} & \text{2}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{2} & \text{-1} & \text{1}  \\ \text{-1} & \text{2} & \text{-1}  \\ \text{1} & \text{-1} & \text{2}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{2} & \text{-1} & \text{1}  \\ \text{-1} & \text{2} & \text{-1}  \\ \text{1} & \text{-1} & \text{2}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{4+1+1} & \text{-2-2-1} & \text{2+1+2}  \\ \text{-2-2-1} & \text{1+4+1} & \text{-1-2-2}  \\ \text{2+1+2} & \text{-1-2-2} & \text{1+1+4}  \\ \end{matrix} \right]\]

\[\therefore {{\text{A}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{6} & \text{-5} & \text{5}  \\ \text{-5} & \text{6} & \text{-5}  \\ \text{5} & \text{-5} & \text{6}  \\ \end{matrix} \right]\]

Similarly,

\[\Rightarrow {{\text{A}}^{\text{3}}}\text{=}{{\text{A}}^{\text{2}}}\text{A=}\left[ \begin{matrix} \text{6} & \text{-5} & \text{5}  \\ \text{-5} & \text{6} & \text{-5}  \\ \text{5} & \text{-5} & \text{6}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{2} & \text{-1} & \text{1}  \\ \text{-1} & \text{2} & \text{-1}  \\ \text{1} & \text{-1} & \text{2}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{3}}}\text{=}\left[ \begin{matrix} \text{12+5+5} & \text{-6-10-5} & \text{6+5+10}  \\ \text{-10-6-5} & \text{5+12+5} & \text{-5-6-10}  \\ \text{10+5+6} & \text{-5-10-6} & \text{5+5+12}  \\ \end{matrix} \right]\]

\[\therefore {{\text{A}}^{\text{3}}}\text{=}\left[ \begin{matrix} \text{22} & \text{-21} & \text{21}  \\ \text{-21} & \text{22} & \text{-21}  \\ \text{21} & \text{-21} & \text{22}  \\ \end{matrix} \right]\]

Now, \[{{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+9A-4I}\] is given by:

\[\Rightarrow {{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+9A-4I=}\left[ \begin{matrix} \text{22} & \text{-21} & \text{21}  \\ \text{-21} & \text{22} & \text{-21}  \\ \text{21} & \text{-21} & \text{22}  \\ \end{matrix} \right]\text{-6}\left[ \begin{matrix} \text{6} & \text{-5} & \text{5}  \\ \text{-5} & \text{6} & \text{-5}  \\ \text{5} & \text{-5} & \text{6}  \\ \end{matrix} \right]\text{+9}\left[ \begin{matrix} \text{2} & \text{-1} & \text{1}  \\ \text{-1} & \text{2} & \text{-1}  \\ \text{1} & \text{-1} & \text{2}  \\ \end{matrix} \right]\text{-4}\left[ \begin{matrix} \text{1} & \text{0} & \text{0}  \\ \text{0} & \text{1} & \text{0}  \\ \text{0} & \text{0} & \text{1}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+9A-4I=}\left[ \begin{matrix} \text{22} & \text{-21} & \text{21}  \\ \text{-21} & \text{22} & \text{-21}  \\ \text{21} & \text{-21} & \text{22}  \\ \end{matrix} \right]\text{-}\left[ \begin{matrix} \text{36} & \text{-30} & \text{30}  \\ \text{-30} & \text{36} & \text{-30}  \\ \text{30} & \text{-30} & \text{36}  \\ \end{matrix} \right]\text{+}\left[ \begin{matrix} \text{18} & \text{-9} & \text{9}  \\ \text{-9} & \text{18} & \text{-9}  \\ \text{9} & \text{-9} & \text{18}  \\ \end{matrix} \right]\text{-}\left[ \begin{matrix} \text{4} & \text{0} & \text{0}  \\ \text{0} & \text{4} & \text{0}  \\ \text{0} & \text{0} & \text{4}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{3}}\text{-6}{{\text{A}}^{\text{2}}}\text{+9A-4I=}\left[ \begin{matrix} \text{40} & \text{-30} & \text{30}  \\ \text{-30} & \text{40} & \text{-30}  \\ \text{30} & \text{-30} & \text{40}  \\ \end{matrix} \right]\text{-}\left[ \begin{matrix} \text{40} & \text{-30} & \text{30}  \\ \text{-30} & \text{40} & \text{-30}  \\ \text{30} & \text{-30} & \text{40}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{0} & \text{0} & \text{0}  \\ \text{0} & \text{0} & \text{0}  \\ \text{0} & \text{0} & \text{0}  \\ \end{matrix} \right]\]

\[\therefore {{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+9A-4I=0}\]

Since, \[{{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+9A-4I=0}\]. 

Multiplying the whole equation by \[{{A}^{-1}}\], we have:

\[\Rightarrow \left( \text{AAA} \right){{\text{A}}^{\text{-1}}}\text{-6}\left( \text{AA} \right){{\text{A}}^{\text{-1}}}\text{+9A}{{\text{A}}^{\text{-1}}}\text{-4I}{{\text{A}}^{\text{-1}}}\text{=0}\] 

\[\Rightarrow \text{AA}\left( \text{A}{{\text{A}}^{\text{-1}}} \right)\text{-6A}\left( \text{A}{{\text{A}}^{\text{-1}}} \right)\text{+9}\left( \text{A}{{\text{A}}^{\text{-1}}} \right)\text{=4}\left( \text{I}{{\text{A}}^{\text{-1}}} \right)\]

\[\Rightarrow \text{AAI-6AI+9I=4}{{\text{A}}^{\text{-1}}}\]

\[\Rightarrow {{\text{A}}^{\text{2}}}\text{-6A+9I=4}{{\text{A}}^{\text{-1}}}\]

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{4}}\left( {{\text{A}}^{\text{2}}}\text{-6A+9I} \right)\]   …... (1)

Now, \[{{\text{A}}^{\text{2}}}\text{-6A+9I}\] is given by:

\[{{\text{A}}^{\text{2}}}\text{-6A+9I=}\left[ \begin{matrix} \text{6} & \text{-5} & \text{5}  \\ \text{-5} & \text{6} & \text{-5}  \\ \text{5} & \text{-5} & \text{6}  \\ \end{matrix} \right]\text{-6}\left[ \begin{matrix} \text{2} & \text{-1} & \text{1}  \\ \text{-1} & \text{2} & \text{-1}  \\ \text{1} & \text{-1} & \text{2}  \\ \end{matrix} \right]\text{+9}\left[ \begin{matrix} \text{0} & \text{0} & \text{0}  \\ \text{0} & \text{0} & \text{0}  \\ \text{0} & \text{0} & \text{0}  \\ \end{matrix} \right]\]

\[{{\text{A}}^{\text{2}}}\text{-6A+9I=}\left[ \begin{matrix} \text{6} & \text{-5} & \text{5}  \\ \text{-5} & \text{6} & \text{-5}  \\ \text{5} & \text{-5} & \text{6}  \\ \end{matrix} \right]\text{-}\left[ \begin{matrix} \text{12} & \text{-6} & \text{6}  \\ \text{-6} & \text{12} & \text{-6}  \\ \text{6} & \text{-6} & \text{12}  \\ \end{matrix} \right]\text{+}\left[ \begin{matrix} \text{9} & \text{0} & \text{0}  \\ \text{0} & \text{9} & \text{0}  \\ \text{0} & \text{0} & \text{9}  \\ \end{matrix} \right]\]

\[\therefore {{\text{A}}^{\text{2}}}\text{-6A+9I=}\left[ \begin{matrix} \text{3} & \text{1} & \text{-1}  \\ \text{1} & \text{3} & \text{1}  \\ \text{-1} & \text{3} & \text{3}  \\ \end{matrix} \right]\]

Substituting for \[{{\text{A}}^{\text{2}}}\text{-6A+9I}\] equation (1), we get

\[{{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{4}}\left[ \begin{matrix} \text{3} & \text{1} & \text{-1}  \\ \text{1} & \text{3} & \text{1}  \\ \text{-1} & \text{3} & \text{3}  \\ \end{matrix} \right]\].


17. Let \[\mathbf{\text{A}}\] be nonsingular square matrix of order \[\mathbf{\text{3 }\!\!\times\!\!\text{ 3}}\] . Then \[\mathbf{\left| \text{adjA} \right|}\] is equal to

  1. \[\mathbf{\left| \text{A} \right|}\]

  2. \[\mathbf{{{\left| \text{A} \right|}^{\text{2}}}}\]

  3. \[\mathbf{{{\left| \text{A} \right|}^{\text{3}}}}\]

  4. \[\mathbf{\text{3}\left| \text{A} \right|}\]

Ans: Given $A$ is a nonsingular square matrix, i.e., it is a square matrix whose determinant is not equal to zero.

The inverse of a matrix is given as \[{{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}adjA\].

$ \Rightarrow {{A}^{-1}}A=\dfrac{1}{\left| A \right|}adjA$ 

$\Rightarrow \left| A \right|I=adjA$

The adjoint of matrix $A$ is given by,

\[\Rightarrow \left( \text{adjA} \right)\text{=A=}\left| \text{A} \right|\text{I=}\left[ \begin{matrix} \left| \text{A} \right| & \text{0} & \text{0}  \\ \text{0} & \left| \text{A} \right| & \text{0}  \\ \text{0} & \text{0} & \left| \text{A} \right|  \\ \end{matrix} \right]\]

\[\Rightarrow \left| \left( \text{adjA} \right)\text{A} \right|\text{=}\left[ \begin{matrix} \left| \text{A} \right| & \text{0} & \text{0}  \\ \text{0} & \left| \text{A} \right| & \text{0}  \\ \text{0} & \text{0} & \left| \text{A} \right|  \\ \end{matrix} \right]\]

\[\Rightarrow \left| \text{adjA} \right|\left| \text{A} \right|\text{=}{{\left| \text{A} \right|}^{\text{3}}}\left| \begin{matrix} \text{1} & \text{0} & \text{0}  \\ \text{0} & \text{1} & \text{0}  \\ \text{0} & \text{0} & \text{1}  \\ \end{matrix} \right|\text{=}{{\left| \text{A} \right|}^{\text{3}}}\left( \text{I} \right)\]

\[\therefore \left| \text{adjA} \right|\text{=}{{\left| \text{A} \right|}^{\text{3}}}\]

Hence, B. \[{{\left| \text{A} \right|}^{\text{2}}}\] is the correct answer.


18. If \[\mathbf{\text{A}}\] is an invertible matrix of order \[\mathbf{\text{2}}\] , then \[\mathbf{\text{det}\left( {{\text{A}}^{\text{-1}}} \right)}\] is equal to

  1. \[\mathbf{\text{det}\left( \text{A} \right)}\]

  2. \[\mathbf{\dfrac{\text{1}}{\text{det}\left( \text{A} \right)}}\]

  3. \[\mathbf{\text{1}}\]

  4. \[\mathbf{\text{0}}\]

Ans: Since \[\text{A}\] is an invertible matrix, thus \[{{\text{A}}^{\text{-1}}}\] exists and it is given by: \[{{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\text{adjA}\].

As matrix \[\text{A}\] is of order \[\text{2}\], 

$\therefore $ Let \[\text{A=}\left[ \begin{matrix} \text{a} & \text{b}  \\ \text{c} & \text{d}  \\ \end{matrix} \right]\] .

Hence, \[\left| \text{A} \right|\text{=ad-bc}\].

The adjoint of \[\text{A}\] would be, \[\text{adjA=}\left[ \begin{matrix} \text{d} & \text{-b}  \\ \text{-c} & \text{a}  \\ \end{matrix} \right]\] .

Now, the inverse of the matrix is given by:

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\text{adjA}\]

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\left[ \begin{matrix} \dfrac{\text{d}}{\left| \text{A} \right|} & \dfrac{\text{-b}}{\left| \text{A} \right|}  \\ \dfrac{\text{-c}}{\left| \text{A} \right|} & \dfrac{\text{a}}{\left| \text{A} \right|}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\left[ \begin{matrix} \dfrac{\text{d}}{\left| \text{A} \right|} & \dfrac{\text{-b}}{\left| \text{A} \right|}  \\ \dfrac{\text{-c}}{\left| \text{A} \right|} & \dfrac{\text{a}}{\left| \text{A} \right|}  \\ \end{matrix} \right]\]

\[\Rightarrow \left| {{\text{A}}^{\text{-1}}} \right|\text{=}\dfrac{\text{1}}{\left| {{\text{A}}^{\text{2}}} \right|}\left| \begin{matrix} \text{d} & \text{-b}  \\ \text{-c} & \text{a}  \\ \end{matrix} \right|\]

\[\Rightarrow \left| {{\text{A}}^{\text{-1}}} \right|=\dfrac{\text{1}}{\left| {{\text{A}}^{\text{2}}} \right|}\left( \text{ad-bc} \right)\]

\[\Rightarrow \left| {{\text{A}}^{\text{-1}}} \right|\text{=}\dfrac{\text{1}}{\left| {{\text{A}}^{\text{2}}} \right|}\text{.}\left| \text{A} \right|\]

\[\therefore \left| {{\text{A}}^{\text{-1}}} \right|\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\]

Thus, \[\text{det}\left( {{\text{A}}^{\text{-1}}} \right)\text{=}\dfrac{\text{1}}{\text{det}\left( \text{A} \right)}\].

Hence, B. \[\dfrac{\text{1}}{\text{det}\left( \text{A} \right)}\] is the correct answer.


Exercise (4.6)

1. Examine the consistency of the system of equations.

\[\mathbf{\text{x+2y=2}}\]

\[\mathbf{\text{2x+3y=3}}\]

Ans: Given equations,

\[\text{x+2y=2}\]

\[\text{2x+3y=3}\]

Let us suppose \[\text{A=}\left[ \begin{matrix} \text{1} & \text{2}  \\ \text{2} & \text{3}  \\ \end{matrix} \right]\], \[\text{X=}\left[ \begin{matrix} x  \\ y  \\ \end{matrix} \right]\] and \[\text{B=}\left[ \begin{matrix} \text{2}  \\ \text{3}  \\ \end{matrix} \right]\] such that, the given system of equations can be written in the form of \[\text{AX=B}\].

Determining the value of $A$, we have:

\[\Rightarrow \left| \text{A} \right|\text{=1}\left( \text{3} \right)-\text{2}\left( \text{2} \right)\text{=3}-\text{4}\]

\[\therefore \left| \text{A} \right|\text{=}-1\ne 0\]

Hence, \[\text{A}\] is non-singular.

Thus, the inverse of $A$, i.e., \[{{\text{A}}^{\text{-1}}}\] exists.

$\therefore $ The given system of equations is consistent.


2. Examine the consistency of the system of equations.

\[\mathbf{\text{2x-y=5}}\]

\[\mathbf{\text{x+y=4}}\]

Ans: Given equations,

\[\text{2x-y=5}\]

\[\text{x+y=4}\]

Let us suppose\[\text{A=}\left[ \begin{matrix} \text{2} & \text{-1}  \\ \text{1} & \text{1}  \\ \end{matrix} \right]\], \[\text{X=}\left[ \begin{matrix} \text{x}  \\ y  \\ \end{matrix} \right]\] and \[\text{B=}\left[ \begin{matrix} \text{5}  \\ \text{4}  \\ \end{matrix} \right]\] such that, the given system of equation can be written in the form of \[\text{AX=B}\].

Determining the value of $A$, we have:

\[\Rightarrow \left| \text{A} \right|\text{=2}\left( \text{1} \right)\text{-}\left( \text{-1} \right)\left( \text{1} \right)\text{=2+1}\]

\[\therefore \left| \text{A} \right|=3\ne 0\]

Hence, \[\text{A}\] is non-singular.

Thus, \[{{\text{A}}^{\text{-1}}}\] exists.

$\therefore $ The given system of equations is consistent.


3. Examine the consistency of the system of equations.

\[\mathbf{\text{x+3y=5}}\]

\[\mathbf{\text{2x+6y=8}}\]

Ans: Given equations,

\[\text{x+3y=5}\]

\[\text{2x+6y=8}\]

We know, that a given system of equations is consistent if it has at least one solution.

Let \[\text{A=}\left[ \begin{matrix} \text{1} & \text{3}  \\ \text{2} & \text{6}  \\ \end{matrix} \right]\] , \[\text{X=}\left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \end{matrix} \right]\] and \[\text{B=}\left[ \begin{matrix} \text{5}  \\ \text{8}  \\ \end{matrix} \right]\] such that, the given system of equation can be written in the form of \[\text{AX=B}\].

Determining the value of $A$, we have:

\[\Rightarrow \left| \text{A} \right|\text{=1}\left( \text{6} \right)\text{-3}\left( \text{2} \right)\]

\[\therefore \left| \text{A} \right|\text{=}0\]

Hence, \[\text{A}\] is a singular matrix.

We know that the adjoint of a square matrix is the transpose of its cofactor matrix.

Determining the adjoint of the matrix $A$, we have:

\[\Rightarrow \left( \text{adjA} \right)\text{=}\left[ \begin{matrix} \text{6} & \text{-3}  \\ \text{-2} & \text{1}  \\ \end{matrix} \right]\]

\[\Rightarrow \left( \text{adjA} \right)\text{B=}\left[ \begin{matrix} \text{6} & \text{-3}  \\ \text{-2} & \text{1}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{5}  \\ \text{8}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{30-24}  \\ \text{-10+8}  \\ \end{matrix} \right]\] \[\therefore \left( \text{adjA} \right)\text{B=}\left[ \begin{matrix} \text{6}  \\ \text{-2}  \\ \end{matrix} \right]\ne 0\]

Thus, the solution of the given system of equations does not exists.

$\therefore $ The given system of equations is inconsistent.


4. Examine the consistency of the system of equations.

\[\mathbf{\text{x+y+z=1}}\]

\[\mathbf{\text{2x+3y+2z=2}}\]

\[\mathbf{\text{ax+ay+2az=4}}\]

Ans: Given equations,

\[\text{x+y+z=1}\]

\[\text{2x+3y+2z=2}\]

\[\text{ax+ay+2az=4}\]

We know, that a given system of equations is consistent if it has at least one solution.

Let \[\text{A=}\left[ \begin{matrix} \text{1} & \text{1} & \text{1}  \\ \text{2} & \text{3} & \text{2}  \\ \text{a} & \text{a} & \text{2a}  \\ \end{matrix} \right]\] , \[\text{X=}\left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\] and \[\text{B=}\left[ \begin{matrix} \text{1}  \\ \text{2}  \\ \text{4}  \\ \end{matrix} \right]\] such that, the given system of equation can be written in the form of \[\text{AX=B}\].

Determining the value of $A$, we have:

\[\Rightarrow \left| \text{A} \right|\text{=1}\left( \text{6a}-\text{2a} \right)-\text{1}\left( \text{4a}-\text{2a} \right)\text{+1}\left( \text{2a}-\text{3a} \right)\]

\[\Rightarrow \left| A \right|\text{=4a}-\text{2a}-a\]

\[\therefore \left| A \right|\text{=}a\ne 0\]

Hence \[\text{A}\] is non-singular matrix.

Thus, \[{{\text{A}}^{\text{-1}}}\] exists.

$\therefore $ The given system of equation is consistent.


5. Examine the consistency of the system of equations.

\[\mathbf{\text{3x-y-2z=2}}\]

\[\mathbf{\text{2y-z=-1}}\]

\[\mathbf{\text{3x-5y=3}}\]

Ans: Given equations,

\[\text{3x-y-2z=2}\]

\[\text{2y-z=-1}\]

\[\text{3x-5y=3}\]

We know, that a given system of equations is consistent if it has at least one solution.

Let \[\text{A=}\left[ \begin{matrix} \text{3} & \text{-1} & \text{-2}  \\ \text{0} & \text{2} & \text{-1}  \\ \text{3} & \text{-5} & \text{0}  \\ \end{matrix} \right]\] , \[\text{X=}\left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\] and \[\text{B=}\left[ \begin{matrix} \text{2}  \\ \text{-1}  \\ \text{3}  \\ \end{matrix} \right]\] such that, this system of equations can be written in the form of \[\text{AX=B}\].

Determining the value of $A$, we have:

\[\Rightarrow \left| \text{A} \right|\text{=3}\left( \text{-5} \right)\text{-0+3}\left( \text{1+4} \right)\]

\[\Rightarrow \left| \text{A} \right|\text{=-15+15}\]

\[\Rightarrow \left| \text{A} \right|\text{=0}\]

\[\therefore \text{A}\] is a singular matrix.

Determining the adjoint of matrix $A$.

Writing the cofactors,

${{A}_{11}}={{(-1)}^{1+1}}\left| \begin{matrix} 2 & -1  \\ -5 & 0  \\ \end{matrix} \right|=-5$

${{A}_{12}}={{(-1)}^{1+2}}\left| \begin{matrix} 0 & -1  \\ 3 & 0  \\ \end{matrix} \right|=-3$

${{A}_{13}}={{(-1)}^{1+3}}\left| \begin{matrix} 0 & 2  \\ 3 & -5  \\ \end{matrix} \right|=-6$

${{A}_{21}}={{(-1)}^{2+1}}\left| \begin{matrix} -1 & -2  \\ -5 & 0  \\ \end{matrix} \right|=10$

${{A}_{22}}={{(-1)}^{2+2}}\left| \begin{matrix} 3 & -2  \\ 3 & 0  \\ \end{matrix} \right|=6$

${{A}_{23}}={{(-1)}^{2+3}}\left| \begin{matrix} 3 & -1  \\ 3 & -5  \\ \end{matrix} \right|=-[-15+3]=12$

${{A}_{31}}={{(-1)}^{3+1}}\left| \begin{matrix} -1 & -2  \\ 2 & -1  \\ \end{matrix} \right|=[1+4]=5$

${{A}_{32}}={{(-1)}^{3+2}}\left| \begin{matrix} 3 & -2  \\ 0 & -1  \\ \end{matrix} \right|=3$

${{A}_{33}}={{(-1)}^{3+3}}\left| \begin{matrix} 3 & -1  \\ 0 & 2  \\ \end{matrix} \right|=6$

Cofactor matrix is \[\left[ \begin{matrix} \text{-5} & \text{-3} & -6  \\ 10 & \text{6} & 12  \\ 5 & 3 & \text{6}  \\ \end{matrix} \right]\]

Taking transpose,

\[\Rightarrow \left( \text{adjA} \right)\text{=}\left[ \begin{matrix} \text{-5} & \text{10} & \text{5}  \\ \text{-3} & \text{6} & \text{3}  \\ \text{-6} & \text{12} & \text{6}  \\ \end{matrix} \right]\]

\[\Rightarrow \left( \text{adjA} \right)\text{B=}\left[ \begin{matrix} \text{-5} & \text{10} & \text{5}  \\ \text{-3} & \text{6} & \text{3}  \\ \text{-6} & \text{12} & \text{6}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{2}  \\ \text{-1}  \\ \text{3}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{-10-10+15}  \\ \text{-6-6+9}  \\ \text{-12-12+18}  \\ \end{matrix} \right]\]

\[\therefore \left( \text{adjA} \right)\text{B=}\left[ \begin{matrix} \text{-5}  \\ \text{-3}  \\ \text{-6}  \\ \end{matrix} \right]\ne 0\]

Thus, the solution of the given system of equation does not exist.

$\therefore $ The system of equations is inconsistent.


6. Examine the consistency of the system of equations.

\[\mathbf{\text{5x-y+4z=5}}\]

\[\mathbf{\text{2x+3y+5z=2}}\]

\[\mathbf{\text{5x-2y+6z=-1}}\]

Ans: Given equations,

\[\text{5x-y+4z=5}\]

\[\text{2x+3y+5z=2}\]

\[\text{5x-2y+6z=-1}\]

Let \[\text{A=}\left[ \begin{matrix} \text{5} & \text{-1} & \text{4}  \\ \text{2} & \text{3} & \text{5}  \\ \text{3} & \text{-2} & \text{6}  \\ \end{matrix} \right]\] , \[\text{X=}\left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\] and \[\text{B=}\left[ \begin{matrix} \text{5}  \\ \text{2}  \\ \text{-1}  \\ \end{matrix} \right]\] such that, the system of equations can be written in the form of \[\text{AX=B}\].

Determining the value of $A$, we have:

\[\Rightarrow \left| \text{A} \right|\text{=5}\left( \text{18+10} \right)\text{+1}\left( \text{12-25} \right)\text{+4}\left( \text{-4-15} \right)\]

\[\Rightarrow \left| \text{A} \right|\text{=5}\left( \text{28} \right)\text{+1}\left( \text{-13} \right)\text{+4}\left( \text{-19} \right)\]

\[\Rightarrow \left| \text{A} \right|\text{=140-13-76}\]

\[\therefore \left| \text{A} \right|\text{=51}\ne \text{0}\]

Hence, \[\text{A}\] is a non-singular matrix.

Thus, \[{{\text{A}}^{\text{-1}}}\] exists.

$\therefore $ The given system of equations is consistent.


7. Solve system of linear equations, using matrix method.

\[\mathbf{\text{5x+2y=4}}\]

\[\mathbf{\text{7x+3y=5}}\]

Ans: Given equations,

\[\text{5x+2y=4}\]

\[\text{7x+3y=5}\]

Let \[\text{A=}\left[ \begin{matrix} \text{5} & \text{2}  \\ \text{7} & \text{3}  \\ \end{matrix} \right]\] , \[\text{X=}\left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \end{matrix} \right]\] and \[\text{B=}\left[ \begin{matrix} \text{4}  \\ \text{5}  \\ \end{matrix} \right]\] such that, this system of equations can be written in the form of \[\text{AX=B}\].

Determining the value of $A$, we have:

\[\Rightarrow \left| \text{A} \right|\text{=15-14}\]

\[\therefore \left| \text{A} \right|=1\ne 0\]

Thus, \[\text{A}\] is a non-singular matrix.

Hence, its inverse exists.

We know that the adjoint of a square matrix is the transpose of its cofactor matrix.

$\therefore adjA=\left[ \begin{matrix} 3 & -2  \\ -7 & 5  \\ \end{matrix} \right]$

Thus,

\[\Rightarrow {{\text{A}}^{\text{-1}}}=\left[ \begin{matrix} \text{3} & \text{-2}  \\ \text{-7} & \text{5}  \\ \end{matrix} \right]\]

\[\Rightarrow \text{X=}{{\text{A}}^{\text{-1}}}\text{B=}\left[ \begin{matrix} \text{3} & \text{-2}  \\ \text{-7} & \text{5}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{4}  \\ \text{5}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{12+10}  \\ \text{-28+52}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{2}  \\ \text{-3}  \\ \end{matrix} \right]\]

Thus, \[\text{x=2}\] and \[\text{y=}-3\].


8. Solve system of linear equations, using matrix method.

\[\mathbf{\text{2x-y=-2}}\]

\[\mathbf{\text{3x+4y=3}}\]

Ans: Given equations,

\[\text{2x-y=-2}\]

\[\text{3x+4y=3}\]

Let \[\text{A=}\left[ \begin{matrix} \text{2} & \text{-1}  \\ \text{3} & \text{4}  \\ \end{matrix} \right]\] , \[\text{X=}\left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \end{matrix} \right]\] and \[\text{B=}\left[ \begin{matrix} \text{-2}  \\ \text{3}  \\ \end{matrix} \right]\] such that, this system of equations can be written in the form of \[\text{AX=B}\].

Determining the value of $A$, we have:

\[\Rightarrow \left| \text{A} \right|\text{=8+3}\]

\[\therefore \left| \text{A} \right|=11\ne 0\]

Thus, \[\text{A}\] is non-singular.

 $\therefore $ It’s inverse exists.

We know that the adjoint of a square matrix is the transpose of its cofactor matrix.

$\therefore adjA=\left[ \begin{matrix} 4 & 1  \\ -3 & 2  \\ \end{matrix} \right]$

Now,

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\left( \text{adjA} \right)\]

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{11}}\left[ \begin{matrix} \text{4} & \text{1}  \\ \text{-3} & \text{2}  \\ \end{matrix} \right]\]

\[\therefore \text{X=}{{\text{A}}^{\text{-1}}}\text{B=}\dfrac{\text{1}}{\text{11}}\left[ \begin{matrix} \text{4} & \text{1}  \\ \text{-3} & \text{2}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{-2}  \\ \text{3}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{11}}\left[ \begin{matrix} \text{-8+3}  \\ \text{6+6}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{11}}\left[ \begin{matrix} \text{-5}  \\ \text{12}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{-}\dfrac{\text{5}}{\text{11}}  \\ \dfrac{\text{12}}{\text{11}}  \\ \end{matrix} \right]\]

Thus, \[x=-\dfrac{5}{11}\] and \[y=\dfrac{12}{11}\].

9. Solve system of linear equations, using matrix method.

\[\mathbf{\text{4x-3y=3}}\]

\[\mathbf{\text{3x-5y=7}}\]

Ans: Given equations,

\[\text{4x-3y=3}\]

\[\text{3x-5y=7}\].

Let \[\text{A=}\left[ \begin{matrix} \text{4} & \text{-3}  \\ \text{3} & \text{-5}  \\ \end{matrix} \right]\] , \[\text{X=}\left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \end{matrix} \right]\] and \[\text{B=}\left[ \begin{matrix} \text{3}  \\ \text{7}  \\ \end{matrix} \right]\] such that, this system of equations can be written in the form of \[\text{AX=B}\].

Determining the value of $A$, we have:

\[\Rightarrow \left| \text{A} \right|\text{=-20+9}\]

\[\therefore \left| \text{A} \right|\text{=-11}\ne \text{0}\]

Thus, \[\text{A}\] is a non-singular matrix.

Hence, its inverse exists.

Formula for inverse is ${{A}^{-1}}=\dfrac{adjA}{\left| A \right|}$.

Finding cofactors,

${{A}_{11}}={{(-1)}^{1+1}}(-5)=-5$

${{A}_{12}}={{(-1)}^{1+2}}(3)=-3$

${{A}_{21}}={{(-1)}^{2+1}}(-3)=3$

${{A}_{22}}={{(-1)}^{2+2}}(4)=4$

Cofactor matrix is $\left[ \begin{matrix} -5 & -3  \\ 3 & 4  \\ \end{matrix} \right]$.

Taking its transpose to get adjoint matrix as $\left[ \begin{matrix} -5 & 3  \\ -3 & 4  \\ \end{matrix} \right]$.

Therefore inverse is

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=-}\dfrac{\text{1}}{\text{11}}\left[ \begin{matrix} \text{-5} & \text{3}  \\ \text{-3} & \text{4}  \\ \end{matrix} \right]\]

\[\therefore \text{X=}{{\text{A}}^{\text{-1}}}\text{B=-}\dfrac{\text{1}}{\text{11}}\left[ \begin{matrix} \text{-5} & \text{3}  \\ \text{-3} & \text{4}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{3}  \\ \text{7}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{11}}\left[ \begin{matrix} \text{5} & \text{-3}  \\ \text{3} & \text{-4}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{3}  \\ \text{7}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{11}}\left[ \begin{matrix} \text{15-21}  \\ \text{9-28}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{-}\dfrac{\text{6}}{\text{11}}  \\ \text{-}\dfrac{\text{19}}{\text{11}}  \\ \end{matrix} \right]\]

Thus, \[x=-\dfrac{6}{11}\] and \[y=-\dfrac{19}{11}\].

10. Solve system of linear equations, using matrix method.

\[\mathbf{\text{5x+2y=3}}\]

\[\mathbf{\text{3x+2y=5}}\]

Ans: Given equations,

\[\text{5x+2y=3}\]

\[\text{5x+2y=3}\]

Let \[\text{A=}\left[ \begin{matrix} \text{5} & \text{2}  \\ \text{3} & \text{2}  \\ \end{matrix} \right]\] , \[\text{X=}\left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \end{matrix} \right]\] and \[\text{B=}\left[ \begin{matrix} \text{3}  \\ \text{5}  \\ \end{matrix} \right]\] such that, this system of equations can be written in the form of \[\text{AX=B}\].

Determining the value of $A$, we have:

\[\Rightarrow \left| \text{A} \right|\text{=10}-6\]

\[\therefore \left| \text{A} \right|\text{=4}\ne \text{0}\]

Thus \[\text{A}\] is non-singular,

Therefore, its inverse exists.

Formula for inverse is ${{A}^{-1}}=\dfrac{adjA}{\left| A \right|}$.

Finding cofactors,

${{A}_{11}}={{(-1)}^{1+1}}(2)=2$

${{A}_{12}}={{(-1)}^{1+2}}(3)=-3$

${{A}_{21}}={{(-1)}^{2+1}}(2)=-2$

${{A}_{22}}={{(-1)}^{2+2}}(5)=5$

Cofactor matrix is $\left[ \begin{matrix} 2 & -3  \\ -2 & 5  \\ \end{matrix} \right]$.

Taking its transpose to get adjoint matrix as $\left[ \begin{matrix} 2 & -2  \\ -3 & 5  \\ \end{matrix} \right]$.

Therefore inverse is

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{4}\left[ \begin{matrix} 2 & -2  \\ -3 & 5  \\ \end{matrix} \right]\]

\[\therefore \text{X=}{{\text{A}}^{\text{-1}}}\text{B=}\dfrac{\text{1}}{4}\left[ \begin{matrix} 2 & -2  \\ -3 & 5  \\ \end{matrix} \right]\left[ \begin{matrix} \text{3}  \\ 5  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{4}\left[ \begin{matrix} 2 & -2  \\ -3 & 5  \\ \end{matrix} \right]\left[ \begin{matrix} \text{3}  \\ 5  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{4}\left[ \begin{matrix} 6-10  \\ -9+25  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} -\dfrac{4}{4}  \\ \dfrac{\text{16}}{4}  \\ \end{matrix} \right]\]

Thus, \[x=-1\] and \[y=4\].

11. Solve system of linear equations, using matrix method.

\[\mathbf{\text{2x+y+z=1}}\]

\[\mathbf{\text{x-2y-z=}\dfrac{\text{3}}{\text{2}}}\]

\[\mathbf{\text{3y-5z=9}}\]

Ans: Given equations,

\[\text{2x+y+z=1}\]

\[\text{x-2y-z=}\dfrac{\text{3}}{\text{2}}\]

\[\text{3y-5z=9}\]

Let \[\text{A=}\left[ \begin{matrix} \text{2} & \text{1} & \text{1}  \\ \text{1} & \text{-2} & \text{-1}  \\ \text{0} & \text{3} & \text{-5}  \\ \end{matrix} \right]\] , \[\text{X=}\left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\] and \[\text{B=}\left[ \begin{matrix} \text{1}  \\ \dfrac{\text{3}}{\text{2}}  \\ \text{9}  \\ \end{matrix} \right]\] such that, this system of equations can be written in the form of \[\text{AX=B}\].

The determinant of $A$ is found by expanding along the first column:

\[\Rightarrow \left| \text{A} \right|\text{=2}\left( \text{10+3} \right)\text{-1}\left( \text{-5-3} \right)\text{+0}\]

\[\Rightarrow \left| \text{A} \right|\text{=2}\left( \text{13} \right)\text{-1}\left( \text{-8} \right)\]

\[\Rightarrow \left| \text{A} \right|=\text{34}\ne \text{0}\]

Thus, \[\text{A}\] is a non-singular matrix.

$\therefore $ Its inverse exists.

Hence,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{A}_{11}}\text{=13}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{A}_{12}}=5\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]

\[\Rightarrow {{A}_{13}}=3\]

Similarly,

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=8}\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=-10\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}-6\]

and

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]

\[\Rightarrow {{\text{A}}_{31}}=1\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]

\[\Rightarrow {{\text{A}}_{32}}=3\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]

\[\Rightarrow {{\text{A}}_{33}}=-5\].

Cofactor matrix is \[\left[ \begin{matrix} 13 & 5 & 3  \\ 8 & -10 & -6  \\ 1 & 3 & -5  \\ \end{matrix} \right]\].

We know that adjoint of a matrix is the transpose of its cofactor matrix.

\[\Rightarrow adjA={{\left[ \begin{matrix} 13 & 5 & 3  \\ 8 & -10 & -6  \\ 1 & 3 & -5  \\ \end{matrix} \right]}^{T}}\]

\[\Rightarrow adjA=\left[ \begin{matrix} \text{13} & \text{8} & \text{1}  \\ \text{5} & \text{-10} & \text{3}  \\ \text{3} & \text{-16} & \text{-5}  \\ \end{matrix} \right]\]

The inverse of a matrix is given by:

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\left( \text{adjA} \right)\]

\[\therefore {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{34}}\left[ \begin{matrix} \text{13} & \text{8} & \text{1}  \\ \text{5} & \text{-10} & \text{3}  \\ \text{3} & \text{-16} & \text{-5}  \\ \end{matrix} \right]\]

\[\therefore \text{X=}{{\text{A}}^{\text{-1}}}\text{B=}\dfrac{\text{1}}{\text{34}}\left[ \begin{matrix} \text{13} & \text{8} & \text{1}  \\ \text{5} & \text{-10} & \text{3}  \\ \text{3} & \text{-16} & \text{-5}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{1}  \\ \dfrac{\text{3}}{\text{2}}  \\ \text{9}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{34}}\left[ \begin{matrix} \text{13+12+9}  \\ \text{5-15+27}  \\ \text{3-9-45}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{34}}\left[ \begin{matrix} \text{34}  \\ \text{17}  \\ \text{-51}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{1}  \\ \dfrac{\text{1}}{\text{2}}  \\ \text{-}\dfrac{\text{3}}{\text{2}}  \\ \end{matrix} \right]\]

Thus, \[x=1\] and \[y=\dfrac{1}{2}\] and \[z=-\dfrac{3}{2}\].

12. Solve system of linear equations, using matrix method.

\[\mathbf{\text{x-y+z=4}}\]

\[\mathbf{\text{2x+y-3z=0}}\]

\[\mathbf{\text{x+y+z=2}}\]

Ans: Given equations,

\[\text{x-y+z=4}\]

\[\text{2x+y-3z=0}\]

\[\text{x+y+z=2}\]

Let \[\text{A=}\left[ \begin{matrix} \text{1} & \text{1} & \text{1}  \\ \text{2} & \text{1} & \text{-3}  \\ \text{1} & \text{1} & \text{1}  \\ \end{matrix} \right]\] , \[\text{X=}\left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\] and \[\text{B=}\left[ \begin{matrix} \text{4}  \\ \text{0}  \\ \text{2}  \\ \end{matrix} \right]\] such that, this system of equations can be written in the form of \[\text{AX=B}\].

Determining the value of $A$, we have:

\[\Rightarrow \left| \text{A} \right|\text{=1}\left( \text{1+3} \right)\text{+1}\left( \text{2+3} \right)\text{+1}\left( \text{2-1} \right)\]

\[\Rightarrow \left| \text{A} \right|\text{=4+5+1}\]

\[\therefore \left| \text{A} \right|\text{=10}\ne \text{0}\]

Thus \[\text{A}\] is non-singular.

$\therefore $ Its inverse exists.

Hence,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{A}_{11}}\text{=4}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{A}_{12}}=-5\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]

\[\Rightarrow {{A}_{13}}=1\]

Similarly,

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=2}\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=0\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}-\text{2}\]

and

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]

\[\Rightarrow {{\text{A}}_{31}}=2\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]

\[\Rightarrow {{\text{A}}_{32}}=5\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]

\[\Rightarrow {{\text{A}}_{33}}=3\].

Cofactor matrix is \[\left[ \begin{matrix} 4 & -5 & 1  \\ 2 & 0 & -2  \\ 2 & 5 & 3  \\ \end{matrix} \right]\].

We know that adjoint of a matrix is the transpose of its cofactor matrix.

\[\Rightarrow adjA={{\left[ \begin{matrix} 4 & -5 & 1  \\ 2 & 0 & -2  \\ 2 & 5 & 3  \\ \end{matrix} \right]}^{T}}\]

The inverse of a matrix is given by:

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\left( \text{adjA} \right)\]

\[\therefore {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{10}}\left[ \begin{matrix} \text{4} & \text{2} & \text{2}  \\ \text{-5} & \text{0} & \text{5}  \\ \text{1} & \text{-2} & \text{3}  \\ \end{matrix} \right]\]

\[\therefore \text{X=}{{\text{A}}^{\text{-1}}}\text{B}\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{10}}\left[ \begin{matrix} \text{4} & \text{2} & \text{2}  \\ \text{-5} & \text{0} & \text{5}  \\ \text{1} & \text{-2} & \text{3}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{4}  \\ \text{0}  \\ \text{2}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{10}}\left[ \begin{matrix} \text{16+0+4}  \\ \text{-20+0+10}  \\ \text{4+0+6}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{10}}\left[ \begin{matrix} \text{20}  \\ \text{-10}  \\ \text{10}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{2}  \\ \text{-1}  \\ \text{1}  \\ \end{matrix} \right]\]

Thus, \[x=2\] , \[y=-1\] and \[z=1\].


13. Solve system of linear equations, using matrix method.

\[\mathbf{\text{2x+3y+3z=5}}\]

\[\mathbf{\text{x-2y+z=-4}}\]

\[\mathbf{\text{3x-y-2z=3}}\]

Ans: Given equations,

\[\text{2x+3y+3z=5}\]

\[\text{x-2y+z=-4}\]

\[\text{3x-y-2z=3}\]

Let \[\text{A=}\left[ \begin{matrix} \text{2} & \text{3} & \text{3}  \\ \text{1} & \text{-2} & \text{1}  \\ \text{3} & \text{-1} & \text{-2}  \\ \end{matrix} \right]\] , \[\text{X=}\left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\] and \[\text{B=}\left[ \begin{matrix} \text{5}  \\ \text{-4}  \\ \text{3}  \\ \end{matrix} \right]\] such that, this system of equations can be written in the form of \[\text{AX=B}\].

Determining the value of $A$, we have:

\[\Rightarrow \left| \text{A} \right|\text{=2}\left( \text{4+1} \right)\text{-3}\left( \text{2-3} \right)\text{+3}\left( \text{-1+6} \right)\]

\[\Rightarrow \left| \text{A} \right|\text{=2}\left( \text{5} \right)\text{-3}\left( \text{-5} \right)\text{+3}\left( \text{5} \right)\]

\[\Rightarrow \left| \text{A} \right|\text{=10+15+15}\]

\[\Rightarrow \left| \text{A} \right|\text{=40}\ne \text{0}\]

Thus, \[\text{A}\] is a non-singular matrix.

$\therefore $ It’s inverse exists.

Hence,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{A}_{11}}\text{=5}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{A}_{12}}=5\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]

\[\Rightarrow {{A}_{13}}=5\]

Similarly,

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=3}\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=-13\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]

\[\Rightarrow {{\text{A}}_{23}}\text{=11}\]

and

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]

\[\Rightarrow {{\text{A}}_{31}}=9\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]

\[\Rightarrow {{\text{A}}_{32}}=1\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]

\[\Rightarrow {{\text{A}}_{33}}=-7\].

Cofactor matrix is \[\left[ \begin{matrix} 5 & 5 & 5  \\ 3 & -13 & 11  \\ 9 & 1 & -7  \\ \end{matrix} \right]\].

We know that adjoint of a matrix is the transpose of its cofactor matrix.

\[\Rightarrow adjA={{\left[ \begin{matrix} 5 & 5 & 5  \\ 3 & -13 & 11  \\ 9 & 1 & -7  \\ \end{matrix} \right]}^{T}}\]

The inverse of a matrix is given by:

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\left( \text{adjA} \right)\]

\[\therefore {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{40}}\left[ \begin{matrix} \text{5} & \text{3} & \text{9}  \\ \text{5} & \text{-13} & \text{1}  \\ \text{5} & \text{11} & \text{-7}  \\ \end{matrix} \right]\]

\[\therefore \text{X=}{{\text{A}}^{\text{-1}}}\text{B}\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{40}}\left[ \begin{matrix} \text{5} & \text{3} & \text{9}  \\ \text{5} & \text{-13} & \text{1}  \\ \text{5} & \text{11} & \text{-7}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{5}  \\ \text{-4}  \\ \text{3}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{40}}\left[ \begin{matrix} \text{25-12+27}  \\ \text{25+52+3}  \\ \text{25-44-21}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{40}}\left[ \begin{matrix} \text{40}  \\ \text{80}  \\ \text{-40}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{1}  \\ \text{2}  \\ \text{-1}  \\ \end{matrix} \right]\]

Thus, \[x=1\], \[y=2\] and \[z=-1\].


14. Solve system of linear equations, using matrix method.

\[\mathbf{\text{x-y+2z=7}}\]

\[\mathbf{\text{3x+4y-5z=-5}}\]

\[\mathbf{\text{2x-y+3z=12}}\]

Ans: Given equations,

\[\text{x-y+2z=7}\]

\[\text{3x+4y-5z=-5}\]

\[\text{2x-y+3z=12}\]

Let \[\text{A=}\left[ \begin{matrix} \text{1} & \text{-1} & \text{2}  \\ \text{3} & \text{4} & \text{-5}  \\ \text{2} & \text{-1} & \text{3}  \\ \end{matrix} \right]\] , \[\text{X=}\left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\] and \[\text{B=}\left[ \begin{matrix} \text{7}  \\ \text{-5}  \\ \text{12}  \\ \end{matrix} \right]\] such that, this system of equations can be written in the form of \[\text{AX=B}\].

Determining the value of $A$, we have:

\[\Rightarrow \left| \text{A} \right|\text{=1}\left( \text{12-5} \right)\text{+1}\left( \text{9+10} \right)\text{+2}\left( \text{-3-8} \right)\]

\[\Rightarrow \left| \text{A} \right|\text{=7+19-22}\]

\[\therefore \left| \text{A} \right|\text{=4}\ne \text{0}\]

Thus, \[\text{A}\] is a non-singular matrix.

$\therefore $ It’s inverse exists.

Now,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{A}_{11}}\text{=7}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{A}_{12}}=-19\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]

\[\Rightarrow {{A}_{13}}=-11\]

Similarly,

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=1}\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=-1\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}-1\]

and

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]

\[\Rightarrow {{\text{A}}_{31}}=-3\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]

\[\Rightarrow {{\text{A}}_{32}}=11\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]

\[\Rightarrow {{\text{A}}_{33}}=7\].

Cofactor matrix is \[\left[ \begin{matrix} 7 & -19 & -11  \\ 1 & -1 & -1  \\ -3 & 11 & 7  \\ \end{matrix} \right]\].

We know that the adjoint of a matrix is the transpose of its cofactor matrix.

\[\Rightarrow adjA={{\left[ \begin{matrix} 7 & -19 & -11  \\ 1 & -1 & -1  \\ -3 & 11 & 7  \\ \end{matrix} \right]}^{T}}\]

\[\Rightarrow adjA=\left[ \begin{matrix} 7 & 1 & -3  \\ -19 & -1 & 11  \\ -11 & -1 & 7  \\ \end{matrix} \right]\]

The inverse of a matrix is given by:

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\left( \text{adjA} \right)\]

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{4}}\left[ \begin{matrix} \text{7} & \text{1} & \text{-3}  \\ \text{-19} & \text{-1} & \text{11}  \\ \text{-11} & \text{-1} & \text{7}  \\ \end{matrix} \right]\]

\[\therefore \text{X=}{{\text{A}}^{\text{-1}}}\text{B}\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{4}}\left[ \begin{matrix} \text{7} & \text{1} & \text{-3}  \\ \text{-19} & \text{-1} & \text{11}  \\ \text{-11} & \text{-1} & \text{7}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{7}  \\ \text{-5}  \\ \text{12}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{4}}\left[ \begin{matrix} \text{49-5-36}  \\ \text{-133+5+132}  \\ \text{-77+5+84}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{4}}\left[ \begin{matrix} \text{8}  \\ \text{4}  \\ \text{12}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{2}  \\ \text{1}  \\ \text{3}  \\ \end{matrix} \right]\]

Thus \[\text{x=2}\] , \[\text{y=1}\] and \[\text{z=3}\]


15. If $\mathbf{A=\left[ \begin{matrix} 2 & -3 & 5  \\ 3 & 2 & -4  \\ 1 & 1 & -2  \\ \end{matrix} \right] }$, find $\mathbf{{{A}^{-1}}}$. Using $\mathbf{{{A}^{-1}}}$ solve the system of equations

$\mathbf{2x-3y+5z=11}$

$\mathbf{3x+2y-4z=-5}$

$\mathbf{x+y-2z=-3}$

Ans: Given equations,

\[2x-3y+5z=11\]

\[3x+2y-4z=-5\]

\[x+y-2z=-3\]

Let \[\text{A=}\left[ \begin{matrix} 2 & -3 & 5  \\ \text{3} & 2 & -4  \\ 1 & \text{1} & -2  \\ \end{matrix} \right]\] , \[\text{X=}\left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\] and \[\text{B=}\left[ \begin{matrix} 11  \\ -5  \\ -3  \\ \end{matrix} \right]\] such that, this system of equations can be written in the form of \[\text{AX=B}\].

Determining the value of $A$, we have:

\[\Rightarrow \left| \text{A} \right|\text{=2}\left( \text{-4+4} \right)\text{+3}\left( \text{-6+4} \right)\text{+5}\left( \text{3-2} \right)\]

\[\Rightarrow \left| \text{A} \right|\text{=0-6+5}\]

\[\therefore \left| \text{A} \right|\text{=-1}\ne \text{0}\]

Thus, \[\text{A}\] is a non-singular matrix.

$\therefore $ It’s inverse exists.

Now,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{A}_{11}}\text{=0}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{A}_{12}}=2\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]

\[\Rightarrow {{A}_{13}}=1\]

Similarly,

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=-1}\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=-9\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}-5\]

and

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]

\[\Rightarrow {{\text{A}}_{31}}=2\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]

\[\Rightarrow {{\text{A}}_{32}}=23\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]

\[\Rightarrow {{\text{A}}_{33}}=13\].

Cofactor matrix is \[\left[ \begin{matrix} 0 & 2 & 1  \\ -1 & -9 & -5  \\ 2 & 23 & 13  \\ \end{matrix} \right]\].

We know that adjoint of a matrix is the transpose of its cofactor matrix.

\[\Rightarrow adjA={{\left[ \begin{matrix} 0 & 2 & 1  \\ -1 & -9 & -5  \\ 2 & 23 & 13  \\ \end{matrix} \right]}^{T}}\]

\[\Rightarrow adjA=\left[ \begin{matrix} 0 & -1 & 2  \\ 2 & -9 & 23  \\ 1 & -5 & 13  \\ \end{matrix} \right]\]

The inverse of a matrix is given by:

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\left( \text{adjA} \right)\]

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{-1}\left[ \begin{matrix} 0 & -1 & 2  \\ 2 & -9 & 23  \\ 1 & -5 & 13  \\ \end{matrix} \right]\]

\[\therefore \text{X=}{{\text{A}}^{\text{-1}}}\text{B}\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=-1}\left[ \begin{matrix} 0 & -1 & 2  \\ 2 & -9 & 23  \\ 1 & -5 & 13  \\ \end{matrix} \right]\left[ \begin{matrix} 11  \\ -5  \\ -3  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=-1}\left[ \begin{matrix} \text{5-6}  \\ \text{22+45-69}  \\ \text{11+25-39}  \\ \end{matrix} \right]\] \[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=-1}\left[ \begin{matrix} -1  \\ -2  \\ -3  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} 1  \\ 2  \\ \text{3}  \\ \end{matrix} \right]\]         

Thus \[\text{x=1}\] , \[\text{y=2}\] and \[\text{z=3}\].


16. The cost of \[\mathbf{\text{4kg}}\] onion, \[\mathbf{\text{3kg}}\] wheat and \[\mathbf{\text{2kg}}\] rice is Rs\[\mathbf{\text{60}}\]. The cost of  \[\mathbf{\text{2kg}}\] onion, \[\mathbf{\text{4kg}}\]wheat and \[\mathbf{\text{6kg}}\] rice is Rs\[\mathbf{\text{90}}\]. The cost of \[\mathbf{\text{6kg}}\] onion \[\mathbf{\text{2kg}}\] wheat and \[\mathbf{\text{3kg}}\] rice is Rs\[\mathbf{\text{70}}\].

Find cost of each item per kg by matrix method.

Ans: Let us suppose that the cost of onions, wheat and rice per kg be Rs \[\text{x}\] , Rs \[\text{y}\] and Rs \[\text{z}\] respectively.

Then, the given situation can be represented by a system of equations as:

\[\Rightarrow \text{4x+3y+2z=60}\]

\[\Rightarrow \text{2x+4y+6z=90}\]

\[\Rightarrow \text{6x+2y+3z=70}\]

Let \[\text{A=}\left[ \begin{matrix} \text{4} & \text{3} & \text{2}  \\ \text{2} & \text{4} & \text{6}  \\ \text{6} & \text{2} & \text{3}  \\ \end{matrix} \right]\] , \[\text{X=}\left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\] and \[\text{B=}\left[ \begin{matrix} \text{60}  \\ \text{90}  \\ \text{70}  \\ \end{matrix} \right]\] such that, this system of equations can be written in the form of \[\text{AX=B}\].

\[\Rightarrow \left| \text{A} \right|\text{=4}\left( \text{12-12} \right)\text{-3}\left( \text{6-36} \right)\text{+2}\left( \text{4-24} \right)\]

\[\Rightarrow \left| \text{A} \right|\text{=0+90-40}\]

\[\Rightarrow \left| \text{A} \right|\text{=50}\ne \text{0}\]

Thus,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{A}_{11}}\text{=0}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{A}_{12}}=30\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]

\[\Rightarrow {{A}_{13}}=-20\]

Similarly,

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}-5\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=0\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]

\[\Rightarrow {{\text{A}}_{23}}\text{=10}\]

And

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]

\[\Rightarrow {{\text{A}}_{31}}=10\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]

\[\Rightarrow {{\text{A}}_{32}}=-20\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]

\[\Rightarrow {{\text{A}}_{33}}=10\].

Cofactor matrix is \[\left[ \begin{matrix} 0 & 30 & -20  \\ -5 & 0 & 10  \\ 10 & -20 & 10  \\ \end{matrix} \right]\].

We know that adjoint of a matrix is the transpose of its cofactor matrix.

\[\Rightarrow adjA={{\left[ \begin{matrix} 0 & 30 & -20  \\ -5 & 0 & 10  \\ 10 & -20 & 10  \\ \end{matrix} \right]}^{T}}\]

\[\therefore \left( \text{adjA} \right)\text{=}\left[ \begin{matrix} \text{0} & \text{-5} & \text{10}  \\ \text{30} & \text{0} & \text{-20}  \\ \text{-20} & \text{10} & \text{10}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\left( \text{adjA} \right)\]

\[\therefore {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{50}}\left[ \begin{matrix} \text{0} & \text{-5} & \text{10}  \\ \text{30} & \text{0} & \text{-20}  \\ \text{-20} & \text{10} & \text{10}  \\ \end{matrix} \right]\]

Since, \[\text{X=}{{\text{A}}^{\text{-1}}}\text{B}\]

\[\Rightarrow \text{X=}\dfrac{\text{1}}{\text{50}}\left[ \begin{matrix} \text{0} & \text{-5} & \text{10}  \\ \text{30} & \text{0} & \text{-20}  \\ \text{-20} & \text{10} & \text{10}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{60}  \\ \text{90}  \\ \text{70}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{50}}\left[ \begin{matrix} \text{0+450+700}  \\ \text{1800+0-1400}  \\ \text{-1200+900+700}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{50}}\left[ \begin{matrix} \text{250}  \\ \text{400}  \\ \text{400}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{5}  \\ \text{8}  \\ \text{8}  \\ \end{matrix} \right]\]

Thus \[\text{x=5}\] , \[\text{y=8}\] , and \[\text{z=8}\]

Hence, the cost of onions is Rs \[\text{5}\] per kg, the cost of wheat is Rs \[\text{8}\] per kg, and the cost of rice is Rs \[\text{8}\] per kg.


Miscellaneous Solutions

1. Prove that the determinant \[\mathbf{\left| \begin{matrix} x & \sin \theta  & \cos \theta   \\ -\sin \theta  & -x & 1  \\ \cos \theta  & 1 & x  \\ \end{matrix} \right|}\] is independent of \[\mathbf{\theta} \].

Ans: Let \[\Delta =\left| \begin{matrix} x & \sin \theta  & \cos \theta   \\ -\sin \theta  & -x & 1  \\ \cos \theta  & 1 & x  \\ \end{matrix} \right|\]

Solving it, we have:

\[\Rightarrow \Delta =x\left( {{x}^{2}}-1 \right)-\sin \theta \left( -x\sin \theta -\cos \theta  \right)+\cos \theta \left( -\sin \theta +x\cos \theta  \right)\]

\[\Rightarrow \Delta ={{x}^{3}}-x+x{{\sin }^{2}}\theta +\sin \theta \cos \theta -\sin \theta \cos \theta +x{{\cos }^{2}}\theta \]

\[\Rightarrow \Delta ={{x}^{3}}-x+x\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta  \right)\]

\[\Rightarrow \Delta \text{=}{{\text{x}}^{\text{3}}}\text{-x+x}\]

\[\therefore \Delta \text{=}{{\text{x}}^{\text{3}}}\]

Hence, \[\text{ }\!\!\Delta\!\!\text{ }\] is independent of \[\theta \] .


2. Without expanding the determinant, prove that \[\mathbf{\left| \begin{matrix} \text{a} & {{\text{a}}^{\text{2}}} & \text{bc}  \\ \text{b} & {{\text{b}}^{\text{2}}} & \text{ca}  \\ \text{c} & {{\text{c}}^{\text{2}}} & \text{ab}  \\ \end{matrix} \right|\text{=}\left| \begin{matrix} \text{1} & {{\text{a}}^{\text{2}}} & {{\text{a}}^{\text{3}}}  \\ \text{1} & {{\text{b}}^{\text{2}}} & {{\text{b}}^{\text{3}}}  \\ \text{1} & {{\text{c}}^{\text{2}}} & {{\text{c}}^{\text{3}}}  \\ \end{matrix} \right|}\].

Ans: Solving the left-hand side determinant.

\[\text{L}\text{.H}\text{.S}\text{.}\] \[\text{=}\left| \begin{matrix} \text{a} & {{\text{a}}^{\text{2}}} & \text{bc}  \\ \text{b} & {{\text{b}}^{\text{2}}} & \text{ca}  \\ \text{c} & {{\text{c}}^{\text{2}}} & \text{ab}  \\ \end{matrix} \right|\]

Applying the row operations, \[{{\text{R}}_{1}}\to \text{a}{{\text{R}}_{1}}\], \[{{\text{R}}_{2}}\to \text{b}{{\text{R}}_{2}}\] and \[{{\text{R}}_{3}}\to \text{c}{{\text{R}}_{3}}\]

\[\Rightarrow \dfrac{\text{1}}{\text{abc}}\left| \begin{matrix} {{\text{a}}^{\text{2}}} & {{\text{a}}^{\text{3}}} & \text{abc}  \\ {{\text{b}}^{\text{2}}} & {{\text{b}}^{\text{3}}} & \text{abc}  \\ {{\text{c}}^{\text{2}}} & {{\text{c}}^{\text{3}}} & \text{abc}  \\ \end{matrix} \right|\]

Taking out \[abc\] common from \[{{\text{C}}_{3}}\]

\[\Rightarrow \dfrac{\text{1}}{\text{abc}}\text{abc}\left| \begin{matrix} {{\text{a}}^{\text{2}}} & {{\text{a}}^{\text{3}}} & \text{1}  \\ {{\text{b}}^{\text{2}}} & {{\text{b}}^{\text{3}}} & \text{1}  \\ {{\text{c}}^{\text{2}}} & {{\text{c}}^{\text{3}}} & \text{1}  \\ \end{matrix} \right|\]

\[\Rightarrow \left| \begin{matrix} {{\text{a}}^{\text{2}}} & {{\text{a}}^{\text{3}}} & \text{1}  \\ {{\text{b}}^{\text{2}}} & {{\text{b}}^{\text{3}}} & \text{1}  \\ {{\text{c}}^{\text{2}}} & {{\text{c}}^{\text{3}}} & \text{1}  \\ \end{matrix} \right|\]

Performing the interchange of columns, \[{{\text{C}}_{\text{1}}}\leftrightarrow {{\text{C}}_{\text{3}}}\] and \[{{\text{C}}_{\text{2}}}\leftrightarrow {{\text{C}}_{\text{3}}}\]

\[\Rightarrow \left| \begin{matrix} \text{1} & {{\text{a}}^{\text{2}}} & {{\text{a}}^{\text{3}}}  \\ \text{1} & {{\text{b}}^{\text{2}}} & {{\text{b}}^{\text{3}}}  \\ \text{1} & {{\text{c}}^{\text{2}}} & {{\text{c}}^{\text{3}}}  \\ \end{matrix} \right|=\] R.H.S

Thus, L.H.S $=$ R.H.S.

\[\left| \begin{matrix} \text{a} & {{\text{a}}^{\text{2}}} & \text{bc}  \\ \text{b} & {{\text{b}}^{\text{2}}} & \text{ca}  \\ \text{c} & {{\text{c}}^{\text{2}}} & \text{ab}  \\ \end{matrix} \right|\text{=}\left| \begin{matrix} \text{1} & {{\text{a}}^{\text{2}}} & {{\text{a}}^{\text{3}}}  \\ \text{1} & {{\text{b}}^{\text{2}}} & {{\text{b}}^{\text{3}}}  \\ \text{1} & {{\text{c}}^{\text{2}}} & {{\text{c}}^{\text{3}}}  \\ \end{matrix} \right|\]

Hence proved.


3. Evaluate \[\mathbf{\left| \begin{matrix} \cos \alpha \cos \beta  & \cos \alpha \sin \beta  & -\sin \alpha   \\ -\sin \beta  & \cos \beta  & 0  \\ \sin \alpha \cos \beta  & \sin \alpha \sin \beta  & \cos \alpha   \\ \end{matrix} \right|}\]

Ans: Let \[\Delta =\left| \begin{matrix} \cos \alpha \cos \beta  & \cos \alpha \sin \beta  & -\sin \alpha   \\ -\sin \beta  & \cos \beta  & 0  \\ \sin \alpha \cos \beta  & \sin \alpha \sin \beta  & \cos \alpha   \\ \end{matrix} \right|\]

Expanding along column \[{{\text{C}}_{3}}\]

\[\Rightarrow \Delta =-\sin \alpha \left( -\sin \alpha {{\sin }^{2}}\beta +{{\cos }^{2}}\beta \sin \alpha  \right)+\cos \alpha \left( \cos \alpha {{\cos }^{2}}\beta +\cos \alpha {{\sin }^{2}}\beta  \right)\]

\[\Rightarrow \Delta ={{\sin }^{2}}\alpha \left( s{{\sin }^{2}}\beta +{{\cos }^{2}}\beta  \right)+{{\cos }^{2}}\alpha \left( {{\cos }^{2}}\beta +{{\sin }^{2}}\beta  \right)\]

\[\Rightarrow \Delta ={{\sin }^{2}}\alpha \left( 1 \right)+{{\cos }^{2}}\alpha \left( 1 \right)\]

\[\therefore \Delta \text{=1}\]


4. If \[\mathbf{\text{a}}\], \[\mathbf{\text{b}}\] and \[\mathbf{\text{c}}\] are real numbers, and \[\mathbf{\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{b+c} & \text{c+a} & \text{a+b}  \\ \text{c+a} & \text{a+b} & \text{b+c}  \\ \text{a+b} & \text{b+c} & \text{c+a}  \\ \end{matrix} \right|\text{=0}}\]

Show that either a+b+c=0 or a=b=c.

Ans: Given, \[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{b+c} & \text{c+a} & \text{a+b}  \\ \text{c+a} & \text{a+b} & \text{b+c}  \\ \text{a+b} & \text{b+c} & \text{c+a}  \\ \end{matrix} \right|\text{=0}\]

Applying the row operation,\[{{R}_{1}}\to {{R}_{1}}\text{+}{{\text{R}}_{2}}\text{+}{{\text{R}}_{3}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{2}\left( \text{a+b+c} \right) & \text{2}\left( \text{a+b+c} \right) & \text{2}\left( \text{a+b+c} \right)  \\ \text{c+a} & \text{a+b} & \text{b+c}  \\ \text{a+b} & \text{b+c} & \text{c+a}  \\ \end{matrix} \right|\]

Taking out \[\text{2}\left( \text{a+b+c} \right)\] common from ${{R}_{1}}$.

\[\Rightarrow \Delta \text{=2}\left( \text{a+b+c} \right)\left| \begin{matrix} \text{1} & \text{1} & \text{1}  \\ \text{c+a} & \text{a+b} & \text{b+c}  \\ \text{a+b} & \text{b+c} & \text{c+a}  \\ \end{matrix} \right|\]

Applying the row operations \[{{\text{C}}_{2}}\to {{C}_{2}}-{{C}_{1}}\] and \[{{C}_{3}}\to {{C}_{3}}-{{C}_{1}}\]

\[\Rightarrow \Delta \text{=2}\left( \text{a+b+c} \right)\left| \begin{matrix} \text{1} & \text{0} & \text{0}  \\ \text{c+a} & \text{b-c} & \text{b-a}  \\ \text{a+b} & \text{c-a} & \text{c-b}  \\ \end{matrix} \right|\]

Expanding along ${{R}_{1}}$

\[\Rightarrow \Delta =2\left( a+b+c \right)\left( 1 \right)\left[ \left( b-c \right)\left( c-b \right)-\left( b-a \right)\left( c-a \right) \right]\]

\[\Rightarrow \Delta =2\left( a+b+c \right)\left[ -{{b}^{2}}-{{c}^{2}}+2bc-bc+ba+ac-{{a}^{2}} \right]\]

\[\Rightarrow \Delta \text{=}2\left( a+b+c \right)\left[ ab+bc+ca-{{a}^{2}}-{{b}^{2}}-{{c}^{2}} \right]\]

It is given that \[\text{ }\!\!\Delta\!\!\text{ =0}\].

Thus, \[\left( \text{a+b+c} \right)\left[ \text{ab+bc+ca-}{{\text{a}}^{\text{2}}}\text{-}{{\text{b}}^{\text{2}}}\text{-}{{\text{c}}^{\text{2}}} \right]\text{=0}\]

$\therefore $ Either \[\text{a+b+c=0}\] , or \[\text{ab+bc+ca-}{{\text{a}}^{\text{2}}}\text{-}{{\text{b}}^{\text{2}}}\text{-}{{\text{c}}^{\text{2}}}\text{=0}\] .

Now, \[\text{ab+bc+ca-}{{\text{a}}^{\text{2}}}\text{-}{{\text{b}}^{\text{2}}}\text{-}{{\text{c}}^{\text{2}}}\text{=0}\]

\[\Rightarrow \text{-2ab-2bc-2ca+2}{{\text{a}}^{\text{3}}}\text{+2}{{\text{b}}^{\text{3}}}\text{+2}{{\text{c}}^{\text{3}}}\text{=0}\]

\[\Rightarrow {{\left( \text{a-b} \right)}^{\text{2}}}\text{+}{{\left( \text{b-c} \right)}^{\text{2}}}\text{+}{{\left( \text{c-a} \right)}^{\text{2}}}\text{=0}\]

\[\Rightarrow \left( \text{a-b} \right)\text{=}{{\left( \text{b-c} \right)}^{\text{2}}}\text{=}{{\left( \text{c-a} \right)}^{\text{2}}}\text{=0}\]                   

Since, \[{{\left( \text{a-b} \right)}^{\text{2}}}\text{,}{{\left( \text{b-c} \right)}^{\text{2}}}\text{,}{{\left( \text{c-a} \right)}^{\text{2}}}\] have non-negative values.

\[\therefore \left( \text{a-b} \right)\text{=}\left( \text{b-c} \right)\text{=}\left( \text{c-a} \right)\text{=0}\]

\[\Rightarrow \text{a=b=c}\]

Thus, if \[\text{ }\!\!\Delta\!\!\text{ =0}\] , then either \[\text{a+b+c=0}\] or \[\text{a=b=c}\] .


5. Solve the equations\[\mathbf{\left| \begin{matrix} \text{x+a} & \text{x} & \text{x}  \\ \text{x} & \text{x+a} & \text{x}  \\ \text{x} & \text{x} & \text{x+a}  \\ \end{matrix} \right|\text{=0}\] , \[\text{a}\ne \text{0}}\].

Ans: Given, \[\left| \begin{matrix} \text{x+a} & \text{x} & \text{x}  \\ \text{x} & \text{x+a} & \text{x}  \\ \text{x} & \text{x} & \text{x+a}  \\ \end{matrix} \right|\text{=0}\]

Applying \[{{R}_{1}}\to {{R}_{1}}\text{+}{{\text{R}}_{2}}\text{+}{{\text{R}}_{3}}\]

\[\Rightarrow \left| \begin{matrix} \text{3x+a} & \text{3x+a} & \text{3x+a}  \\ \text{x} & \text{x+a} & \text{x}  \\ \text{x} & \text{x} & \text{x+a}  \\ \end{matrix} \right|\text{=0}\]

Taking out $\left( 3x+a \right)$ common from ${{R}_{1}}$,

\[\Rightarrow \left( \text{3x+a} \right)\left| \begin{matrix} \text{1} & \text{1} & \text{1}  \\ \text{x} & \text{x+a} & \text{x}  \\ \text{x} & \text{x} & \text{x+a}  \\ \end{matrix} \right|\text{=0}\]

Applying the row operations \[{{\text{C}}_{2}}\to {{C}_{2}}-{{C}_{1}}\] and \[{{C}_{3}}\to {{C}_{3}}-{{C}_{1}}\]

\[\Rightarrow \left( \text{3x+a} \right)\left| \begin{matrix} \text{1} & \text{1} & \text{1}  \\ \text{x} & \text{a} & \text{x}  \\ \text{x} & \text{x} & \text{a}  \\ \end{matrix} \right|\text{=0}\]

Expanding along ${{R}_{1}}$

\[\Rightarrow {{\text{a}}^{\text{2}}}\left( \text{3x+a} \right)\text{=0}\]

But \[\text{a}\ne \text{0}\] .

Hence, \[\text{3x+a=0}\]

\[\Rightarrow \text{x=-}\dfrac{\text{a}}{\text{3}}\]


6. Prove that \[\mathbf{\left| \begin{matrix} {{\text{a}}^{\text{2}}} & \text{bc} & \text{ac+}{{\text{c}}^{\text{2}}}  \\ {{\text{a}}^{\text{2}}}\text{+ab} & {{\text{b}}^{\text{2}}} & \text{ac}  \\ \text{ab} & {{\text{b}}^{\text{2}}}\text{+bc} & {{\text{c}}^{\text{2}}}  \\ \end{matrix} \right|\text{=4}{{\text{a}}^{\text{2}}}{{\text{b}}^{\text{2}}}{{\text{c}}^{\text{2}}}}\].

Ans: Given, \[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} {{\text{a}}^{\text{2}}} & \text{bc} & \text{ac+}{{\text{c}}^{\text{2}}}  \\ {{\text{a}}^{\text{2}}}\text{+ab} & {{\text{b}}^{\text{2}}} & \text{ac}  \\ \text{ab} & {{\text{b}}^{\text{2}}}\text{+bc} & {{\text{c}}^{\text{2}}}  \\ \end{matrix} \right|\]

Taking out \[\text{a}\],\[\text{b}\] and \[\text{c}\] from \[{{\text{C}}_{1}}\], \[{{\text{C}}_{2}}\] and \[{{\text{C}}_{3}}\].

\[\text{ }\!\!\Delta\!\!\text{ =abc}\left| \begin{matrix} \text{a} & \text{c} & \text{a+c}  \\ \text{a+b} & \text{b} & \text{a}  \\ \text{b} & \text{b+c} & \text{c}  \\ \end{matrix} \right|\]

Applying the row operations \[{{\text{R}}_{2}}\to {{\text{R}}_{2}}-{{\text{R}}_{1}}\] and \[{{\text{R}}_{3}}\to {{\text{R}}_{3}}-{{\text{R}}_{1}}\]

\[\text{ }\!\!\Delta\!\!\text{ =abc}\left| \begin{matrix} \text{a} & \text{c} & \text{a+c}  \\ \text{b} & \text{b-c} & \text{-c}  \\ \text{b-a} & \text{b} & \text{-a}  \\ \end{matrix} \right|\]

Applying \[{{\text{R}}_{2}}\to {{\text{R}}_{2}}\text{+}{{\text{R}}_{1}}\]

\[\text{ }\!\!\Delta\!\!\text{ =abc}\left| \begin{matrix} \text{a} & \text{c} & \text{a+c}  \\ \text{a+b} & \text{b} & \text{a}  \\ \text{b-a} & \text{b} & \text{-a}  \\ \end{matrix} \right|\]

Applying \[{{\text{R}}_{3}}\to {{\text{R}}_{3}}\text{+}{{\text{R}}_{2}}\]

\[\text{ }\!\!\Delta\!\!\text{ =abc}\left| \begin{matrix} \text{a} & \text{c} & \text{a+c}  \\ \text{a+b} & \text{b} & \text{a}  \\ \text{2b} & \text{2b} & \text{0}  \\ \end{matrix} \right|\]

\[\Rightarrow \Delta \text{=2a}{{\text{b}}^{\text{2}}}\text{c}\left| \begin{matrix} \text{a} & \text{c} & \text{a+c}  \\ \text{a+b} & \text{b} & \text{a}  \\ \text{1} & \text{1} & \text{0}  \\ \end{matrix} \right|\]

Applying \[{{C}_{2}}\to {{C}_{2}}-{{C}_{1}}\]

Expanding along \[\text{R}_3\],

\[\text{ }\!\!\Delta\!\!\text{ =2a}{{\text{b}}^{\text{2}}}\text{c}\left[ \text{a}\left( \text{c-a} \right)\text{+a}\left( \text{a+c} \right) \right]\]

\[\Rightarrow \Delta \text{=2a}{{\text{b}}^{\text{2}}}\text{c}\left[ \text{ac-}{{\text{a}}^{\text{2}}}\text{+}{{\text{a}}^{\text{2}}}\text{+ac} \right]\]

\[\Rightarrow \Delta \text{=2a}{{\text{b}}^{\text{2}}}\text{c}\left( \text{2ac} \right)\]

\[\Rightarrow \Delta \text{=4}{{\text{a}}^{\text{2}}}{{\text{b}}^{\text{2}}}{{\text{c}}^{\text{2}}}\]

Hence proved.


7. If $\mathbf{{{\mathbf{A}}^{-1}}=\left[ \begin{matrix} 3 & -1 & 1  \\ -15 & 6 & -5  \\ 5 & -2 & 2  \\ \end{matrix} \right]$ and $B=\left[ \begin{matrix} 1 & 2 & -2  \\ -1 & 3 & 0  \\ 0 & -2 & 1  \\ \end{matrix} \right]$, find ${{\left( AB \right)}^{-1}}}$.

Ans: The below result will be used for simplification,

${{\left( AB \right)}^{-1}}={{B}^{-1}}{{A}^{-1}}$

Finding inverse of matrix B. so, the determinant is

$\left| B \right|=1\left( 3-0 \right)-2\left( -1-0 \right)-2\left( 2-0 \right)$

$\Rightarrow \left| B \right|=3+2-4$

$\therefore \left| B \right|=1$

Now finding the cofactors,

${{B}_{11}}={{(-1)}^{1+1}}\left| \begin{matrix} 3 & 0  \\ -2 & 1  \\ \end{matrix} \right|\Rightarrow {{B}_{11}}=3$

${{B}_{12}}={{(-1)}^{1+2}}\left| \begin{matrix} -1 & 0  \\ 0 & 1  \\ \end{matrix} \right|\Rightarrow {{B}_{12}}=1$

${{B}_{13}}={{(-1)}^{1+3}}\left| \begin{matrix} -1 & 3  \\ 0 & -2  \\ \end{matrix} \right|\Rightarrow {{B}_{13}}=2$

${{B}_{21}}={{(-1)}^{2+1}}\left| \begin{matrix} 2 & -2  \\ -2 & 1  \\ \end{matrix} \right|\Rightarrow {{B}_{21}}=-\left( 2-4 \right)=2$

${{B}_{22}}={{(-1)}^{2+2}}\left| \begin{matrix} 1 & -2  \\ 0 & 1  \\ \end{matrix} \right|\Rightarrow {{B}_{22}}=1$

${{B}_{23}}={{(-1)}^{2+3}}\left| \begin{matrix} 1 & 2  \\ 0 & -2  \\ \end{matrix} \right|\Rightarrow {{B}_{23}}=2$

${{B}_{31}}={{(-1)}^{3+1}}\left| \begin{matrix} 2 & -2  \\ 3 & 0  \\ \end{matrix} \right|\Rightarrow {{B}_{31}}=6$

${{B}_{32}}={{(-1)}^{3+2}}\left| \begin{matrix} 1 & -2  \\ -1 & 0  \\ \end{matrix} \right|\Rightarrow {{B}_{32}}=2$

${{B}_{33}}={{(-1)}^{3+3}}\left| \begin{matrix} 1 & 2  \\ -1 & 3  \\ \end{matrix} \right|\Rightarrow {{B}_{33}}=\left( 3+2 \right)=5$

The cofactor matrix is $\left[ \begin{matrix} 3 & 1 & 2  \\ 2 & 1 & 2  \\ 6 & 2 & 5  \\ \end{matrix} \right]$. The adjoint will be the transpose of cofactor matrix.

$adj\left( B \right)=\left[ \begin{matrix} 3 & 2 & 6  \\ 1 & 1 & 2  \\ 2 & 2 & 5  \\ \end{matrix} \right]$

The inverse is given by ${{B}^{-1}}=\frac{adj\left( B \right)}{\left| B \right|}$. So,

${{B}^{-1}}=\left[ \begin{matrix} 3 & 2 & 6  \\ 1 & 1 & 2  \\ 2 & 2 & 5  \\ \end{matrix} \right]$

Now, it is already given that ${{A}^{-1}}=\left[ \begin{matrix} 3 & -1 & 1  \\ -15 & 6 & -5  \\ 5 & -2 & 2  \\ \end{matrix} \right]$.

So, we can compute ${{\left( AB \right)}^{-1}}={{B}^{-1}}{{A}^{-1}}$ as below,

${{\left( AB \right)}^{-1}}=\left[ \begin{matrix} 3 & 2 & 6  \\ 1 & 1 & 2  \\ 2 & 2 & 5  \\ \end{matrix} \right]\left[ \begin{matrix} 3 & -1 & 1  \\ -15 & 6 & -5  \\ 5 & -2 & 2  \\ \end{matrix} \right]$

$\Rightarrow {{\left( AB \right)}^{-1}}=\left[ \begin{matrix} 9-30+30 & -3+12-12 & 3-10+12  \\ 3-15+10 & -1+6-4 & 1-5+4  \\ 6-30+25 & -2+12-10 & 2-10+10  \\ \end{matrix} \right]$

$\therefore {{\left( AB \right)}^{-1}}=\left[ \begin{matrix} 9 & -3 & 5  \\ -2 & 1 & 0  \\ 1 & 0 & 2  \\ \end{matrix} \right]$


8. Let \[\mathbf{\text{A=}\left[ \begin{matrix} \text{1} & \text{2} & \text{1}  \\ \text{2} & \text{3} & \text{1}  \\ \text{1} & \text{1} & \text{5}  \\ \end{matrix} \right]}\] verify that

  1. \[\mathbf{{{\left[ \text{adjA} \right]}^{\text{-1}}}\text{=adj}\left( {{\text{A}}^{\text{-1}}} \right)}\]

Ans: Given, \[\text{A=}\left[ \begin{matrix} \text{1} & \text{2} & \text{1}  \\ \text{2} & \text{3} & \text{1}  \\ \text{1} & \text{1} & \text{5}  \\ \end{matrix} \right]\]

\[\Rightarrow \left| \text{A} \right|\text{=1}\left( \text{15-1} \right)\text{-2}\left( \text{10-1} \right)\text{+1}\left( \text{2-3} \right)\]

\[\Rightarrow \left| A \right|\text{=14-18-1}\]

\[\therefore \left| A \right|\text{=}-5\]

Thus,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{A}_{11}}\text{=14}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{A}_{12}}=-9\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]

\[\Rightarrow {{A}_{13}}=-1\]

Similarly,

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=-9}\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=4\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]

\[\Rightarrow {{\text{A}}_{23}}\text{=1}\]

And

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]

\[\Rightarrow {{\text{A}}_{31}}=-1\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]

\[\Rightarrow {{\text{A}}_{32}}=1\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]

\[\Rightarrow {{\text{A}}_{33}}=-1\].

Cofactor matrix is \[\left[ \begin{matrix} 14 & -9 & -1  \\ -9 & 4 & 1  \\ -1 & 1 & -1  \\ \end{matrix} \right]\].

We know that adjoint of a matrix is the transpose of its cofactor matrix.

\[\Rightarrow adjA={{\left[ \begin{matrix} 14 & -9 & -1  \\ -9 & 4 & 1  \\ -1 & 1 & -1  \\ \end{matrix} \right]}^{T}}\] \[\therefore \text{adjA=}\left[ \begin{matrix} 14 & -9 & -1  \\ -9 & 4 & 1  \\ -1 & 1 & -1  \\ \end{matrix} \right]\]

Let us denote the adjoint of A as B. So, \[B\text{=}\left[ \begin{matrix} 14 & -9 & -1  \\ -9 & 4 & 1  \\ -1 & 1 & -1  \\ \end{matrix} \right]\].

The inverse of A is given by

\[{{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\left( \text{adjA} \right)\]

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=-}\dfrac{\text{1}}{5}\left[ \begin{matrix} 14 & -9 & -1  \\ -9 & 4 & 1  \\ -1 & 1 & -1  \\ \end{matrix} \right]\]

\[\therefore {{A}^{\text{-1}}}\text{=}\dfrac{\text{1}}{5}\left[ \begin{matrix} -14 & 9 & 1  \\ 9 & -4 & -1  \\ 1 & -1 & 1  \\ \end{matrix} \right]\]

Now, we have to verify \[{{\left[ \text{adjA} \right]}^{\text{-1}}}\text{=adj}\left( {{\text{A}}^{\text{-1}}} \right)\].

Let us compute the RHS first, i.e. the adjoint of \[{{A}^{\text{-1}}}\text{=}\dfrac{\text{1}}{5}\left[ \begin{matrix} -14 & 9 & 1  \\ 9 & -4 & -1  \\ 1 & -1 & 1  \\ \end{matrix} \right]\] or \[{{A}^{\text{-1}}}\text{=}\left[ \begin{matrix} -\dfrac{14}{5} & \dfrac{9}{5} & \dfrac{\text{1}}{5}  \\ \dfrac{9}{5} & -\dfrac{4}{5} & -\dfrac{\text{1}}{5}  \\ \dfrac{\text{1}}{5} & -\dfrac{\text{1}}{5} & \dfrac{\text{1}}{5}  \\ \end{matrix} \right]\].

Thus,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{A}_{11}}\text{=-}\dfrac{\text{1}}{5}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{A}_{12}}=-\dfrac{2}{5}\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]

\[\Rightarrow {{A}_{13}}=-\dfrac{\text{1}}{5}\]

Similarly,

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=-}\dfrac{2}{5}\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=-\dfrac{3}{5}\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]

\[\Rightarrow {{\text{A}}_{23}}\text{=-}\dfrac{\text{1}}{5}\]

And

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]

\[\Rightarrow {{\text{A}}_{31}}=-\dfrac{\text{1}}{5}\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]

\[\Rightarrow {{\text{A}}_{32}}=-\dfrac{\text{1}}{5}\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]

\[\Rightarrow {{\text{A}}_{33}}=-1\].

Cofactor matrix is \[\left[ \begin{matrix} -\dfrac{\text{1}}{5} & -\dfrac{2}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{2}{5} & -\dfrac{3}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{\text{1}}{5} & -\dfrac{\text{1}}{5} & -1  \\ \end{matrix} \right]\].  We know that adjoint of a matrix is the transpose of its cofactor matrix.

\[\Rightarrow adj{{A}^{-1}}={{\left[ \begin{matrix} -\dfrac{\text{1}}{5} & -\dfrac{2}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{2}{5} & -\dfrac{3}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{\text{1}}{5} & -\dfrac{\text{1}}{5} & -1  \\ \end{matrix} \right]}^{T}}\]

\[\therefore \text{adj}{{A}^{-1}}\text{=}\left[ \begin{matrix} -\dfrac{\text{1}}{5} & -\dfrac{2}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{2}{5} & -\dfrac{3}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{\text{1}}{5} & -\dfrac{\text{1}}{5} & -1  \\ \end{matrix} \right]\]

Next, moving to the LHS, i.e. the inverse of adjoint of A. We have adjoint of A as matrix B,

\[B\text{=}\left[ \begin{matrix} 14 & -9 & -1  \\ -9 & 4 & 1  \\ -1 & 1 & -1  \\ \end{matrix} \right]\]

Determinant of B is

$\left| B \right|=\left[ 14\left( -4-1 \right)+9\left( 9+1 \right)-1\left( -9+4 \right) \right]$

$\Rightarrow \left| B \right|=\left[ -70+90+5 \right]$

$\therefore \left| B \right|=25$

Now the cofactors,

Thus,

\[\Rightarrow {{B}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{B}_{11}}\text{=-5}\]

\[\Rightarrow {{B}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{B}_{12}}=-10\]

\[\Rightarrow {{B}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]

\[\Rightarrow {{B}_{13}}=-5\]

Similarly,

\[\Rightarrow {{B}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{B}_{21}}\text{=-10}\]

\[\Rightarrow {{B}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{B}_{22}}=-15\]

\[\Rightarrow {{B}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]

\[\Rightarrow {{B}_{23}}\text{=-5}\]

And

\[\Rightarrow {{B}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]

\[\Rightarrow {{B}_{31}}=-5\]

\[\Rightarrow {{B}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]

\[\Rightarrow {{B}_{32}}=-5\]

\[\Rightarrow {{B}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]

\[\Rightarrow {{B}_{33}}=-25\].

Cofactor matrix is \[\left[ \begin{matrix} -5 & -10 & -5  \\ -10 & -15 & -5  \\ -5 & -5 & -25  \\ \end{matrix} \right]\].

We know that adjoint of a matrix is the transpose of its cofactor matrix.

\[\Rightarrow adjB={{\left[ \begin{matrix} -5 & -10 & -5  \\ -10 & -15 & -5  \\ -5 & -5 & -25  \\ \end{matrix} \right]}^{T}}\]

\[\therefore \text{adjB=}\left[ \begin{matrix} -5 & -10 & -5  \\ -10 & -15 & -5  \\ -5 & -5 & -25  \\ \end{matrix} \right]\]

Now the inverse is found as

\[{{B}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| B \right|}\left( \text{adjB} \right)\]

\[\Rightarrow {{B}^{\text{-1}}}\text{=}\dfrac{\text{1}}{25}\left[ \begin{matrix} -5 & -10 & -5  \\ -10 & -15 & -5  \\ -5 & -5 & -25  \\ \end{matrix} \right]\]

\[\therefore {{B}^{\text{-1}}}\text{=}\left[ \begin{matrix} -\dfrac{\text{1}}{5} & -\dfrac{2}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{2}{5} & -\dfrac{3}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{\text{1}}{5} & -\dfrac{\text{1}}{5} & -1  \\ \end{matrix} \right]\]

So, LHS is \[{{\left[ \text{adjA} \right]}^{\text{-1}}}\text{=}\left[ \begin{matrix} -\dfrac{\text{1}}{5} & -\dfrac{2}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{2}{5} & -\dfrac{3}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{\text{1}}{5} & -\dfrac{\text{1}}{5} & -1  \\ \end{matrix} \right]\]. 

Since LHS=RHS, it is verified that \[{{\left[ \text{adjA} \right]}^{\text{-1}}}\text{=adj}\left( {{\text{A}}^{\text{-1}}} \right)\].

  1.  \[\mathbf{{{\left( {{\text{A}}^{\text{-1}}} \right)}^{\text{-1}}}\text{=A}}\]

Ans: Since \[{{A}^{\text{-1}}}\text{=}\left[ \begin{matrix} -\dfrac{14}{5} & \dfrac{9}{5} & \dfrac{\text{1}}{5}  \\ \dfrac{9}{5} & -\dfrac{4}{5} & -\dfrac{\text{1}}{5}  \\ \dfrac{\text{1}}{5} & -\dfrac{\text{1}}{5} & \dfrac{\text{1}}{5}  \\ \end{matrix} \right]\]  and \[\text{adj}{{A}^{-1}}\text{=}\left[ \begin{matrix} -\dfrac{\text{1}}{5} & -\dfrac{2}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{2}{5} & -\dfrac{3}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{\text{1}}{5} & -\dfrac{\text{1}}{5} & -1  \\ \end{matrix} \right]\], thus,

\[\Rightarrow \left| {{\text{A}}^{\text{-1}}} \right|\text{=}\left[ -\dfrac{14}{5}\left( -\dfrac{4}{25}-\dfrac{1}{25} \right)-\dfrac{9}{5}\left( \dfrac{9}{25}+\dfrac{1}{25} \right)+\dfrac{1}{5}\left( -\dfrac{9}{25}+\dfrac{4}{25} \right) \right]\]

\[\Rightarrow \left| {{\text{A}}^{\text{-1}}} \right|\text{=}\left[ \dfrac{70}{125}-\dfrac{90}{125}-\dfrac{5}{125} \right]\]

\[\therefore \left| {{\text{A}}^{\text{-1}}} \right|\text{=-}\dfrac{1}{5}\]

Also, we know that an inverse of a matrix is given by:

\[\Rightarrow {{\left( {{\text{A}}^{\text{-1}}} \right)}^{-1}}\text{=}\dfrac{\text{adj}{{\text{A}}^{\text{-1}}}}{\left| \text{A} \right|}\]

\[\Rightarrow {{\left( {{\text{A}}^{\text{-1}}} \right)}^{-1}}\text{=-5}\left[ \begin{matrix} -\dfrac{\text{1}}{5} & -\dfrac{2}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{2}{5} & -\dfrac{3}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{\text{1}}{5} & -\dfrac{\text{1}}{5} & -1  \\ \end{matrix} \right]\]

\[\Rightarrow {{\left( {{\text{A}}^{\text{-1}}} \right)}^{-1}}\text{=}\left[ \begin{matrix} 1 & 2 & 1  \\ 2 & 3 & 1  \\ 1 & 1 & 1  \\ \end{matrix} \right]\]

\[\Rightarrow {{\left( {{\text{A}}^{\text{-1}}} \right)}^{-1}}\text{=A}\]

Hence verified that \[{{\left( {{\text{A}}^{\text{-1}}} \right)}^{\text{-1}}}\text{=A}\].


9. Evaluate \[\mathbf{\left| \begin{matrix} \text{x} & \text{y} & \text{x+y}  \\ \text{y} & \text{x+y} & \text{x}  \\ \text{x+y} & \text{x} & \text{y}  \\ \end{matrix} \right|}\]

Ans: Given, \[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{x} & \text{y} & \text{x+y}  \\ \text{y} & \text{x+y} & \text{x}  \\ \text{x+y} & \text{x} & \text{y}  \\ \end{matrix} \right|\]

Applying \[{{R}_{1}}\to {{R}_{1}}\text{+}{{\text{R}}_{2}}\text{+}{{\text{R}}_{3}}\]

\[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{2}\left( \text{x+y} \right) & \text{2}\left( \text{x+y} \right) & \text{2}\left( \text{x+y} \right)  \\ \text{y} & \text{x+y} & \text{x}  \\ \text{x+y} & \text{x} & \text{y}  \\ \end{matrix} \right|\]

Taking \[\text{2}\left( \text{x+y} \right)\] common from \[{{\text{R}}_{1}}\]

\[\Rightarrow \Delta \text{=2}\left( \text{x+y} \right)\left| \begin{matrix} \text{1} & \text{1} & \text{1}  \\ \text{y} & \text{x+y} & \text{x}  \\ \text{x+y} & \text{x} & \text{y}  \\ \end{matrix} \right|\]

Applying the row operations \[{{\text{C}}_{2}}\to {{C}_{2}}-{{C}_{1}}\] and \[{{C}_{3}}\to {{C}_{3}}-{{C}_{1}}\]

\[\text{ }\!\!\Delta\!\!\text{ =2}\left( \text{x+y} \right)\left| \begin{matrix} \text{1} & \text{0} & 0  \\ \text{y} & \text{x} & \text{x-y}  \\ \text{x+y} & \text{-y} & \text{-x}  \\ \end{matrix} \right|\]

Expanding along \[{{\text{R}}_{1}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =2}\left( \text{x+y} \right)\left[ 1\left( \left( x\times \left( \text{-x} \right) \right)-\left( -y\left( x-y \right) \right) \right)\text{+0+}0 \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =2}\left( \text{x+y} \right)\left[ 1\left( \left( \text{-}{{\text{x}}^{2}} \right)-\left( -yx+{{y}^{2}} \right) \right) \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =2}\left( \text{x+y} \right)\left[ -{{\text{x}}^{\text{2}}}\text{+xy}-{{\text{y}}^{\text{2}}} \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =-2}\left( \text{x+y} \right)\left[ {{\text{x}}^{\text{2}}}-xy+{{\text{y}}^{\text{2}}} \right]\]

Applying the identity,

\[\therefore \text{ }\!\!\Delta\!\!\text{ =-2}\left[ {{\text{x}}^{3}}\text{+}{{\text{y}}^{3}} \right]\]


10. Evaluate \[\mathbf{\left| \begin{matrix} \text{1} & \text{x} & \text{y}  \\ \text{1} & \text{x+y} & \text{y}  \\ \text{1} & \text{x} & \text{x+y}  \\ \end{matrix} \right|}\]

Ans: Given, \[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{1} & \text{x} & \text{y}  \\ \text{1} & \text{x+y} & \text{y}  \\ \text{1} & \text{x} & \text{x+y}  \\ \end{matrix} \right|\]

Applying the row operations \[{{\text{R}}_{2}}\to {{\text{R}}_{2}}-{{\text{R}}_{1}}\] and \[{{\text{R}}_{3}}\to {{\text{R}}_{3}}-{{\text{R}}_{1}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{1} & \text{x} & \text{y}  \\ \text{0} & \text{y} & \text{0}  \\ \text{0} & \text{0} & \text{x}  \\ \end{matrix} \right|\]

Expanding along \[{{\text{C}}_{1}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =1}\left( \text{xy-0} \right)\]

\[\therefore \text{ }\!\!\Delta\!\!\text{ }=\text{xy}\]


11. Using properties of determinants, prove that: (Not in the current syllabus)

\[\mathbf{\left| \begin{matrix} \text{ }\!\!\alpha\!\!\text{ } & {{\text{ }\!\!\alpha\!\!\text{ }}^{\text{2}}} & \text{ }\!\!\beta\!\!\text{ + }\!\!\gamma\!\!\text{ }  \\ \text{ }\!\!\beta\!\!\text{ } & {{\text{ }\!\!\beta\!\!\text{ }}^{\text{2}}} & \text{ }\!\!\gamma\!\!\text{ + }\!\!\alpha\!\!\text{ }  \\ \text{ }\!\!\gamma\!\!\text{ } & {{\text{ }\!\!\gamma\!\!\text{ }}^{\text{2}}} & \text{ }\!\!\alpha\!\!\text{ + }\!\!\beta\!\!\text{ }  \\ \end{matrix} \right|\text{=}\left( \text{ }\!\!\beta\!\!\text{ - }\!\!\gamma\!\!\text{ } \right)\left( \text{ }\!\!\gamma\!\!\text{ - }\!\!\alpha\!\!\text{ } \right)\left( \text{ }\!\!\alpha\!\!\text{ - }\!\!\beta\!\!\text{ } \right)\left( \text{ }\!\!\alpha\!\!\text{ + }\!\!\beta\!\!\text{ + }\!\!\gamma\!\!\text{ } \right)}\]

Ans:Given, \[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{ }\!\!\alpha\!\!\text{ } & {{\text{ }\!\!\alpha\!\!\text{ }}^{\text{2}}} & \text{ }\!\!\beta\!\!\text{ + }\!\!\gamma\!\!\text{ }  \\ \text{ }\!\!\beta\!\!\text{ } & {{\text{ }\!\!\beta\!\!\text{ }}^{\text{2}}} & \text{ }\!\!\gamma\!\!\text{ + }\!\!\alpha\!\!\text{ }  \\ \text{ }\!\!\gamma\!\!\text{ } & {{\text{ }\!\!\gamma\!\!\text{ }}^{\text{2}}} & \text{ }\!\!\alpha\!\!\text{ + }\!\!\beta\!\!\text{ }  \\ \end{matrix} \right|\]

Applying the row operations \[{{\text{R}}_{2}}\to {{\text{R}}_{2}}-{{\text{R}}_{1}}\] and \[{{\text{R}}_{3}}\to {{\text{R}}_{3}}-{{\text{R}}_{1}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{ }\!\!\alpha\!\!\text{ } & {{\text{ }\!\!\alpha\!\!\text{ }}^{\text{2}}} & \text{ }\!\!\beta\!\!\text{ + }\!\!\gamma\!\!\text{ }  \\ \text{ }\!\!\beta\!\!\text{ - }\!\!\alpha\!\!\text{ } & {{\text{ }\!\!\beta\!\!\text{ }}^{\text{2}}}\text{-}{{\text{ }\!\!\alpha\!\!\text{ }}^{\text{2}}} & \text{ }\!\!\alpha\!\!\text{ - }\!\!\beta\!\!\text{ }  \\ \text{ }\!\!\gamma\!\!\text{ - }\!\!\alpha\!\!\text{ } & {{\text{ }\!\!\gamma\!\!\text{ }}^{\text{2}}}\text{-}{{\text{ }\!\!\alpha\!\!\text{ }}^{\text{2}}} & \text{ }\!\!\alpha\!\!\text{ - }\!\!\gamma\!\!\text{ }  \\ \end{matrix} \right|\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left( \text{ }\!\!\beta\!\!\text{ - }\!\!\alpha\!\!\text{ } \right)\left( \text{ }\!\!\gamma\!\!\text{ - }\!\!\alpha\!\!\text{ } \right)\left| \begin{matrix} \text{ }\!\!\alpha\!\!\text{ } & {{\text{ }\!\!\alpha\!\!\text{ }}^{\text{2}}} & \text{ }\!\!\beta\!\!\text{ + }\!\!\gamma\!\!\text{ }  \\ \text{1} & \text{ }\!\!\beta\!\!\text{ + }\!\!\alpha\!\!\text{ } & \text{-1}  \\ \text{1} & \text{ }\!\!\gamma\!\!\text{ + }\!\!\alpha\!\!\text{ } & \text{-1}  \\ \end{matrix} \right|\]

Applying \[{{\text{R}}_{3}}\to {{\text{R}}_{3}}-{{\text{R}}_{2}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left( \text{ }\!\!\beta\!\!\text{ - }\!\!\alpha\!\!\text{ } \right)\left( \text{ }\!\!\gamma\!\!\text{ - }\!\!\alpha\!\!\text{ } \right)\left| \begin{matrix} \text{ }\!\!\alpha\!\!\text{ } & {{\text{ }\!\!\alpha\!\!\text{ }}^{\text{2}}} & \text{ }\!\!\beta\!\!\text{ + }\!\!\gamma\!\!\text{ }  \\ \text{1} & \text{ }\!\!\beta\!\!\text{ + }\!\!\alpha\!\!\text{ } & \text{-1}  \\ \text{0} & \text{ }\!\!\gamma\!\!\text{ -}\beta  & \text{0}  \\ \end{matrix} \right|\]

Expanding along \[{{\text{R}}_{3}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left( \text{ }\!\!\beta\!\!\text{ - }\!\!\alpha\!\!\text{ } \right)\left( \text{ }\!\!\gamma\!\!\text{ - }\!\!\alpha\!\!\text{ } \right)\left[ \text{-}\left( \text{ }\!\!\gamma\!\!\text{ -}\beta  \right)\left( \text{- }\!\!\alpha\!\!\text{ - }\!\!\beta\!\!\text{ - }\!\!\gamma\!\!\text{ } \right) \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left( \text{ }\!\!\beta\!\!\text{ - }\!\!\alpha\!\!\text{ } \right)\left( \text{ }\!\!\gamma\!\!\text{ - }\!\!\alpha\!\!\text{ } \right)\left( \text{ }\!\!\gamma\!\!\text{ -}\beta  \right)\left( \text{ }\!\!\alpha\!\!\text{ + }\!\!\beta\!\!\text{ + }\!\!\gamma\!\!\text{ } \right)\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left( \text{ }\!\!\beta\!\!\text{ - }\!\!\gamma\!\!\text{ } \right)\left( \text{ }\!\!\gamma\!\!\text{ - }\!\!\alpha\!\!\text{ } \right)\left( \text{ }\!\!\alpha\!\!\text{ - }\!\!\beta\!\!\text{ } \right)\left( \text{ }\!\!\alpha\!\!\text{ + }\!\!\beta\!\!\text{ + }\!\!\gamma\!\!\text{ } \right)\]

Hence proved.

12. Using properties of determinants, prove that:

\[\mathbf{\left| \begin{matrix} \text{x} & {{\text{x}}^{\text{2}}} & \text{1+p}{{\text{x}}^{\text{3}}}  \\ \text{y} & {{\text{y}}^{\text{2}}} & \text{1+p}{{\text{y}}^{\text{3}}}  \\ \text{z} & {{\text{z}}^{\text{2}}} & \text{1+p}{{\text{z}}^{\text{3}}}  \\ \end{matrix} \right|\text{=}\left( \text{1+pxyz} \right)\left( \text{x-y} \right)\left( \text{y-z} \right)\left( \text{z-x} \right)}\]

Ans: Given, \[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{x} & {{\text{x}}^{\text{2}}} & \text{1+p}{{\text{x}}^{\text{3}}}  \\ \text{y} & {{\text{y}}^{\text{2}}} & \text{1+p}{{\text{y}}^{\text{3}}}  \\ \text{z} & {{\text{z}}^{\text{2}}} & \text{1+p}{{\text{z}}^{\text{3}}}  \\ \end{matrix} \right|\]

Applying the row operations \[{{\text{R}}_{2}}\to {{\text{R}}_{2}}-{{\text{R}}_{1}}\] and \[{{\text{R}}_{3}}\to {{\text{R}}_{3}}-{{\text{R}}_{1}}\]

\[\Rightarrow \Delta =\left| \begin{matrix} x & {{x}^{2}} & 1+p{{x}^{3}}  \\ y-x & {{y}^{2}}-x & p\left( {{y}^{3}}-{{x}^{3}} \right)  \\ z-x & {{z}^{2}}-x & p\left( {{z}^{3}}-{{x}^{3}} \right)  \\ \end{matrix} \right|\]

\[\Rightarrow \Delta \text{=}\left( \text{y-x} \right)\left( \text{z-x} \right)\left| \begin{matrix} \text{x} & {{\text{x}}^{\text{2}}} & \text{1+p}{{\text{x}}^{\text{3}}}  \\ \text{1} & \text{y-x} & \text{p}\left( {{\text{y}}^{\text{2}}}\text{-}{{\text{x}}^{\text{2}}}\text{+xy} \right)  \\ \text{1} & \text{z-x} & \text{p}\left( {{\text{z}}^{\text{2}}}\text{-}{{\text{x}}^{\text{2}}}\text{+xz} \right)  \\ \end{matrix} \right|\]

Applying \[{{\text{R}}_{3}}\to {{\text{R}}_{3}}-{{\text{R}}_{2}}\]

\[\Rightarrow \Delta \text{=}\left( \text{y-x} \right)\left( \text{z-x} \right)\left| \begin{matrix} \text{x} & {{\text{x}}^{\text{2}}} & \text{1+p}{{\text{x}}^{\text{3}}}  \\ \text{1} & \text{y+x} & \text{p}\left( {{\text{y}}^{\text{2}}}\text{-}{{\text{x}}^{\text{2}}}\text{+xy} \right)  \\ \text{0} & \text{z-y} & \text{p}\left( \text{z-y} \right)\left( \text{x+y+z} \right)  \\ \end{matrix} \right|\]

Taking $\left( z-y \right)$ from ${{R}_{3}}$

\[\Rightarrow \Delta \text{=}\left( \text{y-x} \right)\left( \text{z-x} \right)\left( \text{z-y} \right)\left| \begin{matrix} \text{x} & {{\text{x}}^{\text{2}}} & \text{1+p}{{\text{x}}^{\text{3}}}  \\ \text{1} & \text{y+x} & \text{p}\left( {{\text{y}}^{\text{2}}}\text{-}{{\text{x}}^{\text{2}}}\text{+xy} \right)  \\ \text{0} & \text{1} & \text{p}\left( \text{x+y+z} \right)  \\ \end{matrix} \right|\]

Expanding along \[{{\text{R}}_{\text{3}}}\]

\[\Rightarrow \Delta \text{=}\left( \text{x-y} \right)\left( \text{y-z} \right)\left( \text{z-x} \right)\left[ -1\left( xp\left( {{\text{y}}^{\text{2}}}\text{-}{{\text{x}}^{2}}\text{+xy} \right)-\left( 1+p{{x}^{3}} \right) \right)\text{+p}\left( \text{x+y+z} \right)\left( \text{x}\left( y+x \right)-{{\text{x}}^{2}} \right) \right]\]

\[\Rightarrow \Delta \text{=}\left( \text{x-y} \right)\left( \text{y-z} \right)\left( \text{z-x} \right)\left[ -1\left( xp{{\text{y}}^{\text{2}}}\text{-p}{{\text{x}}^{3}}\text{+p}{{\text{x}}^{2}}\text{y-1-p}{{\text{x}}^{3}} \right)\text{+p}\left( \text{x+y+z} \right)\left( \text{xy+}{{x}^{2}}-{{\text{x}}^{2}} \right) \right]\]

\[\Rightarrow \Delta \text{=}\left( \text{x-y} \right)\left( \text{y-z} \right)\left( \text{z-x} \right)\left[ -xp{{\text{y}}^{\text{2}}}\text{+p}{{\text{x}}^{3}}\text{-p}{{\text{x}}^{2}}\text{y+1-p}{{\text{x}}^{3}}\text{+pxy}\left( \text{x+y+z} \right) \right]\]

\[\Rightarrow \Delta \text{=}\left( \text{x-y} \right)\left( \text{y-z} \right)\left( \text{z-x} \right)\left[ -xp{{\text{y}}^{\text{2}}}\text{+p}{{\text{x}}^{3}}\text{-p}{{\text{x}}^{2}}\text{y+1-p}{{\text{x}}^{3}}\text{+p}{{\text{x}}^{2}}\text{y+px}{{\text{y}}^{2}}+pxyz \right]\]

\[\Rightarrow \Delta \text{=}\left( \text{x-y} \right)\left( \text{y-z} \right)\left( \text{z-x} \right)\left[ 1+pxyz \right]\]

Hence proved.

13. Using properties of determinants, prove that:

\[\mathbf{\left| \begin{matrix} \text{3a} & \text{-a+b} & \text{-a+c}  \\ \text{-b+a} & \text{3b} & \text{-b+c}  \\ \text{-c+a} & \text{-c+b} & \text{3c}  \\ \end{matrix} \right|\text{=3}\left( \text{a+b+c} \right)\left( \text{ab+ba+ca} \right)}\]

Ans: Given, \[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{3a} & \text{-a+b} & \text{-a+c}  \\ \text{-b+a} & \text{3b} & \text{-b+c}  \\ \text{-c+a} & \text{-c+b} & \text{3c}  \\ \end{matrix} \right|\]

Applying the column operation, \[{{C}_{1}}\to {{C}_{1}}\text{+}{{C}_{2}}\text{+}{{C}_{3}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{a+b+c} & \text{-a+b} & \text{-a+c}  \\ \text{a+b+c} & \text{3b} & \text{-b+c}  \\ \text{a+b+c} & \text{-c+b} & \text{3c}  \\ \end{matrix} \right|\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left( \text{a+b+c} \right)\left| \begin{matrix} \text{1} & \text{-a+b} & \text{-a+c}  \\ \text{1} & \text{3b} & \text{-b+c}  \\ \text{1} & \text{-c+b} & \text{3c}  \\ \end{matrix} \right|\]

Applying \[{{\text{R}}_{2}}\to {{\text{R}}_{2}}-{{\text{R}}_{1}}\] and \[{{\text{R}}_{3}}\to {{\text{R}}_{3}}-{{\text{R}}_{1}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left( \text{a+b+c} \right)\left| \begin{matrix} \text{1} & \text{-a+b} & \text{-a+c}  \\ \text{0} & \text{2b+a} & \text{a-b}  \\ \text{0} & \text{a-c} & \text{2c+a}  \\ \end{matrix} \right|\]

Expanding along \[{{C}_{1}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left( \text{a+b+c} \right)\left[ \left( \text{2b+a} \right)\left( \text{2c+a} \right)\text{-}\left( \text{a-b} \right)\left( \text{a-c} \right) \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left( \text{a+b+c} \right)\left[ \text{4bc+2ab+2ac+}{{\text{a}}^{\text{2}}}\text{-}{{\text{a}}^{\text{2}}}\text{+ac+ba-bc} \right]\].

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left( \text{a+b+c} \right)\left( \text{3ab+3bc+3ac} \right)\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =3}\left( \text{a+b+c} \right)\left( \text{ab+bc+ca} \right)\]

\[\therefore \left| \begin{matrix} \text{3a} & \text{-a+b} & \text{-a+c}  \\ \text{-b+a} & \text{3b} & \text{-b+c}  \\ \text{-c+a} & \text{-c+b} & \text{3c}  \\ \end{matrix} \right|\text{=3}\left( \text{a+b+c} \right)\left( \text{ab+ba+ca} \right)\]

Hence proved.

14. Using properties of determinants, prove this:

\[\mathbf{\left| \begin{matrix} \text{1} & \text{1+p} & \text{1+p+q}  \\ \text{2} & \text{3+2p} & \text{4+3p+2q}  \\ \text{3} & \text{6+3p} & \text{10+6p+3q}  \\ \end{matrix} \right|\text{=1}}\]

Ans: Given, \[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{1} & \text{1+p} & \text{1+p+q}  \\ \text{2} & \text{3+2p} & \text{4+3p+2q}  \\ \text{3} & \text{6+3p} & \text{10+6p+3q}  \\ \end{matrix} \right|\]

Applying \[{{\text{R}}_{2}}\to {{\text{R}}_{2}}-2{{\text{R}}_{1}}\] and \[{{\text{R}}_{3}}\to {{\text{R}}_{3}}-\text{3}{{\text{R}}_{1}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{1} & \text{1+p} & \text{1+p+q}  \\ \text{0} & \text{1} & \text{2+p}  \\ \text{0} & \text{3} & \text{7+3p}  \\ \end{matrix} \right|\]

Applying \[{{\text{R}}_{3}}\to {{\text{R}}_{3}}-\text{3}{{\text{R}}_{2}}\] , we have:

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{1} & \text{1+p} & \text{1+p+q}  \\ \text{0} & \text{1} & \text{2+p}  \\ \text{0} & \text{0} & \text{1}  \\ \end{matrix} \right|\]

Expanding along \[{{C}_{1}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =1}\left| \begin{matrix} \text{1} & \text{2+p}  \\ \text{0} & \text{1}  \\ \end{matrix} \right|\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =1}\left( \text{1-0} \right)\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =1}\]

\[\therefore \left| \begin{matrix} \text{1} & \text{1+p} & \text{1+p+q}  \\ \text{2} & \text{3+2p} & \text{4+3p+2q}  \\ \text{3} & \text{6+3p} & \text{10+6p+3q}  \\ \end{matrix} \right|\text{=1}\]

15. Using properties of determinants, prove that:

\[\mathbf{\left| \begin{matrix} \sin \alpha  & \cos \alpha  & \cos \left( \alpha +\delta  \right)  \\ \sin \beta  & \cos \beta  & \cos \left( \beta +\delta  \right)  \\ \sin \gamma  & \cos \gamma  & cso\left( \gamma +\delta  \right)  \\ \end{matrix} \right|=0}\]

Ans: Given, \[\Delta =\left| \begin{matrix} \sin \alpha  & \cos \alpha  & \cos \left( \alpha +\delta  \right)  \\ \sin \beta  & \cos \beta  & \cos \left( \beta +\delta  \right)  \\ \sin \gamma  & \cos \gamma  & cso\left( \gamma +\delta  \right)  \\ \end{matrix} \right|\]

\[\Rightarrow \Delta =\dfrac{1}{\sin \delta \cos \delta }\left| \begin{matrix} \sin \alpha \sin \delta  & \cos \alpha \cos \delta  & \cos \alpha \cos \delta -\sin \alpha \sin \delta   \\ \sin \beta \sin \delta  & \cos \beta \cos \delta  & \cos \beta \cos \delta -\sin \beta \sin \delta   \\ \sin \gamma \sin \delta  & \cos \gamma \cos \delta  & \cos \gamma \cos \delta -\sin \gamma \sin \delta   \\ \end{matrix} \right|\]

Applying \[{{C}_{1}}\to {{C}_{1}}+{{C}_{3}}\]

\[\Rightarrow \Delta =\dfrac{1}{\sin \delta \cos \delta }\left| \begin{matrix} \cos \alpha \cos \delta  & \cos \alpha \cos \delta  & \cos \alpha \cos \delta -\sin \alpha \sin \delta   \\ \cos \beta \cos \delta  & \cos \beta \cos \delta  & \cos \beta \cos \delta -\sin \beta \sin \delta   \\ \cos \gamma \cos \delta  & \cos \gamma \cos \delta  & \cos \gamma \cos \delta -\sin \gamma \sin \delta   \\ \end{matrix} \right|\]

Since, two columns \[{{\text{C}}_{1}}\] and \[{{C}_{2}}\] are identical.

$\Rightarrow \Delta =0$

\[\therefore \left| \begin{matrix} \sin \alpha  & \cos \alpha  & \cos \left( \alpha +\delta  \right)  \\ \sin \beta  & \cos \beta  & \cos \left( \beta +\delta  \right)  \\ \sin \gamma  & \cos \gamma  & cso\left( \gamma +\delta  \right)  \\ \end{matrix} \right|=0\]

Hence proved.


16. Solve the system of the following equations

\[\mathbf{\dfrac{\text{2}}{\text{x}}\text{+}\dfrac{\text{3}}{\text{y}}\text{+}\dfrac{\text{10}}{\text{z}}\text{=4}}\]

\[\mathbf{\dfrac{\text{4}}{\text{x}}\text{+}\dfrac{\text{6}}{\text{y}}\text{+}\dfrac{\text{5}}{\text{z}}\text{=1}}\]

\[\mathbf{\dfrac{\text{6}}{\text{x}}\text{+}\dfrac{\text{9}}{\text{y}}\text{+}\dfrac{\text{20}}{\text{z}}\text{=2}}\]

Ans: Given equations,

\[\dfrac{\text{2}}{\text{x}}\text{+}\dfrac{\text{3}}{\text{y}}\text{+}\dfrac{\text{10}}{\text{z}}\text{=4}\]

\[\dfrac{\text{4}}{\text{x}}\text{+}\dfrac{\text{6}}{\text{y}}\text{+}\dfrac{\text{5}}{\text{z}}\text{=1}\]

\[\dfrac{\text{6}}{\text{x}}\text{+}\dfrac{\text{9}}{\text{y}}\text{+}\dfrac{\text{20}}{\text{z}}\text{=2}\]

Let \[\dfrac{\text{1}}{\text{x}}\text{=p}\] , \[\dfrac{\text{1}}{\text{y}}\text{=q}\] and \[\dfrac{\text{1}}{\text{z}}\text{=r}\]

Then the given system of equations becomes:

\[\text{2p+3q+10r=4}\]

\[\text{4p-6q+5r=1}\]

\[\text{6p+9q+20r=2}\]

Let \[\text{A=}\left[ \begin{matrix} \text{2} & \text{3} & \text{10}  \\ \text{4} & \text{-6} & \text{5}  \\ \text{6} & \text{9} & \text{-20}  \\ \end{matrix} \right]\] , \[\text{X=}\left[ \begin{matrix} \text{p}  \\ \text{q}  \\ \text{r}  \\ \end{matrix} \right]\] and \[\text{B=}\left[ \begin{matrix} \text{4}  \\ \text{1}  \\ \text{2}  \\ \end{matrix} \right]\] such that, this system can be written in the form of \[\text{AX=B}\].

Now,

\[\Rightarrow \left| A \right|\text{=2}\left( \text{120-45} \right)\text{-3}\left( \text{-80-30} \right)\text{+10}\left( \text{36+36} \right)\]

\[\Rightarrow \left| A \right|\text{=150+330+720}\]

\[\Rightarrow \left| A \right|\text{=1200}\]

Thus, \[\text{A}\] is a non-singular matrix.

$\therefore $ Its inverse exists.

So,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}\left| \begin{matrix} -6 & 5  \\ 9 & -20  \\ \end{matrix} \right|\]

\[\Rightarrow {{A}_{11}}\text{=75}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}\left| \begin{matrix} 4 & 5  \\ 6 & -20  \\ \end{matrix} \right|\]

\[\Rightarrow {{A}_{12}}=110\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}\left| \begin{matrix} 4 & -6  \\ 6 & 9  \\ \end{matrix} \right|\]

\[\Rightarrow {{A}_{13}}=72\]

Similarly,

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}\left| \begin{matrix} 3 & 10  \\ 9 & -20  \\ \end{matrix} \right|\]

\[\Rightarrow {{\text{A}}_{21}}\text{=150}\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}\left| \begin{matrix} 2 & 10  \\ 6 & -20  \\ \end{matrix} \right|\]

\[\Rightarrow {{\text{A}}_{22}}=100\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}\left| \begin{matrix} 2 & 3  \\ 6 & 9  \\ \end{matrix} \right|\]

\[\Rightarrow {{\text{A}}_{23}}\text{=0}\]

and

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}\left| \begin{matrix} 3 & 10  \\ -6 & 5  \\ \end{matrix} \right|\]

\[\Rightarrow {{\text{A}}_{31}}=75\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}\left| \begin{matrix} 2 & 10  \\ 4 & 5  \\ \end{matrix} \right|\]

\[\Rightarrow {{\text{A}}_{32}}=30\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}\left| \begin{matrix} 2 & 3  \\ 4 & -6  \\ \end{matrix} \right|\]

\[\Rightarrow {{\text{A}}_{33}}=-24\].

The inverse of a matrix is given by:

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\left( \text{adjA} \right)\]

\[\therefore {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{1200}}\left[ \begin{matrix} \text{75} & \text{150} & \text{75}  \\ \text{110} & \text{-100} & \text{30}  \\ \text{72} & \text{0} & \text{-24}  \\ \end{matrix} \right]\]

Now, \[\text{X=}{{\text{A}}^{\text{-1}}}\text{B}\]

\[\Rightarrow \left[ \begin{matrix} \text{p}  \\ \text{q}  \\ \text{r}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{1200}}\left[ \begin{matrix} \text{75} & \text{150} & \text{75}  \\ \text{110} & \text{-100} & \text{30}  \\ \text{72} & \text{0} & \text{-24}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{4}  \\ \text{1}  \\ \text{2}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{p}  \\ \text{q}  \\ \text{r}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{1200}}\left[ \begin{matrix} \text{300+150+150}  \\ \text{440-100+60}  \\ \text{288+0-48}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{p}  \\ \text{q}  \\ \text{r}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{1200}}\left[ \begin{matrix} \text{600}  \\ \text{400}  \\ \text{240}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{p}  \\ \text{q}  \\ \text{r}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \dfrac{\text{1}}{\text{2}}  \\ \dfrac{\text{1}}{\text{3}}  \\ \dfrac{\text{1}}{\text{5}}  \\ \end{matrix} \right]\]

\[\therefore \text{P=}\dfrac{\text{1}}{\text{2}}\], \[\text{q=}\dfrac{\text{1}}{\text{3}}\] and \[\text{r=}\dfrac{\text{1}}{\text{5}}\]

Thus \[\text{x=2}\] , \[\text{y=3}\] and \[\text{z=5}\].


17. Choose the correct answer.

If \[\mathbf{\text{a,b,c}}\] are in \[\mathbf{\text{A}\text{.P}}\] , then the determinant

\[\mathbf{\left| \begin{matrix} \text{x+2} & \text{x+3} & \text{x+2a}  \\ \text{x+3} & \text{x+4} & \text{x+2b}  \\ \text{x+4} & \text{x+5} & \text{x+2c}  \\ \end{matrix} \right|}\]

  1. \[\mathbf{\text{0}}\]

  2. \[\mathbf{\text{1}}\]

  3. \[\mathbf{\text{X}}\]

  4. \[\mathbf{\text{2X}}\]

Ans: Given, \[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{x+2} & \text{x+3} & \text{x+2a}  \\ \text{x+3} & \text{x+4} & \text{x+2b}  \\ \text{x+4} & \text{x+5} & \text{x+2c}  \\ \end{matrix} \right|\]

Since \[\text{a}\] , \[\text{b}\], and \[\text{c}\] are in A.P.  \[\text{2b=a+c}\],

\[\Rightarrow \Delta \text{=}\left| \begin{matrix} \text{x+2} & \text{x+3} & \text{x+2a}  \\ \text{x+3} & \text{x+4} & \text{x+}\left( \text{a+c} \right)  \\ \text{x+4} & \text{x+5} & \text{x+2c}  \\ \end{matrix} \right|\]          

Applying $R_1 \rightarrow R_1 - R_2 \, \, \text{and} R_3 \rightarrow R_3 - R_2$

\[\Rightarrow \Delta \text{=}\left| \begin{matrix} \text{-1} & \text{-1} & \text{a-c}  \\ \text{x+3} & \text{x+4} & \text{x+}\left( \text{a+c} \right)  \\ \text{1} & \text{1} & \text{c-a}  \\ \end{matrix} \right|\]

Applying \[{{\text{R}}_{\text{1}}}\to {{\text{R}}_{\text{1}}}\text{-}{{\text{R}}_{\text{3}}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{0} & \text{0} & \text{0}  \\ \text{x+3} & \text{x+4} & \text{x+a+c}  \\ \text{1} & \text{1} & \text{c-a}  \\ \end{matrix} \right|\]

\[\therefore \Delta =0\]


18. Choose the correct answer.If \[\mathbf{\text{X,Y,Z}}\] are nonzero real numbers, then the inverse of matrix \[\mathbf{\text{A=}\left[ \begin{matrix} \text{x} & \text{0} & \text{0}  \\ \text{0} & \text{y} & \text{0}  \\ \text{0} & \text{0} & \text{z}  \\ \end{matrix} \right]}\]

  1. \[\mathbf{\left[ \begin{matrix} {{\text{x}}^{\text{-1}}} & \text{0} & \text{0}  \\ \text{0} & {{\text{y}}^{\text{-1}}} & \text{0}  \\ \text{0} & \text{0} & {{\text{z}}^{\text{-1}}}  \\ \end{matrix} \right]}\]

  2. \[\mathbf{\text{xyz}\left[ \begin{matrix} {{\text{x}}^{\text{-1}}} & \text{0} & \text{0}  \\ \text{0} & {{\text{y}}^{\text{-1}}} & \text{0}  \\ \text{0} & \text{0} & {{\text{z}}^{\text{-1}}}  \\ \end{matrix} \right]}\]

  3. \[\mathbf{\dfrac{\text{1}}{\text{xyz}}\left[ \begin{matrix} \text{x} & \text{0} & \text{0}  \\ \text{0} & \text{y} & \text{0}  \\ \text{0} & \text{0} & \text{z}  \\ \end{matrix} \right]}\]

  4. \[\mathbf{\dfrac{\text{1}}{\text{xyz}}\left[ \begin{matrix} \text{1} & \text{0} & \text{0}  \\ \text{0} & \text{1} & \text{0}  \\ \text{0} & \text{0} & \text{1}  \\ \end{matrix} \right]}\]

Ans: Given, \[\text{A=}\left[ \begin{matrix} \text{x} & \text{0} & \text{0}  \\ \text{0} & \text{y} & \text{0}  \\ \text{0} & \text{0} & \text{z}  \\ \end{matrix} \right]\]

\[\Rightarrow \left| \text{A} \right|\text{=x}\left( \text{yz-0} \right)\]

\[\therefore \left| \text{A} \right|\text{=xyz}\ne \text{0}\]

Thus,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{A}_{11}}\text{=yz}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{A}_{12}}=0\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]

\[\Rightarrow {{A}_{13}}=0\]

Similarly,

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=0}\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=xz\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]

\[\Rightarrow {{\text{A}}_{23}}\text{=0}\]

and

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]

\[\Rightarrow {{\text{A}}_{31}}=0\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]

\[\Rightarrow {{\text{A}}_{32}}=0\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]

\[\Rightarrow {{\text{A}}_{33}}=xy\].

Cofactor matrix is \[\left[ \begin{matrix} yz & 0 & 0  \\ 0 & xz & 0  \\ 0 & 0 & xy  \\ \end{matrix} \right]\].

We know that adjoint of a matrix is the transpose of its cofactor matrix.

\[\Rightarrow adjA={{\left[ \begin{matrix} yz & 0 & 0  \\ 0 & xz & 0  \\ 0 & 0 & xy  \\ \end{matrix} \right]}^{T}}\]

\[\therefore \text{adjA=}\left[ \begin{matrix} \text{yz} & \text{0} & \text{0}  \\ \text{0} & \text{xz} & \text{0}  \\ \text{0} & \text{0} & \text{xy}  \\ \end{matrix} \right]\]

The inverse of a matrix is given by:

\[\Rightarrow {{\text{A}}^{\text{-1}}}=\dfrac{\text{1}}{\left| \text{A} \right|}\left( \text{adjA} \right)\]

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{xyz}}\left[ \begin{matrix} \text{yz} & \text{0} & \text{0}  \\ \text{0} & \text{xz} & \text{0}  \\ \text{0} & \text{0} & \text{xy}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\left[ \begin{matrix} \dfrac{\text{yz}}{\text{xyz}} & \text{0} & \text{0}  \\ \text{0} & \dfrac{\text{xz}}{\text{xyz}} & \text{0}  \\ \text{0} & \text{0} & \dfrac{\text{xy}}{\text{xyz}}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\left[ \begin{matrix} \dfrac{\text{1}}{\text{x}} & \text{0} & \text{0}  \\ \text{0} & \dfrac{\text{1}}{\text{y}} & \text{0}  \\ \text{0} & \text{0} & \dfrac{\text{1}}{\text{z}}  \\ \end{matrix} \right]\]

\[\therefore {{\text{A}}^{\text{-1}}}\text{=}\left[ \begin{matrix} {{\text{x}}^{\text{-1}}} & \text{0} & \text{0}  \\ \text{0} & {{\text{y}}^{\text{-1}}} & \text{0}  \\ \text{0} & \text{0} & {{\text{z}}^{\text{-1}}}  \\ \end{matrix} \right]\]

Thus, A. \[\left[ \begin{matrix} {{\text{x}}^{\text{-1}}} & \text{0} & \text{0}  \\ \text{0} & {{\text{y}}^{\text{-1}}} & \text{0}  \\ \text{0} & \text{0} & {{\text{z}}^{\text{-1}}}  \\ \end{matrix} \right]\] is the correct answer.

19. Choose the correct answer.

Let \[\mathbf{\text{A}=\left[ \begin{matrix} 1 & \sin \theta  & 1  \\ -\sin \theta  & 1 & \sin \theta   \\ -1 & -\sin \theta  & 1  \\ \end{matrix} \right]}\], 

where \[\mathbf{\text{0}\le \text{ }\!\!\theta\!\!\text{ }\le \text{2n}}\] , then

  1. \[\mathbf{\text{Det}\left( \text{A} \right)\text{=0}}\]

  2. \[\mathbf{text{Det}\left( \text{A} \right)\in \left( \text{2,}\infty  \right)}\]

  3. \[\mathbf{\text{Det}\left( \text{A} \right)\in \left( \text{2,4} \right)}\]

  4. \[\mathbf{\text{Det}\left( \text{A} \right)\left[ \text{2,4} \right]}\]

Ans: Given, \[A=\left[ \begin{matrix} 1 & \sin \theta  & 1  \\ -\sin \theta  & 1 & \sin \theta   \\ -1 & -\sin \theta  & 1  \\ \end{matrix} \right]\]

\[\therefore \left| A \right|=1\left( 1+{{\sin }^{2}}\theta  \right)-\sin \theta \left( -\sin \theta +\sin \theta  \right)+1\left( {{\sin }^{2}}\theta +1 \right)\]

\[\Rightarrow \left| A \right|=1+{{\sin }^{2}}\theta +{{\sin }^{2}}\theta +1\]

\[\Rightarrow \left| A \right|=2+2{{\sin }^{2}}\theta \]

\[\Rightarrow \left| A \right|=2\left( 1+{{\sin }^{2}}\theta  \right)\]

We know, \[0\le \sin \theta \le 1\]

\[\Rightarrow 1\le 1+{{\sin }^{2}}\theta \le 2\]

\[\Rightarrow 2\le 2\left( 1+{{\sin }^{2}}\theta  \right)\le 4\]

Thus, D. \[\text{Det}\left( \text{A} \right)\left[ \text{2,4} \right]\] is the correct answer.


NCERT Solutions for Class 12 Maths – Free PDF Download

A student can often feel anxious when they cannot understand the questions. However, students need not panic as they can refer to the solutions for NCERT Solutions for Class 12 Maths Chapter 2 Determinants. Here, they can find the appropriate solutions to Determinants Class 12 which are curated by experienced professionals. These solutions are created and compiled by the subject experts and are available on the website for ready access.


Chapter 4 – Determinants

4.1 Introduction

Class 12 Maths NCERT Solutions Chapter 4 will introduce the learner to Determinants and its basic concepts and theorems. A student will study determinants up to order three only with real entries. One will also study various properties of determinants such as minors, cofactors, and applications of determinants. Topics like the inconsistency of the system of linear equations and solutions of linear equations in two or three variables using the inverse of a matrix are also included.

4.2 Determinant

Under this Class 12 Maths ch 4, students can understand that any matrix inside a mode is read as a determinant of a matrix and not modulus of the matrix. One will also learn how to find the determinant of a matrix of order one, which shall be a little easy to comprehend. Along with this, the section also covers how to determine the matrix of order two and order three. More straightforward calculations on how to expand the determinant along the row or column which contains the maximum number of zeros are also explained. It also makes one understand multiplications in positives instead of negatives.

Exercise 4.1 – 8 Questions (3 Short Questions and 5 Long Questions)

4.3 Properties of Determinants (Not in the current syllabus)

In the NCERT Solutions for Class 12 Maths Chapter 4 Determinants, the focus is on the properties of determinants that simplifies its evaluation by obtaining the maximum number of zeros in a row or column. There’s the section which elaborates on the statement--properties are true for determinants of any order. One can learn more about the properties under this section. Moreover, the lesson deals with understanding six different properties of determinants and the ways to verify and check these properties of the determinants.

Exercise 4.2 – 16 Questions (4 Short Questions 10 Long Questions)

4.4 Area of Triangle

Under Determinants Class 12 PDF, the lesson is about the area of the triangle and its application of vertices and expressions. It touches upon certain topics about how an area in positive quantity and an absolute value of the determinant is taken. It also explains how the positive and negative values of the determinant are used for calculations. It also talks about the area of the triangle formed by three collinear points is zero.

Exercise 4.3 – 5 Questions (2 Short Questions and 3 Long Questions)

4.5 Minors and Cofactors

This section guides students to write the expansion of a determinant in compact form using minors and cofactors. A student will also learn the definition of minors and cofactors. It also elaborates on how the minor of an element of a less order determinant than the order of two is a determinant of order that is less than the order of one. One will also learn five more ways of expansion that can calculate the area of a triangle; the sum of the product of elements of any row or column with their corresponding cofactor of the same.

Exercise 4.4 – 5 Questions (2 Short Questions and 3 Long Questions)

4.6 The Adjoint and Inverse of a Matrix

This section further deals with the inverse and existence of a matrix. The section also contains explanations in terms of finding the adjoint of the matrix. Five definitions are included —the definitions of the adjoint of a square matrix and the theorem to verify the same. A student will learn how to discover five theorems and understand the ways to verify the definitions. The section also includes ways to apply the theorems for verifying the definitions in different kinds of problems.

Exercise 4.5 – 18 Questions (4 Short Questions and 14 Long Questions)

4.7 Applications of Determinants and Matrices

In this Class 12th Maths Chapter 4, the applications of determinants and matrices for solving the system of linear equations in two or three variables are covered. A student will learn to check the consistency of the system of linear equations. The consistent system and inconsistent system are also explained in detail. One will also study the solution system of linear equations using the inverse of a matrix and how to express linear equations as matrix equations and how to solve them using the opposite of the coefficient matrix. The chapter talks about the uniqueness of matrix equations that provide unique solutions for the equation's systems known as the matrix method.

Exercise 4.6 – 16 Questions (3 Short Questions and 13 Long Questions)


Key Features of NCERT Solutions for Class 12 Maths Chapter 4

It is required to go thoroughly through every section of this chapter to gain better knowledge and perspective for the subject. If a learner goes through Determinants Class 12 Solutions that are curated by the expert professionals in this subject, they will get an even better understanding of the chapter and learn the basic and advanced concepts and theorems.

The Key Features are:

  • Attaining more knowledge about the ideas and theories covered in NCERT Solutions for Class 12 Maths Chapter 4.

  • Assessment of one’s knowledge gap with the help of the ch 4 Class 12 Maths PDF.

  • Increase knowledge in main topics.

  • The student will be able to understand assignments based on concepts from the determinants ch 4 Maths Class 12.


Conclusion

The primary goal of NCERT solutions for Class 12 Maths Chapter 4 Determinants is to provide appropriate learning of all core concepts for students to not only score well on their boards but also to develop a deep understanding of all fundamental topics. Students will quickly gain a thorough understanding of this topic if they regularly practice Class 12 Maths NCERT Solutions Chapter 4.


Access Exercise Wise NCERT Solutions for Class 12 Maths Chapter 4


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FAQs on NCERT Solutions for Class 12 Maths Chapter 4 - Determinants

Q1. How many sums are there in the NCERT Solutions for Class 12 Maths Chapter 4?

Ans: There are 8 sums in the first exercise, and Ex.-4.1, 16 sums in the second exercise, Ex.-4.2. There are 5 sums in the third exercise, Ex.-4.3, 5 sums in the fourth exercise, Ex.- 4.4., 18 sums in the fifth exercise Ex.-4.5, and 16 sums in the last exercise, Ex.- 4.6. The last exercise is followed by a miscellaneous exercise, and there are 18 sums in the miscellaneous exercise for this chapter. All the sums are solved and explained in the NCERT Solutions for Class 12 Maths Chapter 4 PDF.

Q2. What are determinants used for?

Ans: Determinants have a number of applications in algebra. The concept of determinants is very useful for solving a set of linear equations. With determinants it becomes easier to understand the change in area, volume, and the change in variables in terms of integrals. Determinants are used for calculating the values of square matrices.

Q3. What type of sums can I expect in the exam from Determinants?

Ans: You can expect sums for solving linear equations from the topic of Determinants. There are various proofs that can be derived using the theory of determinants, so you can expect several ‘prove that’ sums from this topic in your exams. Also, for certain sums, you may have to find the values of some unknown variables using determinants and their resultant values.

Q4. Can I download the NCERT Solutions for Class 12 Maths Chapter 4 for free?

Ans: Yes, you can download the NCERT Solutions for Class 12 Maths Chapter 4 for free from Vedantu. The NCERT Solutions for this chapter are provided in a PDF file on Vedantu. You can refer to these solutions online or download them and practice offline. All you need for downloading the NCERT Solutions of Class 12 Maths Chapter 4, Determinants, is an internet connection and a digital screen. Also, you can print a hardcopy of these NCERT Solutions for your convenience.

Q5. How many problems are there in NCERT Solutions for Class 12 Maths Chapter 4?

Ans: NCERT Solutions for Class 12 Maths Chapter 4 covers an important topic called Determinants. It is an easy chapter but it is quite lengthy and has a few problems to solve. It also has exercises for students to solve. The first exercise has eight questions, the second consists of 16 problems, with a few sub-questions. The third and the fourth exercise consist of five problems each. The fifth one is a bit longer and consists of 18 questions. The sixth exercise consists of 14 questions and the final miscellaneous exercise has 19 questions to solve.

Q6. Why should one read NCERT Books for Class 12 Maths Chapter 4?

Ans: NCERT Class 12 Maths is the prescribed textbook for the Maths syllabus. It is the first book one must cover before jumping into any other reference books. It strictly follows the syllabus of NCERT, and questions will be set based on this in the examination. Vedantu’s NCERT Class 12 Maths Solutions is also based on this book and has solutions for all the problems. Whenever there is any doubt, one can easily refer to these solutions so that no time is wasted searching for the answers.

Q7. What are the uses of determinants according to NCERT Solutions for Class 12 Maths Chapter 4?

Ans: Determinant is an important topic, from the examination point of view and education point of view. This topic is pretty easy and is a very useful one. Students will learn how to use determinants to check the system of linear equations and their consistency. Using determinants, one can express a linear equation and can solve it using the opposite of a coefficient matrix. Students will also learn the method of calculating the area of a triangle using determinants.

Q8. How can I download solutions for Class 12 Math?

Ans: To download solutions for NCERT Class 12 Maths, you can follow these steps:

  1. Click on this NCERT Class 12 Maths Solutions Chapter.

  2. Select the chapter.

  3. This will open to display the solutions of the selected chapter. Click on the Download PDF button to download the PDF for offline use.

Vedantu’s solutions for Class 12 Math are prepared by great experts in the field so you don’t have to worry about the credibility of the solutions. You can also use the official site of  Vedantu or  Vedantu app to get study materials. All the resources are available free of cost.

Q9. How many chapters are there in NCERT Class 12 Maths?

Ans: NCERT Class 12 Maths is one of the most important subjects for a Class 12 student. It builds on some of the most critical concepts for the future of Mathematics and Physics. These topics are essential from the competitive exam point of view as well. There are a total of 13 chapters that are a part of the syllabus for Class 12. Students can use Vedantu’s Class 12 Solutions for Maths to avoid confusion or waste time while studying when stuck at any problem.