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# NCERT Solutions for Class 12 Maths Chapter 4 Determinants

Last updated date: 09th Sep 2024
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## Class 12 Maths Chapter 4 Determinants NCERT Solutions - Free PDF Download

NCERT Solutions for Class 12 Maths Chapter 4 Determinants is an important part of math studies with many real-world applications. They are used to determine whether a system of equations has a unique solution. A determinant is a function that associates a number with any square matrix and has broad applications in engineering, economics, science, and social science. Understanding NCERT Solutions for chapter 4 class 12 Maths Determinants topic is crucial. The expert teachers at Vedantu, with years of experience, have created the Class 12 Maths Chapter 4 Solutions PDF. These solutions are related to determinants and include a PDF file that contains all the answers and explanations.

Table of Content
1. Class 12 Maths Chapter 4 Determinants NCERT Solutions - Free PDF Download
2. Glance on NCERT Solutions for Class 12 Maths Chapter 4 - Determinants
3. Access Exercise Wise NCERT Solutions for Chapter 4 Maths Class 12
4. Exercises Under NCERT Class 12 Maths Chapter 4 Determinants
5. NCERT Solutions for Class 12 Maths Chapter 4 - Determinants
5.1Exercise (4.1)
5.2Exercise (4.2)
5.3Exercise (4.3)
5.4Exercise (4.4)
5.5Exercise (4.5)
5.6Miscellaneous Exercise Solutions
7. Chapter 4 – Determinants
7.14.1 Introduction
7.24.2 Determinant
7.34.3 Area of Triangle
7.44.4 Minors and Cofactors
7.54.5 The Adjoint and Inverse of a Matrix
7.64.6 Applications of Determinants and Matrices
8. The NCERT Solutions for Class 12 Maths Chapter 4 Determinants' Key features
9. Overview of Deleted Syllabus for CBSE Class 12 Determinants 2024-25
10. Class 12 Maths Chapter 4: Exercises Breakdown
11. Other Related Study Material for CBSE Class 12 Maths Chapter 4
12. Chapter-Specific NCERT Solutions for Class 12 Maths
FAQs

## Glance on NCERT Solutions for Class 12 Maths Chapter 4 - Determinants

• Determinants are used in linear algebra that deal with square matrices.

• This article contains important properties that mentions how determinants behave under various operations on matrices like,

• Scaling a row/column by a constant multiplies the determinant by that constant.

• Interchanging rows/columns can change the sign of the determinant.

• Rows/columns with all zeros or identical elements result in a zero determinant.

• The adjoint is a special matrix based on cofactors, and the inverse (if it exists) is closely linked to the determinant.

• There are five exercises and miscellaneous exercises (61 fully solved questions) in class 12th maths chapter 4 Determinants.

## Access Exercise Wise NCERT Solutions for Chapter 4 Maths Class 12

Competitive Exams after 12th Science
More Free Study Material for Determinants
Revision notes
Important questions
Ncert books

## 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. 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 (1,2) and (3,6) 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 (3,1) and (9,3) 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 (4.3)

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 (4.4)

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 (4.5)

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 Exercise 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. 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 ${{\text{C}}_{3}}$

$\Rightarrow \Delta =-\sin \alpha \left( -\sin \alpha {{\sin }^{2}}\beta +{{\cos }^{2}}\beta \sin \alpha \right)+\cos \alpha \left( \cos \alpha {{\cos }^{2}}\beta +\cos \alpha {{\sin }^{2}}\beta \right)$

$\Rightarrow \Delta ={{\sin }^{2}}\alpha \left( s{{\sin }^{2}}\beta +{{\cos }^{2}}\beta \right)+{{\cos }^{2}}\alpha \left( {{\cos }^{2}}\beta +{{\sin }^{2}}\beta \right)$

$\Rightarrow \Delta ={{\sin }^{2}}\alpha \left( 1 \right)+{{\cos }^{2}}\alpha \left( 1 \right)$

$\therefore \Delta \text{=1}$

3. 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,

${{\left( AB \right)}^{-1}}=\left[ \begin{matrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \\ \end{matrix} \right]\left[ \begin{matrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \\ \end{matrix} \right]$

$\Rightarrow {{\left( AB \right)}^{-1}}=\left[ \begin{matrix} 9-30+30 & -3+12-12 & 3-10+12 \\ 3-15+10 & -1+6-4 & 1-5+4 \\ 6-30+25 & -2+12-10 & 2-10+10 \\ \end{matrix} \right]$

$\therefore {{\left( AB \right)}^{-1}}=\left[ \begin{matrix} 9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & 2 \\ \end{matrix} \right]$

4. 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 ${{\left( {{\text{A}}^{\text{-1}}} \right)}^{\text{-1}}}\text{=A}$.

5. 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,

$\therefore \text{ }\!\!\Delta\!\!\text{ =-2}\left[ {{\text{x}}^{3}}\text{+}{{\text{y}}^{3}} \right]$

6. 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 ${{\text{C}}_{1}}$

$\Rightarrow \text{ }\!\!\Delta\!\!\text{ =1}\left( \text{xy-0} \right)$

$\therefore \text{ }\!\!\Delta\!\!\text{ }=\text{xy}$

7. 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 $\text{z=5}$.

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 ${{\text{R}}_{\text{1}}}\to {{\text{R}}_{\text{1}}}\text{-}{{\text{R}}_{\text{3}}}$

$\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{0} & \text{0} & \text{0} \\ \text{x+3} & \text{x+4} & \text{x+a+c} \\ \text{1} & \text{1} & \text{c-a} \\ \end{matrix} \right|$

$\therefore \Delta =0$

9. 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.

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.

## Chapter 4 – Determinants

### 4.1 Introduction

In Chapter 4 of Class 12 Maths NCERT Solutions, you will learn about determinants and their basic concepts and theorems. The chapter covers determinants up to order three with real entries, as well as various properties of determinants such as minors, cofactors, and applications of determinants. Additionally, the chapter includes 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.

### 4.2 Determinant

In Chapter 4 of Class 12 Maths, students can learn how to find the determinant of a matrix. This includes understanding the determinant of a matrix of order one, which is relatively easy to comprehend. The section also covers the determination of the matrix of order two and order three. It explains how to expand the determinant along the row or column containing the maximum number of zeros. Additionally, it helps in understanding positive multiplication rather than negative multiplication.

### 4.3 Area of Triangle

The Determinants Class 12 PDF covers how to calculate the area of a triangle using vertices and expressions. It explains the concept of the area as a positive quantity and the absolute value of the determinant. It also discusses the use of positive and negative values of the determinant in calculations.

### 4.4 Minors and Cofactors

This section teaches students how to simplify the expansion of a determinant using minors and cofactors. It also explains the definitions of minors and cofactors. The text covers five methods to calculate the area of a triangle, including using the sum of the product of elements of any row or column with their corresponding cofactor.

### 4.5 The Adjoint and Inverse of a Matrix

This section is about matrices, focusing on their inverse and existence. It explains how to find the adjoint of a matrix and provides five definitions, including the adjoint of a square matrix and the theorem to verify it. Students will learn how to discover and apply these theorems to verify the definitions of various types of problems.

### 4.6 Applications of Determinants and Matrices

In Chapter 4 maths class 12, students learn about using determinants and matrices to solve systems of linear equations with two or three variables. The chapter covers checking if the system of equations is consistent, explaining both consistent and inconsistent systems in detail. Students also learn how to solve these equations using the inverse of a matrix, express linear equations as matrix equations, and solve them using the inverse of the coefficient matrix. The chapter also discusses the uniqueness of matrix equations in providing unique solutions for systems of equations, known as the matrix method.

## The NCERT Solutions for Class 12 Maths Chapter 4 Determinants' Key features

The following concepts and equations are covered in the NCERT Solutions for chapter 4 Maths class 12 at Vedantu.

• The determinant has zero value if any two rows or any two columns are proportionate or identical.

• If and only if A is not singular, a square matrix A has an inverse.

• When A ≠ 0, the unique solution to the equation AX = B is X = A–1 B.

• A system of equations has a solution if it is consistent.

• There cannot be a solution to an inconsistent set of equations.

• In the matrix equation AX = B, with a square matrix A:

There is a unique solution if | A| ≠ 0.

If (adj A) B ≠ 0 and |A| = 0, then no solution exists.

In case (adj A) B = 0 and | A| = 0, the consistency of the system is uncertain.

## Overview of Deleted Syllabus for CBSE Class 12 Determinants 2024-25

 Chapter Dropped Topics Determinants Exercise 4.3 - Properties of Determinants Page 137-143 Miscellaneous Examples from 30 - 32 and 34 Miscellaneous Exercise Questions 2, 4 - 6, 11 - 15 and 17 Page 144 Summary Points 4–11

### Class 12 Maths Chapter 4: Exercises Breakdown

 Exercise Number of Questions Exercise 4.1 Solutions 8 Questions (2 Long, 5 Short Answers, 1 MCQ) Exercise 4.2 Solutions 5 Questions ( 4 Short Answers, 1 MCQ) Exercise 4.3 Solutions 5 Questions (4 Long, 1 MCQ Exercise 4.4 Solutions 18 Questions (11 Long, 5 Short, 2 MCQs) Exercise 4.5 Solutions 16 Questions (13 Long, 3 Short)

## Conclusion

NCERT Solutions for Class 12 Maths Chapter 4 - Determinants, provided by Vedantu, offers comprehensive guidance for understanding this crucial topic. It is important to focus on grasping the fundamental concepts of determinants, such as properties, operations, and applications in solving equations and matrices. In previous year question papers, around 6-8 questions were typically asked from this chapter. Paying attention to solved examples and practice exercises can enhance your problem-solving skills. Moreover, understanding the theoretical aspects alongside practical applications is essential for a deeper comprehension. By practicing with these solutions, students can build a strong foundation in determinants, which is crucial for higher-level math and various competitive exams.

## Chapter-Specific NCERT Solutions for Class 12 Maths

Given below are the chapter-wise NCERT Solutions for Class 12 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 12 Maths Chapter 4 Determinants

1. How many sums are there in the NCERT Solutions for Class 12 Maths Chapter 4?

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.

2. What are determinants used for?

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.

3. What type of sums can I expect in the exam from Determinants?

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.

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.

5. How many problems are there in NCERT Solutions for Class 12 Maths Chapter 4?

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.

6. Why should one read NCERT Books for Class 12 Maths Chapter 4?

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.

7. What are the uses of determinants according to NCERT Solutions for Class 12 Maths Chapter 4?

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.

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.

9. How many chapters are there in NCERT Class 12 Maths?

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.

10. What are the rules for determinants Class 12?

Please remember the following key rules for determinants in Class 12 Maths:-

• Determinants can only be calculated for square matrices, where the number of rows equals the number of columns.

• Determinants help determine if a system of equations has one unique solution.

• Swapping two rows (or columns) changes the sign of the determinant.

• If any two rows or columns are identical, the determinant becomes zero.

• Multiplying all elements in a row (or column) by a constant multiplies the determinant by that same constant.

11. How many types of determinants are there?

• Scalar or First-Order Determinants: These are single-element matrices or determinants with only one row and one column.

• Minor or Second-Order Determinants: These are determinants obtained by removing either a row or a column from a square matrix.

• Cofactor or Third-Order Determinants: These are determinants obtained by multiplying each element of a row or a column by its cofactor and summing up the products.

12. What is the symbol for determinants?

The symbol used for determinants is denoted by

• Vertical Lines: This is shown as two vertical lines on either side of the matrix. For example, if you have a matrix A, the determinant would be written as |A|.

• Det(A): This is a shorthand way of writing "determinant of A".

13. Is the determinant a scalar?

The determinant is a scalar, which is a single number like 5 or -2. It has a magnitude (how big it is) but no direction. Determinants perfectly fit this definition. They are single numbers calculated from a square matrix, representing a specific property of that matrix.

14. How many exercises are there in chapter 4 class 12 maths?

The number of exercises on determinants class 12 solutions can vary depending on the edition and publisher. These solutions are typically not available in a strictly digital format that allows for easy counting. However, based on what students typically encounter, class 12 determinants NCERT Solutions for determinants likely cover a range of problems to help you practice and understand the concept thoroughly.

15. How to solve a determinant question?

There are two main approaches to solving determinant problems, depending on the size of the matrix:

For 2x2 Matrices: Use a simple formula that multiplies opposite corners together and subtracts the product of the other two elements.

For 3x3 Matrices: This method involves more steps. Here's a simplified breakdown:

• Expand by a Chosen Row or Column: Pick a row or column in the matrix.

• Calculate Minor: For each element in that row/column, calculate a minor which is the determinant of a smaller matrix you get by excluding that element's row and column.

• Multiply by Cofactor: Multiply each minor by its corresponding cofactor, which is either 1 or -1 based on its position (a pattern you'll learn in NCERT solutions).

• Sum the Products: Add up all the products from step c to get the final determinant.

16. What are the uses of determinants according to class 12 determinants NCERT Solutions?

Use of Determinants class 12 solutions have several applications in algebra. They are very useful for solving a set of linear equations, making it easier to understand the change in area, volume, and variables in terms of integrals. Determinants are also used to calculate the values of square matrices.

17. What kind of questions from determinants class 12 solutions may I anticipate seeing on the board exams?

This chapter will likely include problems that require you to use determinants to solve a given set of linear equations. From the standpoint of the exam, proofs that can be obtained from the theory of determinants and "prove that" sums are also significant. In some issues, you should use determinants and their resulting values to find the values of unknown variables.