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# NCERT Solutions for Class 12 Maths Chapter 3 Matrices

Last updated date: 11th Aug 2024
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## Complete Resource of NCERT Class 12 Maths Chapter 3 Matrices - Free PDF Download

NCERT Class 12 Matrices Solutions PDF provided by Vedantu, offer detailed answers and explanations for all the exercises in this chapter. These solutions help students understand the concepts clearly and improve their problem-solving skills. Chapter 3 covers important topics that are crucial for exams, so focusing on understanding each concept is essential.

Table of Content
1. Complete Resource of NCERT Class 12 Maths Chapter 3 Matrices - Free PDF Download
2. Glance of NCERT Solutions for Class 12 Maths Chapter 3  Matrices | Vedantu
3. Access Exercise Wise NCERT Solutions for Chapter 3 Maths Class 12
4. Practice and Master the Concepts of Matrices with NCERT Exemplar for Class 12 Maths Chapter 3
5. Mastering Class 12 Maths Chapter 3: Matrices - MCQs, Question and Answers, and Tips for Success
5.1Exercise 3.1
5.2Exercise 3.2
5.3Exercise 3.3
5.4Miscellaneous Solutions
6. Details of Class 12 Matrices NCERT Solutions – Free PDF Download
6.13.1 Introduction
6.23.2 Matrix
6.33.3 Types of Matrices
6.43.4 Operations on Matrices
6.53.5. Transpose of a Matrix
6.63.6 Symmetric and Skew Symmetric Matrices
7. Some Important Points to remember
8. Summary of NCERT Solutions Class 12 Matrices
9. Overview of Deleted Syllabus for Class 12 Maths Chapter 3
10. Class 12 Maths Chapter 3: Exercise Breakdown
11. Other Study Materials of CBSE Class 12 Maths Chapter 3
12. Chapter-Specific NCERT Solutions for Class 12 Maths
FAQs

In this chapter, key areas include various mathematical or scientific principles (depending on the subject), their applications, and problem-solving techniques. Students should pay special attention to the examples provided and practice the exercises regularly. The solutions by Vedantu are designed to make learning easier and more effective, ensuring students grasp the core ideas thoroughly.

## Glance of NCERT Solutions for Class 12 Maths Chapter 3  Matrices | Vedantu

• The chapter covers topics such as types of matrices (e.g., zero matrix, identity matrix), operations on matrices, properties of determinants, adjoint and inverse of a matrix, and solving systems of linear equations using matrices (e.g., Cramer's rule).

• Important formulas include matrix addition, subtraction, scalar multiplication, matrix multiplication, the transpose of a matrix, and the determinant of a matrix.

• Focus on mastering basic operations, understanding the properties of determinants, and learning methods to find the inverse of a matrix and solve linear equations.

## Access Exercise Wise NCERT Solutions for Chapter 3 Maths Class 12

 S. No Current Syllabus Exercises of Class 12 Maths Chapter 3 1 NCERT Solutions of Class 12 Maths Matrices Exercise 3.1 2 NCERT Solutions of Class 12 Maths Matrices Exercise 3.2 3 NCERT Solutions of Class 12 Maths Matrices Exercise 3.3 4 NCERT Solutions of Class 12 Maths Matrices Exercise 3.4
Competitive Exams after 12th Science
More Free Study Material for Matrices
Revision notes
Important questions
Ncert books

## Practice and Master the Concepts of Matrices with NCERT Exemplar for Class 12 Maths Chapter 3

NCERT Maths Class 12 Chapter 3 Solutions, "Metrices," is based on the concept of Matrix and their properties. The chapter consists of the following exercises:

Exercise 3.1: This exercise introduces types of matrices and covers basic operations like addition, subtraction, and multiplication.

Exercise 3.2: This exercise explores matrix properties like commutative, associative, and identity matrix use.

Exercise 3.3: This exercise focuses on matrix multiplication, its properties, and practicing inverse calculations.

Exercise 3.4: This exercise teaches and applies the concept of matrix transpose and its properties.

Miscellaneous Exercise: Mix of questions testing all matrix concepts and problem-solving skills.

Overall, this chapter is an important topic in linear algebra and covers the fundamental concepts of matrices, including matrix notation, matrix operations, matrix properties, and matrix multiplication.

## Mastering Class 12 Maths Chapter 3: Matrices - MCQs, Question and Answers, and Tips for Success

### Exercise 3.1

1. In the matrix $\mathbf{A=\left[ \begin{matrix} 2 & 5 & 19 & -7 \\ 35 & -2 & \dfrac{5}{2} & 12 \\ \sqrt{3} & 1 & -5 & 17 \\ \end{matrix} \right]}$, write

i. The order of the matrix.

Ans: The order of a matrix is $m\times n$ where $m$ is the number of rows and $n$ is the number of columns. Therefore, here the order is $3\times 4$.

ii. The number of elements.

Ans: Since the order of the given matrix is $3\times 4$ therefore, the number of elements in it is $3\times 4=12$.

iii. Write the elements $\mathbf{{{a}_{13}},{{a}_{21}},{{a}_{33}},{{a}_{24}},{{a}_{23}}}$

Ans: The elements are given as ${{a}_{mn}}$ . Therefore, here ${{a}_{13}}=19$ , ${{a}_{21}}=35$ , ${{a}_{33}}=-5$ , ${{a}_{24}}=12$ , ${{a}_{23}}=\dfrac{5}{2}$.

2. If a matrix has $24$ elements, what are the possible order it can have? What if it has $13$ elements?

Ans: The order of a matrix is $m\times n$ where $m$ is the number of rows and $n$ is the number of columns. To find the possible orders of a matrix, we have to find all the ordered pairs of natural numbers whose product is $24$ .

$\therefore \left( 1\times 24 \right),\left( 24\times 1 \right),\left( 2\times 12 \right),\left( 12\times 2 \right),\left( 3\times 8 \right),\left( 8\times 3 \right),\left( 4\times 6 \right),\left( 6\times 4 \right)$ are all the possible ordered pairs here.

If the matrix had $13$ elements, then the ordered pairs would be $\left( 1\times 13 \right)$ and $\left( 13\times 1 \right)$.

3. If a matrix has $18$ elements, what are the possible orders it can have? What if it has $5$ elements?

Ans: The order of a matrix is $m\times n$ where $m$ is the number of rows and $n$ is the number of columns. To find the possible orders of a matrix, we have to find all the ordered pairs of natural numbers whose product is $18$ .

$\therefore \left( 1\times 18 \right),\left( 18\times 1 \right),\left( 2\times 9 \right),\left( 9\times 2 \right),\left( 3\times 6 \right),\left( 6\times 3 \right)$ are all the possible ordered pairs here.

If the matrix had $5$ elements, then the ordered pairs would be $\left( 1\times 5 \right)$ and $\left( 5\times 1 \right)$.

4. Construct a $3\times 4$ matrix, whose elements are given by

i. $\mathbf{{{a}_{ij}}=\dfrac{1}{2}\left| -3i+j \right|}$

Ans: Given that ${{a}_{ij}}=\dfrac{1}{2}\left| -3i+j \right|$ ,

$\therefore {{a}_{11}}=\dfrac{1}{2}\left| -3\times 1+1 \right|=1$

${{a}_{21}}=\dfrac{1}{2}\left| -3\times 2+1 \right|=\dfrac{5}{2}$

${{a}_{31}}=\dfrac{1}{2}\left| -3\times 3+1 \right|=4$

${{a}_{12}}=\dfrac{1}{2}\left| -3\times 1+2 \right|=\dfrac{1}{2}$

${{a}_{22}}=\dfrac{1}{2}\left| -3\times 2+2 \right|=2$

${{a}_{32}}=\dfrac{1}{2}\left| -3\times 3+2 \right|=\dfrac{7}{2}$

${{a}_{13}}=\dfrac{1}{2}\left| -3\times 1+3 \right|=0$

${{a}_{23}}=\dfrac{1}{2}\left| -3\times 2+3 \right|=\dfrac{3}{2}$

${{a}_{33}}=\dfrac{1}{2}\left| -3\times 3+3 \right|=3$

${{a}_{14}}=\dfrac{1}{2}\left| -3\times 1+4 \right|=\dfrac{1}{2}$

${{a}_{24}}=\dfrac{1}{2}\left| -3\times 2+4 \right|=1$

${{a}_{34}}=\dfrac{1}{2}\left| -3\times 3+4 \right|=\dfrac{5}{2}$

Thus, the required matrix is $A=\left[ \begin{matrix} 1 & \dfrac{1}{2} & 0 & \dfrac{1}{2} \\ \dfrac{5}{2} & 2 & \dfrac{3}{2} & 1 \\ 4 & \dfrac{7}{2} & 3 & \dfrac{5}{2} \\ \end{matrix} \right]$.

ii. $\mathbf{{{a}_{ij}}=2i-j}$

Ans: A $3\times 4$ matrix is given by $A=\left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} & {{a}_{14}} \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} & {{a}_{24}} \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} & {{a}_{34}} \\ \end{matrix} \right]$

Given that ${{a}_{ij}}=2i-j$ ,

$\therefore {{a}_{11}}=2\times 1-1=1$

${{a}_{21}}=2\times 2-1=3$

${{a}_{31}}=2\times 3-1=5$

${{a}_{12}}=2\times 1-2=0$

${{a}_{22}}=2\times 2-2=4$

${{a}_{32}}=2\times 3-2=4$

${{a}_{13}}=2\times 1-3=-1$

${{a}_{23}}=2\times 2-3=1$

${{a}_{33}}=2\times 3-3=3$

${{a}_{14}}=2\times 1-4=-2$

${{a}_{24}}=2\times 2-4=0$

${{a}_{34}}=2\times 3-4=2$

Thus, the required matrix is $A=\left[ \begin{matrix} 1 & 0 & -1 & -2 \\ 3 & 2 & 1 & 0 \\ 5 & 4 & 3 & 2 \\ \end{matrix} \right]$.

5. Find the value of $x,y,z$ from the following equation:

i. $\mathbf{\left[ \begin{matrix} 4 & 3 \\ x & 5 \\ \end{matrix} \right]=\left[ \begin{matrix} y & z \\ 1 & 5 \\ \end{matrix} \right]}$

Ans: Given $\left[ \begin{matrix} 4 & 3 \\ x & 5 \\ \end{matrix} \right]=\left[ \begin{matrix} y & z \\ 1 & 5 \\ \end{matrix} \right]$

Comparing the corresponding elements we get,

$x=1,y=4,z=3$

ii. $\mathbf{\left[ \begin{matrix} x+y & 2 \\ 5+z & xy \\ \end{matrix} \right]=\left[ \begin{matrix} 6 & 2 \\ 5 & 8 \\ \end{matrix} \right]}$

Ans: Given $\left[ \begin{matrix} x+y & 2 \\ 5+z & xy \\ \end{matrix} \right]=\left[ \begin{matrix} 6 & 2 \\ 5 & 8 \\ \end{matrix} \right]$

Comparing the corresponding elements we get,

$x+y=6,xy=8,5+z=5$

Now, $\because 5+z=5$

$\Rightarrow z=0$

We know that, ${{\left( x-y \right)}^{2}}={{\left( x+y \right)}^{2}}-4xy$

$\Rightarrow {{\left( x-y \right)}^{2}}=36-32$

$\Rightarrow \left( x-y \right)=\pm 2$

When $\left( x-y \right)=2$ and $\left( x+y \right)=6$,

We get $x=4,y=2$

When $\left( x-y \right)=-2$ and $\left( x+y \right)=6$,

We get $x=2,y=4$

$\therefore x=4,y=2,z=0$ or $\therefore x=2,y=4,z=0$

iii. $\mathbf{\left[ \begin{matrix} x+y+z \\ x+z \\ y+z \\ \end{matrix} \right]=\left[ \begin{matrix} 9 \\ 5 \\ 7 \\ \end{matrix} \right]}$

Ans: Given $\left[ \begin{matrix} x+y+z \\ x+z \\ y+z \\ \end{matrix} \right]=\left[ \begin{matrix} 9 \\ 5 \\ 7 \\ \end{matrix} \right]$

Comparing the corresponding elements we get,

$x+y+z=9$                      …(1)

$x+z=5$                              …(2)

$y+z=7$                              …(3)

From equation (1) and (2),

$y+5=9$

$\Rightarrow y=4$

From equation (3) we have,

$4+z=7$

$\Rightarrow z=3$

$x+z=5$

$\Rightarrow x=2$

$\therefore x=2,y=4,z=3$

6. Find the value of $\mathbf{a,b,c,d}$ from the equation:

$\mathbf{\left[ \begin{matrix} a-b & 2a+c \\ 2a-b & 3c+d \\ \end{matrix} \right]=\left[ \begin{matrix} -1 & 5 \\ 0 & 13 \\ \end{matrix} \right]}$

Ans: Given $\left[ \begin{matrix} a-b & 2a+c \\ 2a-b & 3c+d \\ \end{matrix} \right]=\left[ \begin{matrix} -1 & 5 \\ 0 & 13 \\ \end{matrix} \right]$

Comparing the corresponding elements we get,

$a-b=-1$                              …(1)

$2a-b=0$                              …(2)

$2a+c=5$                              …(3)

$3c+d=13$                                        …(4)

From equation (2),

$b=2a$

From equation (1),

$a-2a=-1$

$\Rightarrow a=1$

$\Rightarrow b=2$

From equation (3),

$2\times 1+c=5$

$\Rightarrow c=3$

From equation (4),

$3\times 3+d=13$

$\Rightarrow d=4$

$\therefore a=1,b=2,c=3,d=4$

7. $\mathbf{A={{\left[ {{a}_{y}} \right]}_{m\times n}}}$ is a square matrix, if

1. m<n

2. m>n

3. m=n

4. None of these

Ans: A given matrix is said to be a square matrix if the number of rows is equal to the number of columns.

$\therefore A={{\left[ {{a}_{y}} \right]}_{m\times n}}$ is a square matrix if, $m=n$.

Thus, option (C) is correct.

8. Which of the given values of $x$ and $y$ make the following pair of matrices equal $\mathbf{\left[ \begin{matrix} 3x+7 & 5 \\ y+1 & 2-3x \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & y-2 \\ 8 & 4 \\ \end{matrix} \right]}$

1. $\mathbf{x=\dfrac{-1}{3},y=7}$

2. Not possible to find

3. $\mathbf{y=7,x=\dfrac{-2}{3}}$

4. $\mathbf{x=\dfrac{-1}{3},y=\dfrac{-2}{3}}$

Ans: Given $\left[ \begin{matrix} 3x+7 & 5 \\ y+1 & 2-3x \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & y-2 \\ 8 & 4 \\ \end{matrix} \right]$

Comparing the corresponding elements we get,

$3x+7=0$

$\Rightarrow x=-\dfrac{7}{3}$

$y-2=5$

$\Rightarrow y=7$

$y+1=8$

$\Rightarrow y=7$

$2-3x=4$

$\Rightarrow x=-\dfrac{2}{3}$

Since we get two different values of $x$ ,which is not possible. It is not possible to find the values of $x$ and $y$ for which the given matrices are equal.

Thus, the correct option is (B).

9. The number of all possible matrices of order $\mathbf{3\times 3}$ with each entry $0$ or $1$ is:

1. $\mathbf{27}$

2. $\mathbf{18}$

3. $\mathbf{81}$

4. $\mathbf{512}$

Ans: Given a matrix of the order $3\times 3$ has nine elements and each of these elements can be either $0$ or $1$ .

Now, each of the nine elements can be filled in two possible ways.

Therefore, the required number of possible matrices is ${{2}^{9}}=512$.

### Exercise 3.2

1. Let $\mathbf{A=\left[ \begin{matrix} 2 & 4 \\ 3 & 2 \\ \end{matrix} \right],B=\left[ \begin{matrix} 1 & 3 \\ -2 & 5 \\ \end{matrix} \right],C=\left[ \begin{matrix} -2 & 5 \\ 3 & 4 \\ \end{matrix} \right]}$

Find each of the following

i. $\mathbf{A+B}$

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

$\therefore A+B=\left[ \begin{matrix} 2 & 4 \\ 3 & 2 \\ \end{matrix} \right]+\left[ \begin{matrix} 1 & 3 \\ -2 & 5 \\ \end{matrix} \right]$

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

$\Rightarrow A+B=\left[ \begin{matrix} 3 & 7 \\ 1 & 7 \\ \end{matrix} \right]$

ii. $\mathbf{A-B}$

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

$\therefore A-B=\left[ \begin{matrix} 2 & 4 \\ 3 & 2 \\ \end{matrix} \right]-\left[ \begin{matrix} 1 & 3 \\ -2 & 5 \\ \end{matrix} \right]$

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

$\Rightarrow A-B=\left[ \begin{matrix} 1 & 1 \\ 5 & -3 \\ \end{matrix} \right]$

iii. $\mathbf{3A-C}$

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

$\therefore 3A-C=3\left[ \begin{matrix} 2 & 4 \\ 3 & 2 \\ \end{matrix} \right]-\left[ \begin{matrix} -2 & 5 \\ 3 & 4 \\ \end{matrix} \right]$

$\Rightarrow 3A-C=\left[ \begin{matrix} 6 & 12 \\ 9 & 6 \\ \end{matrix} \right]-\left[ \begin{matrix} -2 & 5 \\ 3 & 4 \\ \end{matrix} \right]$

$\Rightarrow 3A-C=\left[ \begin{matrix} 6+2 & 12-5 \\ 9-3 & 6-4 \\ \end{matrix} \right]$

$\Rightarrow 3A-C=\left[ \begin{matrix} 8 & 7 \\ 6 & 2 \\ \end{matrix} \right]$

iv. $\mathbf{AB}$

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

$\therefore AB=\left[ \begin{matrix} 2 & 4 \\ 3 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 3 \\ -2 & 5 \\ \end{matrix} \right]$

$\Rightarrow AB=\left[ \begin{matrix} 2\left( 1 \right)+4\left( -2 \right) & 2\left( 3 \right)+4\left( 5 \right) \\ 3\left( 1 \right)+2\left( -2 \right) & 3\left( 3 \right)+2\left( 5 \right) \\ \end{matrix} \right]$

$\Rightarrow AB=\left[ \begin{matrix} -6 & 26 \\ -1 & 19 \\ \end{matrix} \right]$

v. $\mathbf{BA}$

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

$\therefore BA=\left[ \begin{matrix} 1 & 3 \\ -2 & 5 \\ \end{matrix} \right]\left[ \begin{matrix} 2 & 4 \\ 3 & 2 \\ \end{matrix} \right]$

$\Rightarrow BA=\left[ \begin{matrix} 1\left( 2 \right)+3\left( 3 \right) & 1\left( 4 \right)+3\left( 2 \right) \\ -2\left( 2 \right)+5\left( 3 \right) & -2\left( 4 \right)+5\left( 2 \right) \\ \end{matrix} \right]$

$\Rightarrow BA=\left[ \begin{matrix} 11 & 10 \\ 11 & 2 \\ \end{matrix} \right]$

2. Compute the following:

i. $\mathbf{\left[ \begin{matrix} a & b \\ -b & a \\ \end{matrix} \right]+\left[ \begin{matrix} a & b \\ b & a \\ \end{matrix} \right]}$

Ans: We have to find $\left[ \begin{matrix} a & b \\ -b & a \\ \end{matrix} \right]+\left[ \begin{matrix} a & b \\ b & a \\ \end{matrix} \right]$

$\therefore \left[ \begin{matrix} a & b \\ -b & a \\ \end{matrix} \right]+\left[ \begin{matrix} a & b \\ b & a \\ \end{matrix} \right]=\left[ \begin{matrix} a+a & b+b \\ -b+b & a+a \\ \end{matrix} \right]$

$\Rightarrow \left[ \begin{matrix} a & b \\ -b & a \\ \end{matrix} \right]+\left[ \begin{matrix} a & b \\ b & a \\ \end{matrix} \right]=\left[ \begin{matrix} 2a & 2b \\ 0 & 2a \\ \end{matrix} \right]$

ii. $\mathbf{\left[ \begin{matrix} {{a}^{2}}+{{b}^{2}} & {{b}^{2}}+{{c}^{2}} \\ {{a}^{2}}+{{c}^{2}} & {{a}^{2}}+{{b}^{2}} \\ \end{matrix} \right]+\left[ \begin{matrix} 2ab & 2bc \\ -2ac & -2ab \\ \end{matrix} \right]}$

Ans: We have to find

$\left[ \begin{matrix} {{a}^{2}}+{{b}^{2}} & {{b}^{2}}+{{c}^{2}} \\ {{a}^{2}}+{{c}^{2}} & {{a}^{2}}+{{b}^{2}} \\ \end{matrix} \right]+\left[ \begin{matrix} 2ab & 2bc \\ -2ac & -2ab \\ \end{matrix} \right]$

$\therefore \left[ \begin{matrix} {{a}^{2}}+{{b}^{2}} & {{b}^{2}}+{{c}^{2}} \\ {{a}^{2}}+{{c}^{2}} & {{a}^{2}}+{{b}^{2}} \\ \end{matrix} \right]+\left[ \begin{matrix} 2ab & 2bc \\ -2ac & -2ab \\ \end{matrix} \right]=\left[ \begin{matrix} {{a}^{2}}+{{b}^{2}}+2ab & {{b}^{2}}+{{c}^{2}}+2bc \\ {{a}^{2}}+{{c}^{2}}-2ac & {{a}^{2}}+{{b}^{2}}-2ab \\ \end{matrix} \right]$

$\Rightarrow \left[ \begin{matrix} {{a}^{2}}+{{b}^{2}} & {{b}^{2}}+{{c}^{2}} \\ {{a}^{2}}+{{c}^{2}} & {{a}^{2}}+{{b}^{2}} \\ \end{matrix} \right]+\left[ \begin{matrix} 2ab & 2bc \\ -2ac & -2ab \\ \end{matrix} \right]=\left[ \begin{matrix} {{\left( a+b \right)}^{2}} & {{\left( b+c \right)}^{2}} \\ {{\left( a-c \right)}^{2}} & {{\left( a-b \right)}^{2}} \\ \end{matrix} \right]$

iii. $\left[ \begin{matrix} -1 & 4 & -6 \\ 8 & 5 & 16 \\ 2 & 8 & 5 \\ \end{matrix} \right]+\left[ \begin{matrix} 12 & 7 & 6 \\ 8 & 0 & 5 \\ 3 & 2 & 4 \\ \end{matrix} \right]$

Ans: We have to find $\left[ \begin{matrix} -1 & 4 & -6 \\ 8 & 5 & 16 \\ 2 & 8 & 5 \\ \end{matrix} \right]+\left[ \begin{matrix} 12 & 7 & 6 \\ 8 & 0 & 5 \\ 3 & 2 & 4 \\ \end{matrix} \right]$

$\therefore \left[ \begin{matrix} -1 & 4 & -6 \\ 8 & 5 & 16 \\ 2 & 8 & 5 \\ \end{matrix} \right]+\left[ \begin{matrix} 12 & 7 & 6 \\ 8 & 0 & 5 \\ 3 & 2 & 4 \\ \end{matrix} \right]=\left[ \begin{matrix} -1+12 & 4+7 & -6+6 \\ 8+8 & 5+0 & 16+5 \\ 2+3 & 8+2 & 5+4 \\ \end{matrix} \right]$

$\Rightarrow \left[ \begin{matrix} -1 & 4 & -6 \\ 8 & 5 & 16 \\ 2 & 8 & 5 \\ \end{matrix} \right]+\left[ \begin{matrix} 12 & 7 & 6 \\ 8 & 0 & 5 \\ 3 & 2 & 4 \\ \end{matrix} \right]=\left[ \begin{matrix} 11 & 11 & 0 \\ 16 & 5 & 21 \\ 5 & 10 & 9 \\ \end{matrix} \right]$

iv. $\left[ \begin{matrix} {{\cos }^{2}}x & {{\sin }^{2}}x \\ {{\sin }^{2}}x & {{\cos }^{2}}x \\ \end{matrix} \right]+\left[ \begin{matrix} {{\sin }^{2}}x & {{\cos }^{2}}x \\ {{\cos }^{2}}x & {{\sin }^{2}}x \\ \end{matrix} \right]$

Ans: We have to find $\left[ \begin{matrix} {{\cos }^{2}}x & {{\sin }^{2}}x \\ {{\sin }^{2}}x & {{\cos }^{2}}x \\ \end{matrix} \right]+\left[ \begin{matrix} {{\sin }^{2}}x & {{\cos }^{2}}x \\ {{\cos }^{2}}x & {{\sin }^{2}}x \\ \end{matrix} \right]$

$\therefore \left[ \begin{matrix} {{\cos }^{2}}x & {{\sin }^{2}}x \\ {{\sin }^{2}}x & {{\cos }^{2}}x \\ \end{matrix} \right]+\left[ \begin{matrix} {{\sin }^{2}}x & {{\cos }^{2}}x \\ {{\cos }^{2}}x & {{\sin }^{2}}x \\ \end{matrix} \right]=\left[ \begin{matrix} {{\cos }^{2}}x+{{\sin }^{2}}x & {{\sin }^{2}}x+{{\cos }^{2}}x \\ {{\sin }^{2}}x+{{\cos }^{2}}x & {{\cos }^{2}}x+{{\sin }^{2}}x \\ \end{matrix} \right]$

$\Rightarrow \left[ \begin{matrix} {{\cos }^{2}}x & {{\sin }^{2}}x \\ {{\sin }^{2}}x & {{\cos }^{2}}x \\ \end{matrix} \right]+\left[ \begin{matrix} {{\sin }^{2}}x & {{\cos }^{2}}x \\ {{\cos }^{2}}x & {{\sin }^{2}}x \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right]$

3. Compute the indicated products

i. $\mathbf{\left[ \begin{matrix} a & b \\ -b & a \\ \end{matrix} \right]\left[ \begin{matrix} a & -b \\ b & a \\ \end{matrix} \right]}$

Ans: We have to find $\left[ \begin{matrix} a & b \\ -b & a \\ \end{matrix} \right]\left[ \begin{matrix} a & -b \\ b & a \\ \end{matrix} \right]$

$\therefore \left[ \begin{matrix} a & b \\ -b & a \\ \end{matrix} \right]\left[ \begin{matrix} a & -b \\ b & a \\ \end{matrix} \right]=\left[ \begin{matrix} a\left( a \right)+b\left( b \right) & a\left( -b \right)+b\left( a \right) \\ -b\left( a \right)+a\left( b \right) & -b\left( -b \right)+a\left( a \right) \\ \end{matrix} \right]$

$\Rightarrow \left[ \begin{matrix} a & b \\ -b & a \\ \end{matrix} \right]\left[ \begin{matrix} a & -b \\ b & a \\ \end{matrix} \right]=\left[ \begin{matrix} {{a}^{2}}+{{b}^{2}} & 0 \\ 0 & {{a}^{2}}+{{b}^{2}} \\ \end{matrix} \right]$

ii. $\mathbf{\left[ \begin{matrix} 1 \\ 2 \\ 3 \\ \end{matrix} \right]\left[ \begin{matrix} 2 & 3 & 4 \\ \end{matrix} \right]}$

Ans: We have to find $\left[ \begin{matrix} 1 \\ 2 \\ 3 \\ \end{matrix} \right]\left[ \begin{matrix} 2 & 3 & 4 \\ \end{matrix} \right]$

$\therefore \left[ \begin{matrix} 1 \\ 2 \\ 3 \\ \end{matrix} \right]\left[ \begin{matrix} 2 & 3 & 4 \\ \end{matrix} \right]=\left[ \begin{matrix} 1\left( 2 \right) & 1\left( 3 \right) & 1\left( 4 \right) \\ 2\left( 2 \right) & 2\left( 3 \right) & 2\left( 4 \right) \\ 3\left( 2 \right) & 3\left( 3 \right) & 3\left( 4 \right) \\ \end{matrix} \right]$

$\Rightarrow \left[ \begin{matrix} 1 \\ 2 \\ 3 \\ \end{matrix} \right]\left[ \begin{matrix} 2 & 3 & 4 \\ \end{matrix} \right]=\left[ \begin{matrix} 2 & 3 & 4 \\ 4 & 6 & 8 \\ 6 & 9 & 12 \\ \end{matrix} \right]$

iii. $\mathbf{\left[ \begin{matrix} 1 & -2 \\ 2 & 3 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ \end{matrix} \right]}$

Ans: We have to find $\left[ \begin{matrix} 1 & -2 \\ 2 & 3 \\\end{matrix} \right]\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ \end{matrix} \right]$

$\therefore \left[ \begin{matrix} 1 & -2 \\ 2 & 3 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 1\left( 1 \right)-2\left( 2 \right) & 1\left( 2 \right)-2\left( 3 \right) & 1\left( 3 \right)-2\left( 1 \right) \\ 2\left( 1 \right)+3\left( 2 \right) & 2\left( 2 \right)+3\left( 3 \right) & 2\left( 3 \right)+3\left( 1 \right) \\ \end{matrix} \right]$

$\Rightarrow \left[ \begin{matrix} 1 & -2 \\ 2 & 3 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} -3 & -4 & 1 \\ 8 & 13 & 9 \\ \end{matrix} \right]$

iv. $\mathbf{\left[ \begin{matrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \\ \end{matrix} \right]}$

Ans: We have to find $\left[ \begin{matrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \\ \end{matrix} \right]$

$\therefore \left[ \begin{matrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \\ \end{matrix} \right]=\left[ \begin{matrix} 2\left( 1 \right)+3\left( 0 \right)+4\left( 3 \right) & 2\left( -3 \right)+3\left( 2 \right)+4\left( 0 \right) & 2\left( 5 \right)+3\left( 4 \right)+4\left( 5 \right) \\ 3\left( 1 \right)+4\left( 0 \right)+5\left( 3 \right) & 3\left( -3 \right)+4\left( 2 \right)+5\left( 0 \right) & 3\left( 5 \right)+4\left( 4 \right)+5\left( 5 \right) \\ 4\left( 1 \right)+5\left( 0 \right)+5\left( 3 \right) & 4\left( -3 \right)+5\left( 2 \right)+6\left( 0 \right) & 4\left( 5 \right)+5\left( 4 \right)+6\left( 5 \right) \\ \end{matrix} \right]$

$\Rightarrow \left[ \begin{matrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \\ \end{matrix} \right]=\left[ \begin{matrix} 14 & 0 & 42 \\ 18 & -1 & 56 \\ 22 & -2 & 70 \\ \end{matrix} \right]$

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

Ans: We have to find $\left[ \begin{matrix} 2 & 1 \\ 3 & 2 \\ -1 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 & 1 \\ -1 & 2 & 1 \\ \end{matrix} \right]$

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

$\Rightarrow \left[ \begin{matrix} 2 & 1 \\ 3 & 2 \\ -1 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 & 1 \\ -1 & 2 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 2 & 3 \\ 1 & 4 & 5 \\ -2 & 2 & 0 \\ \end{matrix} \right]$

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

Ans: We have to find $\left[ \begin{matrix} 3 & -1 & 3 \\ -1 & 0 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} 2 & -3 \\ 1 & 0 \\ 3 & 1 \\ \end{matrix} \right]$

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

$\Rightarrow \left[ \begin{matrix} 3 & -1 & 3 \\ -1 & 0 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} 2 & -3 \\ 1 & 0 \\ 3 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 14 & -6 \\ 4 & 5 \\ \end{matrix} \right]$

4.If,

$\mathbf{A=\left[\begin{matrix} 1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1 \\ \end{matrix}\right],B=\left[ \begin{matrix} 3 & -1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3 \\ \end{matrix}\right],C=\left[ \begin{matrix} 4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3 \\ \end{matrix}\right]}$,

then Compute ( A+B ) and ( B-C ).Also, verify that A+( B-C)=( A+B )-C}.

Ans: Given  $A=\left[ \begin{matrix} 1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1 \\ \end{matrix} \right],B=\left[ \begin{matrix} 3 & -1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3 \\ \end{matrix} \right],C=\left[ \begin{matrix} 4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3 \\ \end{matrix} \right]$

$\therefore A+B=\left[ \begin{matrix} 1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1 \\ \end{matrix} \right]+\left[ \begin{matrix} 3 & -1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3 \\ \end{matrix} \right]$

$\Rightarrow A+B=\left[ \begin{matrix} 1+3 & 2-1 & -3+2 \\ 5+4 & 0+2 & 2+5 \\ 1+2 & -1+0 & 1+3 \\ \end{matrix} \right]$

$\Rightarrow A+B=\left[ \begin{matrix} 4 & 1 & -1 \\ 9 & 2 & 7 \\ 3 & -1 & 4 \\ \end{matrix} \right]$

And $B-C=\left[ \begin{matrix} 3 & -1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3 \\ \end{matrix} \right]-\left[ \begin{matrix} 4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3 \\ \end{matrix} \right]$

$\Rightarrow B-C=\left[ \begin{matrix} 3-4 & -1-1 & 2-2 \\ 4-0 & 2-3 & 5-2 \\ 2-1 & 0+2 & 3-3 \\ \end{matrix} \right]$

$\therefore B-C=\left[ \begin{matrix} -1 & -2 & 0 \\ 4 & -1 & 3 \\ 1 & 2 & 0 \\ \end{matrix} \right]$

Now,

$A+\left( B-C \right)=\left[ \begin{matrix} 1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1 \\ \end{matrix} \right]+\left[ \begin{matrix} -1 & -2 & 0 \\ 4 & -1 & 3 \\ 1 & 2 & 0 \\ \end{matrix} \right]$

$\Rightarrow A+\left( B-C \right)=\left[ \begin{matrix} 1-1 & 2-2 & -3+0 \\ 5+4 & 0-1 & 2+3 \\ 1+1 & -1+2 & 1+0 \\ \end{matrix} \right]$

$\therefore A+\left( B-C \right)=\left[ \begin{matrix} 0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1 \\ \end{matrix} \right]$                                                  …(1)

And $\left( A+B \right)-C=\left[ \begin{matrix} 4 & 1 & -1 \\ 9 & 2 & 7 \\ 3 & -1 & 4 \\ \end{matrix} \right]-\left[ \begin{matrix} 4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3 \\ \end{matrix} \right]$

$\Rightarrow \left( A+B \right)-C=\left[ \begin{matrix} 4-4 & 1-1 & -1-2 \\ 9-0 & 2-3 & 7-2 \\ 3-1 & -1+2 & 4-3 \\ \end{matrix} \right]$

$\therefore \left( A+B \right)-C=\left[ \begin{matrix} 0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1 \\ \end{matrix} \right]$                                                  …(2)

Thus, from equation (1) and (2),

$A+\left( B-C \right)=\left( A+B \right)-C$

Hence proved.

5. If $A=\left[ \begin{matrix} \dfrac{2}{3} & 1 & \dfrac{5}{3} \\ \dfrac{1}{3} & \dfrac{2}{3} & \dfrac{4}{3} \\ \dfrac{7}{3} & 2 & \dfrac{2}{3} \\ \end{matrix} \right]$ and $B=\left[ \begin{matrix} \dfrac{2}{5} & \dfrac{3}{5} & 1 \\ \dfrac{1}{5} & \dfrac{2}{5} & \dfrac{4}{5} \\ \dfrac{7}{5} & \dfrac{6}{5} & \dfrac{2}{5} \\ \end{matrix} \right]$ then compute $3A-5B$ .

Ans: Given that $A=\left[ \begin{matrix} \dfrac{2}{3} & 1 & \dfrac{5}{3} \\ \dfrac{1}{3} & \dfrac{2}{3} & \dfrac{4}{3} \\ \dfrac{7}{3} & 2 & \dfrac{2}{3} \\ \end{matrix} \right]$ and $B=\left[ \begin{matrix} \dfrac{2}{5} & \dfrac{3}{5} & 1 \\ \dfrac{1}{5} & \dfrac{2}{5} & \dfrac{4}{5} \\ \dfrac{7}{5} & \dfrac{6}{5} & \dfrac{2}{5} \\ \end{matrix} \right]$

$\therefore 3A-5B=3\left[ \begin{matrix} \dfrac{2}{3} & 1 & \dfrac{5}{3} \\ \dfrac{1}{3} & \dfrac{2}{3} & \dfrac{4}{3} \\ \dfrac{7}{3} & 2 & \dfrac{2}{3} \\ \end{matrix} \right]-5\left[ \begin{matrix} \dfrac{2}{5} & \dfrac{3}{5} & 1 \\ \dfrac{1}{5} & \dfrac{2}{5} & \dfrac{4}{5} \\ \dfrac{7}{5} & \dfrac{6}{5} &\dfrac{2}{5} \\ \end{matrix} \right]$

$\Rightarrow 3A-5B=\left[ \begin{matrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \\ \end{matrix} \right]-\left[ \begin{matrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \\ \end{matrix} \right]$

$\Rightarrow 3A-5B=\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{matrix} \right]$

6. Simplify $\mathbf{\cos \theta \left[ \begin{matrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \\ \end{matrix} \right]+\sin \theta \left[ \begin{matrix} \sin \theta & -\cos \theta \\ \cos \theta & \sin \theta \\ \end{matrix} \right]}$ .

Ans: We have to simplify $\cos \theta \left[ \begin{matrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \\ \end{matrix} \right]+\sin \theta \left[ \begin{matrix} \sin \theta & -\cos \theta \\ \cos \theta & \sin \theta \\ \end{matrix} \right]$

$\therefore \cos \theta \left[ \begin{matrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \\ \end{matrix} \right]+\sin \theta \left[ \begin{matrix} \sin \theta & -\cos \theta \\ \cos \theta & \sin \theta \\ \end{matrix} \right]=\left[ \begin{matrix} {{\cos }^{2}}\theta & \cos \theta \sin \theta \\ -\cos \theta \sin \theta & {{\cos }^{2}}\theta \\ \end{matrix} \right]+\left[ \begin{matrix} {{\sin }^{2}}\theta & -\sin \theta \cos \theta \\ \sin \theta \cos \theta & {{\sin }^{2}}\theta \\ \end{matrix} \right]$

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

$\Rightarrow \cos \theta \left[ \begin{matrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \\ \end{matrix} \right]+\sin \theta \left[ \begin{matrix} \sin \theta & -\cos \theta \\ \cos \theta & \sin \theta \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]$

7. Find $X$ and $Y$ ,if

i. ${X+Y=}\left [ \begin{matrix} 7 & 0 \\ 2 & 5 \\ \end{matrix} \right]$ and $X-Y=\left[ \begin{matrix} 3 & 0 \\ 0 & 3 \\ \end{matrix} \right]$

Ans: Given:

$X+Y=\left[ \begin{matrix} 7 & 0 \\ 2 & 5 \\ \end{matrix} \right]$        …(1)

$X-Y=\left[ \begin{matrix} 3 & 0 \\ 0 & 3 \\ \end{matrix} \right]$   …(2)

Adding equations (1) and (2), we get,

$2X=\left[ \begin{matrix} 7 & 0 \\ 2 & 5 \\ \end{matrix} \right]+\left[ \begin{matrix} 3 & 0 \\ 0 & 3 \\ \end{matrix} \right]$

$\Rightarrow 2X=\left[ \begin{matrix} 10 & 0 \\ 2 & 8 \\ \end{matrix} \right]$

$\Rightarrow X=\dfrac{1}{2}\left[ \begin{matrix} 10 & 0 \\ 2 & 8 \\ \end{matrix} \right]$

$\therefore X=\left[ \begin{matrix} 5 & 0 \\ 1 & 4 \\ \end{matrix} \right]$

Now, since $X+Y=\left[ \begin{matrix} 7 & 0 \\ 2 & 5 \\ \end{matrix} \right]$

$\Rightarrow \left[ \begin{matrix} 5 & 0 \\ 1 & 4 \\ \end{matrix} \right]+Y=\left[ \begin{matrix} 7 & 0 \\ 2 & 5 \\ \end{matrix} \right]$

$\Rightarrow Y=\left[ \begin{matrix} 7 & 0 \\ 2 & 5 \\ \end{matrix} \right]-\left[ \begin{matrix} 5 & 0 \\ 1 & 4 \\ \end{matrix} \right]$

$\therefore Y=\left[ \begin{matrix} 2 & 0 \\ 1 & 1 \\ \end{matrix} \right]$

$\therefore X=\left[ \begin{matrix} 5 & 0 \\ 1 & 4 \\ \end{matrix} \right],Y=\left[ \begin{matrix} 2 & 0 \\ 1 & 1 \\ \end{matrix} \right]$

ii. $\mathbf{2X+3Y=\left[ \begin{matrix} 2 & 3 \\ 4 & 0 \\ \end{matrix} \right]}$ and $\mathbf{3X+2Y=\left[ \begin{matrix} 2 & -2 \\ -1 & 5 \\ \end{matrix} \right]}$

Ans: Given:

$2X+3Y=\left[ \begin{matrix} 2 & 3 \\ 4 & 0 \\ \end{matrix} \right]$                               …(1)

$3X+2Y=\left[ \begin{matrix} 2 & -2 \\ -1 & 5 \\ \end{matrix} \right]$                            …(2)

Multiplying equation (1) with two we get,

$2\left( 2X+3Y \right)=2\left[ \begin{matrix} 2 & 3 \\ 4 & 0 \\ \end{matrix} \right]$

$\Rightarrow \left( 4X+6Y \right)=\left[ \begin{matrix} 4 & 6 \\ 8 & 0 \\ \end{matrix} \right]$                      …(3)

Multiplying equation (2) with three we get,

$3\left( 3X+2Y \right)=3\left[ \begin{matrix} 2 & -2 \\ -1 & 5 \\ \end{matrix} \right]$

$\Rightarrow \left( 9X+6Y \right)=\left[ \begin{matrix} 6 & -6 \\ -3 & 15 \\ \end{matrix} \right]$                 …(4)

From equation (3) and (4),

$\left( 4X+6Y \right)-\left( 9X+6Y \right)=\left[ \begin{matrix} 4 & 6 \\ 8 & 0 \\ \end{matrix} \right]-\left[ \begin{matrix} 6 & -6 \\ -3 & 15 \\ \end{matrix} \right]$

$\Rightarrow -5X=\left[ \begin{matrix} -2 & 12 \\ 11 & -15 \\ \end{matrix} \right]$

$\Rightarrow X=\dfrac{-1}{5}\left[ \begin{matrix} -2 & 12 \\ 11 & -15 \\ \end{matrix} \right]$

$\therefore X=\left[ \begin{matrix} \dfrac{2}{5} & \dfrac{-12}{5} \\ \dfrac{-11}{5} & 3 \\ \end{matrix} \right]$

Now, since $2X+3Y=\left[ \begin{matrix} 2 & 3 \\ 4 & 0 \\ \end{matrix} \right]$

$\therefore 2\left[ \begin{matrix} \dfrac{2}{5} & \dfrac{-12}{5} \\ \dfrac{-11}{5} & 3 \\ \end{matrix} \right]+3Y=\left[ \begin{matrix} 2 & 3 \\ 4 & 0 \\ \end{matrix} \right]$

$\Rightarrow \left[ \begin{matrix} \dfrac{4}{5} & \dfrac{-24}{5} \\ \dfrac{-22}{5} & 6 \\ \end{matrix} \right]+3Y=\left[ \begin{matrix} 2 & 3 \\ 4 & 0 \\ \end{matrix} \right]$

$\Rightarrow 3Y=\left[ \begin{matrix} 2 & 3 \\ 4 & 0 \\ \end{matrix} \right]-\left[ \begin{matrix} \dfrac{4}{5} & \dfrac{-24}{5} \\ \dfrac{-22}{5} & 6 \\ \end{matrix} \right]$

$\Rightarrow 3Y=\left[ \begin{matrix} \dfrac{6}{5} & \dfrac{39}{5} \\ \dfrac{42}{5} & -6 \\ \end{matrix} \right]$

$\therefore Y=\left[ \begin{matrix} \dfrac{2}{5} & \dfrac{13}{5} \\ \dfrac{14}{5} & -2 \\ \end{matrix} \right]$

$\therefore X=\left[ \begin{matrix} \dfrac{2}{5} & \dfrac{-12}{5} \\ \dfrac{-11}{5} & 3 \\ \end{matrix} \right],Y=\left[ \begin{matrix} \dfrac{2}{5} & \dfrac{13}{5} \\ \dfrac{14}{5} & -2 \\ \end{matrix} \right]$

8. Find $\mathbf{X}$ ,if $\mathbf{Y=\left[ \begin{matrix} 3 & 2 \\ 1 & 4 \\ \end{matrix} \right]}$ and $\mathbf{2X+Y=\left[ \begin{matrix} 1 & 0 \\ -3 & 2 \\ \end{matrix} \right]}$ .

Ans: Given $Y=\left[ \begin{matrix} 3 & 2 \\ 1 & 4 \\ \end{matrix} \right]$

And $2X+Y=\left[ \begin{matrix} 1 & 0 \\ -3 & 2 \\ \end{matrix} \right]$

$\Rightarrow 2X+\left[ \begin{matrix} 3 & 2 \\ 1 & 4 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ -3 & 2 \\ \end{matrix} \right]$

$\Rightarrow 2X=\left[ \begin{matrix} 1 & 0 \\ -3 & 2 \\ \end{matrix} \right]-\left[ \begin{matrix} 3 & 2 \\ 1 & 4 \\ \end{matrix} \right]$

$\Rightarrow 2X=\left[ \begin{matrix} -2 & -2 \\ -4 & -2 \\ \end{matrix} \right]$

$\Rightarrow X=\dfrac{1}{2}\left[ \begin{matrix} -2 & -2 \\ -4 & -2 \\ \end{matrix} \right]$

$\therefore X=\left[ \begin{matrix} -1 & -1 \\ -2 & -1 \\ \end{matrix} \right]$

9. Find $X$ and $Y$ ,if $\mathbf{2\left[ \begin{matrix} 1 & 3 \\ 0 & x \\ \end{matrix} \right]+\left[ \begin{matrix} y & 0 \\ 1 & 2 \\ \end{matrix} \right]=\left[ \begin{matrix} 5 & 6 \\ 1 & 8 \\ \end{matrix} \right]}$

Ans: Given: $2\left[ \begin{matrix} 1 & 3 \\ 0 & x \\ \end{matrix} \right]+\left[ \begin{matrix} y & 0 \\ 1 & 2 \\ \end{matrix} \right]=\left[ \begin{matrix} 5 & 6 \\ 1 & 8 \\ \end{matrix} \right]$

$\Rightarrow \left[ \begin{matrix} 2 & 6 \\ 0 & 2x \\ \end{matrix} \right]+\left[ \begin{matrix} y & 0 \\ 1 & 2 \\ \end{matrix} \right]=\left[ \begin{matrix} 5 & 6 \\ 1 & 8 \\ \end{matrix} \right]$

$\Rightarrow \left[ \begin{matrix} 2+y & 6+0 \\ 0+1 & 2x+2 \\ \end{matrix} \right]=\left[ \begin{matrix} 5 & 6 \\ 1 & 8 \\ \end{matrix} \right]$

$\Rightarrow \left[ \begin{matrix} 2+y & 6 \\ 1 & 2x+2 \\ \end{matrix} \right]=\left[ \begin{matrix} 5 & 6 \\ 1 & 8 \\ \end{matrix} \right]$

Comparing the corresponding elements of these two matrices, we get:

$2+y=5$

$\Rightarrow y=3$

$2x+2=8$

$\Rightarrow x=3$

$\therefore x=3,y=3$

10. Solve the equation for $X,Y,Z$ and $t$ if $\mathbf{2\left[ \begin{matrix} x & z \\ y & t \\ \end{matrix} \right]+3\left[ \begin{matrix} 1 & -1 \\ 0 & 2 \\ \end{matrix} \right]=3\left[ \begin{matrix} 3 & 5 \\ 4 & 6 \\ \end{matrix} \right]}$

Ans: Given: $2\left[ \begin{matrix} x & z \\ y & t \\ \end{matrix} \right]+3\left[ \begin{matrix} 1 & -1 \\ 0 & 2 \\ \end{matrix} \right]=3\left[ \begin{matrix} 3 & 5 \\ 4 & 6 \\ \end{matrix} \right]$

$\Rightarrow \left[ \begin{matrix} 2x & 2z \\ 2y & 2t \\ \end{matrix} \right]+\left[ \begin{matrix} 3 & -3 \\ 0 & 6 \\ \end{matrix} \right]=\left[ \begin{matrix} 9 & 15 \\ 12 & 18 \\ \end{matrix} \right]$

$\Rightarrow \left[ \begin{matrix} 2x+3 & 2z-3 \\ 2y & 2t+6 \\ \end{matrix} \right]=\left[ \begin{matrix} 9 & 15 \\ 12 & 18 \\ \end{matrix} \right]$

Equating the corresponding elements of these two matrices, we get:

$2x+3=9$

$\Rightarrow x=3$

$2y=12$

$\Rightarrow y=6$

$2z-3=15$

$\Rightarrow z=9$

$2t+6=18$

$\Rightarrow t=6$

$\therefore x=3,y=6,z=9,t=6$

11. If $\mathbf{x\left[ \begin{matrix} 2 \\ 3 \\ \end{matrix} \right]+y\left[ \begin{matrix} -1 \\ 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 10 \\ 5 \\ \end{matrix} \right]}$ ,find values of $x$ and $y$ .

Ans: Given: $x\left[ \begin{matrix} 2 \\ 3 \\ \end{matrix} \right]+y\left[ \begin{matrix} -1 \\ 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 10 \\ 5 \\ \end{matrix} \right]$

$\Rightarrow \left[ \begin{matrix} 2x \\ 3x \\ \end{matrix} \right]+\left[ \begin{matrix} -y \\ y \\ \end{matrix} \right]=\left[ \begin{matrix} 10 \\ 5 \\ \end{matrix} \right]$

$\Rightarrow \left[ \begin{matrix} 2x-y \\ 3x+y \\ \end{matrix} \right]=\left[ \begin{matrix} 10 \\ 5 \\ \end{matrix} \right]$

Equating the corresponding elements of these two matrices, we get:

$2x-y=10$ and

$3x+y=5$

Adding these two equations, we get:

$5x=15$

$\Rightarrow x=3$

Now, since $3x+y=5$

$\Rightarrow y=5-3x$

$\Rightarrow y=5-9$

$\Rightarrow y=-4$

$\therefore x=3,y=-4$

12. Given $\mathbf{3\left[ \begin{matrix} x & y \\ z & w \\ \end{matrix} \right]=\left[ \begin{matrix} x & 6 \\ -1 & 2w \\ \end{matrix} \right]+\left[ \begin{matrix} 4 & x+y \\ z+w & 2w+3 \\ \end{matrix} \right]}$ ,find the values of $x,y,z$ and $w$ .

Ans: Given: $3\left[ \begin{matrix} x & y \\ z & w \\ \end{matrix} \right]=\left[ \begin{matrix} x & 6 \\ -1 & 2w \\ \end{matrix} \right]+\left[ \begin{matrix} 4 & x+y \\ z+w & 2w+3 \\ \end{matrix} \right]$

$\Rightarrow \left[ \begin{matrix} 3x & 3y \\ 3z & 3w \\ \end{matrix} \right]=\left[ \begin{matrix} x+4 & 6+x+y \\ -1+z+w & 2w+3 \\ \end{matrix} \right]$

Equating the corresponding elements of these two matrices, we get:

$3x=x+4$

$\Rightarrow x=2$

$3y=6+x+y$

$\Rightarrow y=4$

$3w=2w+3$

$\Rightarrow w=3$

$3z=-1+z+w$

$\Rightarrow z=1$

$\therefore x=2,y=4,z=1,w=3$

13. If $\mathbf{F\left( x \right)=\left[ \begin{matrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]}$ , show that $\mathbf{F\left( x \right)F\left( y \right)=F\left( x+y \right)}$ .

Ans: Here, $F\left( x \right)=\left[ \begin{matrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]$

$F\left( y \right)=\left[ \begin{matrix} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]$

$\therefore F\left( x \right)F\left( y \right)=\left[ \begin{matrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]$

$\Rightarrow F\left( x \right)F\left( y \right)=\left[ \begin{matrix} \cos x\cos y-\sin x\sin y+0 & -\cos x\sin y-\sin x\cos y+0 & 0 \\ \sin x\cos y-\cos x\sin y+0 & -\sin x\sin y+\cos x\cos y+0 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]$

$\Rightarrow F\left( x \right)F\left( y \right)=\left[ \begin{matrix} \cos \left( x+y \right) & -\sin \left( x+y \right) & 0 \\ \sin \left( x+y \right) & \cos \left( x+y \right) & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]$                            …(1)

And $F\left( x+y \right)=\left[ \begin{matrix} \cos \left( x+y \right) & -\sin \left( x+y \right) & 0 \\ \sin \left( x+y \right) & \cos \left( x+y \right) & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]$                               …(2)

From equation (1) and (2),

$F\left( x \right)F\left( y \right)=F\left( x+y \right)$

Hence proved.

14. Show that

i. $\mathbf{\left[ \begin{matrix} 5 & -1 \\ 6 & 7 \\ \end{matrix} \right]\left[ \begin{matrix} 2 & 1 \\ 3 & 4 \\ \end{matrix} \right]\ne \left[ \begin{matrix} 2 & 1 \\ 3 & 4 \\ \end{matrix} \right]\left[ \begin{matrix} 5 & -1 \\ 6 & 7 \\ \end{matrix} \right]}$

Ans: LHS: $\left[ \begin{matrix} 5 & -1 \\ 6 & 7 \\ \end{matrix} \right]\left[ \begin{matrix} 2 & 1 \\ 3 & 4 \\ \end{matrix} \right]=\left[ \begin{matrix} 5\left( 2 \right)-1\left( 3 \right) & 5\left( 1 \right)-1\left( 4 \right) \\ 6\left( 2 \right)+7\left( 3 \right) & 6\left( 1 \right)+7\left( 4 \right) \\ \end{matrix} \right]$

$\Rightarrow \left[ \begin{matrix} 5 & -1 \\ 6 & 7 \\ \end{matrix} \right]\left[ \begin{matrix} 2 & 1 \\ 3 & 4 \\ \end{matrix} \right]=\left[ \begin{matrix} 7 & 1 \\ 33 & 34 \\ \end{matrix} \right]$                …(1)

RHS: $\left[ \begin{matrix} 2 & 1 \\ 3 & 4 \\ \end{matrix} \right]\left[ \begin{matrix} 5 & -1 \\ 6 & 7 \\ \end{matrix} \right]=\left[ \begin{matrix} 2\left( 5 \right)+1\left( 6 \right) & 2\left( -1 \right)+1\left( 7 \right) \\ 3\left( 5 \right)+4\left( 6 \right) & 3\left( -1 \right)+4\left( 7 \right) \\ \end{matrix} \right]$

$\Rightarrow \left[ \begin{matrix} 2 & 1 \\ 3 & 4 \\ \end{matrix} \right]\left[ \begin{matrix} 5 & -1 \\ 6 & 7 \\ \end{matrix} \right]=\left[ \begin{matrix} 16 & 5 \\ 39 & 25 \\ \end{matrix} \right]$        …(2)

From equation (1) and (2),

$LHS\ne RHS$

Hence proved.

ii. $\mathbf{\left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \\ \end{matrix} \right]\left[ \begin{matrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \\ \end{matrix} \right]\ne \left[ \begin{matrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \\ \end{matrix} \right]}$

Ans: LHS: $\left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \\ \end{matrix} \right]\left[ \begin{matrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \\ \end{matrix} \right]=\left[ \begin{matrix} 1\left( -1 \right)+2\left( 0 \right)+3\left( 2 \right) & 1\left( 1 \right)+2\left( -1 \right)+3\left( 3 \right) & 1\left( 0 \right)+2\left( 1 \right)+3\left( 4 \right) \\ 0\left( -1 \right)+1\left( 0 \right)+0\left( 2 \right) & 0\left( 1 \right)+1\left( -1 \right)+0\left( 3 \right) & 0\left( 0 \right)+1\left( 1 \right)+0\left( 4 \right) \\ 1\left( -1 \right)+1\left( 0 \right)+0\left( 2 \right) & 1\left( 1 \right)+1\left( -1 \right)+0\left( 3 \right) & 1\left( 0 \right)+1\left( 1 \right)+0\left( 4 \right) \\ \end{matrix} \right]$

$\Rightarrow \left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \\ \end{matrix} \right]\left[ \begin{matrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \\ \end{matrix} \right]=\left[ \begin{matrix} 5 & 8 & 14 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \\ \end{matrix} \right]$                            …(1)

RHS: $\left[ \begin{matrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \\ \end{matrix} \right]=\left[ \begin{matrix} -1\left( 1 \right)+1\left( 0 \right)+0\left( 1 \right) & -1\left( 2 \right)+1\left( 1 \right)+0\left( 1 \right) & -1\left( 3 \right)+1\left( 0 \right)+0\left( 0 \right) \\ 0\left( 1 \right)-1\left( 0 \right)+1\left( 1 \right) & 0\left( 2 \right)-1\left( 1 \right)+1\left( 1 \right) & 0\left( 3 \right)-1\left( 0 \right)+1\left( 0 \right) \\ 2\left( 1 \right)+3\left( 0 \right)+4\left( 1 \right) & 2\left( 2 \right)+3\left( 1 \right)+4\left( 1 \right) & 2\left( 3 \right)+3\left( 0 \right)+4\left( 0 \right) \\ \end{matrix} \right]$

$\Rightarrow \left[ \begin{matrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \\ \end{matrix} \right]=\left[ \begin{matrix} -1 & -1 & 3 \\ 1 & 0 & 0 \\ 6 & 11 & 6 \\ \end{matrix} \right]$                          …(2)

From equation (1) and (2),

$LHS\ne RHS$

Hence proved.

15. Find $\mathbf{{{A}^{2}}-5A+6I}$ if $\mathbf{A=\left[ \begin{matrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \\ \end{matrix} \right]}$ .

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

${{A}^{2}}=AA$

$\therefore {{A}^{2}}=\left[ \begin{matrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \\ \end{matrix} \right]\left[ \begin{matrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \\ \end{matrix} \right]$

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

$\Rightarrow {{A}^{2}}=\left[ \begin{matrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \\ \end{matrix} \right]$

$\therefore {{A}^{2}}-5A+6I=\left[ \begin{matrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \\ \end{matrix} \right]-5\left[ \begin{matrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \\ \end{matrix} \right]+6\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]$

$\Rightarrow {{A}^{2}}-5A+6I=\left[ \begin{matrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \\ \end{matrix} \right]-\left[ \begin{matrix} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \\ \end{matrix} \right]+\left[ \begin{matrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \\ \end{matrix} \right]$

$\Rightarrow {{A}^{2}}-5A+6I=\left[ \begin{matrix} 5-10 & -1-0 & 2-5 \\ 9-10 & -2-5 & 5-15 \\ 0-5 & -1+5 & -2-0 \\ \end{matrix} \right]+\left[ \begin{matrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \\ \end{matrix} \right]$

$\Rightarrow {{A}^{2}}-5A+6I=\left[ \begin{matrix} -5 & -1 & -3 \\ -1 & -7 & -10 \\ -5 & 4 & -2 \\ \end{matrix} \right]+\left[ \begin{matrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \\ \end{matrix} \right]$

$\Rightarrow {{A}^{2}}-5A+6I=\left[ \begin{matrix} 1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \\ \end{matrix} \right]$

16. If $\mathbf{A=\left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \\ \end{matrix} \right]}$ , prove that $\mathbf{{{A}^{3}}-6{{A}^{2}}+7A+21=0}$ .

Ans: Given: $A=\left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \\ \end{matrix} \right]$

${{A}^{2}}=AA$

$\therefore {{A}^{2}}=\left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \\ \end{matrix} \right]$

$\Rightarrow {{A}^{2}}=\left[ \begin{matrix} 1+0+4 & 0+0+0 & 2+0+6 \\ 0+0+2 & 0+4+0 & 0+2+3 \\ 2+0+6 & 0+0+0 & 4+0+9 \\ \end{matrix} \right]$

$\Rightarrow {{A}^{2}}=\left[ \begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \\ \end{matrix} \right]$

Now, ${{A}^{3}}={{A}^{2}}\cdot A$

$\therefore {{A}^{3}}=\left[ \begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \\ \end{matrix} \right]$

$\Rightarrow {{A}^{3}}=\left[ \begin{matrix} 5+0+16 & 0+0+0 & 10+0+24 \\ 2+0+10 & 0+8+0 & 4+4+15 \\ 8+0+26 & 0+0+0 & 16+0+39 \\ \end{matrix} \right]$

$\Rightarrow {{A}^{3}}=\left[ \begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \\ \end{matrix} \right]$

$\therefore {{A}^{3}}-6{{A}^{2}}+7A+21=\left[ \begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \\ \end{matrix} \right]-6\left[ \begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \\ \end{matrix} \right]+7\left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \\ \end{matrix} \right]+2\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]$

$\Rightarrow {{A}^{3}}-6{{A}^{2}}+7A+21=\left[ \begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \\ \end{matrix} \right]-\left[ \begin{matrix} 30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \\ \end{matrix} \right]+\left[ \begin{matrix} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \\ \end{matrix} \right]+\left[ \begin{matrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \\ \end{matrix} \right]$

$\Rightarrow {{A}^{3}}-6{{A}^{2}}+7A+21=\left[ \begin{matrix} 21-30+7+2 & 0-0+0+0 & 34-48+14+0 \\ 12-12+0+0 & 8-24+14+2 & 23-30+7+0 \\ 34-48+14+0 & 0-0+0+0 & 55-78+21+2 \\ \end{matrix} \right]$

$\Rightarrow {{A}^{3}}-6{{A}^{2}}+7A+21=\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{matrix} \right]$

$\therefore {{A}^{3}}-6{{A}^{2}}+7A+21=0$

Hence proved.

17.  If $\mathbf{A=\left[ \begin{matrix} 3 & -2 \\ 4 & -2 \\ \end{matrix} \right]}$ and $\mathbf{I=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]}$ , find $\mathbf{k}$ so that $\mathbf{{{A}^{2}}=kA-21}$ .

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

${{A}^{2}}=AA$

$\Rightarrow {{A}^{2}}=\left[ \begin{matrix} 3 & -2 \\ 4 & -2 \\ \end{matrix} \right]\left[ \begin{matrix} 3 & -2 \\ 4 & -2 \\ \end{matrix} \right]$

$\Rightarrow {{A}^{2}}=\left[ \begin{matrix} 9-8 & -6+4 \\ 12-8 & -8+4 \\ \end{matrix} \right]$

$\Rightarrow {{A}^{2}}=\left[ \begin{matrix} 1 & -2 \\ 4 & -4 \\ \end{matrix} \right]$

Now, ${{A}^{2}}=kA-21$

$\Rightarrow \left[ \begin{matrix} 1 & -2 \\ 4 & -4 \\ \end{matrix} \right]=k\left[ \begin{matrix} 3 & -2 \\ 4 & -2 \\ \end{matrix} \right]-2\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]$

$\Rightarrow \left[ \begin{matrix} 1 & -2 \\ 4 & -4 \\ \end{matrix} \right]=\left[ \begin{matrix} 3k & -2k \\ 4k & -2k \\ \end{matrix} \right]-\left[ \begin{matrix} 2 & 0 \\ 0 & 2 \\ \end{matrix} \right]$

$\Rightarrow \left[ \begin{matrix} 1 & -2 \\ 4 & -4 \\ \end{matrix} \right]=\left[ \begin{matrix} 3k-2 & -2k \\ 4k & -2k-2 \\ \end{matrix} \right]$

Equating the corresponding elements, we have:

$3k-2=1$

$\Rightarrow k=1$

Thus, the value of $k$ is $1$.

18. If $\mathbf{A=\left[ \begin{matrix} 0 & -\tan \dfrac{\alpha }{2} \\ \tan \dfrac{\alpha }{2} & 0 \\ \end{matrix} \right]}$ and $I$ is the identity matrix of order $2$ , show that $\mathbf{I+A=\left( I-A \right)\left[ \begin{matrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \\ \end{matrix} \right]}$

Ans: Given: $A=\left[ \begin{matrix} 0 & -\tan \dfrac{\alpha }{2} \\ \tan \dfrac{\alpha }{2} & 0 \\ \end{matrix} \right]$

LHS: $I+A=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]+\left[ \begin{matrix} 0 & -\tan \dfrac{\alpha }{2} \\ \tan \dfrac{\alpha }{2} & 0 \\ \end{matrix} \right]$

$\Rightarrow I+A=\left[ \begin{matrix} 1 & -\tan \dfrac{\alpha }{2} \\ \tan \dfrac{\alpha }{2} & 1 \\ \end{matrix} \right]$                                           …(1)

RHS: $\left( I-A \right)\left[ \begin{matrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \\ \end{matrix} \right]=\left( \left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]-\left[ \begin{matrix} 0 & -\tan \dfrac{\alpha }{2} \\ \tan \dfrac{\alpha }{2} & 0 \\ \end{matrix} \right] \right)\left[ \begin{matrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \\ \end{matrix} \right]$

$\Rightarrow \left( I-A \right)\left[ \begin{matrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & \tan \dfrac{\alpha }{2} \\ -\tan \dfrac{\alpha }{2} & 1 \\ \end{matrix} \right]\left[ \begin{matrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \\ \end{matrix} \right]$

$\Rightarrow \left( I-A \right)\left[ \begin{matrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \\ \end{matrix} \right]=\left[ \begin{matrix} \cos \alpha +\sin \alpha \tan \dfrac{\alpha }{2} & -\sin \alpha +\cos \alpha \tan \dfrac{\alpha }{2} \\ -\cos \alpha \tan \dfrac{\alpha }{2}+\sin \alpha & \sin \alpha \tan \dfrac{\alpha }{2}+\cos \alpha \\ \end{matrix} \right]$

$\Rightarrow \left( I-A \right)\left[ \begin{matrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \\ \end{matrix} \right]=\left[ \begin{matrix} 1-2{{\sin }^{2}}\dfrac{\alpha }{2}+2\sin \dfrac{\alpha }{2}-\cos \dfrac{\alpha }{2}\tan \dfrac{\alpha }{2} & -2\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2}+\left( 2{{\cos }^{2}}\dfrac{\alpha }{2}-1 \right)\tan \dfrac{\alpha }{2} \\ -\left( 2{{\cos }^{2}}\dfrac{\alpha }{2}-1 \right)\tan \dfrac{\alpha }{2}+2\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2} & 2\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2}\tan \dfrac{\alpha }{2}+1-2{{\sin }^{2}}\dfrac{\alpha }{2} \\ \end{matrix} \right]$

$\Rightarrow \left( I-A \right)\left[ \begin{matrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \\ \end{matrix} \right]=\left[ \begin{matrix} 1-2{{\sin }^{2}}\dfrac{\alpha }{2}+2{{\sin }^{2}}\dfrac{\alpha }{2} & -2\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2}+2\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2}-\tan \dfrac{\alpha }{2} \\ -\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2}+\tan \dfrac{\alpha }{2}+2\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2} & 2{{\sin }^{2}}\dfrac{\alpha }{2}+1-2{{\sin }^{2}}\dfrac{\alpha }{2} \\ \end{matrix} \right]$

$\Rightarrow \left( I-A \right)\left[ \begin{matrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & -\tan \dfrac{\alpha }{2} \\ \tan \dfrac{\alpha }{2} & 1 \\ \end{matrix} \right]$                                        …(2)

Thus, from equation (1) and (2).

$LHS=RHS$

Hence proved.

19. A trust fund has Rs $30,000$ that must be invested in two different types of bonds. The first bond pays $5%$ interest per year, and the second bond pays $7%$ interest per year. Using matrix multiplication, determine how to divide Rs $30,000$ among the two types of bonds. If the trust fund must obtain an annual total interest of:

i. Rs $\mathbf{1,800}$

Ans: Let Rs $x$ be invested in the first round.

Then, the sum of money invested in the second bond pays Rs $\left( 30000-x \right)$

It is given that the first bond pays $5%$ interest per year, and the second bond pays $7%$ interest per year.

We know that,

Simple interest for one year is $\dfrac{principle\times rate}{100}$ .

Therefore, in order to obtain an annual total interest of Rs $1,800$ ,

$\left[ \begin{matrix} x & \left( 30000-x \right) \\ \end{matrix} \right]\left[ \begin{matrix} \dfrac{5}{100} \\ \dfrac{7}{100} \\ \end{matrix} \right]=1800$

$\Rightarrow \dfrac{5x}{100}+\dfrac{7\left( 30000-x \right)}{100}=1800$

$\Rightarrow 210000-2x=180000$

$\Rightarrow x=15000$

Thus, in order to obtain an annual total interest of Rs $1,800$ , the trust fund should invest Rs $15000$ in the first bond and the remaining Rs $15000$ in the second bond.

ii. Rs $\mathbf{2,000}$

Ans: Let Rs $x$ be invested in the first round.

Then, the sum of money invested in the second bond pays Rs $\left( 30000-x \right)$

It is given that the first bond pays $5%$ interest per year, and the second bond pays $7%$ interest per year.

We know that,

Simple interest for one year is $\dfrac{principle\times rate}{100}$ .

Therefore, in order to obtain an annual total interest of Rs $2,000$ ,

$\left[ \begin{matrix} x & \left( 30000-x \right) \\ \end{matrix} \right]\left[ \begin{matrix} \dfrac{5}{100} \\ \dfrac{7}{100} \\ \end{matrix} \right]=2000$

$\Rightarrow \dfrac{5x}{100}+\dfrac{7\left( 30000-x \right)}{100}=2000$

$\Rightarrow 210000-2x=200000$

$\Rightarrow x=5000$

Thus, in order to obtain an annual total interest of Rs $2,000$ , the trust fund should invest Rs $5000$ in the first bond and the remaining Rs $25000$ in the second bond.

20. The bookshop of a particular school has $10$ dozen chemistry books, $8$ dozen physics books, $10$ dozen economics books. Their selling prices are Rs $80$, Rs $60$ and Rs $40$ each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.

Ans: The total amount of money that will be received from the sale of all these books can be represented in the form of a matrix as:

$12\left[ \begin{matrix} 10 & 8 & 10 \\ \end{matrix} \right]\left[ \begin{matrix} 80 \\ 60 \\ 40 \\ \end{matrix} \right]$

$\Rightarrow 12\left[ 10\times 80+8\times 60+10\times 40 \right]$

$\Rightarrow 12\left[ 1680 \right]$

$\Rightarrow 20160$

Therefore, the bookshop will receive Rs $20160$ from the sale.

21. Assume $X,Y,Z,W$ and $P$ are the matrices of order $\mathbf{2\times n,3\times k,2\times p,n\times 3}$ and $\mathbf{p\times k}$ respectively. The restriction on $\mathbf{n,k,p}$ so that $\mathbf{PY+WY}$ will be defined are:

1. $\mathbf{k=3,p=n}$

2. $\mathbf{k}$ is arbitrary, $\mathbf{p=2}$

3. $\mathbf{p}$ is arbitrary, $\mathbf{k=3}$

4. $\mathbf{k=2,p=3}$

Ans: Matrices $P$ and $Y$ are of the orders $p\times k$ and $3\times k$ respectively.

Therefore, the matrix $PY$ will be defined if $k=3$ .

Also, $PY$ will be of order $p\times k$ .

Since the number of columns in matrix $W$ is equal to the number of rows in matrix $Y$ .

Therefore, matrix $WY$ is well-defined and is of the order $n\times k$ .

Moreover, Matrices $PY$ and $WY$ can be added only when their orders are the same.

But $PY$ is of the order $p\times k$ and $WY$ is of the order $n\times k$ .

Therefore, we must have $p=n$ .

$\therefore p=n,k=3$ are the restrictions on $n,k,p$ so that $PY+WY$ will be defined.

Thus, option (A) is correct.

22. Assume $\mathbf{X,Y,Z,W}$ and $\mathbf{P}$ are matrices of order $\mathbf{2\times n,3\times k,2\times p,n\times 3}$ and $\mathbf{p\times k}$ respectively. If $\mathbf{p=n}$ , then the order of the matrix $\mathbf{7X-5Z}$ is:

1. $\mathbf{p\times 2}$

2. $\mathbf{2\times n}$

3. $\mathbf{n\times 3}$

4. $\mathbf{p\times n}$

Ans: Matrix $X$ is of the order $2\times n$ .

Thus, matrix $7X$ is also of the same order.

Since, $p=n$

Matrix $Z$ is of the order $2\times p$ or $2\times n$ .

Thus, matrix $5Z$ is also of the same order.

Since, both the matrices $7X$ and $5Z$ are of the same order $2\times n$.

Therefore, $7X-5Z$ is well defined and is of the order $2\times n$ .

Thus, option (B) is correct.

### Exercise 3.3

1. Find the transpose of each of the following matrices:

i. $\left[ \begin{matrix} 5 \\ \dfrac{1}{2} \\ -1 \\ \end{matrix} \right]$

Ans: The transpose of a matrix is obtained by changing its rows into columns and its columns into rows.

Thus, if $A=\left[ \begin{matrix} 5 \\ \dfrac{1}{2} \\ -1 \\ \end{matrix} \right]$ , then ${{A}^{T}}=\left[ \begin{matrix} 5 & \dfrac{1}{2} & -1 \\ \end{matrix} \right]$

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

Ans: The transpose of a matrix is obtained by changing its rows into columns and its columns into rows.

Thus, if $A=\left[ \begin{matrix} 1 & -1 \\ 2 & 3 \\ \end{matrix} \right]$ , then ${{A}^{T}}=\left[ \begin{matrix} 1 & 2 \\ -1 & 3 \\ \end{matrix} \right]$

iii. $\left[ \begin{matrix} -1 & 5 & 6 \\ \sqrt{3} & 5 & 6 \\ 2 & 3 & -1 \\ \end{matrix} \right]$

Ans: The transpose of a matrix is obtained by changing its rows into columns and its columns into rows.

Thus, if $A=\left[ \begin{matrix} -1 & 5 & 6 \\ \sqrt{3} & 5 & 6 \\ 2 & 3 & -1 \\ \end{matrix} \right]$ , then ${{A}^{T}}=\left[ \begin{matrix} -1 & \sqrt{3} & 2 \\ 5 & 5 & 3 \\ 6 & 6 & -1 \\ \end{matrix} \right]$

2. If $A=\left[ \begin{matrix} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \\ \end{matrix} \right]$ and $B=\left[ \begin{matrix} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \\ \end{matrix} \right]$ , then verify that

i. $\left( A+B \right)'=A'+B'$

Ans: Given that: $A=\left[ \begin{matrix} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \\ \end{matrix} \right]$ and $B=\left[ \begin{matrix} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \\ \end{matrix} \right]$

Thus, we have $A'=\left[ \begin{matrix} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \\ \end{matrix} \right]$ and $B'=\left[ \begin{matrix} -4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1 \\ \end{matrix} \right]$

$\left( A+B \right)=\left[ \begin{matrix} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \\ \end{matrix} \right]+\left[ \begin{matrix} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \\ \end{matrix} \right]$

$\therefore \left( A+B \right)=\left[ \begin{matrix} -5 & 3 & -2 \\ 6 & 9 & 9 \\ -1 & 4 & 2 \\ \end{matrix} \right]$

$\Rightarrow \left( A+B \right)'=\left[ \begin{matrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \\ \end{matrix} \right]$                                   …(1)

And $A'+B'=\left[ \begin{matrix} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \\ \end{matrix} \right]+\left[ \begin{matrix} -4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1 \\ \end{matrix} \right]$

$\therefore A'+B'=\left[ \begin{matrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \\ \end{matrix} \right]$                                         …(2)

Thus, from equation (1) and (2),

$\left( A+B \right)'=A'+B'$

Hence proved.

ii. $\left( A-B \right)'=A'-B'$

Ans: Given that: $A=\left[ \begin{matrix} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \\ \end{matrix} \right]$ and $B=\left[ \begin{matrix} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \\ \end{matrix} \right]$

Thus, we have $A'=\left[ \begin{matrix} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \\ \end{matrix} \right]$ and $B'=\left[ \begin{matrix} -4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1 \\ \end{matrix} \right]$

$\left( A-B \right)=\left[ \begin{matrix} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \\ \end{matrix} \right]-\left[ \begin{matrix} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \\ \end{matrix} \right]$

$\therefore \left( A-B \right)=\left[ \begin{matrix} 3 & 1 & 8 \\ 4 & 5 & 9 \\ -3 & -2 & 0 \\ \end{matrix} \right]$

$\Rightarrow \left( A-B \right)'=\left[ \begin{matrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \\ \end{matrix} \right]$                                   …(1)

And $A'-B'=\left[ \begin{matrix} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \\ \end{matrix} \right]-\left[ \begin{matrix} -4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1 \\ \end{matrix} \right]$

$\therefore A'-B'=\left[ \begin{matrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \\ \end{matrix} \right]$                                         …(2)

Thus, from equation (1) and (2),

$\left( A-B \right)'=A'-B'$

Hence proved.

3. If $A'=\left[ \begin{matrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \\ \end{matrix} \right]$ and $B=\left[ \begin{matrix} -1 & 2 & 1 \\ 1 & 2 & 3 \\ \end{matrix} \right]$ , then verify that

i. $\left( A+B \right)'=A'+B'$

Ans: We know that, $A=\left( A' \right)'$

Thus, we have: $A=\left[ \begin{matrix} 3 & -1 & 0 \\ 4 & 2 & 1 \\ \end{matrix} \right]$

And $B'=\left[ \begin{matrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \\ \end{matrix} \right]$

$\therefore A+B=\left[ \begin{matrix} 3 & -1 & 0 \\ 4 & 2 & 1 \\ \end{matrix} \right]+\left[ \begin{matrix} -1 & 2 & 1 \\ 1 & 2 & 3 \\ \end{matrix} \right]$

$\Rightarrow A+B=\left[ \begin{matrix} 2 & 1 & 1 \\ 5 & 4 & 4 \\ \end{matrix} \right]$

$\therefore \left( A+B \right)'=\left[ \begin{matrix} 2 & 5 \\ 1 & 4 \\ 1 & 4 \\ \end{matrix} \right]$                                …(1)

$A'+B'=\left[ \begin{matrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \\ \end{matrix} \right]+\left[ \begin{matrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \\ \end{matrix} \right]$

$\therefore A'+B'=\left[ \begin{matrix} 2 & 5 \\ 1 & 4 \\ 1 & 4 \\ \end{matrix} \right]$                                      …(2)

Thus, from equation (1) and (2),

$\left( A+B \right)'=A'+B'$

Hence proved.

ii. $\left( A-B \right)'=A'-B'$

Ans: We know that, $A=\left( A' \right)'$

Thus, we have: $A=\left[ \begin{matrix} 3 & -1 & 0 \\ 4 & 2 & 1 \\ \end{matrix} \right]$

And $B'=\left[ \begin{matrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \\ \end{matrix} \right]$

$\therefore A-B=\left[ \begin{matrix} 3 & -1 & 0 \\ 4 & 2 & 1 \\ \end{matrix} \right]-\left[ \begin{matrix} -1 & 2 & 1 \\ 1 & 2 & 3 \\ \end{matrix} \right]$

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

$\therefore \left( A-B \right)'=\left[ \begin{matrix} 4 & 3 \\ -3 & 0 \\ -1 & -2 \\ \end{matrix} \right]$                                …(1)

$A'-B'=\left[ \begin{matrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \\ \end{matrix} \right]-\left[ \begin{matrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \\ \end{matrix} \right]$

$\therefore A'-B'=\left[ \begin{matrix} 4 & 3 \\ -3 & 0 \\ -1 & -2 \\ \end{matrix} \right]$                                      …(2)

Thus, from equation (1) and (2),

$\left( A-B \right)'=A'-B'$

Hence proved.

4. If $A'=\left[ \begin{matrix} -2 & 3 \\ 1 & 2 \\ \end{matrix} \right]$ and $B=\left[ \begin{matrix} -1 & 0 \\ 1 & 2 \\ \end{matrix} \right]$ , then ( A+2B \right)' .

Ans: We know that, $A=\left( A' \right)'$

$\therefore A=\left[ \begin{matrix} -2 & 1 \\ 3 & 2 \\ \end{matrix} \right]$

$\therefore A+2B=\left[ \begin{matrix} -2 & 1 \\ 3 & 2 \\ \end{matrix} \right]+2\left[ \begin{matrix} -1 & 0 \\ 1 & 2 \\ \end{matrix} \right]$

$\Rightarrow A+2B=\left[ \begin{matrix} -2 & 1 \\ 3 & 2 \\ \end{matrix} \right]+\left[ \begin{matrix} -2 & 0 \\ 2 & 4 \\ \end{matrix} \right]$

$\Rightarrow A+2B=\left[ \begin{matrix} -4 & 1 \\ 5 & 6 \\ \end{matrix} \right]$

$\therefore \left( A+2B \right)'=\left[ \begin{matrix} -4 & 5 \\ 1 & 6 \\ \end{matrix} \right]$

5. For the matrices $A$ and $B$ , verify that $\left( AB \right)'=B'A'$ where

i. $A=\left[ \begin{matrix} 1 \\ -4 \\ 3 \\ \end{matrix} \right],B=\left[ \begin{matrix} -1 & 2 & 1 \\ \end{matrix} \right]$

Ans: LHS: $AB=\left[ \begin{matrix} 1 \\ -4 \\ 3 \\ \end{matrix} \right]\left[ \begin{matrix} -1 & 2 & 1 \\ \end{matrix} \right]$

$\Rightarrow AB=\left[ \begin{matrix} -1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3 \\ \end{matrix} \right]$

$\therefore \left( AB \right)'=\left[ \begin{matrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \\ \end{matrix} \right]$                                 …(1)

Now, $A'=\left[ \begin{matrix} 1 & -4 & 3 \\ \end{matrix} \right]$

And $B'=\left[ \begin{matrix} -1 \\ 2 \\ 1 \\ \end{matrix} \right]$

RHS: $\therefore B'A'=\left[ \begin{matrix} -1 \\ 2 \\ 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & -4 & 3 \\ \end{matrix} \right]$

$\therefore B'A'=\left[ \begin{matrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \\ \end{matrix} \right]$                                     …(2)

Thus, from equation (1) and (2),

$\left( AB \right)'=B'A'$

Hence proved.

ii. $A=\left[ \begin{matrix} 0 \\ 1 \\ 2 \\ \end{matrix} \right], B=\left[ \begin{matrix} 1 & 5 & 7 \\ \end{matrix} \right]$

Ans: LHS: $AB=\left[ \begin{matrix} 0 \\ 1 \\ 2 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 5 & 7 \\ \end{matrix} \right]$

$\Rightarrow AB=\left[ \begin{matrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \\ \end{matrix} \right]$

$\therefore \left( AB \right)'=\left[ \begin{matrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \\ \end{matrix} \right]$                                 …(1)

Now, $A'=\left[ \begin{matrix} 0 & 1 & 2 \\ \end{matrix} \right]$

And $B'=\left[ \begin{matrix} 1 \\ 5 \\ 7 \\ \end{matrix} \right]$

RHS: $\therefore B'A'=\left[ \begin{matrix} 1 \\ 5 \\ 7 \\ \end{matrix} \right]\left[ \begin{matrix} 0 & 1 & 2 \\ \end{matrix} \right]$

$\therefore B'A'=\left[ \begin{matrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \\ \end{matrix} \right]$                                  …(2)

Thus, from equation (1) and (2),

$\left( AB \right)'=B'A'$

Hence proved.

6. If

i. $A=\left[ \begin{matrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \\ \end{matrix} \right]$ , then verify that $A'A=I$

Ans: Here, $A'=\left[ \begin{matrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \\ \end{matrix} \right]$

$\therefore A'A=\left[ \begin{matrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \\ \end{matrix} \right]\left[ \begin{matrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \\ \end{matrix} \right]$

$\Rightarrow A'A=\left[ \begin{matrix} {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha & \sin \alpha \cos \alpha -\sin \alpha \cos \alpha \\ \sin \alpha \cos \alpha -\sin \alpha \cos \alpha & {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha \\ \end{matrix} \right]$

$\therefore A'A=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]$

$\therefore A'A=I$

Hence proved.

ii. $A=\left[ \begin{matrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \\ \end{matrix} \right]$ , then verify that $A'A=I$

Ans: Here, $A'=\left[ \begin{matrix} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \\ \end{matrix} \right]$

$\therefore A'A=\left[ \begin{matrix} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \\ \end{matrix} \right]\left[ \begin{matrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \\ \end{matrix} \right]$

$\Rightarrow A'A=\left[ \begin{matrix} {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha & \sin \alpha \cos \alpha -\sin \alpha \cos \alpha \\ \sin \alpha \cos \alpha -\sin \alpha \cos \alpha & {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha \\ \end{matrix} \right]$

$\therefore A'A=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]$

$\therefore A'A=I$

Hence proved.

7. Show that

i. The matrix $A=\left[ \begin{matrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \\ \end{matrix} \right]$ is a symmetric matrix.

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

$\therefore A'=\left[ \begin{matrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \\ \end{matrix} \right]$

Here, we have $A'=A$

Thus, $A$ is a symmetric matrix.

ii. The matrix $A=\left[ \begin{matrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right]$ is a skew symmetric matrix.

Ans: Given $A=\left[ \begin{matrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right]$

$\therefore A'=\left[ \begin{matrix} 0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0 \\ \end{matrix} \right]$

$\Rightarrow A'=-\left[ \begin{matrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right]$

Here, we have $A'=-A$

Thus, $A$ is a skew symmetric matrix.

8. For the matrix $A=\left[ \begin{matrix} 1 & 5 \\ 6 & 7 \\ \end{matrix} \right]$ , verify that

i. $\left( A+A' \right)$ is a symmetric matrix.

Ans: Given $A=\left[ \begin{matrix} 1 & 5 \\ 6 & 7 \\ \end{matrix} \right]$

$\therefore A'=\left[ \begin{matrix} 1 & 6 \\ 5 & 7 \\ \end{matrix} \right]$

$A+A'=\left[ \begin{matrix} 1 & 5 \\ 6 & 7 \\ \end{matrix} \right]+\left[ \begin{matrix} 1 & 6 \\ 5 & 7 \\ \end{matrix} \right]$

$\Rightarrow A+A'=\left[ \begin{matrix} 2 & 11 \\ 11 & 14 \\ \end{matrix} \right]$

$\left( A+A' \right)'=\left[ \begin{matrix} 2 & 11 \\ 11 & 14 \\ \end{matrix} \right]$

Here, we have $A+A'=\left( A+A' \right)'$

Thus, $A+A'$ is a symmetric matrix.

ii. $\left( A-A' \right)$ is a skew symmetric matrix.

Ans: Given $A=\left[ \begin{matrix} 1 & 5 \\ 6 & 7 \\ \end{matrix} \right]$

$\therefore A'=\left[ \begin{matrix} 1 & 6 \\ 5 & 7 \\ \end{matrix} \right]$

$A-A'=\left[ \begin{matrix} 1 & 5 \\ 6 & 7 \\ \end{matrix} \right]-\left[ \begin{matrix} 1 & 6 \\ 5 & 7 \\ \end{matrix} \right]$

$\Rightarrow A-A'=\left[ \begin{matrix} 0 & -1 \\ 1 & 0 \\ \end{matrix} \right]$

$\left( A-A' \right)'=\left[ \begin{matrix} 0 & 1 \\ -1 & 0 \\ \end{matrix} \right]$

$\Rightarrow \left( A-A' \right)'=-\left[ \begin{matrix} 0 & -1 \\ 1 & 0 \\ \end{matrix} \right]$

Here, we have $A-A'=-\left( A-A' \right)'$

Thus, $A-A'$ is a skew symmetric matrix.

9. Find $\dfrac{1}{2}\left( A+A' \right)$ and $\dfrac{1}{2}\left( A-A' \right)$ , when $A=\left[ \begin{matrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \\ \end{matrix} \right]$ .

Ans: Given $A=\left[ \begin{matrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \\ \end{matrix} \right]$

$\therefore A'=\left[ \begin{matrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \\ \end{matrix} \right]$

$A+A'=\left[ \begin{matrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \\ \end{matrix} \right]+\left[ \begin{matrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \\ \end{matrix} \right]$

$\Rightarrow A+A'=\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{matrix} \right]$

$\therefore \dfrac{1}{2}\left( A+A' \right)=\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{matrix} \right]$

Now, $A-A'=\left[ \begin{matrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \\ \end{matrix} \right]-\left[ \begin{matrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \\ \end{matrix} \right]$

$\left( A-A' \right)=\left[ \begin{matrix} 0 & 2a & 2b \\ -2a & 0 & 2c \\ -2b & -2c & 0 \\ \end{matrix} \right]$

$\therefore \dfrac{1}{2}\left( A-A' \right)=\left[ \begin{matrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \\ \end{matrix} \right]$

10. Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

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

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

Now, $A+A'=\left[ \begin{matrix} 3 & 5 \\ 1 & -1 \\ \end{matrix} \right]+\left[ \begin{matrix} 3 & 1 \\ 5 & -1 \\ \end{matrix} \right]$

$\Rightarrow A+A'=\left[ \begin{matrix} 6 & 6 \\ 6 & -2 \\ \end{matrix} \right]$

Let $P=\dfrac{1}{2}\left( A+A' \right)$

$\Rightarrow P=\dfrac{1}{2}\left[ \begin{matrix} 6 & 6 \\ 6 & -2 \\ \end{matrix} \right]$

$\Rightarrow P=\left[ \begin{matrix} 3 & 3 \\ 3 & -1 \\ \end{matrix} \right]$

Now, $P'=\left[ \begin{matrix} 3 & 3 \\ 3 & -1 \\ \end{matrix} \right]$

Here, $P=P'$

$\therefore P=\dfrac{1}{2}\left( A+A' \right)$ is a symmetric matrix.

Now, $A-A'=\left[ \begin{matrix} 3 & 5 \\ 1 & -1 \\ \end{matrix} \right]-\left[ \begin{matrix} 3 & 1 \\ 5 & -1 \\ \end{matrix} \right]$

$\Rightarrow A-A'=\left[ \begin{matrix} 0 & 4 \\ -4 & 0 \\ \end{matrix} \right]$

Let $Q=\dfrac{1}{2}\left( A-A' \right)$

$\Rightarrow Q=\dfrac{1}{2}\left[ \begin{matrix} 0 & 4 \\ -4 & 0 \\ \end{matrix} \right]$

$\Rightarrow Q=\left[ \begin{matrix} 0 & 2 \\ -2 & 0 \\ \end{matrix} \right]$

Now, $Q'=\left[ \begin{matrix} 0 & 2 \\ -2 & 0 \\ \end{matrix} \right]$

Here, $Q=-Q'$

$\therefore Q=\dfrac{1}{2}\left( A-A' \right)$ is a skew symmetric matrix.

Thus, $A$is the sum of matrices $P=\left[ \begin{matrix} 3 & 3 \\ 3 & -1 \\ \end{matrix} \right]$ and $Q=\left[ \begin{matrix} 0 & 2 \\ -2 & 0 \\ \end{matrix} \right]$.

ii. $\left[ \begin{matrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \\ \end{matrix} \right]$

Ans: Let $A=\left[ \begin{matrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \\ \end{matrix} \right]$ and $A'=\left[ \begin{matrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \\ \end{matrix} \right]$

Now, $A+A'=\left[ \begin{matrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \\ \end{matrix} \right]+\left[ \begin{matrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \\ \end{matrix} \right]$

Let $P=\dfrac{1}{2}\left( A+A' \right)$

$\Rightarrow P=\dfrac{1}{2}\left[ \begin{matrix} 12 & -4 & 4 \\ -4 & 6 & -2 \\ 4 & -2 & 6 \\ \end{matrix} \right]$

$\Rightarrow P=\left[ \begin{matrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \\ \end{matrix} \right]$

Now, $P'=\left[ \begin{matrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \\ \end{matrix} \right]$

Here, $P=P'$

$\therefore P=\dfrac{1}{2}\left( A+A' \right)$ is a symmetric matrix.

Now, $A-A'=\left[ \begin{matrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \\ \end{matrix} \right]-\left[ \begin{matrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \\ \end{matrix} \right]$

$\Rightarrow A-A'=\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{matrix} \right]$

Let $Q=\dfrac{1}{2}\left( A-A' \right)$

$\Rightarrow Q=\dfrac{1}{2}\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{matrix} \right]$

$\Rightarrow Q=\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{matrix} \right]$

Now, $Q'=\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{matrix} \right]$

Here, $Q=-Q'$

$\therefore Q=\dfrac{1}{2}\left( A-A' \right)$ is a skew symmetric matrix.

Thus, $A$is the sum of matrices $P=\left[ \begin{matrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \\ \end{matrix} \right]$ and $Q=\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{matrix} \right]$.

iii. $\left[ \begin{matrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \\ \end{matrix} \right]$

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

Now, $A+A'=\left[ \begin{matrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \\ \end{matrix} \right]+\left[ \begin{matrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \\ \end{matrix} \right]$

$\Rightarrow A+A'=\left[ \begin{matrix} 6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4 \\ \end{matrix} \right]$

Let $P=\dfrac{1}{2}\left( A+A' \right)$

$\Rightarrow P=\dfrac{1}{2}\left[ \begin{matrix} 6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4 \\ \end{matrix} \right]$

$\Rightarrow P=\left[ \begin{matrix} 3 & \dfrac{1}{2} & \dfrac{-5}{2} \\ \dfrac{1}{2} & -2 & -2 \\ \dfrac{-5}{2} & -2 & 2 \\ \end{matrix} \right]$

Now, $P'=\left[ \begin{matrix} 3 & \dfrac{1}{2} & \dfrac{-5}{2} \\ \dfrac{1}{2} & -2 & -2 \\ \dfrac{-5}{2} & -2 & 2 \\ \end{matrix} \right]$

Here, $P=P'$

$\therefore P=\dfrac{1}{2}\left( A+A' \right)$ is a symmetric matrix.

Now, $A-A'=\left[ \begin{matrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \\ \end{matrix} \right]-\left[ \begin{matrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \\ \end{matrix} \right]$

$\Rightarrow A-A'=\left[ \begin{matrix} 0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0 \\ \end{matrix} \right]$

Let $Q=\dfrac{1}{2}\left( A-A' \right)$

$\Rightarrow Q=\dfrac{1}{2}\left[ \begin{matrix} 0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0 \\ \end{matrix} \right]$

$\Rightarrow Q=\left[ \begin{matrix} 0 & \dfrac{5}{2} & \dfrac{3}{2} \\ \dfrac{-5}{2} & 0 & 3 \\ \dfrac{-3}{2} & -3 & 0 \\ \end{matrix} \right]$

Now, $Q'=\left[ \begin{matrix} 0 & -\dfrac{5}{2} & -\dfrac{3}{2} \\ \dfrac{5}{2} & 0 & -3 \\ \dfrac{3}{2} & 3 & 0 \\ \end{matrix} \right]$

Here, $Q=-Q'$

$\therefore Q=\dfrac{1}{2}\left( A-A' \right)$ is a skew symmetric matrix.

Thus, $A$is the sum of matrices $P=\left[ \begin{matrix} 3 & \dfrac{1}{2} & \dfrac{-5}{2} \\ \dfrac{1}{2} & -2 & -2 \\ \dfrac{-5}{2} & -2 & 2 \\ \end{matrix} \right]$ and $Q=\left[ \begin{matrix} 0 & \dfrac{5}{2} & \dfrac{3}{2} \\ \dfrac{-5}{2} & 0 & 3 \\ \dfrac{-3}{2} & -3 & 0 \\ \end{matrix} \right]$.

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

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

Now, $A+A'=\left[ \begin{matrix} 1 & 5 \\ -1 & 2 \\ \end{matrix} \right]+\left[ \begin{matrix} 1 & -1 \\ 5 & 2 \\ \end{matrix} \right]$

$\Rightarrow A+A'=\left[ \begin{matrix} 2 & 4 \\ 4 & 4 \\ \end{matrix} \right]$

Let $P=\dfrac{1}{2}\left( A+A' \right)$

$\Rightarrow P=\dfrac{1}{2}\left[ \begin{matrix} 2 & 4 \\ 4 & 4 \\ \end{matrix} \right]$

$\Rightarrow P=\left[ \begin{matrix} 1 & 2 \\ 2 & 2 \\ \end{matrix} \right]$

Now, $P'=\left[ \begin{matrix} 1 & 2 \\ 2 & 2 \\ \end{matrix} \right]$

Here, $P=P'$

$\therefore P=\dfrac{1}{2}\left( A+A' \right)$ is a symmetric matrix.

Now, $A-A'=\left[ \begin{matrix} 1 & 5 \\ -1 & 2 \\ \end{matrix} \right]-\left[ \begin{matrix} 1 & -1 \\ 5 & 2 \\ \end{matrix} \right]$

$\Rightarrow A-A'=\left[ \begin{matrix} 0 & 6 \\ -6 & 0 \\ \end{matrix} \right]$

Let $Q=\dfrac{1}{2}\left( A-A' \right)$

$\Rightarrow Q=\dfrac{1}{2}\left[ \begin{matrix} 0 & 6 \\ -6 & 0 \\ \end{matrix} \right]$

$\Rightarrow Q=\left[ \begin{matrix} 0 & 3 \\ -3 & 0 \\ \end{matrix} \right]$

Now, $Q'=\left[ \begin{matrix} 0 & -3 \\ 3 & 0 \\ \end{matrix} \right]$

Here, $Q=-Q'$

$\therefore Q=\dfrac{1}{2}\left( A-A' \right)$ is a skew symmetric matrix.

Thus, $A$is the sum of matrices $P=\left[ \begin{matrix} 1 & 2 \\ 2 & 2 \\ \end{matrix} \right]$ and $Q=\left[ \begin{matrix} 0 & 3 \\ -3 & 0 \\ \end{matrix} \right]$.

11. If $A,B$ are symmetric matrices of same order, then $AB-BA$ is a:

1. Skew symmetric matrix

2. Symmetric matrix

3. Zero matrix

4. Identity matrix

Ans: $A$ and $B$ are symmetric matrix, therefore, we have:

$A'=A$ and $B'=B$                                                    …(1)

Here, $\left( AB-BA \right)'=\left( AB \right)'-\left( BA \right)'$

$\Rightarrow \left( AB-BA \right)'=B'A'-A'B'$

From (1),

$\Rightarrow \left( AB-BA \right)'=BA-AB$

$\Rightarrow \left( AB-BA \right)'=-\left( AB-BA \right)$                           …(2)

From equation (2),

$AB-BA$ is a skew symmetric matrix.

12. If $A=\left[ \begin{matrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \\ \end{matrix} \right]$ , then $\mathbf{A+A'=I}$ , if the value of $\mathbf{\alpha}$ is

1. $\dfrac{\pi }{6}$

2. $\dfrac{\pi }{3}$

3. $n$

4. $\dfrac{3\pi }{2}$

Ans: Given $A=\left[ \begin{matrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \\ \end{matrix} \right]$

$\Rightarrow A'=\left[ \begin{matrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \\ \end{matrix} \right]$

Now, $\because A+A'=I$

$\therefore \left[ \begin{matrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \\ \end{matrix} \right]+\left[ \begin{matrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]$

$\Rightarrow \left[ \begin{matrix} 2\cos \alpha & 0 \\ 0 & 2\cos \alpha \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]$

Equating the corresponding elements of the two matrices, we have:

$\cos \alpha =\dfrac{1}{2}$

$\Rightarrow \alpha ={{\cos }^{-1}}\left( \dfrac{1}{2} \right)$

$\Rightarrow \alpha =\dfrac{\pi }{3}$

Thus, option (B) is correct.

### Miscellaneous Solutions

1. Let $A=\left[ \begin{matrix} 0 & 1 \\ 0 & 0 \\ \end{matrix} \right]$ , show that ${{\left( aI+bA \right)}^{n}}={{a}^{n}}I+n{{a}^{n-1}}bA$ , where $I$ is the identity matrix of order $2}$ and $n\in N$.

Ans: Given $A=\left[ \begin{matrix} 0 & 1 \\ 0 & 0 \\ \end{matrix} \right]$

By using the principle of mathematical induction.

For $n=1$ ,

$P\left( 1 \right):{{\left( aI+bA \right)}^{1}}={{a}^{1}}I+{{a}^{0}}bA$

$\Rightarrow P\left( 1 \right):\left( aI+bA \right)=aI+bA$

Therefore, the result is true for $n=1$ .

Let the result be true for $n=k$ .

That is, $P\left( k \right):{{\left( aI+bA \right)}^{k}}={{a}^{k}}I+k{{a}^{k-1}}bA$

Now, we have to prove that the result is true for $n=k+1$ .

${{\left( aI+bA \right)}^{k-1}}={{\left( aI+bA \right)}^{k}}\left( aI+bA \right)$

$\Rightarrow {{\left( aI+bA \right)}^{k-1}}=\left( {{a}^{k}}I+k{{a}^{k-1}}bA \right)\left( aI+bA \right)$

$\Rightarrow {{\left( aI+bA \right)}^{k-1}}={{a}^{k}}I+\left( k+1 \right){{a}^{k}}bA+k{{a}^{k-1}}{{b}^{2}}{{A}^{2}}$                        …(1)

Now,

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

${{A}^{2}}=\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \\ \end{matrix} \right]$

${{A}^{2}}=0$

$\Rightarrow {{\left( aI+bA \right)}^{k-1}}={{a}^{k}}I+\left( k+1 \right){{a}^{k}}bA$

Thus, the result is true for $n=k+1$

Therefore, by the principal of mathematical induction, we have:

${{\left( aI+bA \right)}^{n}}={{a}^{n}}I+n{{a}^{n-1}}bA$ where $A=\left[ \begin{matrix} 0 & 1 \\ 0 & 0 \\ \end{matrix} \right]$ , $n\in N$.

2. If $A=\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{matrix} \right]$ , prove that ${{A}^{n}}=\left[ \begin{matrix} {{3}^{n-1}} & {{3}^{n-1}} & {{3}^{n-1}} \\ {{3}^{n-1}} & {{3}^{n-1}} & {{3}^{n-1}} \\ {{3}^{n-1}} & {{3}^{n-1}} & {{3}^{n-1}} \\ \end{matrix} \right]$ , $n\in N$.

Ans: Given $A=\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{matrix} \right]$

By using the principles of mathematical induction.

For $n=1$ , we have

$P\left( 1 \right)=\left[ \begin{matrix} {{3}^{1-1}} & {{3}^{1-1}} & {{3}^{1-1}} \\ {{3}^{1-1}} & {{3}^{1-1}} & {{3}^{1-1}} \\ {{3}^{1-1}} & {{3}^{1-1}} & {{3}^{1-1}} \\ \end{matrix} \right]$

$\Rightarrow P\left( 1 \right)=\left[ \begin{matrix} {{3}^{0}} & {{3}^{0}} & {{3}^{0}} \\ {{3}^{0}} & {{3}^{0}} & {{3}^{0}} \\ {{3}^{0}} & {{3}^{0}} & {{3}^{0}} \\ \end{matrix} \right]$

$\Rightarrow P\left( 1 \right)=\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{matrix} \right]$

$\Rightarrow P\left( 1 \right)=A$

Therefore, the result is true for $n=1$ .

Let the result be true for $n=k$ .

$P\left( k \right):{{A}^{k}}=\left[ \begin{matrix} {{3}^{k-1}} & {{3}^{k-1}} & {{3}^{k-1}} \\ {{3}^{k-1}} & {{3}^{k-1}} & {{3}^{k-1}} \\ {{3}^{k-1}} & {{3}^{k-1}} & {{3}^{k-1}} \\ \end{matrix} \right]$

Now, we have to prove that the result is true for $n=k+1$ .

Now, ${{A}^{k+1}}=A\cdot {{A}^{k}}$

$\Rightarrow {{A}^{k+1}}=\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} {{3}^{k-1}} & {{3}^{k-1}} & {{3}^{k-1}} \\ {{3}^{k-1}} & {{3}^{k-1}} & {{3}^{k-1}} \\ {{3}^{k-1}} & {{3}^{k-1}} & {{3}^{k-1}} \\ \end{matrix} \right]$

$\Rightarrow {{A}^{k+1}}=\left[ \begin{matrix} 3\cdot {{3}^{k-1}} & 3\cdot {{3}^{k-1}} & 3\cdot {{3}^{k-1}} \\ 3\cdot {{3}^{k-1}} & 3\cdot {{3}^{k-1}} & 3\cdot {{3}^{k-1}} \\ 3\cdot {{3}^{k-1}} & 3\cdot {{3}^{k-1}} & 3\cdot {{3}^{k-1}} \\ \end{matrix} \right]$

$\Rightarrow {{A}^{k+1}}=\left[ \begin{matrix} {{3}^{\left( k+1 \right)-1}} & {{3}^{\left( k+1 \right)-1}} & {{3}^{\left( k+1 \right)-1}} \\ {{3}^{\left( k+1 \right)-1}} & {{3}^{\left( k+1 \right)-1}} & {{3}^{\left( k+1 \right)-1}} \\ {{3}^{\left( k+1 \right)-1}} & {{3}^{\left( k+1 \right)-1}} & {{3}^{\left( k+1 \right)-1}} \\ \end{matrix} \right]$

Thus, the result is true for $n=k+1$

Therefore, by the principal of mathematical induction, we have:

${{A}^{n}}=\left[ \begin{matrix} {{3}^{n-1}} & {{3}^{n-1}} & {{3}^{n-1}} \\ {{3}^{n-1}} & {{3}^{n-1}} & {{3}^{n-1}} \\ {{3}^{n-1}} & {{3}^{n-1}} & {{3}^{n-1}} \\ \end{matrix} \right]$ where $A=\left[ \begin{matrix} 0 & 1 \\ 0 & 0 \\ \end{matrix} \right]$ , $n\in N$.

3. If $A=\left[ \begin{matrix} 3 & -4 \\ 1 & -1 \\ \end{matrix} \right]$ , then prove ${{A}^{n}}=\left[ \begin{matrix} 1+2n & -4n \\ n & 1-2n \\ \end{matrix} \right]$ where $n$ is any positive integer.

Ans: Given $A=\left[ \begin{matrix} 3 & -4 \\ 1 & -1 \\ \end{matrix} \right]$

By using the principle of mathematical induction.

For $n=1$ ,

$P\left( 1 \right):A=\left[ \begin{matrix} 3 & -4 \\ 1 & -1 \\ \end{matrix} \right]$

Therefore, the result is true for $n=1$ .

Let the result be true for $n=k$ .

That is, $P\left( k \right):{{A}^{k}}=\left[ \begin{matrix} 1+2k & -4k \\ k & 1-2k \\ \end{matrix} \right]$ , $n\in N$

Now, we have to prove that the result is true for $n=k+1$ .

${{A}^{k+1}}=A\cdot {{A}^{k}}$

$\Rightarrow {{A}^{k+1}}=\left[ \begin{matrix} 1+2k & -4k \\ k & 1-2k \\ \end{matrix} \right]\left[ \begin{matrix} 3 & -4 \\ 1 & -1 \\ \end{matrix} \right]$

$\Rightarrow {{A}^{k+1}}=\left[ \begin{matrix} 3+2k & -4-4k \\ 1+k & -1-2k \\ \end{matrix} \right]$

$\Rightarrow {{A}^{k+1}}=\left[ \begin{matrix} 1+2\left( k+1 \right) & -4\left( k+1 \right) \\ 1+k & 1-2\left( k+1 \right) \\ \end{matrix} \right]$

Thus, the result is true for $n=k+1$

Therefore, by the principal of mathematical induction, we have:

${{A}^{n}}=\left[ \begin{matrix} 1+2n & -4n \\ n & 1-2n \\ \end{matrix} \right]$ , $n\in N$.

4. If $A$ and $B$ are symmetric matrices, prove that $AB-BA$ is a skew symmetric matrix.

Ans: $A$ and $B$ are symmetric matrix, therefore, we have:

$A'=A$ and $B'=B$                                                    …(1)

Here, $\left( AB-BA \right)'=\left( AB \right)'-\left( BA \right)'$

$\Rightarrow \left( AB-BA \right)'=B'A'-A'B'$

From (1),

$\Rightarrow \left( AB-BA \right)'=BA-AB$

$\Rightarrow \left( AB-BA \right)'=-\left( AB-BA \right)$                           …(2)

From equation (2),

$AB-BA$ is a skew symmetric matrix.

5. Show that the matrix $B'AB$ is symmetric or skew symmetric accordingly as $A$ is symmetric or skew symmetric.

Ans: Let $A$ be a symmetric matrix, then $A'=A$               …(1)

$\left( B'AB \right)'=\left\{ B'\left( AB \right) \right\}'$

$\Rightarrow \left( B'AB \right)'=\left( AB \right)'B'$

$\Rightarrow \left( B'AB \right)'=B'\left( A'B \right)$

From (1),

$\Rightarrow \left( B'AB \right)'=B'\left( AB \right)$

Thus, if $A$ is a symmetric matrix, then $B'AB$ is a symmetric matrix.

Let $A$ be a skew symmetric matrix, then $A'=-A$               …(2)

$\left( B'AB \right)'=\left\{ B'\left( AB \right) \right\}'$

$\Rightarrow \left( B'AB \right)'=\left( AB \right)'B'$

$\Rightarrow \left( B'AB \right)'=B'\left( A'B \right)$

From (2),

$\Rightarrow \left( B'AB \right)'=B'\left( -AB \right)$

$\Rightarrow \left( B'AB \right)'=-B'AB$

Thus, if $A$ is a skew-symmetric matrix, then $B'AB$ is a skew-symmetric matrix.

Therefore, the matrix $B'AB$ is symmetric or skew symmetric accordingly as $A$ is symmetric or skew symmetric.

6. Solve the system of linear equations, using matrix method.

$2x-y=-2$

$3x+4y=3$

Ans: The given system of equation can be written in the form of $AX=B$ where,

$A=\left[ \begin{matrix} 2 & -1 \\ 3 & 4 \\ \end{matrix} \right],X=\left[ \begin{matrix} x \\ y \\ \end{matrix} \right],B=\left[ \begin{matrix} -2 \\ 3 \\ \end{matrix} \right]$

Now, $\left| A \right|=8+3=11$

$\because \left| A \right|\ne 0$ ,therefore its inverse exists.

We know that,

${{A}^{-1}}=\dfrac{1}{\left| A \right|}adjA$

$\therefore {{A}^{-1}}=\dfrac{1}{11}adj\left[ \begin{matrix} 4 & 1 \\ -3 & 2 \\ \end{matrix} \right]$

We can write $X={{A}^{-1}}B$

$\therefore X=\dfrac{1}{11}\left[ \begin{matrix} 4 & 1 \\ -3 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} -2 \\ 3 \\ \end{matrix} \right]$

$\Rightarrow \left[ \begin{matrix} x \\ y \\ \end{matrix} \right]=\dfrac{1}{11}\left[ \begin{matrix} -5 \\ 12 \\ \end{matrix} \right]$

$\Rightarrow \left[ \begin{matrix} x \\ y \\ \end{matrix} \right]=\left[ \begin{matrix} \dfrac{-5}{11} \\ \dfrac{12}{11} \\ \end{matrix} \right]$

Thus, $x=\dfrac{-5}{11}$ and $y=\dfrac{12}{11}$.

7. For what values of $x$ , $\left[ \begin{matrix} 1 & 2 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} 0 \\ 2 \\ x \\ \end{matrix} \right]=O$ ?

Ans: Given $\left[ \begin{matrix} 1 & 2 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} 0 \\ 2 \\ x \\ \end{matrix} \right]=O$

$\Rightarrow \left[ \begin{matrix} 1+4+1 & 2+0+0 & 0+2+2 \\ \end{matrix} \right]\left[ \begin{matrix} 0 \\ 2 \\ x \\ \end{matrix} \right]=O$

$\Rightarrow \left[ \begin{matrix} 6 & 2 & 4 \\\end{matrix} \right]\left[ \begin{matrix} 0 \\ 2 \\ x \\ \end{matrix} \right]=O$

$\Rightarrow \left[ 6\left( 0 \right)+2\left( 2 \right)+4\left( x \right) \right]=O$

$\Rightarrow \left[ 4+4x \right]=\left[ 0 \right]$

$\Rightarrow 4+4x=0$

$\therefore x=-1$

Thus, the required value of $x$ is $-1$ .

8. If $A=\left[ \begin{matrix} 3 & 1 \\ -1 & 2 \\ \end{matrix} \right]$ , show that ${{A}^{2}}-5A+7I=O$ .

Ans: Given $A=\left[ \begin{matrix} 3 & 1 \\ -1 & 2 \\ \end{matrix} \right]$

$\therefore {{A}^{2}}=A\cdot A$

$\Rightarrow {{A}^{2}}=\left[ \begin{matrix} 3 & 1 \\ -1 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} 3 & 1 \\ -1 & 2 \\ \end{matrix} \right]$

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

$\Rightarrow {{A}^{2}}=\left[ \begin{matrix} 9-1 & 3+2 \\ -3-2 & -1+4 \\ \end{matrix} \right]$

$\Rightarrow {{A}^{2}}=\left[ \begin{matrix} 8 & 5 \\ -5 & 3 \\ \end{matrix} \right]$

LHS: ${{A}^{2}}-5A+7I$

$\therefore \left[ \begin{matrix} 8 & 5 \\ -5 & 3 \\ \end{matrix} \right]-5\left[ \begin{matrix} 3 & 1 \\ -1 & 2 \\ \end{matrix} \right]+7\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]$

$\Rightarrow \left[ \begin{matrix} 8 & 5 \\ -5 & 3 \\ \end{matrix} \right]-\left[ \begin{matrix} 15 & 5 \\ -5 & 10 \\ \end{matrix} \right]+\left[ \begin{matrix} 7 & 0 \\ 0 & 7 \\ \end{matrix} \right]$

$\Rightarrow \left[ \begin{matrix} 0 & 0 \\ 0 & 0 \\ \end{matrix} \right]$

$\Rightarrow O$

RHS: $\Rightarrow O$

$LHS=RHS$

$\therefore {{A}^{2}}-5A+7I=O$ hence proved.

9. Find $X$ , if $\left[ \begin{matrix} x & -5 & -1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ 4 \\ 1 \\ \end{matrix} \right]=O$ .

Ans: Given $\left[ \begin{matrix} x & -5 & -1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ 4 \\ 1 \\ \end{matrix} \right]=O$

$\Rightarrow \left[ \begin{matrix} x-2 & -10 & 2x-8 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ 4 \\ 1 \\ \end{matrix} \right]=O$

$\Rightarrow \left[ x\left( x-2 \right)-40+2x-8 \right]=O$

$\Rightarrow \left[ {{x}^{2}}-48 \right]=\left[ 0 \right]$

$\Rightarrow {{x}^{2}}-48=0$

$\therefore x=\pm 4\sqrt{3}$

Thus, the required value of $x$ is $\pm 4\sqrt{3}$ .

10. A manufacture produces three products $X,Y,Z$ which he sells in two markets.

Annual sales are indicated below:

 Market Products $I$ $10000$ $2000$ $18000$ $II$ $6000$ $20000$ $8000$
1. If unit sale prices of $X,Y,Z$ are Rs $2.50$ , Rs $1.50$ and Rs $1.00$ , respectively, find the total revenue in each market with the help of matrix algebra.

Ans: Here the total revenue in market $I$ can be represented in the form of a matrix as:

$\left[ \begin{matrix} 10000 & 2000 & 18000 \\ \end{matrix} \right]\left[ \begin{matrix} 2.50 \\ 1.50 \\ 1.00 \\ \end{matrix} \right]$

$\Rightarrow 1000\times 2.50+2000\times 1.50+18000\times 1.00$

$\Rightarrow 46000$

And, the total revenue in market $II$ can be represented in the form of a matrix as:

$\left[ \begin{matrix} 6000 & 20000 & 8000 \\ \end{matrix} \right]\left[ \begin{matrix} 2.50 \\ 1.50 \\ 1.00 \\ \end{matrix} \right]$

$\Rightarrow 6000\times 2.50+20000\times 1.50+8000\times 1.00$

$\Rightarrow 53000$

Thus, the total revenue in market $I$ is Rs $46000$ and the same in market $II$ is Rs $53000$.

1. If the unit costs of the above three commodities are Rs $2.00$ , Rs $1.00$ and $50$ paise respectively. Find the gross profit.

Ans: Here, the total cost prices of all the products in the market $I$ can be represented in the form of a matrix as:

$\left[ \begin{matrix} 10000 & 2000 & 18000 \\ \end{matrix} \right]\left[ \begin{matrix} 2.00 \\ 1.00 \\ 0.50 \\ \end{matrix} \right]$

$\Rightarrow 1000\times 2.00+2000\times 1.00+18000\times 0.50$

$\Rightarrow 31000$

As the total revenue in market $I$ is Rs $46000$ , the gross profit in this market is $Rs46000-Rs31000=Rs15000$ .

And, the total revenue in market $II$ can be represented in the form of a matrix as:

$\left[ \begin{matrix} 6000 & 20000 & 8000 \\ \end{matrix} \right]\left[ \begin{matrix} 2.00 \\ 1.00 \\ 0.50 \\ \end{matrix} \right]$

$\Rightarrow 6000\times 2.00+20000\times 1.00+8000\times 0.50$

$\Rightarrow 36000$

As the total revenue in market $II$ is Rs $53000$ , the gross profit in this market is $Rs53000-Rs36000=Rs17000$ .

11. Find the matrix $X$ so that $X\left[ \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ \end{matrix} \right]=\left[ \begin{matrix} -7 & -8 & -9 \\ 2 & 4 & 6 \\ \end{matrix} \right]$.

Ans: Given $X\left[ \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ \end{matrix} \right]=\left[ \begin{matrix} -7 & -8 & -9 \\ 2 & 4 & 6 \\ \end{matrix} \right]$

Here, $X$ has to be a $2\times 2$ matrix.

Now, let $X=\left[ \begin{matrix} a & c \\ b & d \\ \end{matrix} \right]$

Thus, we have:

$\left[ \begin{matrix} a & c \\ b & d \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ \end{matrix} \right]=\left[ \begin{matrix} -7 & -8 & -9 \\ 2 & 4 & 6 \\ \end{matrix} \right]$

$\Rightarrow \left[ \begin{matrix} a+4c & 2a+5c & 3a+6c \\ b+4d & 2b+5d & 3b+6d \\ \end{matrix} \right]=\left[ \begin{matrix} -7 & -8 & -9 \\ 2 & 4 & 6 \\ \end{matrix} \right]$

Comparing the corresponding elements of two matrices, we have:

$a+4c=-7$

$2a+5c=-8$

$3a+6c=-9$

$b+4d=2$

$2b+5d=4$

$3b+6d=6$

Now, solving the above equations we get,

$\therefore a=1,b=2,c=-2,d=0$

Thus, the required matrix $X$ is $\left[ \begin{matrix} 1 & -2 \\ 2 & 0 \\ \end{matrix} \right]$.

12. If $A$ and $B$ are square matrices of the same order such that $AB=BA$ , then prove by induction that $A{{B}^{n}}={{B}^{n}}A$ . Further, prove that ${{\left( AB \right)}^{n}}={{B}^{n}}{{A}^{n}}$ for all $n\in N$.

Ans: Given, $A$ and $B$ are square matrices of the same order such that $AB=BA$ .

For $n=1$ , we have:

$P\left( 1 \right):AB=BA$                          (Given)

$\Rightarrow A{{B}^{1}}={{B}^{1}}A$

Therefore, the result is true for $n=1$ .

Let the result be true for $n=k$ .

$P\left( k \right):A{{B}^{k}}={{B}^{k}}A$                                        …(1)

Now, we have to prove that the result is true for $n=k+1$ .

$A{{B}^{k+1}}=A{{B}^{k}}\cdot B$

$\Rightarrow A{{B}^{k+1}}=\left( {{B}^{k}}A \right)B$

$\Rightarrow A{{B}^{k+1}}={{B}^{k}}\left( AB \right)$

$\Rightarrow A{{B}^{k+1}}={{B}^{k}}\left( BA \right)$

$\Rightarrow A{{B}^{k+1}}=\left( {{B}^{k}}B \right)A$

$\Rightarrow A{{B}^{k+1}}={{B}^{k+1}}A$

Therefore, the result is true for $n=k+1$ .

By the principle of mathematical induction, we have $A{{B}^{n}}={{B}^{n}}A$ , $n\in N$.

13. Choose the correct answer in the following questions:

If $A=\left[ \begin{matrix} \alpha & \beta \\ \gamma & -\alpha \\ \end{matrix} \right]$ is such that ${{A}^{2}}=I$ then

1. $1+{{\alpha }^{2}}+\beta \gamma =0$

2. $1-{{\alpha }^{2}}+\beta \gamma =0$

3. $1-{{\alpha }^{2}}-\beta \gamma =0$

4. $1+{{\alpha }^{2}}-\beta \gamma =0$

Ans: Given $A=\left[ \begin{matrix} \alpha & \beta \\ \gamma & -\alpha \\ \end{matrix} \right]$

$\therefore {{A}^{2}}=A\cdot A$

$\Rightarrow {{A}^{2}}=\left[ \begin{matrix} \alpha & \beta \\ \gamma & -\alpha \\ \end{matrix} \right]\left[ \begin{matrix} \alpha & \beta \\ \gamma & -\alpha \\ \end{matrix} \right]$

$\Rightarrow {{A}^{2}}=\left[ \begin{matrix} {{\alpha }^{2}}+\beta \gamma & \alpha \beta -\alpha \beta \\ \alpha \gamma -\alpha \gamma & \beta \gamma +{{\alpha }^{2}} \\ \end{matrix} \right]$

Now, ${{A}^{2}}=I$

$\Rightarrow \left[ \begin{matrix} {{\alpha }^{2}}+\gamma & 0 \\ 0 & \beta \gamma +{{\alpha }^{2}} \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]$

Equating the corresponding elements, we get:

$\beta \gamma +{{\alpha }^{2}}=1$

$\Rightarrow {{\alpha }^{2}}+\beta \gamma -1=0$

$\Rightarrow 1-{{\alpha }^{2}}-\beta \gamma =0$

14. If the matrix $A$ is both symmetric and skew symmetric, then

1. $A$ is a diagonal matrix

2. $A$ is a zero matrix

3. $A$ is a square matrix

4. None of these

Ans: If a matrix $A$ is both symmetric and skew symmetric, then

$A'=A$ and $A'=-A$

$\Rightarrow A=-A$

$\Rightarrow A+A=O$

$\Rightarrow A=O$

Thus, option (B) is correct.

15. If $A$ is a square matrix such that ${{A}^{2}}=A$ , then ${{\left( I+A \right)}^{3}}-7A$ is equal to

1. $A$

2. $I-A$

3. $I$

4. $3A$

Ans: ${{\left( I+A \right)}^{3}}-7A={{I}^{3}}+{{A}^{3}}+3A+3{{A}^{2}}-7A$

$\Rightarrow {{\left( I+A \right)}^{3}}-7A=I+{{A}^{3}}+3A+3{{A}^{2}}-7A$

Given that ${{A}^{2}}=A$ ,

$\Rightarrow {{\left( I+A \right)}^{3}}-7A=I+{{A}^{2}}\cdot A+3A+3{{A}^{2}}-7A$

$\Rightarrow {{\left( I+A \right)}^{3}}-7A=I+A\cdot A-A$

$\Rightarrow {{\left( I+A \right)}^{3}}-7A=I$

Thus, option (C) is correct.

## Details of Class 12 Matrices NCERT Solutions – Free PDF Download

### 3.1 Introduction

In NCERT Matrices Class 12 Maths Chapter 3 PDF, students will explore concepts of matrices. The learners can clearly understand the necessary application of matrices in various branches of mathematics. This mathematical tool simplifies the work to a great extent as compared to other straight forward methods.

The concept of matrices in Class 12 develops from the need to simplify solving systems of linear equations. This chapter introduces students to the basics of matrices and matrix algebra. Matrices are incredibly useful, not just for solving equations but also in various real-world applications. They are used in electronic spreadsheets for budgeting, sales projections, and cost estimates, as well as in science for analyzing experimental results. Matrices also help in physical operations like magnification and reflection and are used in cryptography. Understanding matrices is essential across different fields such as genetics, economics, sociology, psychology, and industrial management.

### 3.2 Matrix

In this section of the chapter,  the fundamental principles of matrices are discussed. With the help of examples cited by established academicians, students can understand the whole concept. Get the proper definition of Matrix Class 12 and learn the order of a matrix considering only those matrices whose elements are real numbers or functions taking actual values. Students will get to know that the horizontal and vertical lines of entries in a matrix are called rows and columns respectively. This section also talks about the size of a matrix by the number of rows and columns it contains.

### 3.3 Types of Matrices

This section of the chapter expalins various types of matrices, such as column, row, square, diagonal, scalar, identity, and zero matrices, using clear examples. It also examines when two matrices are considered equal, providing examples and solutions to illustrate this.

The questions included are designed to deepen students' understanding of these concepts and encourage a thorough exploration of the Matrix Class 12 NCERT Solutions. This part helps students get a better grasp of different matrix forms and their properties, enhancing their mathematical skills and knowledge.

### 3.4 Operations on Matrices

Matrix Class 12 NCERT Solutions introduces certain operations on matrices, namely, the addition of matrices, multiplication of a matrix by a scalar, differences and multiplication of matrices. Highlighting properties of matrix addition, scalar multiplication of a matrix, multiplication of matrices, etc., students can get a profound understanding of how matrices operate.

### 3.5. Transpose of a Matrix

This section will focus on transposing a matrix and unique types of matrices such as symmetric and skew-symmetric matrices. It will establish the foundations of the transpose of the matrix with compelling examples and properties. This chapter teaches the basics using clear examples and solutions.

### 3.6 Symmetric and Skew Symmetric Matrices

In this section, the chapter will assist in learning symmetric and skew-symmetric matrices. One can also comprehend the difference between a symmetric matrix and a skew-symmetric matrix. It is imperative to know about the transpose of a matrix and how to find it. Moreover, it is essential to remember all the parts of Chapter 3 that have been covered up until now. This section also explores theorems proving the nature of the symmetric matrix and skew-symmetric matrix.

## Some Important Points to remember

1. A matrix is said to be an ordered rectangular array of numbers or functions. These numbers or functions in the array are called the elements or the entries of the matrix.

2. Order of a Matrix

The order of the matrix determines the dimension of the matrix and the number of rows and columns in the matrix. The general representation of matrix order is Amxn, where m is the number of rows and n is the number of columns in the matrix.

3. Types of Matrices

• Column Matrix: A column matrix is a matrix with only one column.

• Row Matrix: A row matrix is a matrix with only one row.

• Square Matrix: A square matrix is one that has an equal number of rows and columns.

• Diagonal Matrix: A square matrix where only the diagonal elements are non-zero and all other elements are zero.

• Scalar Matrix: A diagonal matrix where all the diagonal elements are the same non-zero number.

• Identity Matrix: A diagonal matrix where all diagonal elements are '1' and all other elements are zero, often symbolized as 'I'.

• Zero or Null Matrix: A matrix where all elements are zero.

4. Equality of Matrices: Let A and B be two matrices. These matrices will be equal, if

(i) orders of A and B will be the same

(ii) corresponding elements of two matrices are the same

5. Operations on Matrices

• Addition of Matrices: You can add two matrices by adding the numbers that are in the same position in each matrix, as long as both matrices are the same size.

• Subtraction of Matrices: Subtracting matrices works like addition; you subtract the numbers in the same positions, but only if the matrices are the same size.

• Multiplication of Matrices: To multiply matrices, the number of columns in the first matrix must match the number of rows in the second; you then multiply and sum specific pairs of numbers.

6. Properties of Multiplication of Matrices

• Non-commutativity:

Matrix multiplication will be not commutative i.e. if AB & BA are both defined, then it is not mandatory that AB ≠ BA.

• Associative law:

For three matrices A, B, and C, if multiplication is defined, then we can write it as A (BC) = (AB) C.

• Multiplicative identity:

For any square matrix A, there will be an identity matrix of the same order in which  IA = AI = A.

## Summary of NCERT Solutions Class 12 Matrices

• A matrix is an ordered rectangular array of numbers or functions.

• A matrix having $m$ rows and $n$ columns is called a matrix of order $m \times n$.

• $\left[a_y\right]_{m \times 1}$ is a column matrix.

• $\left[a_h\right]_{\mathrm{b} x}$ is a row matrix.

• An $m \times n$ matrix is a square matrix if $m=n$.

• $A=\left[a_{i j}\right]_{\text {nex }}$ is a diagonal matrix if $a_{i j}=0$, when $i \neq j$.

• $A_y=\left[a_{i j}\right]_{n \times n}$ is a scalar matrix if $a_{i j}=0$, when $i \neq j, a_{i j}=k,(k$ is some constant $)$, when $i=j$.

• $A=\left[a_y\right]_{m \times n}$ is an identity matrix, if $a_{i j}=1$. when $i=j, a_{i j}=0$, when $i \neq j$.

• A zero matrix has all its elements as zero.

• $A=\left[a_i\right]-\left[b_y\right]-B$ if

1. A and B are of asme order,

2. $a_{i j}-b_{i j}$ for all possible values of $i$ and $j$.

• $k a=k\left[a_y\right]_{m \times n}-\left[k\left(a_y\right)\right]_{m \times n}$

• $-A=(-1) A$

• $A-B=A+(-1) B$

• $A+B=B+A$

• $(A+B)+C=A+(B+C)$, where $A, B$ and $C$ are of aame order.

• $k(A+B)=k A+k B$, where $\mathrm{A}$ and $\mathrm{B}$ of asme order, $k$ is constant.

• $(k+l) A=k A+l A$, where $k$ and $l$ are constant.

• If $A=\left[a_{j j}\right]_{m \beta n}$ and $B=\left[b_{i j}\right]_{N \times p}$, then $A B=C-\left[C_{j k}\right]_{w \times p}$, where $c_{j k}=\sum_{j=1}^n a_j b_{j k}$

1. $A(B C)=(A B) C$

2. $A(B+C)=A B+A C$

3. $(A+B) C=A C+B C$

•  If $A=\left[a_W\right]_{n \times N}$, then $A$ or $A^T=\left[a_N\right]_{N \times m}$

1. $\left(A^{\prime}\right)^{\prime}=A$,

2. $(k t)^{\prime}=k A^{\prime}$,

3. $(A+B)^{\prime}=A^{\prime}+B^{\prime}$;

4. $(A B)^{\prime}=B \cdot A^{\prime}$

• $A$ is a symmetric matrix if $A^{\prime}=A$.

• A is a skew symmetric matrix if $A^{\prime}=-A^{\prime}$.

• Any aquare matrix can be represented as the sum of a symmetric and a skew symmetric matrix.

• Elementary operations of a matrix are as follows:

1. $R_i \leftrightarrow R_j$ or $C_i \leftrightarrow C_i$

2. $R_l \rightarrow k R_i$ or $C_l \rightarrow k C_i$

3. $R_i \rightarrow R_i+k R_j$ or $C_i \rightarrow C_i+k C_j$

• If $A$ and $B$ re two square matrices such that $A B-B A=I$, then $B$ is the inverse matrix of $A$ and is denoted by $A^{-1}$ and $A$ is the inverse of $\mathrm{B}$.

## Overview of Deleted Syllabus for Class 12 Maths Chapter 3

 Chapter Dropped Topics Metrices 3.7 Elementary Operations (Transformation) of a Matrix 3.8.1 Inverse of Matrices by Elementary Operations (Retain Question 18 of Exercise 3.4) Page 98 Example 26 Page Number 100-101:Miscellaneous Exercise Questions 1, 2, 3 and 12 Page 102 Third Last Point of Summary

## Class 12 Maths Chapter 3: Exercise Breakdown

 Exercise Number of Questions Exercise 3.1 Solutions 10 Questions (7 Short Answers, 3 MCQs) Exercise 3.2 Solutions 22 Questions (14 Long, 6 Short, 2 MCQs) Exercise 3.3 Solutions 12 Questions (10 Short Answers, 2 MCQs) Exercise 3.4 Solutions 18 Questions (4 Long, 13 Short, 1 MCQ) Miscellaneous Exercise Solutions 15 Questions (7 Long, 5 Short, 3 MCQs)

## Conclusion

NCERT Solutions for Class 12 Chapter 4 - Matrices offer a comprehensive understanding of the topic. It's crucial to focus on understanding the fundamental concepts of matrices, such as types of matrices, operations, and applications. Previous year question papers have typically included around 5-7 questions from this chapter, covering various aspects like matrix multiplication, determinants, and their properties. Therefore, it's important to practice solving different types of problems to gain proficiency. Vedantu's solutions provide clear explanations and step-by-step approaches, aiding in grasping the concepts effectively. By utilizing these resources and practicing regularly, students can enhance their understanding and excel in this topic.

## Other Study Materials of CBSE Class 12 Maths Chapter 3

 S. No Important Links for Chapter 3 Matrices 1 Matrices Revision Notes 2 Matrices Important Questions 3 Matrices Formula 4 Matrices NCERT Exemplar Solution 5 Matrices RS Aggarwal Solutions

## 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 3 Matrices

1. What are the main topics and subtopics of the chapter Matrices?

Matrices are one of the easiest chapters in maths, which, when understood, would become fun to solve. The concepts are as follows: Applications, Matrices, Order of matrices, Types of matrices includes Column matrix, Row matrix, Square matrix, Diagonal matrix, Scalar matrix, Identity matrix, Zero matrix and Equal matrix. You will also read about Operations of matrices which includes Addition of matrices, Subtraction of matrices and Multiplication of matrices.

The chapter also explains about various laws which are: Commutative law, Associative law, Distributive law, Existence of multiplicative identity and Cancellation law. You will also get an idea of Transpose of a matrix, Properties of transpose, Symmetric and skew matrices and Invertible matrices.

2. Give me a glimpse of the chapter Matrices?

Matrices are majorly rectangular arrays of numbers which are represented in rows and columns. Basically, matrices are used to perform various mathematical operations such as addition, multiplication, subtraction and division. Representing data related to population, infant mortality rate, etc. are some of the widely used areas where matrices are implemented to simplify the calculations and ease the representation of data.

Matrices have substantial use in plotting graphs, statistics and various scientific research purposes. The most common and popular application of matrices is in solving linear equations. Matrices are even used to represent the coefficients of a linear equation. Other than that, matrices even find application in 3D maths, where they are used to define the relationship between two coordinate spaces.

3. How do our solutions help you in scoring good marks in the examination?

Start solving easier problems and slowly move to the medium level and then to the complex level. After each and every topic try to solve the questions and check where you are lagging behind. Practice the weaker area of that particular concept as many times as you can. This also helps you in knowing your strength along with the weakness.

Maths is always a part of our life and it will also be used in our day to day life. The short-cut technique to score well in maths is to practice. Do not mug up the solutions, try to understand them and solve it in your own way and then check where you have gone wrong.

4. Why should we learn about Matrices in NCERT Solutions for Class 12 Maths Chapter 3?

Matrices are a powerful and essential tool in Mathematics. They are rectangular arrays of numbers represented in the form of rows and columns. It is used to perform several mathematical operations like addition, subtraction, multiplication, and division. It is widely used in various areas to simplify complex operations like plotting graphs, representing the population, statistics, and in various research papers. It also simplifies the method of solving complex linear equations.

5. How many exercises are there in the Matrix?

There are four exercises and one miscellaneous exercise in the chapter Matrix. You can refer to the Maths Solutions for Class 12 by Vedantu, where you will find the detailed solution of every question in your Maths book. The solutions are explained in a simple and detailed manner and will clear your doubts. It also includes tips and important points that will help you score high in your exams.

6. Is Class 12 maths tough?

No, Class 12 Maths is not tough. You can score high by regular practice and strong concepts. Solve every question in your NCERT Maths book, including the solved questions. Refer to Vedantu’s Maths Solution for Class 12. It focuses on strengthening your concepts so that you can solve a variety of questions. Practice some questions daily to improve your solving skills. You can refer to Vedantu’s Revision Class 12 Notes which includes all the important concepts and formulas compiled in one place.

7. What are Matrices in Maths Class 12?

Matrices are rectangular arrays where data is represented in the form of rows and columns. It is a very easy and scoring topic. This chapter includes:

• Types of Matrices

• Operations on Matrices

• Transpose of Matrices

• Symmetric and Skew-symmetric Matrices

• Elementary Operations (Transformation) on Matrices

• Invertible Matrices

These topics are explained in easy language so that the students can score high. You can practice a variety of questions in this chapter since it is relatively easier.

8. Why should you refer to Vedantu’s Solutions for Class 12 Maths?

Vedantu’s Solutions for Class 12 Maths are prepared by experienced subject-matter experts and include answers to every question in your NCERT Maths book. It includes all the important formulas, concepts, and theorems that are very important from the exam point of view. The solutions are updated according to the latest guidelines of the CBSE board. This is also very useful for revision before your exam. It includes step-wise solutions that will help you to understand the concepts well. The solution PDFs and other study materials such as important questions and revision notes can also be downloaded from the Vedantu app as well for free of cost.

9. Why understanding Metrices is crucial for exams in Class 12?

Mastery of matrices is essential for solving complex mathematical problems and is frequently tested in board exams.

10. What should I focus on in Chapter 3 Maths Class 12?

The NCERT Solutions Class 12 Maths Chapter 3 covers :

• The order of a matrix

• Types of matrices

• Mathematical operation

• Transpose of a matrix

• Symmetric matrices

• Skew-symmetric matrices

• Simple operations

• Invertible matrices.

11. How many questions from matrices appear in the board exams?

This chapter has 62 questions in 4 exercises along with 15 more provided in a miscellaneous exercise. Out of these 41 questions are short answer types, 11 multiple choice questions, and 25 long answer type questions.

12. Where can I find step-by-step process for matrices class 12 ncert solutions?

Vedantu provides comprehensive solutions that break down complex matrix problems into simple steps.

13. What are the trickiest topics in Chapter 3 Maths Class 12?

Focus more on matrix multiplication and finding inverses as these areas are often challenging for students.

14. How can I effectively practice matrix operations from Chapter 3?

Regular practice using NCERT exercises and additional problems on platforms like Vedantu is recommended.

15. Are matrix concepts important for other competitive exams too?

Yes, understanding matrices is useful for various competitive exams, especially in technical fields.