Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# NCERT Solutions for Exercise 3.2 Class 12 Maths Chapter 3 Matrices

Last updated date: 13th Jul 2024
Total views: 600.3k
Views today: 17.00k

## NCERT Solutions for Class 12 Maths Chapter 3 Matrices Exercise 3.2 - Free PDF Download

The NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.2 Matrices provides complete solutions to the problems in the Exercise. These NCERT Solutions are intended to assist students with the CBSE Class 12 board examination.

Table of Content
1. NCERT Solutions for Class 12 Maths Chapter 3 Matrices Exercise 3.2 - Free PDF Download
2. Glance on NCERT Solutions for Exercise 3.2 Class 12 Maths Chapter 3
3. Topics Covered in  Class 12 Maths Chapter 3 Exercise 3.2
4. Access NCERT Solutions for Class 12 Maths Chapter 3 - Matrices
4.1Exercise 3.2
5. NCERT Solutions for Class 12 Maths Chapter 3 Exercises
6. CBSE Class 12 Maths Chapter 3 Other Study Materials
7. Chapter-Specific NCERT Solutions for Class 12 Maths
FAQs

Students should thoroughly study this NCERT solution in order to solve all types of questions based on arithmetic progression. By completing these practice questions with the NCERT Maths Solutions Chapter 3 Exercise 3.2 Class 12, you will be better prepared to understand all of the different types of questions that may be asked in the Class 12 board exams.

## Glance on NCERT Solutions for Exercise 3.2 Class 12 Maths Chapter 3

• The NCERT Solutions for Maths Class 12 Ex 3.2 focus on solidifying your understanding of arithmetic operations on matrices.

• In Class 12 Ex 3.2 there are overall 22 Questions with 3 Short Answers and 19 Long Answers

• This exercise likely covers addition, subtraction, multiplication by a scalar, and potentially matrix multiplication.

• Remember the properties that hold true for these operations, such as commutativity, associativity, and distributive property (for matrix addition and scalar multiplication).

• There are problems related to finding the transpose of a matrix and solving systems of linear equations using matrices

## Topics Covered in  Class 12 Maths Chapter 3 Exercise 3.2

• Operations on Matrices

• Addition and subtraction of matrices

• Multiplication of a matrix by scalar

• Properties of matrix addition and scalar multiplication

• Commutativity

• Associativity

• Distributive property

• Existence of additive identity and inverse.

• Multiplication of matrices

• Properties of multiplication of matrices

Competitive Exams after 12th Science

## Access NCERT Solutions for Class 12 Maths Chapter 3 - Matrices

### Exercise 3.2

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

Find each of the following

i. $\mathbf{A}+\mathbf{B}$

$\mathrm{A}+\mathrm{B}=\left[\begin{array}{ll} 2 & 4 \\3 & 2\end{array}\right]+\left[\begin{array}{cc} 1 & 3 \\-2 & 5\end{array}\right]=\left[\begin{array}{ll} 3 & 7 \\1 & 7 \end{array}\right]$

ii. $\mathbf{A}-\mathbf{B}$

Ans: Subtracting the matrices

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

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

Ans: Subtracting the matrices

\begin{aligned} &3 \mathrm{~A}-\mathrm{C}=3\left[\begin{array}{cc} 2 & 4 \\ 3 & 2 \end{array}\right]-\left[\begin{array}{cc} -2 & 5 \\ 3 & 4 \end{array}\right] \\ &{\left[\begin{array}{cc} 6 & 12 \\ 9 & 6 \end{array}\right]-\left[\begin{array}{cc} -2 & 5 \\3 & 4 \end{array}\right]=\left[\begin{array}{cc} 8 & 7 \\ 6 & 2 \end{array}\right]} \end{aligned}

iv. AB

Ans: Matrix A has 2 columns. This number is equal to the number of rows in matrix B. Therefore, AB is defined as:

\begin{aligned} &\mathrm{AB}=\left[\begin{array}{cc} 2 & 4 \\ 3 & 2\end{array}\right]\left[\begin{array}{cc} 1 & 3 \\ -2 & 5 \end{array}\right] \\ &{\left[\begin{array}{ll} 2(1)+4(-2) & 2(3)+4(5) \\ 3(1)+2(-2) & 3(3)+2(5) \end{array}\right]=\left[\begin{array}{cc} -6 & 26 \\ 1 & 19\end{array}\right]} \end{aligned}

2. Compute the following:

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

Ans: Computing,

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

ii. $\left[\begin{array}{cc}a^{2}+b^{2} & b^{2}+c^{2} \\ a^{2}+c^{2} & a^{2}+b^{2}\end{array}\right]+\left[\begin{array}{cc}2 a b & 2 b c \\ -2 a c & -2 a b\end{array}\right]$

Ans: Computing,

\begin{aligned} &{\left[\begin{array}{cc} a^{2}+b^{2} & b^{2}+c^{2} \\ a^{2}+c^{2} & a^{2}+b^{2} \end{array}\right]+\left[\begin{array}{cc} 2 a b & 2 b c \\ -2 a c & -2 a b \end{array}\right]=\left[\begin{array}{cc} a^{2}+b^{2}+2 a b & b^{2}+c^{2}+2 b c \\ a^{2}+c^{2}-2 a c & a^{2}+b^{2}-2 a b \end{array}\right]} \\ &{\left[\begin{array}{ll} a^{2}+b^{2}+2 a b & b^{2}+c^{2}+2 b c \\ a^{2}+c^{2}-2 a c & a^{2}+b^{2}-2 a b \end{array}\right]=\left[\begin{array}{ll} (a+b)^{2} & (b+c)^{2} \\ (a-c)^{2} & (a-b)^{2} \end{array}\right]} \end{aligned}

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

Ans: Computing,

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

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

Ans: Computing,

\begin{aligned} &{\left[\begin{array}{cc} \cos ^{2} x & \sin ^{2} x \\ \sin ^{2} x & \cos ^{2} x \end{array}\right]+\left[\begin{array}{cc} \sin ^{2} x & \cos ^{2} x \\ \cos ^{2} x & \sin ^{2} x \end{array}\right]=\left[\begin{array}{cc} \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{array}\right]} \\ &{\left[\begin{array}{ll} \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{array}\right]=\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right],\left(\because \sin ^{2} x+\cos ^{2} x=1\right)} \end{aligned}

3. Compute the indicated products

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

Ans: Given: $\left[\begin{array}{cc}\mathrm{a} & \mathrm{b} \\ -\mathrm{b} & \mathrm{a}\end{array}\right]\left[\begin{array}{cc}\mathrm{a} & -\mathrm{b} \\ \mathrm{b} & \mathrm{a}\end{array}\right]$

\begin{aligned} {\left[\begin{array}{cc} a & b \\ -b & a \end{array}\right]\left[\begin{array}{cc} a & -b \\ b & a \end{array}\right] } &=\left[\begin{array}{cc} a(a)+b(b) & a(-b)+a(b) \\ -b(a)+b(a) & -b(-b)+a(a) \end{array}\right] \\ &=\left[\begin{array}{cc} a^{2}+b^{2} & 0 \\ 0 & a^{2}+b^{2} \end{array}\right] \end{aligned}

ii. $\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]\left[\begin{array}{lll}2 & 3 & 4\end{array}\right]$

Ans: Given: $\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]\left[\begin{array}{lll}2 & 3 & 4\end{array}\right]$

$\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]\left[\begin{array}{lll}2 & 3 & 4\end{array}\right]=\left[\begin{array}{lll}1(2) & 1(3) & 1(4) \\ 2(2) & 2(3) & 2(4) \\ 3(2) & 3(3) & 3(4)\end{array}\right]$

$=\left[\begin{array}{ccc} 2 & 3 & 4 \\ 4 & 6 & 8 \\ 6 & 9 & 12 \end{array}\right]$

iii. $\left[\begin{array}{cc}1 & -2 \\ 2 & 3\end{array}\right]\left[\begin{array}{lll}1 & 2 & 3 \\ 2 & 3 & 1\end{array}\right]$

Ans: Given: $\left[\begin{array}{cc}1 & -2 \\ 2 & 3\end{array}\right]\left[\begin{array}{ccc}1 & 2 & 3 \\ 2 & 3 & 1\end{array}\right]$

\begin{aligned} {\left[\begin{array}{cc} 1 & -2 \\ 2 & 3 \end{array}\right]\left[\begin{array}{lll} 1 & 2 & 3 \\ 2 & 3 & 1 \end{array}\right] } &=\left[\begin{array}{lll} 1(1)-2(2) & 1(2)-2(3) & 1(3)-2(1) \\ 2(1)+3(2) & 2(2)+3(3) & 2(3)+3(1) \end{array}\right] \\ &=\left[\begin{array}{ccc} -3 & -4 & 1 \\ 8 & 13 & 9 \end{array}\right] \end{aligned}

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

Ans: Given: $\left[\begin{array}{lll}2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6\end{array}\right]\left[\begin{array}{ccc}1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5\end{array}\right]$

\begin{aligned} {\left[\begin{array}{lll} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{array}\right]\left[\begin{array}{ccc} 1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \end{array}\right] } &=\left[\begin{array}{ccc} 2(1)+3(0)+4(3) & 2(-3)+3(2)+4(0) & 2(5)+3(4)+4(5) \\ 3(1)+4(0)+5(3) & 3(-3)+4(2)+5(0) & 3(5)+4(4)+5(5) \\ 4(1)+5(0)+6(3) & 4(-3)+5(2)+6(0) & 4(5)+5(4)+6(5) \end{array}\right] \\ &=\left[\begin{array}{ccc} 14 & 0 & 42 \\ 18 & -1 & 56 \\ 22 & -2 & 70 \end{array}\right] \end{aligned}

v. $\left[\begin{array}{cc}2 & 1 \\ 3 & 2 \\ -1 & 1\end{array}\right]\left[\begin{array}{ccc}1 & 0 & 1 \\ -1 & 2 & 1\end{array}\right]$

Ans: Given: $\left[\begin{array}{cc}2 & 1 \\ 3 & 2 \\ -1 & 1\end{array}\right]\left[\begin{array}{ccc}1 & 0 & 1 \\ -1 & 2 & 1\end{array}\right]$

\begin{aligned} {\left[\begin{array}{cc} 2 & 1 \\ 3 & 2 \\ -1 & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 0 & 1 \\ -1 & 2 & 1 \end{array}\right] } &=\left[\begin{array}{ccc} 2(1)+1(-1) & 2(0)+1(2) & 2(1)+1(1) \\ 3(1)+2(-1) & 3(0)+2(2) & 3(1)+2(1) \\ -1(1)+1(-1) & -1(0)+1(2) & -1(1)+1(1) \end{array}\right] \\ &=\left[\begin{array}{ccc} 1 & 2 & 3 \\ 1 & 4 & 5 \\ -2 & 2 & 0 \end{array}\right] \end{aligned}

vi. $\left[\begin{array}{ccc}3 & -1 & 3 \\ -1 & 0 & 2\end{array}\right]\left[\begin{array}{cc}2 & -3 \\ 1 & 0 \\ 3 & 1\end{array}\right]$

Ans: Given: $\left[\begin{array}{ccc}3 & -1 & 3 \\ -1 & 0 & 2\end{array}\right]\left[\begin{array}{cc}2 & -3 \\ 1 & 0 \\ 3 & 1\end{array}\right]$

\begin{aligned} {\left[\begin{array}{ccc} 3 & -1 & 3 \\ -1 & 0 & 2 \end{array}\right]\left[\begin{array}{cc} 2 & -3 \\ 1 & 0 \\ 3 & 1 \end{array}\right] } &=\left[\begin{array}{cc} 3(2)-1(1)+3(3) & 3(-3)-1(0)+3(1) \\ -1(2)+0(1)+2(3) & -1(-3)+0(0)+2(1) \end{array}\right] \\ &=\left[\begin{array}{cc} 14 & -6 \\ 4 & 5 \end{array}\right] \end{aligned}

4. If $A=\left[\begin{array}{ccc}1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1\end{array}\right], B=\left[\begin{array}{ccc}3 & 1 & 2 \\ 4 & 3 & 5 \\ 2 & -2 & 3\end{array}\right]$ and $C=\left[\begin{array}{ccc}4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3\end{array}\right]$, then Compute $(A+B)$ and $(B-C)$. Also verify that $A+(B-C)=(A+B)-C$

Ans:

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

Computing $(B-C)=\left[\begin{array}{ccc}3 & 1 & 2 \\ 4 & 3 & 5 \\ 2 & -2 & 3\end{array}\right]-\left[\begin{array}{ccc}4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3\end{array}\right]=\left[\begin{array}{ccc}-1 & -2 & 0 \\ 4 & -1 & 3 \\ 1 & 2 & 0\end{array}\right]$

To verify $\mathrm{A}+(\mathrm{B}-\mathrm{C})=(\mathrm{A}+\mathrm{B})-\mathrm{C}$,

We calculate $\mathrm{A}+(\mathrm{B}-\mathrm{C})$ and $(\mathrm{A}+\mathrm{B})-\mathrm{C}$ first, and then check that $\mathrm{LHS}=\mathrm{RHS}$

Computing $A+(B-C)$

$\left[\begin{array}{ccc} 1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1 \end{array}\right]+\left[\begin{array}{ccc} -1 & -2 & 0 \\ 4 & -1 & 3 \\ 1 & 2 & 0 \end{array}\right]=\left[\begin{array}{ccc} 0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1 \end{array}\right] \text {, and }$ Computing $(A+B)-C$ $\left[\begin{array}{ccc} 4 & 1 & -1 \\ 9 & 2 & 7 \\ 3 & -1 & 4 \end{array}\right]-\left[\begin{array}{ccc} 4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3 \end{array}\right]=\left[\begin{array}{ccc} 0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1 \end{array}\right]$

Hence, we have verified that $A+(B-C)=(A+B)-C$.

5. If $\mathbf{A}=\left[\begin{array}{ccc}\frac{2}{3} & 1 & \frac{5}{3} \\ \frac{1}{3} & \frac{2}{3} & \frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3}\end{array}\right]$ and $\mathbf{B}=\left[\begin{array}{ccc}\frac{2}{5} & \frac{3}{5} & 1 \\ \frac{1}{5} & \frac{2}{5} & \frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{5}\end{array}\right]$ then calculate $\mathbf{3 A}-\mathbf{5}$

Ans: Evaluating $3 \mathrm{~A}-5 \mathrm{~B}$, ${\left[\begin{array}{ccc}\frac{2}{3} & 1 & \frac{5}{3} \\ \frac{1}{3} & \frac{2}{3} & \frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3}\end{array}\right]-5\left[\begin{array}{ccc}\frac{2}{5} & \frac{3}{5} & 1 \\ \frac{1}{5} & \frac{2}{5} & \frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{5}\end{array}\right]=\left[\begin{array}{lll}2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2\end{array}\right]-\left[\begin{array}{lll}2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2\end{array}\right] }$

$=\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]$

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

Ans: Simplifying $\cos \theta\left[\begin{array}{cc}\cos \theta & \sin \theta \\ \sin \theta & \cos \theta\end{array}\right]+\sin \theta\left[\begin{array}{cc}\sin \theta & -\cos \theta \\ \cos \theta & \sin \theta\end{array}\right]$

\begin{aligned} &\Rightarrow\left[\begin{array}{cc} \cos ^{2} \theta & \cos \theta \sin \theta \\ -\sin \theta \cos \theta & \cos ^{2} \theta \end{array}\right]+\left[\begin{array}{cc} \sin ^{2} \theta & -\sin \theta \cos \theta \\ \sin \theta \cos \theta & \sin ^{2} \theta \end{array}\right] \\ &\Rightarrow\left[\begin{array}{cc} \cos ^{2} \theta+\sin ^{2} \theta & \cos \theta \sin \theta-\sin \theta \cos \theta \\ -\sin \theta \cos \theta+\sin \theta \cos \theta & \cos ^{2} \theta+\sin ^{2} \theta \end{array}\right] \end{aligned}

$\Rightarrow\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \quad\left(\because \cos ^{2} \theta+\sin ^{2} \theta=1\right)$

7. Find $\mathrm{X}$ and $\mathrm{Y}$, if

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

Ans: Given:

\begin{aligned} &X+Y=\left[\begin{array}{ll} 7 & 0 \\ 2 & 5 \end{array}\right] \ldots(1) \\ &X-Y=\left[\begin{array}{ll} 3 & 0 \\ 0 & 3 \end{array}\right] \ldots(2) \end{aligned} Adding these two equations, we get $2 X=\left[\begin{array}{ll} 7 & 0 \\ 2 & 5 \end{array}\right]+\left[\begin{array}{ll} 3 & 0 \\ 0 & 3 \end{array}\right]$ \begin{aligned} &2 X=\left[\begin{array}{cc} 10 & 0 \\ 2 & 8 \end{array}\right] \\ &X=\left[\begin{array}{ll} 5 & 0 \\ 1 & 4 \end{array}\right] \end{aligned} Putting $X$ in (1) \begin{aligned} &{\left[\begin{array}{ll} 5 & 0 \\ 1 & 4 \end{array}\right]+Y=\left[\begin{array}{ll} 7 & 0 \\ 2 & 5 \end{array}\right] Y=\left[\begin{array}{ll} 7-5 & 0-0 \\ 2-1 & 5-4 \end{array}\right]} \\ &Y=\left[\begin{array}{ll} 2 & 0 \\ 1 & 1 \end{array}\right] \end{aligned}

ii. $2 \mathrm{X}+3 \mathrm{Y}=\left[\begin{array}{ll}2 & 3 \\ 4 & 0\end{array}\right]$ and $3 \mathrm{X}+2 \mathrm{Y}=\left[\begin{array}{cc}2 & -2 \\ -1 & 5\end{array}\right]:$

Ans: Given:

\begin{aligned} &2 X+3 Y=\left[\begin{array}{cc} 2 & 3 \\ 4 & 0 \end{array}\right] \ldots(1) \\ &3 X+2 Y=\left[\begin{array}{cc} 2 & -2 \\ -1 & 5 \end{array}\right] \ldots \end{aligned} Multiplying (1) with 2, we have $4 X+6 Y=\left[\begin{array}{ll}4 & 6 \\ 8 & 0\end{array}\right] \ldots(3)$ Multiplying (2) with 3, we have $9 X+6 Y=\left[\begin{array}{cc} 6 & -6 \\ -3 & 15 \end{array}\right] \ldots(4)$ From (3) and (4), we have $-5 X=\left[\begin{array}{cc}4-6 & 6-(-6) \\ 8-(-3) & 0-15\end{array}\right] \therefore X=\left[\begin{array}{cc}\frac{2}{5} & -\frac{12}{5} \\ -\frac{11}{5} & 3\end{array}\right]$ Putting $\mathrm{X}$ in (1) $2\left[\begin{array}{cc} \frac{2}{5} & -\frac{12}{5} \\ -\frac{11}{5} & 3 \end{array}\right]+3 \mathrm{Y}=\left[\begin{array}{ll} 2 & 3 \\ 4 & 0 \end{array}\right] 3 \mathrm{Y}=\left[\begin{array}{cc} \frac{6}{5} & \frac{39}{5} \\ \frac{42}{5} & -6 \end{array}\right]$ $\therefore \mathrm{Y}=\left[\begin{array}{cc} \frac{2}{5} & \frac{13}{5} \\ \frac{14}{5} & -2 \end{array}\right]$

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

Ans: Given:

$2 \mathrm{X}+\mathrm{Y}=\left[\begin{array}{cc}1 & 0 \\ -3 & 2\end{array}\right]$, where $\mathrm{Y}=\left[\begin{array}{ll}3 & 2 \\ 1 & 4\end{array}\right]$

$2 \mathrm{X}+\left[\begin{array}{ll}3 & 2 \\ 1 & 4\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ -3 & 2\end{array}\right] 2 \mathrm{X}=\left[\begin{array}{cc}1-3 & 0-2 \\ -3-1 & 2-4\end{array}\right]$

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

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

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

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

\begin{aligned} &2+y=5 \Rightarrow y=3 \\ &2 x+2=8 \Rightarrow x=3 \\ &\therefore x=3 \text { and } y=3 \end{aligned}

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

Ans: Given:

$2\left[\begin{array}{cc} x & z \\ y & t \end{array}\right]+3\left[\begin{array}{cc} 1 & -1 \\ 0 & 2 \end{array}\right]=3\left[\begin{array}{ll} 3 & 5 \\ 4 & 6 \end{array}\right]$ \begin{aligned} &{\left[\begin{array}{cc} 2 x & 2 z \\ 2 y & 2 t \end{array}\right]+\left[\begin{array}{cc} 3 & -3 \\ 0 & 6 \end{array}\right]=\left[\begin{array}{cc} 9 & 15 \\ 12 & 18 \end{array}\right]} \\ &{\left[\begin{array}{cc} 2 x+3 & 2 z-3 \\ 2 y & 2 t+6 \end{array}\right]=\left[\begin{array}{cc} 9 & 15 \\ 12 & 18 \end{array}\right]} \end{aligned}

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

\begin{aligned} &2 x+3=9 \\ &\Rightarrow x=3 \end{aligned} \begin{aligned} &2 \mathrm{y}=12 \\ &\Rightarrow \mathrm{y}=6 \end{aligned} $2 z-3=15$ $\Rightarrow \mathrm{z}=9$ $2 t+6=18$ $\Rightarrow \mathrm{t}=6$ $\therefore \mathrm{x}=3, \mathrm{y}=6, \mathrm{z}=9 \text { and } \mathrm{t}=6 \text {. }$

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

Ans: Given:

\begin{aligned} &x\left[\begin{array}{l} 2 \\ 3 \end{array}\right]+y\left[\begin{array}{c} -1 \\ 1 \end{array}\right]=\left[\begin{array}{c} 10 \\ 5 \end{array}\right] \\ &{\left[\begin{array}{l} 2 x \\ 3 x \end{array}\right]+\left[\begin{array}{c} -y \\ y \end{array}\right]=\left[\begin{array}{c} 10 \\ 5 \end{array}\right]} \\ &{\left[\begin{array}{l} 2 x-y \\ 3 x+y \end{array}\right]=\left[\begin{array}{c} 10 \\ 5 \end{array}\right]} \end{aligned} Comparing the corresponding elements of these two matrices, we get $2 x-y=10$ and $3 x+y=5$

Adding these two equations, we have:

\begin{aligned} &5 \mathrm{x}=15 \\ &\Rightarrow \mathrm{x}=3 \end{aligned} Now, putting $x=3$ in any equation, we have: \begin{aligned} &\mathrm{y}=5-3 \mathrm{x} \\ &\Rightarrow \mathrm{y}=-4 \\ &\therefore \mathrm{x}=3 \text { and } \mathrm{y}=-4 \end{aligned}

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

Ans: Simplifying

$\left[\begin{array}{cc}3 x & 3 y \\ 3 z & 3 w\end{array}\right]=\left[\begin{array}{cc}x+4 & 6+y+x \\ -1+z+w & 2 w+3\end{array}\right]$ Comparing the corresponding elements of these two matrices, we get: $3 x=x+4$ $\Rightarrow x=2$ \begin{aligned} &3 y=6+x+y \\ &\Rightarrow y=4 \end{aligned} \begin{aligned} &3 w=2 w+3 \\ &\Rightarrow w=3 \end{aligned} $3 z=-1+z+w$ \begin{aligned} &\Rightarrow \mathrm{z}=1 \\ &\therefore \mathrm{x}=2, \mathrm{y}=4, \mathrm{z}=1 \text { and } \mathrm{w}=3 \end{aligned}

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

Ans: To show $\mathrm{F}(\mathrm{x}) \mathrm{F}(\mathrm{y})=\mathrm{F}(\mathrm{x}+\mathrm{y})$,

We first calculate $\mathrm{F}(\mathrm{x}) \mathrm{F}(\mathrm{y})$ and $\mathrm{F}(\mathrm{x}+\mathrm{y})$, and check that both are equal LHS

$F(x)=\left[\begin{array}{ccc}\cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right]$ and $F(y)=\left[\begin{array}{ccc}\cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1\end{array}\right]$ $F(x) F(y)=\left[\begin{array}{cccc}\cos x \cos y-\sin x \sin y & -\cos x \sin y-\sin x \cos y & 0 \\ \sin x \cos y+\cos x \sin y & -\sin x \sin y+\cos x \cos y & 0 \\ 0 & 0 & 0 & 1\end{array}\right]$ $F(x) F(y)=\left[\begin{array}{ccc}\cos (x+y) & -\sin (x+y) & 0 \\ \sin (x+y) & \operatorname{cox}(x+y) & 0 \\ 0 & 0 & 1\end{array}\right]$

\begin{aligned} &F(x) F(y)=\left[\begin{array}{ccc} \cos x \cos y-\sin x \sin y & -\cos x \sin y-\sin x \cos y & 0 \\ \sin x \cos y+\cos x \sin y & -\sin x \sin y+\cos x \cos y & 0 \\ 0 & 0 & 1 \end{array}\right] \\ &F(x) F(y)=\left[\begin{array}{ccc} \cos (x+y) & -\sin (x+y) & 0 \\ \sin (x+y) & \operatorname{cox}(x+y) & 0 \\ 0 & 0 & 1 \end{array}\right] \end{aligned}

RHS

$F(x+y)=\left[\begin{array}{ccc} \cos (x+y) & -\sin (x+y) & 0 \\ \sin (x+y) & \operatorname{cox}(x+y) & 0 \\ 0 & 0 & 1 \end{array}\right]$

$\mathrm{LHS}=\mathrm{RHS}$, proved.

14. Show that:

i. $\left[\begin{array}{cc}5 & -1 \\ 6 & 7\end{array}\right]\left[\begin{array}{ll}2 & 1 \\ 3 & 4\end{array}\right] \neq\left[\begin{array}{cc}2 & 1 \\ 3 & 4\end{array}\right]\left[\begin{array}{cc}5 & -1 \\ 6 & 7\end{array}\right]$

Ans: To show, we first calculate LHS and RHS

LHS

\begin{aligned} &{\left[\begin{array}{cc} 5 & -1 \\ 6 & 7 \end{array}\right]\left[\begin{array}{ll} 2 & 1 \\ 3 & 4 \end{array}\right]=\left[\begin{array}{cc} 5(2)-1(3) & 5(1)-1(4) \\ 6(2)+7(3) & 6(1)+7(4) \end{array}\right]} \\ &{\left[\begin{array}{cc} 10-3 & 5-4 \\ 12+21 & 6+28 \end{array}\right]=\left[\begin{array}{cc} 7 & 1 \\ 33 & 34 \end{array}\right]} \end{aligned} RHS \begin{aligned} &{\left[\begin{array}{cc} 2 & 1 \\ 3 & 4 \end{array}\right]\left[\begin{array}{cc} 5 & -1 \\ 6 & 7 \end{array}\right]=\left[\begin{array}{ll} 2(5)+1(6) & 2(-1)+1(7) \\ 3(5)+4(6) & 3(-1)+4(7) \end{array}\right]} \\ &{\left[\begin{array}{cc} 10+6 & -2+7 \\ 15+24 & -3+28 \end{array}\right]=\left[\begin{array}{cc} 16 & 5 \\ 39 & 25 \end{array}\right]} \end{aligned}

${\left[\begin{array}{cc}2 & 1 \\ 3 & 4\end{array}\right]\left[\begin{array}{cc}5 & -1 \\ 6 & 7\end{array}\right] }$ ${\left[\begin{array}{cc}10+6 & -2+ \\ 15+24 & -3+\end{array}\right.}$ $\therefore$ LHS $\neq$ RHS

ii. $\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0\end{array}\right]\left[\begin{array}{ccc}-1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4\end{array}\right] \neq\left[\begin{array}{ccc}-1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4\end{array}\right]\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0\end{array}\right]$

Ans: To show, we first calculate LHS and RHS

LHS

$\left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{array}\right]\left[\begin{array}{ccc} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{array}\right]=\left[\begin{array}{lll} 1(-1)+2(0)+3(2) & 1(1)+2(-1)+3(3) & 1(0)+2(1)+3(4) \\ 0(-1)+1(0)+0(2) & 0(1)+1(-1)+0(3) & 0(0)+1(1)+0(4) \\ 1(-1)+1(0)+0(2) & 1(1)+1(-1)+0(3) & 1(0)+1(1)+0(4) \end{array}\right]$

$=\left[\begin{array}{ccc} 5 & 8 & 14 \\ 0 & -1 & 1 \\ -1 & 0 & 1 \end{array}\right]$

RHS

\begin{aligned} {\left[\begin{array}{ccc} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{array}\right]\left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{array}\right]}&\\=\left[\begin{array}{ccc} -1(1)+1(0)+0(1) & -1(2)+1(1)+0(1) & -1(3)+1(0)+0(0) \\ 0(1)-1(0)+1(1) & 0(2)-1(1)+1(1) & 0(3)-1(0)+1(0) \\ 2(1)+3(0)+4(1) & 2(2)+3(1)+4(1) & 2(3)+3(0)+4(0) \end{array}\right] \\ &\\=\left[\begin{array}{ccc} -1 & -1 & -3 \\ 1 & 0 & 0 \\ 6 & 11 & 6 \end{array}\right] \end{aligned}

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

Ans: We know that, $\mathrm{A}^{2}=\mathrm{A} \times \mathrm{A}$

\begin{aligned} &{\left[\begin{array}{ccc} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right]\left[\begin{array}{ccc} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right]=\left[\begin{array}{ll} 2(2)+0(2)+1(1) & 2(0)+0(1)+1(-1) & 2(1)+0(3)+1(0) \\ 2(2)+1(2)+3(1) & 2(0)+1(1)+3(-1) & 2(1)+1(3)+3(0) \\ 1(2)-1(2)+0(1) & 1(0)-1(1)+0(-1) & 1(1)-1(3)+0(0) \end{array}\right]} \\ &{\left[\begin{array}{lll} 4+0+1 & 0+0-1 & 2+0+0 \\ 4+2+3 & 0+1-3 & 2+3+0 \\ 2-2+0 & 0-1+0 & 1-3+0 \end{array}\right]=\left[\begin{array}{ccc} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{array}\right]} \end{aligned}

Substituting value of $\mathrm{A}^{2}, \mathrm{~A}, \mathrm{I}$ in $\mathrm{A}^{2}-5 \mathrm{~A}+6 \mathrm{I}$

$\Rightarrow\left[\begin{array}{ccc} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{array}\right]-5\left[\begin{array}{ccc} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right]+6\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$

$\Rightarrow \begin{bmatrix} 5-10+6 &-1 &2-5 \\ 9-10&-2-5+6 & 5-15\\ -5+0 & -1+5 & -2-0+6 \end{bmatrix}\Rightarrow \begin{bmatrix} 1 & -1 &-3 \\ -1 &-1 & -10\\ -5& 4 & 4 \end{bmatrix}$

16. If  $\mathbf{A=\begin{bmatrix} 1 &0 & 2\\ 0&2 & 1\\ 2 & 0& 3 \end{bmatrix}},\text{Prove that}\; A^3-6A^2+7A+2I=0$

Ans: We know that, $\mathrm{A}^{2}=\mathrm{A} \times \mathrm{A}$ and $\mathrm{A}^{3}=\mathrm{A}^{2} \times \mathrm{A}$ For $\mathrm{A}^{2}$,

\begin{aligned} &{\left[\begin{array}{lll} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}\right]\left[\begin{array}{lll} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}\right]=\left[\begin{array}{ll} 1(1)+0(0)+2(2) & 1(0)+0(2)+2(0) & 1(2)+0(1)+2(3) \\ 0(1)+2(0)+1(2) & 0(0)+2(2)+1(0) & 0(2)+2(1)+1(3) \\ 2(1)+0(0)+3(2) & 2(0)+0(2)+3(0) & 2(2)+0(1)+3(3) \end{array}\right]} \\ &{\left[\begin{array}{lll} 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{array}\right]=\left[\begin{array}{ccc} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{array}\right]} \end{aligned}

For $\mathrm{A}^{3}$,

\begin{aligned} &{\left[\begin{array}{ccc} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{array}\right]\left[\begin{array}{lll} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}\right]=\left[\begin{array}{cc} 5(1)+0(0)+8(2) & 5(0)+0(2)+8(0) & 5(2)+0(1)+8(3) \\ 2(1)+4(0)+5(2) & 2(0)+4(3)+5(0) & 2(2)+4(1)+5(3) \\ 8(1)+0(0)+13(2) & 8(0)+0(2)+13(0) & 8(2)+0(1)+13(3) \end{array}\right]} \\ &{\left[\begin{array}{ccc} 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{array}\right]=\left[\begin{array}{ccc} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{array}\right]} \end{aligned}

Substituting value of $\mathrm{A}^{3}, \mathrm{~A}^{2}, \mathrm{~A}, \mathrm{I}$ in $\mathrm{A}^{3}-6 \mathrm{~A}^{2}+7 \mathrm{~A}+2 \mathrm{I}=0$

\begin{aligned} &\Rightarrow\left[\begin{array}{ccc} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{array}\right]-6\left[\begin{array}{ccc} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{array}\right]+7\left[\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}\right]+2\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ &\Rightarrow\left[\begin{array}{lll} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{array}\right]-\left[\begin{array}{ccc} 30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \end{array}\right]+\left[\begin{array}{ccc} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{array}\right]+\left[\begin{array}{lll} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array}\right] \\ &\Rightarrow\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \\ &\therefore \mathrm{A}^{3}-6 \mathrm{~A}^{2}+7 \mathrm{~A}+2 \mathrm{I}=0 \end{aligned}

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

Ans: We know that, $\mathrm{A}^{2}=\mathrm{A} \times \mathrm{A}$

\begin{aligned} {\left[\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right]\left[\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right] } &=\left[\begin{array}{ll} 3(3)-2(4) & 3(-2)-2(-2) \\ 4(3)-2(4) & 4(-2)-2(-2) \end{array}\right] \\ &=\left[\begin{array}{ll} 1 & -2 \\ 4 & -4 \end{array}\right] \end{aligned} Now $\mathrm{A}^{2}=\mathrm{k} \mathrm{A}-2 \mathrm{I}$ \begin{aligned} &\Rightarrow\left[\begin{array}{cc} 1 & -2 \\ 4 & -4 \end{array}\right]=\mathrm{k}\left[\begin{array}{cc} 3 & -2 \\ 4 & -2 \end{array}\right]-2\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ &\Rightarrow\left[\begin{array}{cc} 1 & -2 \\ 4 & -4 \end{array}\right]=\left[\begin{array}{cc} 3 \mathrm{k}-2 & -2 \mathrm{k} \\ 4 \mathrm{k} & -2 \mathrm{k}-2 \end{array}\right] \end{aligned}

Comparing the corresponding elements, we have:

Comparing the corresponding elements, we have:

\begin{aligned} &\Rightarrow 3 \mathrm{k}-2=1 \\ &\Rightarrow \mathrm{k}+1 \end{aligned}

Thus, the value of $\mathrm{k}$ is 1 .

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

Ans: LHS

$\mathrm{I}+\mathrm{A}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]+\left[\begin{array}{cc} 0 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 0 \end{array}\right]$ $=\left[\begin{array}{cc} 1 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 1 \end{array}\right]$

RHS

\begin{aligned} &(\mathrm{I}-\mathrm{A})\left[\begin{array}{cc} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]=\left[\begin{array}{cc} 1 & \tan \frac{\alpha}{2} \\ -\tan \frac{\alpha}{2} & 1 \end{array}\right]\left[\begin{array}{cc} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right] \\ &\Rightarrow\left[\begin{array}{cc} \cos \alpha+\sin \alpha \tan \frac{\alpha}{2} & -\sin \alpha+\cos \alpha \tan \frac{\alpha}{2} \\ -\cos \alpha \tan \frac{\alpha}{2}+\sin \alpha & \sin \alpha \tan \frac{\alpha}{2}+\cos \alpha \end{array}\right] \end{aligned}

\begin{aligned} &\Rightarrow\left[\begin{array}{cc} 1-2 \sin ^{2} \frac{\alpha}{2}+2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} \tan \frac{\alpha}{2} & -2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}+\left(2 \cos ^{2} \frac{\alpha}{2}-1\right) \tan \frac{\alpha}{2} \\ -\left(2 \cos ^{2} \frac{\alpha}{2}-1\right) \tan \frac{\alpha}{2}+2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} & 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} \tan \frac{\alpha}{2}+1-2 \sin ^{2} \frac{\alpha}{2} \end{array}\right] \\ &\Rightarrow\left[\begin{array}{cc} 1-2 \sin ^{2} \frac{\alpha}{2}+2 \sin ^{2} \frac{\alpha}{2} & -2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}+2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}-\tan \frac{\alpha}{2} \\ -2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}+\tan \frac{\alpha}{2}+2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} & 2 \sin ^{2} \frac{\alpha}{2}+1-2 \sin ^{2} \frac{\alpha}{2} \end{array}\right] \\ &\Rightarrow\left[\begin{array}{cc} 1 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 1 \end{array}\right] \end{aligned}

We get LHS = RHS.

19. A trust fund has Rs30,000 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 :

(a) Rs 1,800

Ans: Let Rs $x$ be invested in the first bond. Then, the sum of money invested in the second bond pays Rs $(30,000-\mathrm{x})$

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

Therefore, in order to obtain an annual total interest of Rs 1,800 , we have:

$\Rightarrow\left[\begin{array}{ll}x & 30,000-x\end{array}\right]\left[\begin{array}{c}\frac{5}{100} \\ \frac{7}{100}\end{array}\right]=1,800$

\begin{aligned} &\Rightarrow \frac{5 x}{100}+\frac{7(30,000-x)}{100}=1,800 \\ &\Rightarrow 5 x+2,10,000-7 x=1,80,000 \\ &\Rightarrow 2 x=30,000 \\ &\Rightarrow x=15,000 \end{aligned}

Thus, in order to obtain an annual total interest of Rs 1,800 , the trust fund should invest Rs 15,000 in the first bond and the remaining Rs 15,000 in the second bond.

(b) Rs 2,000

Ans: Let Rs $x$ be invested in the first bond. Then, the sum of money invested in the second bond pays Rs $(30,000-\mathrm{x})$

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

Therefore, in order to obtain an annual total interest of Rs 2,000 , we have:

$\Rightarrow\left[\begin{array}{ll}x & 30,000-x\end{array}\right]\left[\begin{array}{c}\frac{5}{100} \\ \frac{7}{100}\end{array}\right]=2,000$

\begin{aligned} &\Rightarrow \frac{5 x}{100}+\frac{7(30,000-x)}{100}=2,000 \\ &\Rightarrow 5 x+2,10,000-7 x=2,00,000 \\ &\Rightarrow 2 x=10,000 \\ &\Rightarrow x=5,000 \end{aligned}

Thus, in order to obtain an annual total interest of Rs 2,000 , the trust fund should invest Rs 5,000 in the first bond and the remaining Rs 25,000 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 $\mathbf{4 0}$ each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.

Ans: The bookshop has 10 dozen chemistry books, 8 dozen physics books, and 10 dozen economics books. The selling prices of a chemistry book, a physics book, and an economics book are respectively given as Rs 80, Rs 60 and Rs 40 . 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:

\begin{aligned} &\Rightarrow 12\left[\begin{array}{lll} 10 & 8 & 10 \end{array}\right]\left[\begin{array}{l} 80 \\ 60 \\ 40 \end{array}\right] \\ &\Rightarrow 12[10 \times 80+8 \times 60+10 \times 40] \\ &\Rightarrow 12(800+480+400) \\ &\Rightarrow 12(1680) \\ &\Rightarrow 20160 \end{aligned}

Thus, the bookshop will receive Rs 20160 from the sale of all these books.

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

A. $k=3, p=n$

B. $k$ is arbitrary, $p=2$

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

D. $k=2, p=3$

Ans: The correct option is Option(A)

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

Therefore, matrix PY will be defined if $\mathrm{k}=3$.

Consequently, $P Y$ will be of the order $p \times k$.

Matrices $\mathrm{W}$ and $\mathrm{Y}$ are of the orders $\mathrm{n} \times 3$ and $3 \times \mathrm{k}$ respectively.

Since the number of columns in $\mathrm{W}$ is equal to the number of rows in $\mathrm{Y}$, matrix $\mathrm{WY}$ is well-defined and is of the order $\mathrm{n} \times \mathrm{k}$.

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

However, PY is of the order $\mathrm{p} \times \mathrm{k}$ and $\mathrm{WY}$ is of the order $\mathrm{n} \times \mathrm{k}$. Therefore. we must have $\mathrm{p}=\mathrm{n}$.

Thus, $\mathrm{k}=3$ and $\mathrm{p}=\mathrm{n}$, are the restrictions on $\mathrm{n}, \mathrm{k}$ and $\mathrm{p}$ so that $\mathrm{PY}+\mathrm{WY}$ will be defined.

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

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

B. $2 \times n$

C. $\mathrm{n} \times 3$

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

Ans: The correct answer is B.

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

Therefore, matrix $7 \mathrm{X}$ is also of the same order.

Matrix $Z$ is of the order $2 \times p$ or $2 \times n$ (Because $n=p$ )

Therefore, matrix $5 \mathrm{Z}$ is also of the same order.

Now, both the matrices $7 \mathrm{X}$ and $5 \mathrm{Z}$ are of the order $2 \times \mathrm{n}$.

Thus, matrix $7 \mathrm{X}-5 \mathrm{Z}$ is well-defined and is of the order $2 \times \mathrm{n}$.

## Conclusion

In Exercise 3.2 of Class 12 Maths Chapter 3 on Matrices, covere important concepts like types of matrices, operations on matrices, and properties associated with them. It's crucial to understand the addition, subtraction, and multiplication of matrices, as well as the properties such as commutativity and associativity that apply to these operations. Previous year question papers often include 2-3 questions from this section, making it vital to master these problems to score well in exams. Additionally, focusing on the types of matrices like row matrix, column matrix, square matrix, and identity matrix is essential for a clear understanding. Mastering Class 12 maths Ex 3.2 will strengthen your foundation in matrix algebra, which is fundamental in various fields including computer science, engineering, and physics. Ensure you grasp the operations thoroughly and can identify different types of matrices easily. This will pave the way for tackling more complex problems in the future.

## NCERT Solutions for Class 12 Maths Chapter 3 Exercises

 Chapter 3 - Matrices All Exercises in PDF Format Exercise 3.1 10 Questions & Solutions (5 Short Answers, 5 Long Answers) Exercise 3.3 12 Questions & Solutions (4 Short Answers, 8 Long Answers) Exercise 3.4 18 Questions & Solutions (18 Short Answers)

## CBSE Class 12 Maths Chapter 3 Other Study Materials

 Sr.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 Exercise 3.2 Class 12 Maths Chapter 3 Matrices

1. How many questions are there in Class 12 Maths Chapter 3 Exercise 3.2?

Exercise 3.2 of Class 12 Maths Chapter 3 named Matrices contains various types of 22 questions in total. Our solutions contain answers to all of these questions. These solutions are carefully created by our in-house subject matter experts.

2. What are the topics and sub-topics contained in Class 12 Maths Chapter 3?

The topics and sub-topics covered in Class 12 Chapter 3 Matrices are given below in tabular format.

3.1 Introduction

3.2 Matrix

• 3.2.1 Order of a matrix

3.3 Types of Matrices

• 3.3.1 Equality of matrices

3.4 Operations on Matrices

• 3.4.2 Multiplication of a matrix by a scalar

• 3.4.3 Properties of matrix addition

• 3.4.4 Properties of scalar multiplication of a matrix

• 3.4.5 Multiplication of matrices

• 3.4.6 Properties of multiplication of matrices

3.5. Transpose of a Matrix

• 3.5.1 Properties of transpose of the matrices

3.6 Symmetric and Skew Symmetric Matrices

3.7 Elementary Operation (Transformation) of a Matrix (not in current syllabus)

3.8 Invertible Matrices

• 3.8.1 Inverse of a matrix by elementary operations.

Yes, you can download the Class 12 Maths Chapter 3 Exercise 3.2 NCERT Solutions directly from Vedantu website and mobile app at absolutely free of cost. These are created by the experienced teachers and subject matter experts in a clear and concise manner so that students can easily hold a grasp over all the concepts.

4. How can I find NCERT Solutions for Chapter 3 of Class 12 Maths to prepare for exams?

Students can download NCERT Solutions for Chapter 3 Class 12 Maths from Vedantu. All Solutions for Chapter 3 Matrices of Class 12 Maths are given in a simple method so that students can understand the concepts given in Chapter 3 of Class 12 Maths. They can practice all NCERT questions to revise the main concepts given in Chapter 3 of Class 12 Maths to score high marks. All NCERT Solutions are given by expert Maths professionals so that students can understand the main concepts.

5. How many exercises are there in Chapter 3 Matrices of Class 12 Maths?

There are a total of four exercises in Chapter 3 Matrices of Class 12 Maths. The Chapter covers a wide range of topics related to matrices in four exercises. Students will learn different operations on matrices. They should practice questions given in each exercise for a proper understanding of the Chapter. Students can also refer to the NCERT Solutions for Chapter 3 of Class 12 Maths to learn the basic concepts of the chapter.

6. What are the different types of Matrices students will study in Chapter 3 of Class 12 Maths?

Students will study different types of matrices in Chapter 3 of Class 12 Maths. They will study column matrix, square matrix, row matrix, diagonal matrix, zero matrices, identity matrix, and scalar matrix. Different types of matrices are explained in different exercises with examples. Students can also understand the concept of different matrices from NCERT Solutions given on Vedantu website and on the Vedantu app at free of cost. NCERT Solutions can help students to have deeper knowledge and understanding of the different topics related to matrices.

7. Is it sufficient to prepare NCERT Solutions for Chapter 3 of Class 12 Maths for exams?

Students can practice all NCERT Solutions for Chapter 3 of Class 12 Maths to prepare for their exams. They can download the NCERT Solutions for Chapter 3 of  Chapter Class 12 Maths for free of cost. They can save the solutions and can refer to all NCERT Solutions of Chapter 3 of Class 12 Maths to revise for the exams. NCERT Solutions can help students to get a deep understanding of matrices and score high marks in exams.

8. What is the best way to score high marks in Chapter 3 Matrices of Class 12 Maths?

Students should practice all NCERT Solutions to score high marks in Chapter 3 Matrices of Class 12 Maths. NCERT Solutions are solved by expert professional Maths teachers so that students can understand the concepts of the chapter. All solutions are given in a simple way for proper understanding. Most questions asked in the board papers are based on the NCERT Solutions that can help students to get an idea about the type of questions they can get on matrices.