Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

NCERT Solutions for Class 12 Maths Chapter 3: Matrices - Exercise 3.1

ffImage
Last updated date: 28th Mar 2024
Total views: 567.6k
Views today: 10.67k
MVSAT 2024

NCERT Solutions for Class 12 Maths Chapter 3 (Ex 3.2)

Free PDF download of NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.2 (Ex 3.2) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 12 Maths Chapter 3 Matrices Exercise 3.2 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.


Class:

NCERT Solutions for Class 12

Subject:

Class 12 Maths

Chapter Name:

Chapter 3 - Matrices

Exercise:

Exercise - 3.2

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2023-24

Medium:

English and Hindi

Available Materials:

  • Chapter Wise

  • Exercise Wise

Other Materials

  • Important Questions

  • Revision Notes



NCERT Solutions for Class 12 Maths of Exercise 3.2 are given in this article. This article contains solutions for all the questions asked in exercise 3.2 – Matrices. These solutions are provided by our subject experts in easy to understand language to clarify the doubts that students face while solving the questions from the NCERT maths book. Students can also download the NCERT Maths Solutions of Class 12 Chapter 3 Exercise 3.2 free pdf from the link below and practice the questions offline.

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

Ans: Adding the matrices

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

NCERT Solutions for Class 12 Maths Chapter 3 All Other Exercises

Chapter 3 - Matrix Other 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)


Topics Covered In Class 12 Maths Chapter 3 Exercise 3.2

The questions asked in this exercise are based on the following topics:

  • Addition of matrices

  • Multiplication of a matrix by scalar

  • Properties of matrix addition

  • Properties of scalar multiplication of a matrix

  • Multiplication of matrices

  • Properties of multiplication of matrices


What is a Matrix?

In Mathematics, a matrix is defined as the set of numbers arranged in rows and columns to form a rectangular array. The numbers in the matrix are termed as the elements or entries of the matrix.


What are the Operations of Matrices?

The operation of matrices includes addition, subtraction, and multiplication of matrices. The matrix operation helps to combine two or more matrices in a single matrix.


NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.2

Opting for the NCERT solutions for Ex 3.2 Class 12 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 3.2 Class 12 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 12 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 12 Maths Chapter 3 Exercise 3.2 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 12 Maths Chapter 3 Exercise 3.2, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

Do not delay any more. Download the NCERT solutions for Class 12 Maths Chapter 3 Exercise 3.2 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.

FAQs on NCERT Solutions for Class 12 Maths Chapter 3: Matrices - Exercise 3.1

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

  • 3.4.1 Addition of 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

  • 3.8 Invertible Matrices

  • 3.8.1 Inverse of a matrix by elementary operations.

3. Can I download Class 12 Maths Chapter 3 Exercise 3.2 NCERT Solutions for free?

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.