NCERT Solutions for Exercise 3.2 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:

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

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.

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