2 Marks Questions

Question 12 Marks

Three car dealers, say A, B and C deals in three cars, Sedan, cars Hatchback cars and SUV cars. The sales figure of 220 and 2021 showed that dealer A sold 120 Hatch back, 50 Sedan and 10 SUV cars in 2020 and 300 Hatchback, 150 Sedan and 20 SUV in 2021. Dealer B sold 100 Hatchback, 30 Sedan, 5 SUV cars in 2020 and 200 Hatchback, 50 Sedan, 6 SUV cars in 2021. Dealer C sold 90 Hatchback, 40 Sedan and 2 SUV cars in 2020 and 100 Hatchback, 60 Sedan and 5 SUV cars in 2021

Q.1. Find the total number of cars sold in two given years, by each dealer.
Q.2. If each dealer receive profit of ₹ $ 50,000$ on sale of a Hatch car, ₹ $ 1,00,000$ on sale of a sedan car and ₹ $ 1,50,000$ on a sale of a SUV car, then find the amount of profit received in year 2021 by each dealer.

Answer

(1) Let matrix P denotes that dealer A sold 120 Hatchback, 50 sedan and 10 SUV; dealer B sold 100 Hatchback, 30 sedan and 5 SUV cars and dealer C sold 90 Hatchback, 40 sedan and 2 SUV cars in year 2020.

Now, total number of cars sold in two given years, by each dealer, is given by

(2) The amount of profit in year 2021 received by each dealer is given by the matrix.

$=\begin{array}{l}A \\ B \\ C\end{array}\left[\begin{array}{c}15000000+15000000+3000000 \\ 10000000+5000000+900000 \\ 5000000+6000000+750000\end{array}\right]$
$=\begin{array}{r}A \\ B \\ C\end{array}\left[\begin{array}{l}33000000 \\ 15900000 \\ 11750000\end{array}\right]$

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Question 22 Marks

In a city there are two factories $A$ and $B$. Each factory produces sports clothes for boys and girls. There are three types of clothes produced in both the factories type I, type II and type III. For boys the number of units of types I, II and III respectively are 80, 70 and 65 in factory A and 85, 65 and 72 are in factory B. For girls the number of units of types I, II and III respectively are $80,75,90$ in factory A and 50, 55, 80 are in factory B.

Q.1. Write the matrices $P$ and $Q$, if $P$ represents the matrix of number of units of each type produced by factory $A$ for both boys and girls; and $Q$ represents the matrix of number of units of each type produced by factory B for both boys are girls.
Q.2. Find the total production of sports clothes of each type for boys and girls.

Answer

(1) In factory A, number of units of type I, II and III for boys are $80,70,65$ respectively and for girls number of units of type I, II and III are $80,75,90$ respectively.

(2) In factory B, number of units of type I, II and III for boys are $85,65,72$ respectively and for girls number of units of type I, II and III are $50,55,80$ respectively.

(2) Let matrix $X$ represent the number of units of each type produced by factory A for boys and matrix Y represents the number of units of each type produced by factory B for boys.

Now, total production of sports clothes of each type for boys $=X+Y$
$\begin{array}{l}=\left[\begin{array}{lll}80 & 70 & 65\end{array}\right]+\left[\begin{array}{lll}85 & 65 & 72\end{array}\right] \\=\left[\begin{array}{lll}165 & 135 & 137\end{array}\right]\end{array}$
Similarly, for girls, let matrix $S$ represents the number of units of each type produced by factory A and matrix T represents the number of units of each type produced by factory B.

$\begin{array}{l}\text{Now, required matrix}\quad = S + T \\ =\left[\begin{array}{lll}80 & 75 & 90\end{array}\right]+\left[\begin{array}{lll}50 & 55 & 80\end{array}\right] \\ =\left[\begin{array}{lll}130 & 130 & 170\end{array}\right]\end{array}$

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Question 32 Marks

Prove that the diagonal elements of a skew symmetric matrix are all zero.

Answer

Let $A$ be a skew-symmetric matrix. Then by definition $A^{\prime}=-A$
or the $(i, j)^{\text {th }}$ element of $A^{\prime}=$ the $(i, j)^{\text {th }}$ element of $(-A)$
or the $(j, i)^{\text {th }}$ element of $A=-$ the $(i, j)^{\text {th }}$ element of $A$
For the diagonal elements $i=j$ or the $(i, j)$ the element of $A=-$ the $(i, j)^{\text {th }}$ element of $A$
or the $(i, i)^{\text {th }}$ element of $A=0$
Hence the diagonal elements are all zero.

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Question 42 Marks

Show that $A^{\prime} A$ and $A A^{\prime}$ are both symmetric matrices for any matrix $A$.

Answer

$\text{Let}\quad P= A'A$
$\therefore\quad P'=(A'A)'$
$=A'(A')' \quad [\because (AB)'=B'A']$
$=A'A=P$
So, $A'A$ is symmetric matrix for any matrix $A.$
Similarly,
$\text{Let}\quad Q=AA'$
$Q'=(AA')'$
$=(A')'A'$
$=AA'$
$=A$
So, $AA'$ is symmetric matrix for any matrix $A.$

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Question 52 Marks

If $A=\left[\begin{array}{cc}4 & 2 \\ -1 & 1\end{array}\right]$, show that $(A-2 I)(A-3 I)=0$.

Answer

$\begin{array}{l}\text { Given, } A=\left[\begin{array}{cc}4 & 2 \\ -1 & 1\end{array}\right] \\ \begin{aligned} \therefore A-2 I & =\left[\begin{array}{cc}4 & 2 \\ -1 & 1\end{array}\right]\end{aligned}-2\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right] \\ \\ =\left[\begin{array}{cc}4 & 2 \\ -1 & 1\end{array}\right]+\left[\begin{array}{cc}-2 & 0 \\ 0 & -2\end{array}\right] \\ \\ =\left[\begin{array}{cc}4-2 & 2+0 \\ -1+0 & 1-2\end{array}\right] \\ \\ =\left[\begin{array}{cc}2 & 2 \\ -1 & -1\end{array}\right]\end{array}$
$\begin{aligned}\text { and }A-3 I & =\left[\begin{array}{cc}4 & 2 \\-1 & 1\end{array}\right]-3\left[\begin{array}{ll}1 & 0 \\0 & 1\end{array}\right] \\& =\left[\begin{array}{cc}4 & 2 \\-1 & 1\end{array}\right]+\left[\begin{array}{cc}-3 & 0 \\0 & -3\end{array}\right] \\& =\left[\begin{array}{cc}4-3 & 2+0 \\-1+0 & 1-3\end{array}\right] \\& =\left[\begin{array}{cc}1 & 2 \\-1 & -2\end{array}\right]\end{aligned}$
Now, LHS $=(A-2 I)(A-3 I)$
$\begin{array}{l}=\left[\begin{array}{cc}2 & 2 \\
-1 & -1\end{array}\right]\left[\begin{array}{cc}1 & 2 \\-1 & -2\end{array}\right] \\=\left[\begin{array}{cc}2-2 & 4-4 \\-1+1 & -2+2\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\0 & 0\end{array}\right] \\=0=\text { RHS }\end{array}$

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Question 62 Marks

If $A=\left[\begin{array}{ccc}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0\end{array}\right]$, then find $\left(A^2-5 A\right)$.

Answer

$\begin{array}{l}\text { Given, } A=\left[\begin{array}{ccc}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0\end{array}\right] \\ \therefore A^2=\left[\begin{array}{ccc}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0\end{array}\right]\left[\begin{array}{ccc}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0\end{array}\right]\end{array}$
$\begin{array}{l}=\left[\begin{array}{ccc}4+0+1 & 0+0-1 & 2+0+0 \\ 4+2+3 & 0+1-3 & 2+3+0 \\ 2-2+0 & 0-1+0 & 1-3+0\end{array}\right] \\ =\left[\begin{array}{ccc}5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2\end{array}\right]\end{array}$
$\text{Now,}\quad A^2-5 A=\left[\begin{array}{ccc}5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2\end{array}\right]-5\left[\begin{array}{ccc}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0\end{array}\right]$
$\begin{array}{l}=\left[\begin{array}{ccc}5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2\end{array}\right]-\left[\begin{array}{ccc}10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0\end{array}\right] \\ =\left[\begin{array}{ccc}-5 & -1 & -3 \\ -1 & -7 & -10 \\ -5 & 4 & -2\end{array}\right]\end{array}$

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Question 72 Marks

Find a matrix $A$ such that $2 A-3 B+5 C=0$, where
$B=\left[\begin{array}{ccc}-2 & 2 & 0 \\3 & 1 & 4\end{array}\right] \text { and } C=\left[\begin{array}{ccc}2 & 0 & -2 \\7 & 1 & 6\end{array}\right]$

Answer

Given, $B=\left[\begin{array}{ccc}-2 & 2 & 0 \\ 3 & 1 & 4\end{array}\right]$ and $C=\left[\begin{array}{ccc}2 & 0 & -2 \\ 7 & 1 & 6\end{array}\right]$
Also, given $2A-3B+5C=0$
$\Rightarrow 2A=3B-5C$
$\Rightarrow 2 A=3\left[\begin{array}{ccc}-2 & 2 & 0 \\ 3 & 1 & 4\end{array}\right]-5\left[\begin{array}{ccc}2 & 0 & -2 \\ 7 & 1 & 6\end{array}\right]$
$\Rightarrow 2 A=\left[\begin{array}{ccc}-6 & 6 & 0 \\ 9 & 3 & 12\end{array}\right]+\left[\begin{array}{ccc}-10 & 0 & 10 \\ -35 & -5 & -30\end{array}\right]$
$\Rightarrow 2 A=\left[\begin{array}{ccc}-16 & 6 & 10 \\ -26 & -2 & -18\end{array}\right]$
$\Rightarrow A=\frac{1}{2}\left[\begin{array}{ccc}-16 & 6 & 10 \\ -26 & -2 & -18\end{array}\right]$
$\Rightarrow A=\left[\begin{array}{ccc}-8 & 3 & 5 \\ -13 & -1 & -9\end{array}\right]$

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Question 82 Marks

If $A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]$ and $I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$, find $k$ so that $A^2=5 A+k I$.

Answer

$A^2=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]$
$\begin{aligned} A^2 & =\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right] \\ 5 A & =\left[\begin{array}{cc}15 & 5 \\ -5 & 10\end{array}\right], \\ k I & =\left[\begin{array}{ll}k & 0 \\ 0 & k\end{array}\right]\end{aligned}$
$\begin{aligned}\because A^2-5 A & =k I \\ {\left[\begin{array}{cc}-7 & 0 \\ 0 & -7\end{array}\right] } & =\left[\begin{array}{ll}k & 0 \\ 0 & k\end{array}\right] \Rightarrow k=-7\end{aligned}$

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Question 92 Marks

If $\left(\begin{array}{lll}2 & 1 & 3\end{array}\right)\left(\begin{array}{ccc}-1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1\end{array}\right)\left(\begin{array}{c}1 \\ 0 \\ -1\end{array}\right)=A$, then write the order of matrix $A$.

Answer

We have,
$A=\left(\begin{array}{lll}2 & 1 & 3\end{array}\right)\left(\begin{array}{ccc}-1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1\end{array}\right)\left(\begin{array}{c}1 \\ 0 \\ -1\end{array}\right)$
$=\left(\begin{array}{lll}-2-1 & 1+3 & -2+3\end{array}\right)\left(\begin{array}{c}1 \\ 0 \\ -1\end{array}\right)$
$=\left(\begin{array}{lll}-3 & 4 & 1\end{array}\right)\left(\begin{array}{c}1 \\ 0 \\ -1\end{array}\right)$
$=(-3-1)=(-4)_{1 \times 1}$
$\therefore$ Order of matrix $A$ is $1 \times 1.$
OR
Let $P=\left(\begin{array}{lll}2 & 1 & 3\end{array}\right), Q=\left(\begin{array}{ccc}-1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1\end{array}\right)$ and $R=\left(\begin{array}{c}1 \\ 0 \\ -1\end{array}\right)$
order of $PQ =1 \times 3$
order of $PQR(A)=1\times1$

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Question 102 Marks

If $A=\left(\begin{array}{ccc}1 & 2 & 2 \\ 2 & 1 & x \\ -2 & 2 & -1\end{array}\right)$ is a matrix satisfying $A A^{\prime}=9 I$, find $x$.

Answer

$\begin{array}{c}\text { Given, } A=\left(\begin{array}{ccc}1 & 2 & 2 \\2 & 1 & x \\-2 & 2 & -1\end{array}\right) \\A^{\prime}=\left(\begin{array}{ccc}1 & 2 & -2 \\2 & 1 & 2 \\2 & x & -1\end{array}\right)\end{array}$
Since, $A A^{\prime}=9 I$
$\begin{array}{l}\left(\begin{array}{ccc}1 & 2 & 2 \\ 2 & 1 & x \\ -2 & 2 & -1\end{array}\right)\left(\begin{array}{ccc}1 & 2 & -2 \\ 2 & 1 & 2 \\ 2 & x & -1\end{array}\right)=9\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right) \\ \left(\begin{array}{ccc}1+4+4 & 2+2+2 x & -2+4-2 \\ 2+2+2 x & 4+1+x^2 & -4+2-x \\ -2+4-2 & -4+2-x & 4+4+1\end{array}\right)=\left(\begin{array}{lll}9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9\end{array}\right)\end{array}$
$\left(\begin{array}{ccc}9 & 2 x+4 & 0 \\2 x+4 & x^2+5 & -x-2 \\0 & -x-2 & 9\end{array}\right)=\left(\begin{array}{lll}9 & 0 & 0 \\0 & 9 & 0 \\0 & 0 & 9\end{array}\right)$
$\begin{aligned}\text{Now,}\quad x^2+5 & =9 \\ x^2 & =9-5 \\ x^2 & =4 \\ x & =\sqrt{4} \\ x & = \pm 2\end{aligned}$
$\begin{aligned}\text{Also,}\quad 2 x+4 & =0 \\ 2 x & =-4 \\ x & =-\frac{4}{2}\end{aligned}$
$\text{So,}\quad x=-2$

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Question 112 Marks

Write a $2 \times 2$ matrix which is both symmetric and skew symmetric.

Answer

$\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$ is $a$ $2 \times 2$ symmetric as well as skew symmetric matrix.

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Question 122 Marks

If the matrix $A=\left[\begin{array}{ccc}0 & a & -3 \\ 2 & 0 & -1 \\ b & 1 & 0\end{array}\right]$ is skew symmetric. Find the values of ' $a$ ' and ' $b$ '.

Answer

A is a skew symmetric matrix
$\begin{array}{l}\Rightarrow A^{\prime}=-A \\A=\left[\begin{array}{ccc}0 & a & -3 \\2 & 0 & -1 \\b & 1 & 0\end{array}\right] \\A^{\prime}=\left[\begin{array}{ccc}0 & 2 & b \\a & 0 & 1 \\-3 & -1 & 0\end{array}\right] \\A^{\prime}=-A\end{array}$
Now, $\left[\begin{array}{ccc}0 & 2 & b \\ a & 0 & 1 \\ -3 & -1 & 0\end{array}\right]=-\left[\begin{array}{ccc}0 & a & -3 \\ 2 & 0 & -1 \\ b & 1 & 0\end{array}\right]$
$\left[\begin{array}{ccc}0 & 2 & b \\a & 0 & 1 \\-3 & -1 & 0\end{array}\right]=\left[\begin{array}{ccc}0 & -a & 3 \\-2 & 0 & 1 \\-b & -1 & 0\end{array}\right]$
On comparing we get,
$\begin{array}{l}a=-2 \\b=3\end{array}$

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Question 132 Marks

Matrix $A=\left[\begin{array}{ccc}0 & 2 b & -2 \\ 3 & 1 & 3 \\ 3 a & 3 & -1\end{array}\right]$ is given to be symmetric, find values of $a$ and $b$.

Answer

As $A$ is a symmetric matrix,
$\text{or}\quad A'=A$
$\therefore\left[\begin{array}{ccc}0 & 3 & 3 a \\ 2 b & 1 & 3 \\ -2 & 3 & -1\end{array}\right]=\left[\begin{array}{ccc}0 & 2 b & -2 \\ 3 & 1 & 3 \\ 3 a & 3 & -1\end{array}\right]$
$\therefore$ By equality of matrices, $a= \frac {-2}{3}$ and $b= \frac{3}{2}.$

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Question 142 Marks

If $\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{c}x \\ -1 \\ z\end{array}\right]=\left[\begin{array}{l}1 \\ 2 \\ 1\end{array}\right]$, find $x+y+z.$

Answer

$\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1 \\ 2 \\ 1\end{array}\right]$
$\begin{array}{l}x=1, y=-2, z=1 \\\text { or } x+y+z=0\end{array}$

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Question 152 Marks

If $3 A-B=\left[\begin{array}{ll}5 & 0 \\ 1 & 1\end{array}\right]$ and $B=\left[\begin{array}{ll}4 & 3 \\ 2 & 5\end{array}\right]$, then find the matrix $A.$

Answer

Given $3 A-B=\left[\begin{array}{ll}5 & 0 \\ 1 & 1\end{array}\right], B=\left[\begin{array}{ll}4 & 3 \\ 2 & 5\end{array}\right]$
$\text { or } 3 A-\left[\begin{array}{ll}4 & 3 \\2 & 5\end{array}\right]=\left[\begin{array}{ll}5 & 0 \\1 & 1\end{array}\right]$
$\begin{array}{l}\text { or } 3 A=\left[\begin{array}{ll}5 & 0 \\ 1 & 1\end{array}\right]+\left[\begin{array}{ll}4 & 3 \\ 2 & 5\end{array}\right] \\ \text { or } 3 A=\left[\begin{array}{ll}9 & 3 \\ 3 & 6\end{array}\right] \\ \text { or } A=\left[\begin{array}{ll}3 & 1 \\ 1 & 2\end{array}\right]\end{array}$

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Question 162 Marks

If $\left[\begin{array}{cc}a+4 & 3 b \\ 8 & -6\end{array}\right]=\left[\begin{array}{cc}2 a+2 & b+2 \\ 8 & a-8 b\end{array}\right],$ write the value of $a-2 b$.

Answer

$\left[\begin{array}{cc}a+4 & 3 b \\8 & -6\end{array}\right]=\left[\begin{array}{cc}2 a+2 & b+2 \\8 & a-8 b\end{array}\right]$
By equating, $a+4=2 a+2$ or $a=2$
$3 b=b+2 \text { or } b=1$
$\therefore a-2b=2-2(1)$
$=2-2$
$=0$

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Question 172 Marks

If $\left(\begin{array}{ll}2 x & 4\end{array}\right)\binom{x}{-8}=0$, find the positive value of $x$.

Answer

$\left(\begin{array}{ll}2 x & 4\end{array}\right)\binom{x}{-8}=0$
$\text{or}\quad (2x^{2}-32)=0$
By equating,
$2x^{2}-32=0$
$\text{or}\quad 2x^{2}=32$
$\text{or}\quad x^{2}=16$
$x= \pm 4$
$\text{or}\quad x=+4$
$\therefore$ Positive value of $x=4$

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Question 182 Marks

If $\left[\begin{array}{cc}x y & 4 \\ z+6 & x+y\end{array}\right]=\left[\begin{array}{cc}8 & w \\ 0 & 6\end{array}\right]$, write the value of $x+y+z$.

Answer

$\left[\begin{array}{cc}x y & 4 \\ z+6 & x+y\end{array}\right]=\left[\begin{array}{cc}8 & w \\ 0 & 6\end{array}\right]$
By equating $\quad z+6=0$ and $x+y=6$
or $z=-6, x+y=6$
$x+y+z=6-6$
$\therefore x+y+z=0$

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Question 192 Marks

If $A$ is a square matrix such that $A^2=I$, then find the simplified value of $(A-I)^3+(A+I)^3-7 A$.

Answer

Given $A^2=I$
$\begin{array}{l}\therefore(A-I)^3+(A+I)^3-7 A \\=A^3-I^3-3 A^2+3 A I^2+A^3+I^3+3 A^2 I+3 A I^2-7 A \\=A I-I-3 I+3 A+A I+I+3 I+3 A-7 A \\=A+3 A+A+3 A-7 A=A\end{array}$

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Question 202 Marks

Write the number of all possible matrices of order $2 \times 2$ with each entry 1,2 or 3 .

Answer

Number of elements of $2 \times 2$ matrix $=4$
$\therefore$ Number of ways to write 1, 2 or 3 at 4 places
$=3^4$
$=81$
$\therefore$ 81 matrices of order $2 \times 2$ are possible with each entry 1, 2 or 3.

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Question 212 Marks

If $A=\left(\begin{array}{cc}0 & 3 \\ 2 & -5\end{array}\right)$ and $k A=\left(\begin{array}{cc}0 & 4 a \\ -8 & 5 b\end{array}\right)$, find the values of $k$ and $a$.

Answer

$\begin{array}{l}A=\left[\begin{array}{cc}0 & 3 \\2 & -5\end{array}\right] \\ k A=\left[\begin{array}{cc}0 & 3 k \\2 k & -5 k
\end{array}\right] \\\text { But given } k A=\left[\begin{array}{cc}0 & 4 a \\-8 & 5 b\end{array}\right] \\\therefore\left[\begin{array}{cc}0 & 3 k \\2 k & -5 k\end{array}\right]=\left[\begin{array}{cc}0 & 4 a \\-8 & 5 b\end{array}\right]\end{array}$
Equating individual terms,
$\begin{aligned}2 k & =-8 \Rightarrow k=-4 \\3 k & =4 a \\3 \times(-4) & =4 a \Rightarrow a=-3\end{aligned}$

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