ML Aggarwal Solution Class 10 Chapter 8 Matrices Exercise MCQs

 MCQs

Question 1

If $A=\left[a_{ij}\right]_{2 \times 2}$ where $a_{ij}=i+j,$ then $A$ is equal to

(a) $\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$

(b) $\left[\begin{array}{ll}2 & 3 \\ 3 & 4\end{array}\right]$

(c) $\left[\begin{array}{ll}1 & 2 \\ 1 & 2\end{array}\right]$

(d) $\left[\begin{array}{ll}1 & 1 \\ 2 & 2\end{array}\right]$

Sol :

$A=\left[a_{i j}\right]_{2 \times 2}$ where $a_{i j}=i+j,$ then $A$ is equal to

$\left[\begin{array}{ll}2 & 3 \\ 3 & 4\end{array}\right]$

Ans (b)

Question 2

If $\left[\begin{array}{cc}x+3 & 4 \\ y-4 & x+y\end{array}\right]=\left[\begin{array}{ll}5 & 4 \\ 3 & 9\end{array}\right]$  then the values of x and y are

(a) x = 2, y = 7

(b) x = 7, y = 2

(c) x = 3, y = 6

(d) x = – 2, y = 7

Sol :

$\left[\begin{array}{cc}x+3 & 4 \\ y-4 & x+y\end{array}\right]=\left[\begin{array}{ll}5 & 4 \\ 3 & 9\end{array}\right]$

Comparing we get

x + 3 = 5

⇒ x = 5 – 3 = 2

and y – 4 = 3

⇒ y = 3 + 4 = 7

x = 2, y = 7 

Ans (a)

Question 3

If $\left[\begin{array}{cc}x+2 y & -y \\ 3 x & 7\end{array}\right]=\left[\begin{array}{cc}-4 & 3 \\ 6 & 4\end{array}\right]$ then the values of x and y are

(a) x = 2, y = 3

(b) x = 2, y = – 3

(c) x = – 2, y = 3

(d) x = 3, y = 2

Sol :

$\left[\begin{array}{cc}x+2 y & -y \\ 3 x & 7\end{array}\right]=\left[\begin{array}{cc}-4 & 3 \\ 6 & 4\end{array}\right]$

Comparing, we get

3x = 6

$\Rightarrow x=\frac{6}{3}=2$⇒ -y = 3

⇒ y = – 3

x = 2, y = -3 (b)

Question 4

If $\left[\begin{array}{cc}x-2 y & 5 \\ 3 & y\end{array}\right]=\left[\begin{array}{cc}6 & 5 \\ 3 & -2\end{array}\right]$  then the value of x is

(a) – 2

(b) 0

(c) 1

(d) 2

Sol :

$\left[\begin{array}{cc}x-2 y & 5 \\ 3 & y\end{array}\right]=\left[\begin{array}{cc}6 & 5 \\ 3 & -2\end{array}\right]$

Comparing, we get

y = -2

and x – 2y = 6

⇒ x – 2 x (-2) = 6

⇒ x + 4 = 6

⇒ x = 6 – 4 = 2

Ans (d)

Question 5

If $\left[\begin{array}{cc}x+2 y & 3 y \\ 4 x & 2\end{array}\right]=\left[\begin{array}{cc}0 & -3 \\ 8 & 2\end{array}\right]$ then the value of x – y is

(a) – 3

(b) 1

(c) 3

(d) 5

Sol :

$\left[\begin{array}{cc}x+2 y & 3 y \\ 4 x & 2\end{array}\right]=\left[\begin{array}{cc}0 & -3 \\ 8 & 2\end{array}\right]$

Comparing, we get

3y = -3

$\Rightarrow y=\frac{-3}{3}=-1$

4x=8

$\Rightarrow x=\frac{8}{4}=2$

x – y = 2 – (-1) = 2 + 1 = 3

Ans (c)

Question 6

If $x\left[\begin{array}{l}2 \\ 3\end{array}\right]+y\left[\begin{array}{c}-1 \\ 0\end{array}\right]=\left[\begin{array}{c}10 \\ 6\end{array}\right]$ then the values of x and y are

(a) x = 2, y = 6

(b) x = 2, y = – 6

(c) x = 3, y = – 4

(d) x = 3, y = – 6

Sol :

Given :

$x\left[\begin{array}{l}2 \\ 3\end{array}\right]+y\left[\begin{array}{c}-1 \\ 0\end{array}\right]=\left[\begin{array}{c}10 \\ 6\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}2 x \\ 3 x\end{array}\right]+\left[\begin{array}{c}-y \\ 0\end{array}\right]=\left[\begin{array}{c}10 \\ 6\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}2 x-y \\ 3 x+0\end{array}\right]=\left[\begin{array}{c}10 \\ 6\end{array}\right]$

$\Rightarrow\left[\begin{array}{c}2 x-y \\ 3 x\end{array}\right]=\left[\begin{array}{c}10 \\ 6\end{array}\right]$

Comparing , we get

$3 x=6 \Rightarrow x=\frac{6}{3}=2$

and 2x-y=10

$2 \times 2-y=10$

$\Rightarrow 4-y=10$

$\Rightarrow-y=10-4=6$

$\Rightarrow y=-6$

$\therefore x=2, y=-6$

Ans (b)

Question 7

If $B=\left[\begin{array}{cc}-1 & 5 \\ 0 & 3\end{array}\right]$ and $A-2 B=\left[\begin{array}{cc}0 & 4 \\ -7 & 5\end{array}\right]$

then the matrix A is equal to

(a) $\left[\begin{array}{cc}2 & 14 \\ -7 & 11\end{array}\right]$

(b) $\left[\begin{array}{cc}-2 & 14 \\ 7 & 11\end{array}\right]$

(c) $\left[\begin{array}{cc}2 & -14 \\ 7 & 11\end{array}\right]$

(d) $\left[\begin{array}{ll}-2 & 14 \\ -7 & 11\end{array}\right]$

Sol :

Given :

$B=\left[\begin{array}{cc}-1 & 5 \\ 0 & 3\end{array}\right]$

$A-2 B=\left[\begin{array}{cc}0 & 4 \\ -7 & 5\end{array}\right]$

$2 B=2 \times\left[\begin{array}{cc}-1 & 5 \\ 0 & 3\end{array}\right]=\left[\begin{array}{cc}-2 & 10 \\ 0 & 6\end{array}\right]$

$A-2 B=\left[\begin{array}{cc}0 & 4 \\ -7 & 5\end{array}\right]$

$\Rightarrow A=\left[\begin{array}{cc}0 & 4 \\ -7 & 5\end{array}\right]+2 B$

$\Rightarrow \mathrm{A}=\left[\begin{array}{cc}0 & 4 \\ -7 & 5\end{array}\right]+\left[\begin{array}{cc}-2 & 10 \\ 0 & 6\end{array}\right]$

$=\left[\begin{array}{cc}0-2 & 4+10 \\ -7+0 & 5+6\end{array}\right]=\left[\begin{array}{cc}-2 & 14 \\ -7 & 11\end{array}\right]$

Ans (d)

Question 8

If $A+B=\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right]$ and $A-2 B=\left[\begin{array}{cc}-1 & 1 \\ 0 & -1\end{array}\right]$

then A is equal to

(a) $\frac{1}{3}\left[\begin{array}{ll}1 & 1 \\ 2 & 1\end{array}\right]$

(b) $\frac{1}{3}\left[\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right]$

(c) $\left[\begin{array}{ll}1 & 1 \\ 2 & 1\end{array}\right]$

(d) $\left[\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right]$

Sol :

$A+B=\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right]$ and

$A-2 B=\left[\begin{array}{cc}-1 & 1 \\ 0 & -1\end{array}\right]$

$\Rightarrow 2 \mathrm{~A}+2 \mathrm{~B}=\left[\begin{array}{ll}2 & 0 \\ 2 & 2\end{array}\right]$ 

(multiplying by 2)...(i)

$A-2 B=\left[\begin{array}{cc}-1 & 1 \\ 0 & -1\end{array}\right]$...(ii)

Adding (i), and (ii) , we get

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

$\therefore \mathrm{A}=\frac{1}{3}\left[\begin{array}{ll}1 & 1 \\ 2 & 1\end{array}\right]$

Ans (a)

Question 9

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

(a) $\left[\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right]$

(b) $\left[\begin{array}{ll}0 & 0 \\ 1 & 1\end{array}\right]$

(c) $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

(d) $\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$

Sol :

Given :

$A=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

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

$=\left[\begin{array}{ll}1+0 & 0+0 \\ 0+0 & 0+1\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

Ans (d)

Question 10

If $A=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right],$ then $A^{2}=$

(a) $\left[\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right]$

(b) $\left[\begin{array}{ll}0 & 0 \\ 1 & 1\end{array}\right]$

(c) $\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$

(d) $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

Sol :

Given :

$A=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$

$\mathrm{A}^{2}=\mathrm{A} \times \mathrm{A}=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right] \times\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$

$=\left[\begin{array}{ll}0+1 & 0+0 \\ 0+0 & 1+0\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

Ans (d)

Question 11

If $A=\left[\begin{array}{ll}0 & 0 \\ 1 & 0\end{array}\right],$ then $A^{2}=$

(a) A

(b) O

(c) I

(d) $2 \mathrm{~A}$

Sol :

given :

$A=\left[\begin{array}{ll}0 & 0 \\ 1 & 0\end{array}\right]$

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

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

Ans (b)

Question 12

If $A=\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right],$ then $A^{2}=$

(a) $\left[\begin{array}{ll}2 & 0 \\ 1 & 1\end{array}\right]$

(b) $\left[\begin{array}{ll}1 & 0 \\ 1 & 2\end{array}\right]$

(c) $\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right]$

(d) none of these

Sol :

Given :

$A=\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right]$

$\mathrm{A}^{2}=\mathrm{A} \times \mathrm{A}=\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right] \times\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right]$

$=\left[\begin{array}{ll}1+0 & 0+0 \\ 1+1 & 0+1\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right]$

Ans (c)

Question 13

If $A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right],$ then $A^{2}=$

(a) $\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]$

(b) $\left[\begin{array}{cc}8 & -5 \\ 5 & 3\end{array}\right]$

(c) $\left[\begin{array}{cc}8 & -5 \\ -5 & -3\end{array}\right]$

(d) $\left[\begin{array}{cc}8 & -5 \\ -5 & 3\end{array}\right]$

Sol :

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

$A^{2}=A \times A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]$

$=\left[\begin{array}{cc}9+(-1) & 3+2 \\ -3-2 & -1+4\end{array}\right]=\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]$

Ans (a)

Question 14

If $A=\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right],$ then $A^{2}=p A$, then the value of p is 

(a) 2

(b) 4

(c) -2

(d) -4

Sol :

$A=\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right]$

and $A^{2}=p A$

$A^{2}=A \times A=\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right] \times\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right]$

$=\left[\begin{array}{cc}4+4 & -4-4 \\ -4-4 & 4+4\end{array}\right]=\left[\begin{array}{cc}8 & -8 \\ -8 & 8\end{array}\right]$

$p \mathrm{~A}=p\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right]=\left[\begin{array}{cc}2 p & -2 p \\ -2 p & 2 p\end{array}\right]$

$\because A^{2}=p A$

$\therefore\left[\begin{array}{cc}8 & -8 \\ -8 & 8\end{array}\right]=\left[\begin{array}{cc}2 p & -2 p \\ -2 p & 2 p\end{array}\right]$

Comparing , we get

$8=2 p \Rightarrow p=4$

Ans (b)