ML Aggarwal Solution Class 10 Chapter 8 Matrices Exercise Test

 Test 

Question 1

Find the values of a and below

$\left[\begin{array}{cc}a+3 & b^{2}+2 \\ 0 & -6\end{array}\right]=\left[\begin{array}{cc}2 a+1 & 3 b \\ 0 & b^{2}-5 b\end{array}\right]$

Sol :

$\left[\begin{array}{cc}a+3 & b^{2}+2 \\ 0 & -6\end{array}\right]=\left[\begin{array}{cc}2 a+1 & 3 b \\ 0 & b^{2}-5 b\end{array}\right]$

comparing the corresponding elements

a + 3 = 2a + 1

⇒ 2a – a =3 – 1

⇒ a = 2

b² + 2 = 3b

⇒ b² – 3b + 2 = 0

⇒ b² – b – 2b + 2 = 0

⇒ b (b – 1) – 2 (b – 1) = 0

⇒ (b – 1) (b – 2) = 0.

Either b – 1 = 0, then b = 1

or b – 2 = 0, then b = 2

Hence a = 2, b = 2 or 1


Question 2

Find a, b, c and d if $3\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]=\left[\begin{array}{cc}4 & a+b \\ c+d & 3\end{array}\right]+\left[\begin{array}{cc}a & 6 \\ -1 & 2 d\end{array}\right]$

Sol :

Given :

$3\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]=\left[\begin{array}{cc}4 & a+b \\ c+d & 3\end{array}\right]+\left[\begin{array}{cc}a & 6 \\ -1 & 2 d\end{array}\right]$

$\Rightarrow\left[\begin{array}{ll}3 a & 3 b \\ 3 c & 3 d\end{array}\right]=\left[\begin{array}{ll}4+a & a+b+6 \\ c+d-1 & 3+2 d\end{array}\right]$

Comparing the corresponding elements:

3a=4+a 

$\Rightarrow 3 a-a=4$

$ \Rightarrow 2 a=4$

$\therefore a=2$

$3 b=a+b+6 \Rightarrow 3 b-b=2+6$

$\Rightarrow 2 b=8$

$\therefore b=4$

3d=3+2d 

$\Rightarrow 3 d-2 d=3$

$\therefore d=3$

3c=c+d-1 

$\Rightarrow 3 c-c=3-1$

2c=2 

$\Rightarrow c=1$

Hence a=2, b=4, c=1, d=3


Question 3

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

Sol :

Given :

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

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

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

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

$X=\frac{1}{2}\left[\begin{array}{ll}-2 & -2 \\ -4 & -2\end{array}\right]=\left[\begin{array}{ll}-1 & -1 \\ -2 & -1\end{array}\right]$


Question 4

Determine the matrices A and B when

$A+2 B=\left[\begin{array}{cc}1 & 2 \\ 6 & -3\end{array}\right]$ and $2 A-B=\left[\begin{array}{cc}2 & -1 \\ 2 & -1\end{array}\right]$
Sol :

$A+2 B=\left[\begin{array}{cc}1 & 2 \\ 6 & -3\end{array}\right]$...(i)

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

Multiplying (i) by 1 and (ii) by 2

$A+2 B=\left[\begin{array}{rr}1 & 2 \\ 6 & -3\end{array}\right]$

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

Adding we get 

$5 \mathrm{~A}=\left[\begin{array}{rr}1 & 2 \\ 6 & -3\end{array}\right]+\left[\begin{array}{rr}4 & -2 \\ 4 & -2\end{array}\right]=\left[\begin{array}{rr}5 & 0 \\ 10 & -5\end{array}\right]$

$\mathbf{A}=\frac{1}{5}\left[\begin{array}{rr}5 & 0 \\ 10 & -5\end{array}\right]=\left[\begin{array}{rr}1 & 0 \\ 2 & -1\end{array}\right]$

From
(i) $A+2 B=\left[\begin{array}{rr}1 & 2 \\ 6 & -3\end{array}\right]$

$=\left[\begin{array}{rr}1 & 0 \\ 2 & -1\end{array}\right]+2 \mathrm{~B}=\left[\begin{array}{rr}1 & 2 \\ 6 & -3\end{array}\right]$

$2 \mathrm{~B}=\left[\begin{array}{rr}1 & 2 \\ 6 & -3\end{array}\right]-\left[\begin{array}{rr}1 & 0 \\ 2 & -1\end{array}\right]=\left[\begin{array}{rr}0 & 2 \\ 4 & -2\end{array}\right]$

$\therefore B=\frac{1}{2}\left[\begin{array}{rr}0 & 2 \\ 4 & -2\end{array}\right]=\left[\begin{array}{rr}0 & 1 \\ 2 & -1\end{array}\right]$

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

Question 5

(i) Find the matrix B if A = $\left[\begin{array}{ll}4 & 1 \\ 2 & 3\end{array}\right]$ and $A^{2}=A+2 B$

(ii) If $A=\left[\begin{array}{cc}1 & 2 \\ -3 & 4\end{array}\right], B=\left[\begin{array}{cc}0 & 1 \\ -2 & 5\end{array}\right]$
and $C=\left[\begin{array}{cc}-2 & 0 \\ -1 & 1\end{array}\right]$ find $A(4 B-3 C)$

Sol :

$A=\left[\begin{array}{ll}4 & 1 \\ 2 & 3\end{array}\right]$
let $B=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$

$A^{2}=A \times A=\left[\begin{array}{ll}4 & 1 \\ 2 & 3\end{array}\right]\left[\begin{array}{ll}4 & 1 \\ 2 & 3\end{array}\right]$
$=\left[\begin{array}{rr}16+2 & 4+3 \\ 8+6 & 2+9\end{array}\right]$
$=\left[\begin{array}{rr}18 & 7 \\ 14 & 11\end{array}\right]$

$A+2 B=\left[\begin{array}{ll}4 & 1 \\ 2 & 3\end{array}\right]+2\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$

$=\left[\begin{array}{ll}4 & 1 \\ 2 & 3\end{array}\right]+\left[\begin{array}{ll}2 a & 2 b \\ 2 c & 2 d\end{array}\right]=\left[\begin{array}{ll}4+2 a & 1+2 b \\ 2+2 c & 3+2 d\end{array}\right]$

$\therefore A^{2}=A+2 B$
$\therefore\left[\begin{array}{ll}18 & 7 \\ 14 & 11\end{array}\right]=\left[\begin{array}{ll}4+2 a & 1+2 b \\ 2+2 c & 3+2 d\end{array}\right]$

Comparing the corresponding elements

$4+2 a=18 \Rightarrow 2 a=18-4=14$
$\therefore a=7$
$1+2 b=7 \Rightarrow 2 b=7-1=6$
$\therefore b=3$
$2+2 c=14 \Rightarrow 2 c=14-2=12$
$\therefore c=6$
$3+2 d=11 \Rightarrow 2 d=11-3=8$
$\therefore d=4$

Hence a=7, b=3, c=6, d=4
$\therefore B=\left[\begin{array}{ll}7 & 3 \\ 6 & 4\end{array}\right]$

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

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

$=\left[\begin{array}{rr}0 & 4 \\ -8 & 20\end{array}\right]-\left[\begin{array}{rr}-6 & 0 \\ -3 & 3\end{array}\right]$

$=\left[\begin{array}{rr}0-(-6) & 4-0 \\ -8-(-3) & 20-3\end{array}\right]=\left[\begin{array}{rr}0+6 & 4-0 \\ -8+3 & 20-3\end{array}\right]$

$=\left[\begin{array}{rr}6 & 4 \\ -5 & 17\end{array}\right]$

Now $A(4 B-3 C)=\left[\begin{array}{rr}1 & 2 \\ -3 & 4\end{array}\right]\left[\begin{array}{rr}6 & 4 \\ -5 & 17\end{array}\right]$

$=\left[\begin{array}{rr}1 \times 6+2(-5) & 1 \times 4+2 \times 17 \\ -3 \times 6+4 \times(-5) & -3 \times 4+4 \times 17\end{array}\right]$

$=\left[\begin{array}{rr}6-10 & 4+34 \\ -18-20 & -12+68\end{array}\right]=\left[\begin{array}{rr}-4 & 38 \\ -38 & 56\end{array}\right]$


Question 6

If $A=\left[\begin{array}{ll}1 & 4 \\ 1 & 0\end{array}\right], B=\left[\begin{array}{cc}2 & 1 \\ 3 & -1\end{array}\right]$ and $C=\left[\begin{array}{ll}2 & 3 \\ 0 & 5\end{array}\right]$ compute (AB)C = (CB)A ?

Sol :
Given :
$\mathrm{A}=\left[\begin{array}{ll}1 & 4 \\ 1 & 0\end{array}\right]$
$\mathrm{B}=\left[\begin{array}{cc}2 & 1 \\ 3 & -1\end{array}\right]$
$\mathrm{C}=\left[\begin{array}{ll}2 & 3 \\ 0 & 5\end{array}\right]$

(AB) $\mathrm{C}=\left[\left[\begin{array}{ll}1 & 4 \\ 1 & 0\end{array}\right] \times\left[\begin{array}{ll}2 & 1 \\ 3 & -1\end{array}\right]\right]\left[\begin{array}{ll}2 & 3 \\ 0 & 5\end{array}\right]$

$=\left[\begin{array}{ll}2+12 & 1-4 \\ 2+0 & 1+0\end{array}\right]\left[\begin{array}{ll}2 & 3 \\ 0 & 5\end{array}\right]$

$=\left[\begin{array}{rr}14 & -3 \\ 2 & 1\end{array}\right]\left[\begin{array}{ll}2 & 3 \\ 0 & 5\end{array}\right]$

$=\left[\begin{array}{rr}28+0 & 42-15 \\ 4+0 & 6+5\end{array}\right]=\left[\begin{array}{rr}28 & 27 \\ 4 & 11\end{array}\right]$


(CB) $\mathrm{A}=\left[\left[\begin{array}{ll}2 & 3 \\ 0 & 5\end{array}\right]\left[\begin{array}{rr}2 & 1 \\ 3 & -1\end{array}\right]\right]\left[\begin{array}{ll}1 & 4 \\ 1 & 0\end{array}\right]$

$=\left[\begin{array}{ll}4+9 & 2-3 \\ 0+15 & 0-5\end{array}\right]\left[\begin{array}{ll}1 & 4 \\ 1 & 0\end{array}\right]$

$=\left[\begin{array}{ll}13 & -1 \\ 15 & -5\end{array}\right]\left[\begin{array}{ll}1 & 4 \\ 1 & 0\end{array}\right]=\left[\begin{array}{ll}13-1 & 52+0 \\ 15-5 & 60+0\end{array}\right]$

$=\left[\begin{array}{ll}12 & 52 \\ 10 & 60\end{array}\right]$

It is clear from above that

(AB)≠(CB)A


Question 7

If $A=\left[\begin{array}{ll}3 & 2 \\ 0 & 5\end{array}\right]$ and $B=\left[\begin{array}{ll}1 & 0 \\ 1 & 2\end{array}\right]$ find the each of the following and state it

they are equal:

(i) $(A+B)(A-B)$

(ii) $A^{2}-B^{2}$

Sol :

Given :

$A=\left[\begin{array}{ll}3 & 2 \\ 0 & 5\end{array}\right]$

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

(i) (A+B)(A-B)

$=\left\{\left[\begin{array}{ll}3 & 2 \\ 0 & 5\end{array}\right]+\left[\begin{array}{ll}1 & 0 \\ 1 & 2\end{array}\right]\right\}\times\left\{\left[\begin{array}{ll}3 & 2 \\ 0 & 5\end{array}\right]-\left[\begin{array}{ll}1 & 0 \\ 1 & 2\end{array}\right]\right\}$

$=\left[\begin{array}{ll}3+1 & 2+0 \\ 0+1 & 5+2\end{array}\right] \times\left[\begin{array}{ll}3-1 & 2-0 \\ 0-1 & 5-2\end{array}\right]$

$=\left[\begin{array}{ll}4 & 2 \\ 1 & 7\end{array}\right] \times\left[\begin{array}{cc}2 & 2 \\ -1 & 3\end{array}\right]$

$=\left[\begin{array}{ll}8-2 & 8+6 \\ 2-7 & 2+21\end{array}\right]=\left[\begin{array}{cc}6 & 14 \\ -5 & 23\end{array}\right]$


(ii) $\mathrm{A}^{2}-\mathrm{B}^{2}$

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

$=\left[\begin{array}{ll}9+0 & 6+10 \\ 0+0 & 0+25\end{array}\right]-\left[\begin{array}{ll}1+0 & 0+0 \\ 1+2 & 0+4\end{array}\right]$

$=\left[\begin{array}{ll}9 & 16 \\ 0 & 25\end{array}\right]-\left[\begin{array}{ll}1 & 0 \\ 3 & 4\end{array}\right]=\left[\begin{array}{ll}9-1 & 16-0 \\ 0-3 & 25-4\end{array}\right]$

$=\left[\begin{array}{cc}8 & 16 \\ -3 & 21\end{array}\right]$

We see that $(\mathrm{A}+\mathrm{B})(\mathrm{A}-\mathrm{B}) \neq \mathrm{A}^{2}-\mathrm{B}^{2}$


Question 8

If $A=\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]$ find $A^{2}-5 A-141$

Where I is unit matrix of order 2 x 2

Sol :

Given

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

$A^{2}=A \times A=\left[\begin{array}{rr}3 & -5 \\ -4 & 2\end{array}\right]\left[\begin{array}{rr}3 & -5 \\ -4 & 2\end{array}\right]$

$=\left[\begin{array}{rr}9+20 & -15-10 \\ -12-8 & 20+4\end{array}\right]=\left[\begin{array}{rr}29 & -25 \\ -20 & 24\end{array}\right]$

$5 A=5\left[\begin{array}{rr}3 & -5 \\ -4 & 2\end{array}\right]=\left[\begin{array}{rr}15 . & -25 \\ -20 & 10\end{array}\right]$

$\therefore \mathrm{A}^{2}-5 \mathrm{~A}-14 \mathrm{I}=\left[\begin{array}{rr}29 & -25 \\ -20 & 24\end{array}\right]$

$-\left[\begin{array}{rr}15 \cdot & -25 \\ -20 & 10\end{array}\right]-14\left[\begin{array}{rr}1 & 0 \\ 0 & 1\end{array}\right]$

$=\left[\begin{array}{rr}29 & -25 \\ -20 & 24\end{array}\right]-\left[\begin{array}{rr}15 & -20 \\ -20 & 10\end{array}\right]-\left[\begin{array}{rr}14 & 0 \\ 0 & 14\end{array}\right]$

$=\left[\begin{array}{ll}29-15-14 & -25+25-0 \\ -20+20+0 & 24-10-14\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$


Question 9

If $A=\left[\begin{array}{ll}3 & 3 \\ p & q\end{array}\right]$ and $A^{2}=0$ find $p$ and $q$

Sol :

Given :

$A=\left[\begin{array}{ll}3 & 3 \\ p & q\end{array}\right]$

$A^{2}=A \times A=\left[\begin{array}{ll}3 & 3 \\ p & q\end{array}\right]\left[\begin{array}{ll}3 & 3 \\ p & q\end{array}\right]$

$=\left[\begin{array}{ll}9+3 p & 9+3 q \\ 3 p+p q & 3 p+q^{2}\end{array}\right]$

But $A^{2}=0$

$\therefore\left[\begin{array}{ll}9+3 p & 9+3 q \\ 3 p+p q & 3 p+q^{2}\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$

Comparing the corresponding elements

9+3p=0

$ \Rightarrow 3 p=-9 \Rightarrow p=-3$

9+3q=0 

$\Rightarrow 3 q=-9 \Rightarrow q=-3$

Hence p=-3, q=-3


Question 10

If $A=\left[\begin{array}{cc}\frac{3}{5} & \frac{2}{5} \\ x & y\end{array}\right]$ and $A^{2}=1,$ find x, y

Sol :

Given :

$A=\left[\begin{array}{ll}\frac{3}{5} & \frac{2}{5} \\ x & y\end{array}\right]$

$A^{2}=A \times A=\left[\begin{array}{ll}\frac{3}{5} & \frac{2}{5} \\ x & y\end{array}\right]\left[\begin{array}{ll}\frac{3}{5} & \frac{2}{5} \\ x & y\end{array}\right]$

$=\left[\begin{array}{ll}\frac{9}{25}+\frac{2}{5} x & \frac{6}{25}+\frac{2}{5} y \\ \frac{3}{5} x+x y & \frac{2}{5} x+y^{2}\end{array}\right]$

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

$\left[\begin{array}{ll}\frac{9}{25}+\frac{2}{5} x & \frac{6}{25}+\frac{2}{5} y \\ \frac{3}{5} x+x y & \frac{2}{5} x+y^{2}\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

Comparing the corresponding elements,

$\frac{9}{25}+\frac{2}{5} x=1$

$ \Rightarrow \frac{2}{5} x=1-\frac{9}{25}=\frac{16}{25}$

$x=\frac{16}{25} \times \frac{5}{2}=\frac{8}{5}$

$\frac{6}{25}+\frac{2}{5} y=0 $

$\Rightarrow \frac{2}{5} y=\frac{-6}{25}$

$y=\frac{-6}{25} \times \frac{5}{2}=\frac{-3}{5}$

Hence $x=\frac{8}{5}, y=\frac{-3}{5}$


Question 11

If $\left[\begin{array}{cc}-1 & 0 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ 0 & -1\end{array}\right]$ find a, b, c and d

Sol :

$\left[\begin{array}{cc}-1 & 0 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ 0 & -1\end{array}\right]$

$\Rightarrow\left[\begin{array}{rr}-a+0 & -b+0 \\ 0+c & 0+d\end{array}\right]=\left[\begin{array}{rr}1 & 0 \\ 0 & -1\end{array}\right]$

$\Rightarrow\left[\begin{array}{rr}-a & -b \\ c & d\end{array}\right]=\left[\begin{array}{rr}1 & 0 \\ 0 & -1\end{array}\right]$

Comparing the corresponding elements

$-a=1 \Rightarrow a=-1$

$-b=0 \Rightarrow b=0$

$c=0$ and $d=-1$

Hence a=-1, b=0, c=0, d=-1


Question 12

Find a and b if

$\left[\begin{array}{cc}a-b & b-4 \\ b+4 & a-2\end{array}\right]\left[\begin{array}{ll}2 & 0 \\ 0 & 2\end{array}\right]=\left[\begin{array}{cc}-2 & -2 \\ 14 & 0\end{array}\right]$

Sol :

Given :

$\left[\begin{array}{ll}a-b & b-4 \\ b+4 & a-2\end{array}\right]\left[\begin{array}{ll}2 & 0 \\ 0 & 2\end{array}\right]=\left[\begin{array}{cc}-2 & -2 \\ 14 & 0\end{array}\right]$

$\Rightarrow\left[\begin{array}{ll}2 a-2 b+0 & 0+2 a-2 b \\ 2 b+8+0 & 0+2 a-4\end{array}\right]=\left[\begin{array}{rr}-2 & -2 \\ 14 & 0\end{array}\right]$

$\Rightarrow\left[\begin{array}{ll}2 a-2 b & 2 a-2 b \\ 2 b+8 & 2 a-4\end{array}\right]=\left[\begin{array}{rr}-2 & -2 \\ 14 & 0\end{array}\right]$

Comparing the corresponding elements

$2 a-4=0

$ \Rightarrow 2 a=4 \Rightarrow a=2$

2 a-2 b=-2 

$\Rightarrow 2 \times 2-2 b=-2$

$\Rightarrow 4-2 b=-2 $

$\Rightarrow-2 b=-2-4=-6$

$\Rightarrow b=3$

Hence a=2, b=3


Question 13

If $A=\left[\begin{array}{cc}\sec 60^{\circ} & \cos 90^{\circ} \\ -3 \tan A 5^{\circ} & \sin 90^{\circ}\end{array}\right]$ and $B=\left[\begin{array}{cc}0 & \cos 45^{\circ} \\ -2 & 3 \sin 90^{\circ}\end{array}\right]$

Find (i) 2A – 3B (ii) A² (iii) BA

Sol :

Given

$A=\left[\begin{array}{cc}\sec 60^{\circ} & \cos 90^{\circ} \\ -3 \tan A 5^{\circ} & \sin 90^{\circ}\end{array}\right]$ and
$B=\left[\begin{array}{cc}0 & \cos 45^{\circ} \\ -2 & 3 \sin 90^{\circ}\end{array}\right]$

$A=\left[\begin{array}{ll}\sec 60^{\circ} & \cos 90^{\circ} \\ -3 \tan 45^{\circ} & \sin 90^{\circ}\end{array}\right]=\left[\begin{array}{rr}2 & 0 \\ -3 & 1\end{array}\right]$

$\left(\because \sec 60^{\circ}\right)=2, \cos 90^{\circ}=0, \tan 45^{\circ}=1 \left.\sin 90^{\circ}=1\right)$

$B=\left[\begin{array}{rr}0 & \cot 45^{\circ} \\ -2 & 3 \sin 90^{\circ}\end{array}\right]=\left[\begin{array}{rr}0 & 1 \\ -2 & 3\end{array}\right]$

$\left(\because \cot 45^{\circ}=1\right)$

(i) 2A-3B

$=2\left[\begin{array}{rr}2 & 0 \\ -3 & 1\end{array}\right]-3\left[\begin{array}{rr}0 & 1 \\ -2 & 3\end{array}\right]$

$=\left[\begin{array}{rr}4 & 0 \\ -6 & 2\end{array}\right]-\left[\begin{array}{rr}0 & 3 \\ -6 & 9\end{array}\right]$

$=\left[\begin{array}{rr}4-0 & 0-3 \\ -6+6 & 2-9\end{array}\right]=\left[\begin{array}{ll}4 & -3 \\ 0 & -7\end{array}\right]$


(ii) $A^{2}=A \times A=\left[\begin{array}{rr}2 & 0 \\ -3 & 1\end{array}\right]\left[\begin{array}{rr}2 & 0 \\ -3 & 1\end{array}\right]$

$=\left[\begin{array}{rr}4+0 & 0+0 \\ -6-3 & 0+1\end{array}\right]=\left[\begin{array}{rr}4 & 0 \\ -9 & 1\end{array}\right]$


(iii) $\mathrm{BA}=\left[\begin{array}{rr}0 & 1 \\ -2 & 3\end{array}\right]\left[\begin{array}{rr}2 & 0 \\ -3 & 1\end{array}\right]$

$=\left[\begin{array}{rr}0-3 & 0+1 \\ -4-9 & 0+3\end{array}\right]=\left[\begin{array}{ll}-3 & 1 \\ -13 & 3\end{array}\right]$

Comments

Post a Comment

Popular posts from this blog

ML Aggarwal Solution Class 10 Chapter 15 Circles Exercise 15.1

Image
 Exercise 15.1 Question 1 Using the given information, find the value of x in each of the following figures : Sol : (i) ∠ADB and ∠ACB are in the same segment. ∠ADB = ∠ACB = 50° Now in ∆ADB, ∠DAB + X + ∠ADB = 180° (Angles of a triangles) $\Rightarrow 42^{\circ}+x+50^{\circ}=180^{\circ}$ $\Rightarrow x+92^{\circ}=180^{\circ}$ $\Rightarrow x=180^{\circ}-92^{\circ}=88^{\circ}$ (ii) ∠ABD=∠ACD (Angles in the same segment) But ∠ACD=32° ∴∠ABD=32° Now in ΔABD ∠ABD+∠ADB+∠DAB=180° ⇒32°+45°+x=180° ⇒77°+x=180° ⇒x=180°-77°=103° (iii) ∠BAD=∠BCD (Angles in the same segement) But ∠BAD=20° $\therefore \angle B C D=20^{\circ}$ $\because \angle \mathrm{CEA}=90^{\circ}$ $\therefore \angle \mathrm{CED}=90^{\circ}$ Now in $\Delta$ CED, $\angle C E D+\angle B C D+\angle C D E=180^{\circ}$ $\Rightarrow 90^{\circ}+20^{\circ}+x=180^{\circ}$ $\Rightarrow 110^{\circ}+x=180^{\circ}$ $\Rightarrow x=180^{\circ}-110^{\circ}=70^{\circ}$ (iv) In ΔABC ∠ABC+∠ACB+∠BAC=180° (∵Sum of angles of a triangle) $\Rightarrow 69...
Read more

ML Aggarwal Solution Class 9 Chapter 20 Statistics Exercise 20.2

  Exercise 20.2 Question 1 Sol : (i) discrete  (ii) continuous  (iii) discrete  (iv) continuous  (v) continuous  Question 2 Sol : Given data  13, 6, 10 ,5 ,11 ,14 ,2 ,8 ,15 ,16 ,9 ,13 ,17 ,11, 19 ,5 ,7 ,12 ,20 ,21 ,18 ,1 ,8 ,12 ,18 Classes 0-4 5-9 10-14 15-9 20-24 Frequency 2 7 8 6 2 Question 3 Sol : (i)  Classes 1-10 11-20 21-30 31-40 Frequency 8 7 6 6 (ii) Range = Maximum value - Minimum value  =40-2=38 (iii) Third class of frequency table $=\frac{21+30}{2}=\frac{51}{2}$ =25.5 Question 4 Sol : Variate : A particular value of a variable is called variate Class size :  The difference be...
Read more

ML Aggarwal Solution Class 9 Chapter 3 Expansions Exercise 3.2

Exercise 3.2 Question 1 Sol : Given : x-y=8..(i) xy=5...(ii) Squaring both sides in equation (i) ∴(x-y) 2 =8 2 ⇒x 2 -2.xy+y 2 =64 ⇒x 2 -2(5)+y 2 =64 ⇒x 2 +y 2 =64+10 ⇒x 2 +y 2 =74 Question 2 Sol : ⇒Given : x+y=10...(i) ⇒xy=21....(ii) Squaring both sides in equation (i) ⇒(x+y) 2 =10 2 ⇒x 2 +2xy+y 2 =100 ⇒x 2 +2(21)+y 2 =100 ⇒x 2 +4 2 +y 2 =100 ⇒x 2 +y 2 =100-42 ⇒x 2 +y 2 =58 2(x 2 +y 2 )=2×58=116 Question 3 Sol : ⇒Given : 2a+3b=7..(i) ⇒ab=2....(ii) Squaring both sides in equal (i) ⇒(2a+3b) 2 =7 2 ⇒(2a) 2 +2.2a+3b+(3b) 2 =49 ⇒4a 2 +12ab+9b 2 =49 ⇒4a 2 +12(2)+9b 2 =49 ⇒4a 2 +24+9b 2 =49 ⇒4a 2 +9b 2 =49-24 ⇒4a 2 +9b 2 =25 Question 4 Sol : ⇒Given : 3x-4y=16...(i) ⇒xy=4..(ii) Squaring on both sides in equation (i) ⇒(3x-4y) 2 =16 2 ⇒(3x) 2 -2.3x.4y+(4y) 2 =256 ⇒9x 2 -24xy+16y 2 =256 ⇒9x 2 -24(4)+16y 2 =256 ⇒9x 2 -96+16y 2 =256 ⇒9x 2 +16y 2 =256+96 ⇒9x 2 +16y 2 =352 Question 5 Sol : ⇒Given : x+y=8...(i) ⇒x-y=2...(ii) Since we know that ⇒2(x) 2 +2(y) 2 =(x+y) 2 +(x-y) 2 From (i) ⇒S...
Read more