ML Aggarwal Solution Class 10 Chapter 8 Matrices Exercise 8.3

 Exercise 8.3

Question 1

If $A=\left[\begin{array}{cc}3 & 5 \\ 4 & -2\end{array}\right]$ and $B=\left[\begin{array}{l}2 \\ 4\end{array}\right]$ , is the product AB possible ? Give a reason if yes, find AB

Sol :

Yes, the product is possible because of

number of column in A = number of row in B

i.e., (2 x 2). (2 x 1) = (2 x 1) is the order of the matrix.

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

$=\left[\begin{array}{c}3 \times 2+5 \times 4 \\ 4 \times 2+(-2) \times 4\end{array}\right]$

$=\left[\begin{array}{c}6+20 \\ 8-8\end{array}\right]=\left[\begin{array}{c}26 \\ 0\end{array}\right]$

Question 2

If $A=\left[\begin{array}{ll}2 & 5 \\ 1 & 3\end{array}\right], B=\left[\begin{array}{cc}1 & -1 \\ -3 & 2\end{array}\right]$, find AB and BA, Is AB = BA ?

Sol :

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

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

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

$=\left[\begin{array}{rr}2-15 & -2+10 \\ 1-9 & -1+6\end{array}\right]=\left[\begin{array}{rr}-13 & 8 \\ -8 & 5\end{array}\right]$

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

$=\left[\begin{array}{rr}2-1 & 5-3 \\ -6+2 & -15+6\end{array}\right]$

$=\left[\begin{array}{rr}1 & 2 \\ -4 & -9\end{array}\right]$

Hence AB≠BA

Question 3

If $\mathrm{P}=\left[\begin{array}{cc}4 & 6 \\ 2 & -8\end{array}\right], \mathrm{Q}=\left[\begin{array}{cc}2 & -3 \\ -1 & 1\end{array}\right]$

Find 2PQ

Sol :

$P=\left[\begin{array}{cc}4 & 6 \\ 2 & -8\end{array}\right]$

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

$2 P Q=2\left[\begin{array}{rr}4 & 6 \\ 2 & -8\end{array}\right] \times\left[\begin{array}{lr}2 & -3 \\ -1 & 1\end{array}\right]$

$=2\left[\begin{array}{cc}8-6 & -12+6 \\ 4+8 & -6-8\end{array}\right]$

$=2\left[\begin{array}{cc}8 & -6 \\ 12 & -14\end{array}\right]$

$=\left[\begin{array}{cc}4 & -12 \\ 24 & -28\end{array}\right]$

Question 4

Given $A=\left[\begin{array}{ll}1 & 1 \\ 8 & 3\end{array}\right]$ , evaluate $A^{2}-4 A$

Sol :

$A=\left[\begin{array}{ll}1 & 1 \\ 8 & 3\end{array}\right]$

$A^{2}-4 A=\left[\begin{array}{ll}1 & 1 \\ 8 & 3\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 8 & 3\end{array}\right]-4\left[\begin{array}{ll}1 & 1 \\ 8 & 3\end{array}\right]$

$=\left[\begin{array}{cc}1+8 & 1+3 \\ 8+24 & 8+9\end{array}\right]-\left[\begin{array}{cc}4 & 4 \\ 32 & 12\end{array}\right]$

$=\left[\begin{array}{cc}9 & 4 \\ 32 & 17\end{array}\right]-\left[\begin{array}{cc}4 & 4 \\ 32 & 12\end{array}\right]$

$=\left[\begin{array}{cccc}9 & -4 & 4 & -4 \\ 32 & -32 & 17 & -12\end{array}\right]$

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

Question 5

If $A=\left[\begin{array}{ll}3 & 7 \\ 2 & 4\end{array}\right], B=\left[\begin{array}{ll}0 & 2 \\ 5 & 3\end{array}\right]$ and $C=\left[\begin{array}{cc}1 & -5 \\ -4 & 6\end{array}\right]$

Find AB – 5C

Sol :

$A=\left[\begin{array}{ll}3 & 7 \\ 2 & 4\end{array}\right], B=\left[\begin{array}{ll}0 & 2 \\ 5 & 3\end{array}\right]$ and $C=\left[\begin{array}{cc}1 & -5 \\ -4 & 6\end{array}\right]$

$A B=\left[\begin{array}{ll}3 & 7 \\ 2 & 4\end{array}\right]\left[\begin{array}{ll}0 & 2 \\ 5 & 3\end{array}\right]$

$=\left[\begin{array}{ll}3 \times 0+7 \times 5 & 3 \times 2+7 \times 3 \\ 2 \times 0+4 \times 5 & 2 \times 2+4 \times 3\end{array}\right]$

$=\left[\begin{array}{ll}0+35 & 6+21 \\ 0+20 & 4+12\end{array}\right]=\left[\begin{array}{ll}35 & 27 \\ 20 & 16\end{array}\right]$

$5 \mathrm{C}=5\left[\begin{array}{cc}1 & -5 \\ -4 & 6\end{array}\right]=\left[\begin{array}{cc}5 & -25 \\ -20 & 30\end{array}\right]$

$\mathrm{AB}-5 \mathrm{C}=\left[\begin{array}{cc}35 & 27 \\ 20 & 16\end{array}\right]-\left[\begin{array}{cc}5 & -25 \\ -20 & 30\end{array}\right]$

$=\left[\begin{array}{cc}30 & 52 \\ 40 & -14\end{array}\right]$

Question 6

If $A=\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right]$ and $B=\left[\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right]$ , find A(BA)

Sol :

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

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

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

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

$\mathrm{A}(\mathrm{BA})=\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right] \times\left[\begin{array}{ll}4 & 5 \\ 5 & 4\end{array}\right]$

$=\left[\begin{array}{cc}4+10 & 5+8 \\ 8+5 & 10+4\end{array}\right]=\left[\begin{array}{cc}14 & 13 \\ 13 & 14\end{array}\right]$

Question 7

Given matrices:

$A=\left[\begin{array}{ll}2 & 1 \\ 4 & 2\end{array}\right]$ and $B=\left[\begin{array}{cc}3 & 4 \\ -1 & -2\end{array}\right], C=\left[\begin{array}{cc}-3 & 1 \\ 0 & -2\end{array}\right]$

Find the products of (i) ABC (ii) ACB and state whether they are equal.

Sol :

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

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

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

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

$=\left[\begin{array}{rr}6-1 & 8-2 \\ 12-2 & 16-4\end{array}\right]\left[\begin{array}{rr}-3 & 1 \\ 0 & -2\end{array}\right]$

$=\left[\begin{array}{rr}5 & 6 \\ 10 & 12\end{array}\right] \times\left[\begin{array}{rr}-3 & 1 \\ 0 & -2\end{array}\right]$

$=\left[\begin{array}{rr}-15+0 & 5-12 \\ -30+0 & 10-24\end{array}\right]=\left[\begin{array}{rr}-15 & -7 \\ -30 & -14\end{array}\right]$

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

$=\left[\begin{array}{rr}-6+0 & 2-2 \\ -12+0 & 4-4\end{array}\right] \times\left[\begin{array}{rr}3 & 4 \\ -1 & -2\end{array}\right]$

$=\left[\begin{array}{rr}-6 & 0 \\ -12 & 0\end{array}\right] \times\left[\begin{array}{rr}3 & 4 \\ -1 & -2\end{array}\right]$

$=\left[\begin{array}{ll}-18+0 & -24+0 \\ -36+0 & -48+0\end{array}\right]=\left[\begin{array}{ll}-18 & -24 \\ -36 & -48\end{array}\right]$

$\therefore \mathrm{ABC} \neq \mathrm{ACB}$

Question 8

Evaluate: $\left[\begin{array}{cc}4 \sin 30^{\circ} & 2 \cos 60^{\circ} \\ \sin 90^{\circ} & 2 \cos 0^{\circ}\end{array}\right]\left[\begin{array}{ll}4 & 5 \\ 5 & 4\end{array}\right]$

Sol :

$\left[\begin{array}{cc}4 \sin 30^{\circ} & 2 \cos 60^{\circ} \\ \sin 90^{\circ} & 2 \cos 0^{\circ}\end{array}\right]\left[\begin{array}{ll}4 & 5 \\ 5 & 4\end{array}\right]$

$\sin 30^{\circ}=\frac{1}{2}, \cos 60^{\circ}=\frac{1}{2}$

$\therefore\left[\begin{array}{cc}4 \times \frac{1}{2} & 2 \times \frac{1}{2} \\ 1 & 2 \times 1\end{array}\right]\left[\begin{array}{cc}4 & 5 \\ 5 & 4\end{array}\right]$

$=\left[\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right]\left[\begin{array}{ll}4 & 5 \\ 5 & 4\end{array}\right]$

$=\left[\begin{array}{ll}2 \times 4+1 \times 5 & 2 \times 5+1 \times 4 \\ 1 \times 4+2 \times 5 & 1 \times 5+2 \times 4\end{array}\right]$

$=\left[\begin{array}{cc}8+5 & 10+4 \\ 4+10 & 5+8\end{array}\right]=\left[\begin{array}{cc}13 & 14 \\ 14 & 13\end{array}\right]$

Question 9

If $A=\left[\begin{array}{cc}-1 & 3 \\ 2 & 4\end{array}\right], B=\left[\begin{array}{cc}2 & -3 \\ -4 & -6\end{array}\right]$ find the matrix AB + BA

Sol :

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

$B=\left[\begin{array}{cc}2 & -3 \\ -4 & -6\end{array}\right]$

$A B=\left[\begin{array}{cc}-1 & 3 \\ 2 & 4\end{array}\right] \times\left[\begin{array}{cc}2 & -3 \\ -4 & -6\end{array}\right]$

$=\left[\begin{array}{rr}-2-12 & 3-18 \\ 4-16 & -6-24\end{array}\right]=\left[\begin{array}{ll}-14 & -15 \\ +12 & -30\end{array}\right]$

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

$=\left[\begin{array}{rr}-2-6 & 6-12 \\ 4-12 & -12-24\end{array}\right]=\left[\begin{array}{rr}-8 & -6 \\ -8 & -36\end{array}\right]$

$\therefore A B+B A$

$=\left[\begin{array}{ll}-14 & -15 \\ -12 & -30\end{array}\right]+\left[\begin{array}{ll}-8 & -6 \\ -8 & -36\end{array}\right]$

$=\left[\begin{array}{rr}-14-8 & -15-6 \\ -12-8 & -30-36\end{array}\right]=\left[\begin{array}{ll}-22 & -21 \\ -20 & -66\end{array}\right]$

Question 10

$A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$ and $B=\left[\begin{array}{ll}6 & 1 \\ 1 & 1\end{array}\right], C=\left[\begin{array}{cc}-2 & -3 \\ 0 & 1\end{array}\right]$

find each of the following and state if they are equal.

(i) CA + B

(ii) A + CB

Sol :

(i) CA + B

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

$=\left[\begin{array}{rr}-11+6 & -16+1 \\ 3+1 & 4+1\end{array}\right]=\left[\begin{array}{rr}-5 & -15 \\ 4 & 5\end{array}\right]$

(ii) A+CB

$=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]+\left[\begin{array}{rr}-2 & -3 \\ 0 & 1\end{array}\right] \times\left[\begin{array}{ll}6 & 1 \\ 1 & 1\end{array}\right]$

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

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

$=\left[\begin{array}{rr}1-15 & 2-5 \\ 3+1 & 4+1\end{array}\right]=\left[\begin{array}{rr}-14 & -3 \\ 4 & 5\end{array}\right]$

We can say that CA+B≠A+CB

Question 11

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

Find 2B – A²

Sol :

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

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

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

$=\left[\begin{array}{cc}6 & 4 \\ -4 & 2\end{array}\right]$

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

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

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

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

$=\left[\begin{array}{rr}9 & 4 \\ -4 & 5\end{array}\right]$

Question 12

If $\mathrm{A}=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{ll}2 & 1 \\ 4 & 2\end{array}\right], \mathrm{C}=\left[\begin{array}{ll}5 & 1 \\ 7 & 4\end{array}\right],$ compute

(i) A(B + C)

(ii) (B + C)A

Sol :

(i) A(B + C)

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

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

$C=\left[\begin{array}{ll}5 & 1 \\ 7 & 4\end{array}\right]$

$A(B+C)=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]\left[\begin{array}{ll}2 & 1 \\ 4 & 2\end{array}\right]+\left[\begin{array}{ll}5 & 1 \\ 7 & 4\end{array}\right]$

$=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]\left[\begin{array}{ll}2+5 & 1+1 \\ 4+7 & 2+4\end{array}\right]$

$=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]\left[\begin{array}{rr}7 & 2 \\ 11 & 6\end{array}\right]$

$=\left[\begin{array}{rr}7+22 & 2+12 \\ 21+44 & 6+24\end{array}\right]=\left[\begin{array}{ll}29 & 14 \\ 65 & 30\end{array}\right]$

$(B+C) A=\left[\begin{array}{rr}7 & 2 \\ 11 & 6\end{array}\right] \times\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$

$=\left[\begin{array}{rr}7+6 & 14+8 \\ 11+18 & 22+24\end{array}\right]=\left[\begin{array}{rr}13 & 22 \\ 29 & 46\end{array}\right]$

Question 13

If $\mathrm{A}=\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{ll}2 & 1 \\ 3 & 2\end{array}\right] ,\mathrm{C}=\left[\begin{array}{ll}1 & 3 \\ 3 & 1\end{array}\right]$

find the matrix C(B – A)

Sol :

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

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

$\mathrm{C}=\left[\begin{array}{ll}1 & 3 \\ 3 & 1\end{array}\right]$

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

$C(B-A)=\left[\begin{array}{ll}1 & 3 \\ 3 & 1\end{array}\right] \times\left[\begin{array}{ll}1 & -1 \\ 1 & -1\end{array}\right]$

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

Question 14

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

Find A² + AB + B²

Sol :

Given that

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

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

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

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

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

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

$=\left[\begin{array}{ll}1 \times 2+0 \times-1 & 1 \times 3+0 \times 0 \\ 2 \times 2+1 \times-1 & 2 \times 3+1 \times 0\end{array}\right]=\left[\begin{array}{ll}2 & 3 \\ 3 & 6\end{array}\right]$

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

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

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

$A^{2}+A B+B^{2}=\left[\begin{array}{ll}1 & 0 \\ 4 & 1\end{array}\right]+\left[\begin{array}{ll}2 & 3 \\ 3 & 6\end{array}\right]+\left[\begin{array}{cc}1 & 6 \\ -2 & -3\end{array}\right]$

$=\left[\begin{array}{cc}1+2+1 & 0+3+6 \\ 4+3-2 & 1+6+-3\end{array}\right]=\left[\begin{array}{ll}4 & 9 \\ 5 & 4\end{array}\right]$

Question 15

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

Find A² + AC – 5B

Sol :

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

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

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

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

(Subsitituting the values from question)

$=\left[\begin{array}{cc}4+0 & 2-2 \\ 0+0 & 0+4\end{array}\right]+\left[\begin{array}{cc}-6-1 & 4+4 \\ 0+2 & 0-8\end{array}\right]-5\left[\begin{array}{cc}4 & 1 \\ -3 & -2\end{array}\right]$

$=\left[\begin{array}{ll}4 & 0 \\ 0 & 4\end{array}\right]+\left[\begin{array}{cc}-7 & 8 \\ 2 & -8\end{array}\right]-\left[\begin{array}{cc}20 & 5 \\ -15 & -10\end{array}\right]$

$=\left[\begin{array}{cc}4-7-20 & 0+8-5 \\ 0+2+15 & 4-8+10\end{array}\right]=\left[\begin{array}{cc}-23 & 3 \\ 17 & 6\end{array}\right]$

Question 16

If $A=\left[\begin{array}{cc}1 & 0 \\ 0 & -1\end{array}\right]$ find $A^{2}$ and $A^{3}$ Also state that which of these is equal to A

Sol :

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

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

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

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

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

From above, it is clear that $A^{3}=A$

Question 17

If $X=\left[\begin{array}{cc}4 & 1 \\ -1 & 2\end{array}\right]$, show that 6X – X² = 9I Where I is the unit matrix.

Sol :

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

$x^{2}=x \times x=\left[\begin{array}{rr}4 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{rr}4 & 1 \\ -1 & 2\end{array}\right]$

$=\left[\begin{array}{rr}16-1 & 4+2 \\ -4-2 & -1+4\end{array}\right]=\left[\begin{array}{rr}15 & 6 \\ -6 & 3\end{array}\right]$

L.H.S. $6 \mathrm{X}-\mathrm{X}^{2}=6\left[\begin{array}{rr}4 & 1 \\ -1 & 2\end{array}\right]-\left[\begin{array}{rr}15 & 6 \\ -6 & 3\end{array}\right]$

$=\left[\begin{array}{rr}24 & 6 \\ -6 & 12\end{array}\right]-\left[\begin{array}{rr}15 & 6 \\ -6 & 3\end{array}\right]$

$=\left[\begin{array}{rr}24-15 & 6-6 \\ -6 & -6 & 12-3\end{array}\right]=\left[\begin{array}{ll}9 & 0 \\ 0 & 9\end{array}\right]$

$=9\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=9 \mathrm{L}=\mathrm{R} . \mathrm{H.S}$

Hence proved

Question 18

Show that $\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right]$  is a solution of the matrix equation X² – 2X – 3I = 0,Where I is the unit matrix of order 2

Sol :

X² – 2X – 3I = 0

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

or 

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

$\therefore X^{2}=\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right]$

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

Now $X^{2}-2 X-3 I$

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

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

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

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

$\therefore X^{2}-2 X-3 I=0$

Hence proved

Question 19

Find the matrix X of order 2 × 2 which satisfies the equation

$\left[\begin{array}{ll}3 & 7 \\ 2 & 4\end{array}\right]\left[\begin{array}{ll}0 & 2 \\ 5 & 3\end{array}\right]+2 X=\left[\begin{array}{cc}1 & -5 \\ -4 & 6\end{array}\right]$

Sol :

$\left[\begin{array}{ll}3 & 7 \\ 2 & 4\end{array}\right]\left[\begin{array}{ll}0 & 2 \\ 5 & 3\end{array}\right]+2 X=\left[\begin{array}{cc}1 & -5 \\ -4 & 6\end{array}\right]$

$\Rightarrow\left[\begin{array}{ll}0+35 & 6+21 \\ 0+20 & 4+12\end{array}\right]+2 X=\left[\begin{array}{rr}1 & -5 \\ -4 & 6\end{array}\right]$

$\Rightarrow\left[\begin{array}{ll}35 & 27 \\ 20 & 16\end{array}\right]+2 X=\left[\begin{array}{rr}1 & -5 \\ -4 & 6\end{array}\right]$

$2 X=-\left[\begin{array}{ll}35 & 27 \\ 20 & 16\end{array}\right]+\left[\begin{array}{rr}1 & -5 \\ -4 & -6\end{array}\right]=\left[\begin{array}{ll}-34 & -32 \\ -24 & -10\end{array}\right]$

$X=\frac{1}{2}\left[\begin{array}{ll}-34 & -32 \\ -24 & -10\end{array}\right]=\left[\begin{array}{rr}-17 & -16 \\ -12 & -5\end{array}\right]$

Question 20

If $A=\left[\begin{array}{ll}1 & 1 \\ x & x\end{array}\right]$, find the value of x, so that A² – 0

Sol :

Given :

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

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

$=\left[\begin{array}{rr}1+x & 1+x \\ x+x^{2} & x+x^{2}\end{array}\right]$

$\because \quad A^{2}=0$

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

Comparing $1+x=0 \Rightarrow x=-1$

Question 21

If $\left[\begin{array}{ll}1 & 3 \\ 0 & 0\end{array}\right]\left[\begin{array}{c}2 \\ -1\end{array}\right]=\left[\begin{array}{l}x \\ 0\end{array}\right]$ Find the value of $x$

Sol :

$\left[\begin{array}{ll}1 & 3 \\ 0 & 0\end{array}\right]\left[\begin{array}{c}2 \\ -1\end{array}\right]=\left[\begin{array}{l}x \\ 0\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}2 & -3 \\ 0 & 0\end{array}\right]=\left[\begin{array}{l}x \\ 0\end{array}\right]$

$\Rightarrow\left[\begin{array}{c}-1 \\ 0\end{array}\right]=\left[\begin{array}{l}x \\ 0\end{array}\right]$

Comparing the corresponding elements

x=-1

Question 22

(i) Find x and y if $\left[\begin{array}{cc}-3 & 2 \\ 0 & -5\end{array}\right]\left[\begin{array}{l}x \\ 2\end{array}\right]=\left[\begin{array}{c}-5 \\ y\end{array}\right]$

(ii) Find x and y if $\left[\begin{array}{cc}2 x & x \\ y & 3 y\end{array}\right]\left[\begin{array}{l}3 \\ 2\end{array}\right]=\left[\begin{array}{c}16 \\ 9\end{array}\right]$

Sol :

(i) $\left[\begin{array}{cc}-3 & 2 \\ 0 & -5\end{array}\right]\left[\begin{array}{l}x \\ 2\end{array}\right]=\left[\begin{array}{c}-5 \\ y\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}-3 x & 4 \\ 0 & -10\end{array}\right]=\left[\begin{array}{c}-5 \\ y\end{array}\right]$

Comparing the corresponding elements 

-3x+4=-5 

$\Rightarrow-3 x=-5-4=-9$

$\therefore x=\frac{-9}{-3}=3$

-10=y 

$\Rightarrow y=-10$

Hence , x=3, y=-10

(ii) $\left[\begin{array}{cc}2 x & x \\ y & 3 y\end{array}\right]\left[\begin{array}{l}3 \\ 2\end{array}\right]=\left[\begin{array}{c}16 \\ 9\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}2 x \times 3+x \times 2 \\ y \times 3+3 y \times 2\end{array}\right]=\left[\begin{array}{c}16 \\ 9\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}6 x+2 x \\ 3 y+6 y\end{array}\right]=\left[\begin{array}{c}16 \\ 9\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}8 x \\ 9 y\end{array}\right]=\left[\begin{array}{c}16 \\ 9\end{array}\right]$

Comparing, we get

8x=16 

$\Rightarrow x=\frac{16}{8}=2$

and 9y=9 

$\Rightarrow y=\frac{9}{9}=1$

Here x=2, y=1

Question 23

Find x and y if

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

Sol :

Given

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

$\Rightarrow \left[\begin{array}{rr}2 x+2 y & -y \\ 4 x & -x+y\end{array}\right]=\left[\begin{array}{l}3 \\ 2\end{array}\right]$

$\Rightarrow \left[\begin{array}{ll}2 x & +y \\ 3 x & +y\end{array}\right]=\left[\begin{array}{l}3 \\ 2\end{array}\right]$

Question 24

If $\left[\begin{array}{ll}1 & 2 \\ 3 & 3\end{array}\right]\left[\begin{array}{ll}x & 0 \\ 0 & y\end{array}\right]=\left[\begin{array}{ll}x & 0 \\ 9 & 0\end{array}\right]$  find the values of x and y

Sol :

Given :

$\left[\begin{array}{ll}1 & 2 \\ 3 & 3\end{array}\right]\left[\begin{array}{ll}x & 0 \\ 0 & y\end{array}\right]=\left[\begin{array}{ll}x & 0 \\ 9 & 0\end{array}\right]$

$\Rightarrow \left[\begin{array}{rr}x+0 & 0+2 y \\ 3 x+0 & 0+3 y\end{array}\right]=\left[\begin{array}{ll}x & 0 \\ 9 & 0\end{array}\right]$

$\Rightarrow \left[\begin{array}{rr}x & 2 y \\ 3 x & 3 y\end{array}\right]=\left[\begin{array}{ll}x & 0 \\ 9 & 0\end{array}\right]$

Question 25

If $\left[\begin{array}{ll}3 & 4 \\ 2 & 5\end{array}\right]=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$ write down the values of a, b, c and d

Sol :

Given

$\left[\begin{array}{ll}3 & 4 \\ 2 & 5\end{array}\right]=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

$\Rightarrow \left[\begin{array}{ll}3 & 4 \\ 2 & 5\end{array}\right]=\left[\begin{array}{ll}a+0 & 0+b \\ c+0 & 0+d\end{array}\right]$

$\Rightarrow \left[\begin{array}{ll}3 & 4 \\ 2 & 5\end{array}\right]=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$

Comparing the corresponding elements

a = 3, b = 4, c = 2, d = 5

Question 26

Find the value of x given that A² = B

Where $A=\left[\begin{array}{cc}2 & 12 \\ 0 & 1\end{array}\right]$ and

$B=\left[\begin{array}{ll}4 & x \\ 0 & 1\end{array}\right]$

Sol :

$A=\left[\begin{array}{ll}2 & 12 \\ 0 & 1\end{array}\right]$ and

$B=\left[\begin{array}{ll}4 & x \\ 0 & 1\end{array}\right]$

$A^{2}=B$

$\Rightarrow\left[\begin{array}{ll}2 & 12 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}2 & 12 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}4 & x \\ 0 & 1\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}2 \times 2+12 \times 0 & 2 \times 12+12 \times 1 \\ 0 \times 2+1 \times 0 & 0 \times 12+1 \times 1\end{array}\right]=\left[\begin{array}{ll}4 & x \\ 0 & 1\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}4+0 & 24+12 \\ 0+0 & 0+1\end{array}\right]=\left[\begin{array}{ll}4 & x \\ 0 & 1\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}4 & 36 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}4 & x \\ 0 & 1\end{array}\right]$

Comparing the corresponding elements of two equal matrices, x=36

Question 27

If $A=\left[\begin{array}{ll}2 & x \\ 0 & 1\end{array}\right]$ and $B=\left[\begin{array}{cc}4 & 36 \\ 0 & 1\end{array}\right],$ find the value of $x,$ given that $A^{2}-B$

Sol :

Given :

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

$=\left[\begin{array}{rr}4+0 & 2 x+x \\ 0+0 & 0+1\end{array}\right]=\left[\begin{array}{rr}4 & 3 x \\ 0 & 1\end{array}\right]$

$\because A^{2}=B$

$\therefore\left[\begin{array}{rr}4 & 3 x \\ 0 & 1\end{array}\right]=\left[\begin{array}{rr}4 & 36 \\ 0 & 1\end{array}\right]$

Corresponding the corresponding elements

3 x=36 

$\Rightarrow x=12$

Hence x=12

Question 28

If $A=\left[\begin{array}{ll}3 & x \\ 0 & 1\end{array}\right]$ and $B=\left[\begin{array}{cc}9 & 16 \\ 0 & -y\end{array}\right]$ find $x$ and $y$ when $A^{2}=B$

Sol :

Given : 

$A=\left[\begin{array}{ll}3 & x \\ 0 & 1\end{array}\right]$ and $B=\left[\begin{array}{ll}9 & 16 \\ 0 & -y\end{array}\right]$  find x and y when A² = B

Now, $A^{2}=A \times A$

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

$=\left[\begin{array}{cc}9 & 3 x+x \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}9 & 4 x \\ 0 & 1\end{array}\right]$

We have $A^{2}=B$ 

Two matrices are equal if each and every corresponding element is equal.

Thus, $\left[\begin{array}{cc}9 & 4 x \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}9 & 16 \\ 0 & -y\end{array}\right]$

$\Rightarrow 4 x=16$ and $1=-y$

$\Rightarrow x=4$ and $y=-1$

Question 29

Find x, y if $\left[\begin{array}{cc}-2 & 0 \\ 3 & 1\end{array}\right]\left[\begin{array}{c}-1 \\ 2 x\end{array}\right]+3\left[\begin{array}{c}-2 \\ 1\end{array}\right]=2\left[\begin{array}{l}y \\ 3\end{array}\right]$

Sol :

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

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

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

$\Rightarrow\left[\begin{array}{c}-4 \\ 2 x\end{array}\right]=\left[\begin{array}{c}2 y \\ 6\end{array}\right]$

$\Rightarrow 2 x=6$ and $2 y=-4$

$\Rightarrow x=\frac{6}{2}$ and $y=-\frac{4}{2}$

$\Rightarrow x=3$ and $\Rightarrow y=-2$

Question 30

If $\left[\begin{array}{ll}a & 1 \\ 1 & 0\end{array}\right]\left[\begin{array}{cc}4 & 3 \\ -3 & 2\end{array}\right]=\left[\begin{array}{cc}b & 11 \\ 4 & c\end{array}\right]$ find a,b and c

Sol :

$\left[\begin{array}{ll}a & 1 \\ 1 & 0\end{array}\right]\left[\begin{array}{cc}4 & 3 \\ -3 & 2\end{array}\right]=\left[\begin{array}{cc}b & 11 \\ 4 & c\end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}4 a-3 & 3 a+2 \\ 4+0 & 3+0\end{array}\right]=\left[\begin{array}{cc}b & 11 \\ 4 & c\end{array}\right]$

Comparing the corresponding elements

3a+2=11 

$\Rightarrow 3 a=11-2=9$

$\therefore a=\frac{9}{3}=3$

4a-3=b 

$\Rightarrow b=4 \times 3-3=12-3=9$

3=c

Question 31

If $A=\left[\begin{array}{cc}1 & 4 \\ 0 & -1\end{array}\right], B=\left[\begin{array}{cc}2 & x \\ 0 & -\frac{1}{2}\end{array}\right]$  find the value of x if AB = BA

Sol :

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

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

$=\left[\begin{array}{ll}2+0 & 8-x \\ 0+0 & 0+\frac{1}{2}\end{array}\right]=\left[\begin{array}{lr}2 & 8-x \\ 0 & \frac{1}{2}\end{array}\right]$

$\because \mathrm{AB}=\mathrm{BA}$

$\therefore\left[\begin{array}{rr}2 & x-2 \\ 0 & \frac{1}{2}\end{array}\right]=\left[\begin{array}{lr}2 & 8-x \\ 0 & \frac{1}{2}\end{array}\right]$

Comparing the corresponding elements

x-2=8-x

$\Rightarrow x+x=8+2 \Rightarrow 2 x=10$

$\therefore x=\frac{10}{2}=5$

Question 32

If $A=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]$ find x and y so that A² – xA + yI

Sol :

Given :

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

$=\left[\begin{array}{ll}4+3 & 6+6 \\ 2+2 & 3+4\end{array}\right]=\left[\begin{array}{rr}7 & 12 \\ 4 & 7\end{array}\right]$

$\because \mathrm{A}^{2}=x \mathrm{~A}+y \mathrm{I}$

$\Rightarrow\left[\begin{array}{ll}7 & 12 \\ 4 & 7\end{array}\right]=x\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]+y\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

$\Rightarrow\left[\begin{array}{ll}7 & 12 \\ 4 & 7\end{array}\right]=\left[\begin{array}{rr}2 x & 3 x \\ x & 2 x\end{array}\right]+\left[\begin{array}{ll}y & 0 \\ 0 & y\end{array}\right]$

$=\left[\begin{array}{rr}2 x+y & 3 x \\ x & 2 x+y\end{array}\right]$

Comparing the corresponding elements $3 x=12 \Rightarrow x=4$

2x+y=7

$\Rightarrow 2 \times 4+y=7$

$\Rightarrow 8+y=7 \Rightarrow y=7-8=-1$

Hence x=4, y=-1

Question 33

If $P=\left[\begin{array}{ll}2 & 6 \\ 3 & 9\end{array}\right], Q=\left[\begin{array}{ll}3 & x \\ y & 2\end{array}\right]$

find x and y such that PQ = 0

Sol :

Given :

$P=\left[\begin{array}{ll}2 & 6 \\ 3 & 9\end{array}\right]$

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

$\mathrm{PQ}=\left[\begin{array}{ll}2 & 6 \\ 3 & 9\end{array}\right]\left[\begin{array}{ll}3 & x \\ y & 2\end{array}\right]$

$=\left[\begin{array}{ll}6+6 y & 2 x+12 \\ 9+9 y & 3 x+18\end{array}\right]$

∵PQ=0

$\therefore\left[\begin{array}{ll}6+6 y & 2 x+12 \\ 9+9 y & 3 x+18\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$

Comparing the corresponding elements 

6+6 y=0 

$\Rightarrow 6 y=-6 \Rightarrow \quad y=-1$

2x+12=0 

$\Rightarrow 2x=-12 \Rightarrow x=-6$

Hence x=-6, y=-1

Question 34

Let $M \times\left[\begin{array}{ll}1 & 1 \\ 0 & 2\end{array}\right]=\left[\begin{array}{ll}1 & 2\end{array}\right]$  where M is a matrix

(i) State the order of matrix M

(ii) Find the matrix M

Sol :

(i) M is the order of 1 x 2

let M = [x y]

$\therefore\left[\begin{array}{ll}x & y\end{array}\right] \times\left[\begin{array}{ll}1 & 1 \\ 0 & 2\end{array}\right]=\left[\begin{array}{ll}1 & 2\end{array}\right]$

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

Comparing the corresponding elements x=1 and 

x+2y=2 

$\Rightarrow 1+2 y=2$

$\Rightarrow 2 y=2-1=1 \Rightarrow y=\frac{1}{2}$

Hence $x=1, y=\frac{1}{2}$

$\therefore M=\left[1 \quad \frac{1}{2}\right]$

Question 35

Given $\left[\begin{array}{cc}2 & 1 \\ -3 & 4\end{array}\right] \quad X=\left[\begin{array}{l}7 \\ 6\end{array}\right]$

(i) the order of the matrix X

(ii) the matrix X

Sol :

We have

$\left[\begin{array}{cc}2 & 1 \\ -3 & 4\end{array}\right], X=\left[\begin{array}{l}7 \\ 6\end{array}\right]$

(i) We see that 2×2×2×1=2×1

The order of X is 2×1

(ii) Let $X=\left[\begin{array}{l}x \\ y\end{array}\right]$

So, $\left[\begin{array}{cc}2 & 1 \\ -3 & 4\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}7 \\ 6\end{array}\right] \Rightarrow\left[\begin{array}{c}2 x+y \\ -3 x+4 y\end{array}\right]=\left[\begin{array}{l}7 \\ 6\end{array}\right]$

2x+y=7..(i)

-3x+4y=6...(ii)

Multiplying (i) by 3 and (ii) by 2, and adding we get :

$\begin{array}{r}6 x+3 y=21 \\-6 x+8 y=12 \\\hline 11 y=33\end{array}$

y=3

From (i) , 2x=7-3=4

x=2

So, $X=\left[\begin{array}{l}2 \\ 3\end{array}\right]$

Question 36

Solve the matrix equation : $\left[\begin{array}{l}4 \\ 1\end{array}\right] , \mathrm{X}=\left[\begin{array}{ll}-4 & 8 \\ -1 & 2\end{array}\right]$

Sol :

$\left[\begin{array}{l}4 \\ 1\end{array}\right], X=\left[\begin{array}{ll}-4 & 8 \\ -1 & 2\end{array}\right]$

Let matrix X = [x y]

$\therefore\left[\begin{array}{l}4 \\ 1\end{array}\right][x, y]=\left[\begin{array}{ll}-4 & 8 \\ -1 & 2\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}4 x & 4 y \\ x & y\end{array}\right]=\left[\begin{array}{ll}-4 & 8 \\ -1 & 2\end{array}\right]$

Comparing the corresponding elements. 

4x=-4 

$\Rightarrow \quad x=-1$

4y=8 

$\Rightarrow \quad y=2$

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

Question 37

(i) If $A=\left[\begin{array}{cc}2 & -1 \\ -4 & 5\end{array}\right]$ and $B=\left[\begin{array}{c}-3 \\ 2\end{array}\right]$  find the matrix C such that CA=B

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

Sol :

(i) given

$A=\left[\begin{array}{cc}2 & -1 \\ -4 & 5\end{array}\right]$

$B=\left[\begin{array}{c}-3 \\ 2\end{array}\right]$

Let Matrix $\mathrm{C}=\left[\begin{array}{l}x \\ y\end{array}\right]$

$\therefore \mathrm{AC}=\left[\begin{array}{rr}2 & -1 \\ -4 & 5\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}2 x-y \\ -4 x+5 y\end{array}\right]$

But AC=B

$\therefore\left[\begin{array}{c}2 x-y \\ -4 x+5 y\end{array}\right]=\left[\begin{array}{r}-3 \\ 2\end{array}\right]$

Comparing the corresponding elements.

2 x-y=-3...(i)

-4 x+5 y=2..(ii)

Multiplying (i) by 5 and (ii) by 1

$\begin{array}{l}

10 x-5 y=-15 \\

-4 x+5 y=2 \\

\hline 6 x=-13

\end{array}$

$x=\frac{-13}{6}$

Substituting the value of x in (i)

$2\left(\frac{-13}{6}\right)-y=-3 \Rightarrow \frac{-13}{3}-y=-3$

$-y=-3+\frac{13}{3}=\frac{-9+13}{3}=\frac{4}{3}$

$\therefore y=-\frac{4}{3}$

$\therefore$ Matrix $C=\left[\begin{array}{c}\frac{-13}{6} \\ \frac{-4}{3}\end{array}\right]$

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

Let Matrix C=[x y]

Since the matrix $A$ is $2 \times 2$ and $B=7 \times 2$

$\because \mathrm{CA}=\mathrm{B}$

$\therefore(x y)\left[\begin{array}{cc}2 & -1 \\ -4 & 5\end{array}\right]=\left[\begin{array}{cc}0 & -3\end{array}\right]$

=(2 x-4 y-x+5 y)=[0-3]

Comparing 

2x-4 y=0

$\Rightarrow x-2 y=0$

$\therefore x=2 y$

and $-x+5 y=-3 \Rightarrow-2 y+5 y=-3$

$\Rightarrow 3 y=-3 \Rightarrow y=-1$

$\therefore x=2 y=2 \times(-1)=-2$

Hence , $C=\left[\begin{array}{ll}x & y\end{array}\right]=\left[\begin{array}{ll}-2 & -1\end{array}\right]$

Question 38

If $A=\left[\begin{array}{cc}3 & -4 \\ -1 & 2\end{array}\right]$ , find matrix B such that BA = I,where I is unity matrix of order 2

Sol :

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

BA = I, where I is unity matrix of order 2

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

Let $\mathbf{B}=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$

$\therefore \mathrm{BA}$

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

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

$\therefore\left[\begin{array}{ll}3 a-b & -4 a+2 b \\ 3 c-d & -4 c+2 d\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

Comparing the corresponding terms, we get

$3 a-b=1,-4 a+2 b=0 \Rightarrow 2 b=4 a \Rightarrow b=2 a$

$\therefore 3 a-b=1 \Rightarrow 3 a-2 a=1 \Rightarrow a=1$

$\quad$ and $b=2 a \Rightarrow b=2 \times 1=2$

$\therefore a=1, b=2$

$\quad$ and $3 c-d=0 \Rightarrow d=3 c$

$\quad-4 c+2 d=1 \Rightarrow-4 c+2 \times 3 c=1$

$\Rightarrow-4 c+6 c=1 \Rightarrow 2 c=1 \quad \Rightarrow c=\frac{1}{2}$

and $d=3 c=3 \times \frac{1}{2}=\frac{3}{2}$

Hence $a=1, b=2, c=\frac{1}{2}, d=\frac{3}{2}$

$\therefore$ Matrix $B=\left[\begin{array}{cc}1 & 2 \\ \frac{1}{2} & \frac{3}{2}\end{array}\right]$

Question 39

If $B=\left[\begin{array}{cc}-4 & 2 \\ 5 & -1\end{array}\right]$ and $C=\left[\begin{array}{cc}17 & -1 \\ 47 & -13\end{array}\right]$

find the matrix A such that AB = C

Sol :

$B=\left[\begin{array}{cc}-4 & 2 \\ 5 & -1\end{array}\right]$

$C=\left[\begin{array}{cc}17 & -1 \\ 47 & -13\end{array}\right]$

and AB = C

Let $\mathrm{A}=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$

Then $\mathrm{AB}=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right] \times\left[\begin{array}{cc}-4 & 2 \\ 5 & -1\end{array}\right]$

$=\left[\begin{array}{ll}-4 a+5 b & 2 a-b \\ -4 c+5 d & 2 c-d\end{array}\right]$

$\because A B=C$

$\therefore\left[\begin{array}{ll}-4 a+5 b & 2 a-b \\ -4 c+5 d & 2 c-d\end{array}\right]=\left[\begin{array}{cc}17 & -1 \\ 47 & -13\end{array}\right]$

Comparing corresponding elements, we get

$\because-4 a+5 b=17$...(i)

$2 a-b=-1$...(ii)

$-4 c+5 d=47$...(iii)

$2 c-d=-13$...(iv)

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

-4a+5b=17

4a-2b=-2

Adding 

$3 b=15 \Rightarrow b=\frac{15}{3}=5$

$2 a-b=-1 \Rightarrow 2 a-5=-1 \Rightarrow 2 a=-1+5$

$\quad=4 \Rightarrow a=\frac{4}{2}=2$

$\therefore a=2, b=5$

Again multiplying (iii) by 1 and (iv) by 2 ,

-4 c+5 d=47

4 c-2 d=-26

Adding  $3 d=21 \Rightarrow d=\frac{21}{3}=7$

and $2 c-d=-13 \Rightarrow 2 c-7=-13$

$\Rightarrow 2 c=-13+7=-6 \Rightarrow c=\frac{-6}{2}=-3$

$\therefore c=-3, d=7$

Now matrix $\mathrm{A}=\left[\begin{array}{cc}2 & 5 \\ -3 & 7\end{array}\right]$