Exercise 8 C
Question 1
Ans: Given,
$A=\left[\begin{array}{ll}1 & 2 \\3 & 4\end{array}\right], B=\left[\begin{array}{ll}6 & 1 \\1 & 1\end{array}\right], C=\left[\begin{array}{cc}-2 & -3 \\0 & 1\end{array}\right]$
(a) $A B=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]\left[\begin{array}{ll}6 & 1 \\ 1 & 1\end{array}\right]$
$=\left[\begin{array}{ll}1 \times 6+ 2 \times1 & 1\times1+2\times1 \\ 3 \times 6+4 \times1 & 3\times 1+4\times 1\end{array}\right]$
$=\left[\begin{array}{ll}6+2 & 1+2 \\ 18+y & 3+4\end{array}\right]$
$=\left[\begin{array}{ll}8 & 3 \\ 22 & 7\end{array}\right]$
$B A=\left[\begin{array}{ll}6 & 1 \\ 1 & 1\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$
$=\left[\begin{array}{ll}6 \times 1+1 \times 3 & 6 \times 2+1 \times 4 \\ 1 \times 1+1 \times 3 & 1 \times 2+1 \times 4\end{array}\right]$
$=\left[\begin{array}{ll}6+3 & 12+4 \\ 1+3 & 2+4\end{array}\right]$
$=\left[\begin{array}{ll}9 & 16 \\ 4 & 6\end{array}\right]$
$A C=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]\left[\begin{array}{cc}-2 & -3 \\ 0 & 1\end{array}\right]$
$=\left[\begin{array}{ll}1 \times (-2)+2\times0 & 1 \times(-3)+2 \times 1 \\ 3 \times(-2)+4 \times 0 & 3 \times(-3)+4 \times 1\end{array}\right]$
$\begin{aligned} &=\left[\begin{array}{cc}-2+0 & -3+2 \\ -6+0 & -9+4\end{array}\right] \\ &=\left[\begin{array}{cc}-2 & -1 \\ -6 & -5\end{array}\right] \\ C A &=\left[\begin{array}{cc}-2 & -3 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right] \end{aligned}$
$=\left[\begin{array}{cc}-2 \times 1+(-3) \times 3 & -2 \times 2+(-3) \times 4 \\ 0 \times 1+1 \times 3 & 0 \times 2+1 \times 4\end{array}\right]$
$=\left[\begin{array}{cc}-2-9 & -4-12 \\ 0+3 & 0+4\end{array}\right]$
$=\left[\begin{array}{cc}-11 & -16 \\ 3 & 4\end{array}\right]$
$B C=\left[\begin{array}{cc}6 & 1 \\ 1 & 1\end{array}\right]\left[\begin{array}{cc}-2 & -3 \\ 0 & 1\end{array}\right]$
$=\left[\begin{array}{cc}6 \times(-2)+1 \times 0 & 6 \times(-3)+1 \times 1 \\ 1 \times(-2)+1 \times 0 & 1 \times(-3)+1 \times 1\end{array}\right]$
$=\left[\begin{array}{cc}-12+0 & -18+1 \\ -2+0 & -371\end{array}\right]$
$=\left[\begin{array}{cc}-12 & -17 \\ -2 & -2\end{array}\right]$
$C B=\left[\begin{array}{cc}-2 & -3 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}6 & 1 \\ 1 & 1\end{array}\right] .$
$=\left[\begin{array}{cc}-2 \times 6+(-3) \times 1 & -2 \times 1+(-3) \times 1 \\ 0 \times 6+1 \times 1 & 0 \times 1+1 \times 1\end{array}\right]$
$=\left[\begin{array}{cc}-12-3 & -2-3 \\ 0+1 & 0+1\end{array}\right]$
$=\left[\begin{array}{cc}-15 & -5 \\ 1 & 1\end{array}\right]$
(b) $A^{2}=A \times A$
$=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$
$=\left[\begin{array}{ll}1 \times 1+2 \times 3 & 1 \times 2+2 \times 4 \\ 3 \times 1+4 \times 3 & 3 \times 2+4 \times 4\end{array}\right]$
$=\left[\begin{array}{ll}1+6 & 2+8 \\ 3+12 & 6+16\end{array}\right]$
$=\left[\begin{array}{cc}7 & 10 \\ 15 & 22\end{array}\right]$
$\begin{aligned} B^{2} &=B \times B \\ &=\left[\begin{array}{ll}6 & 1 \\ 1 & 1\end{array}\right]\left[\begin{array}{ll}6 & 1 \\ 1 & 1\end{array}\right] \\ &=\left[\begin{array}{ll}6 \times 6+\mid \times 1 & 6 \times 1+1 \times 1 \\ 1 \times 6+1 \times 1 & 1 \times 1+1 \times 1\end{array}\right] \\ &=\left[\begin{array}{ll}36+1 & 6+1 \\ 6+1 & 1+1\end{array}\right] \\ &=\left[\begin{array}{ll}37 & 7 \\ 7 & 2\end{array}\right] \end{aligned}$
$\begin{aligned} C^{2} &=C \times C \\ &=\left[\begin{array}{cc}-2 & -3 \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}-2 & -3 \\ 0 & 1\end{array}\right] \end{aligned}$
$=\left[\begin{array}{cc}-2 \times(-2)+(-3) \times 0 & -2 \times-3)+(-3) \times 1 \\ 0 \times(-2)+1 \times 0 & 0 \times(-3)+1 \times 1\end{array}\right]$
$=\left[\begin{array}{cc}4+0 & 6+3 \\ 0+0 & 0+1\end{array}\right]$
$=\left[\begin{array}{ll}4 & 3 \\ 0 & 1\end{array}\right] $
(C) $A(B C)=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]\left[\begin{array}{cc}-12 & -17 \\ -2 & -2\end{array}\right]$
$=\left[\begin{array}{cc}-12-4 & -17-4 \\ -36-8 & -51-8\end{array}\right]$
$=\left[\begin{array}{ll}-16 & -21 \\ -44 & -59\end{array}\right]$
$(A B) C=\left[\begin{array}{cc}8 & 3 \\ 22 & 7\end{array}\right]\left[\begin{array}{cc}-2 & -3 \\ 0 & 1\end{array}\right]$
$=\left[\begin{array}{lr}8 \times(-2)+3 \times 0 & 8 \times(-3)+3 \times 1 \\ 22 \times(-2)+7 \times 0 & 22 \times(-3)+7 \times 1\end{array}\right]$
$\begin{aligned} &=\left[\begin{array}{ll}-16+0 & -24+3 \\ -44+0 & -66+7\end{array}\right] \\ &=\left[\begin{array}{ll}-16 & -21 \\ -44 & -59\end{array}\right] \\ B(C A) &=\left[\begin{array}{ll}6 & 1 \\ 1 & 1\end{array}\right]\left[\begin{array}{cc}-11 & -16 \\ 3 & 4\end{array}\right] \end{aligned}$
$\begin{aligned}=&\left[\begin{array}{ll}6 \times(-11)+1 \times 3 & 6 \times(-16)+1 \times 4 \\ 1 \times-11+1 \times 3 & 1 \times(-16)+1 \times 4\end{array}\right] \\ &\end{aligned}$
$=\left[\begin{array}{ll}-66+3 & -96+4 \\ -11+3 & -16+4\end{array}\right]$
$=\left[\begin{array}{ll}-63 & -92 \\ -8 & -12\end{array}\right]$
(BC) A = $\left[\begin{array}{cc}-12 & -17 \\ -2 & -2\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$
$=\left[\begin{array}{ll}-12 \times 1+(-17) \times 3 & -12 \times 2+(-17) \times 4 \\ -2 \times 1+(-2) \times 3 & -2 \times 2+(-2) \times 4\end{array}\right]$
$=\left[\begin{array}{ll}-12-51 & -24-68 \\ -2-6 & -4-8\end{array}\right]$
$=\left[\begin{array}{ll}-63 & -92 \\ -8 & -12\end{array}\right]$
Question 2
Ans:(i) False
(ii) True
Question 3
Ans: $\left[\begin{array}{cc}a & 3 \\ 1 & 2\end{array}\right]\left[\begin{array}{c}2 \\ -1\end{array}\right]=\left[\begin{array}{l}5 \\ 0\end{array}\right]$
$\left[\begin{array}{c}a \times 2+3 \times(-1) \\ 1 \times 2+2 \times 1)\end{array}\right]=\left[\begin{array}{l}5 \\ 0\end{array}\right]$
$\left[\begin{array}{c}2 a-3 \\ 2-2\end{array}\right]=\left[\begin{array}{l}5 \\ 0\end{array}\right]$
$\left[\begin{array}{c}2 a-3 \\ 0\end{array}\right]=\left[\begin{array}{l}5 \\ 0\end{array}\right]$
On comparing their corresponding elements
$\begin{gathered}2 a-3=5 \\2 a=5+3 \\2 a=8 \\a=\frac{8}{2} \\a=4 .\end{gathered}$
Hence, a=4
Question 4
Ans:
$\begin{aligned} A^{2} &=A \times A \\ &=\left[\begin{array}{ll}1 & 4 \\ 2 & 1\end{array}\right]\left[\begin{array}{ll}1 & 4 \\ 2 & 1\end{array}\right] \end{aligned}$
$=\left[\begin{array}{ll}1\times 1+4 \times 2 & 1 \times y+4 \times 1 \\ 2 \times 1+ 1 \times 2 & 2 \times 4+1\times 1\end{array}\right]$
$\begin{aligned} &=\left[\begin{array}{ll}1+8 & 4+4 \\ 2+2 & 8+1\end{array}\right] \\ &=\left[\begin{array}{ll}9 & 8 \\ 4 & 9\end{array}\right] \\ B C &=\left[\begin{array}{cc}-3 & 2 \\ 4 & 0\end{array}\right]\left[\begin{array}{ll}1 & 0 \\ 0 & 2\end{array}\right] \end{aligned}$
$=\left[\begin{array}{cc}-3 \times 1+2 \times 0 & -3 \times 0+2 \times 2 \\ 4 \times 1+0 \times 0 & 4 \times 0+0 \times 2\end{array}\right]$
$\begin{aligned} &=\left[\begin{array}{cc}-3+0 & 0+4 \\ 4+0 & 0+0\end{array}\right] \\ &=\left[\begin{array}{cc}-3 & 4 \\ 4 & 0\end{array}\right] \\ \therefore & A^{2}+B C \\ &\left[\begin{array}{ll}9 & 8 \\ 4 & 9\end{array}\right]+\left[\begin{array}{cc}-3 & 4 \\ 4 & 0\end{array}\right] \\ &\left[\begin{array}{cc}9-3 & 8+4 \\ 4+4 & 9+0\end{array}\right] \\ &\left[\begin{array}{ll}6 & 12 \\ 8 & 9\end{array}\right] \end{aligned}$
Question 5
Ans: $\left[\begin{array}{cc}1 & 0 \\ 0 & -1\end{array}\right]\left[\begin{array}{l}3 \\ 4\end{array}\right]=\left[\begin{array}{l}x \\ y\end{array}\right]$
$\left[\begin{array}{l}1 \times 3+0 \times 4 \\ 0 \times 3+(-1) \times 4\end{array}\right]=\left[\begin{array}{l}x \\ y\end{array}\right]$
$\left[\begin{array}{c}3+0 \\ 0-4\end{array}\right]=\left[\begin{array}{l}x \\ y\end{array}\right]$
$\left[\begin{array}{c}3 \\ -4\end{array}\right]=\left[\begin{array}{l}x \\ y\end{array}\right]$
On comparing their corresponding elements,
x= 3 and , y = -4
Question 6
Ans: (i) The order of matrix M will be 1 $ \times$ 2
(ii) Let M = (x y)
$\begin{aligned} \therefore &[x, y] \times\left[\begin{array}{ll}1 & 1 \\ 0 & 2\end{array}\right]=\left[\begin{array}{ll}1 & 2\end{array}\right] \\ &[x+1+y \times 0 \cdot x \times 1+y+2]=\left[\begin{array}{ll}1 & 2\end{array}\right] \\ &[x+0 \quad x+2 y]=\left[\begin{array}{ll}1 & 2\end{array}\right] \end{aligned}$
$\left[\begin{array}{ll}x & x+2 y\end{array}\right]=\left[\begin{array}{ll}1 & 2\end{array}\right]$
On comparing their corresponding elements,
$x=1 .$ and $x+2 y=2$
$x=1$ and 1 +2y=2
$\begin{array}{lll}x=1 & \text { and } & 2 y=2-1 \\ & x=1 & \text {and } & y=\frac{1}{2}\end{array}$
$\begin{aligned} \therefore M &=\left[\begin{array}{ll}x & y\end{array}\right] \\ &=\left[\begin{array}{ll}1 & \frac{1}{2}\end{array}\right] . \end{aligned}$
Question 7
Ans: Given,
A =$\left[\begin{array}{cc}4 & -2 \\ 6 & -3\end{array}\right], \quad B=\left[\begin{array}{cc}0 & 2 \\ 1 & -1\end{array}\right], \quad C=\left[\begin{array}{cc}-2 & 3 \\ 1 & -3\end{array}\right]$
(i)
$\begin{aligned} A^{2} &=A x A \\ &=\left[\begin{array}{ll}4 & -2 \\ R& -3\end{array}\right]\left[\begin{array}{ll}4 & -2 \\ 6 & -3\end{array}\right] \end{aligned}$
$=\left[\begin{array}{ll}4 \times 4+(-2)-66 & 4 \times(-2)+(-2) \times(-3) \\ 6 \times 4+(-3) \times 6 & 6 \times(-2)+(-3) \times(-3)\end{array}\right]$
$=\left[\begin{array}{cc}16-12 & -8+6 \\ 24-18 & -12+9\end{array}\right]$
$=\left[\begin{array}{cc}4 & -2 \\ 6 & -3\end{array}\right]$
(ii) $\begin{aligned} B C &=\left[\begin{array}{cc}0 & 2 \\ 1 & -1\end{array}\right]\left[\begin{array}{cc}-2 & 3 \\ 1 & -3\end{array}\right] \\ &=\left[\begin{array}{cc}0 \times(-2)+2 \times 1 & 0 \times 3+2 \times(-3) \\ 1 \times(-2)+(-1) \times 1 & 1 \times 3+(-1) \times(3)\end{array}\right] \end{aligned}$
$=\left[\begin{array}{cc}0+2 & 0-6 \\ -2-1 & 3+3\end{array}\right]$
$=\left[\begin{array}{cc}2 & -6 \\ -3 & 6\end{array}\right]$
(iii) $A^{2}-A+B C$
$\left[\begin{array}{cc}4 & -2 \\ 6 & -3\end{array}\right]-\left[\begin{array}{cc}4 & -2 \\ 6 & -3\end{array}\right]+\left[\begin{array}{cc}2 & -6 \\ -3 & 6\end{array}\right]$
$\left[\begin{array}{cc}4-4+2 & -2+2-6 \\ 6-6-3 & -3+3+6\end{array}\right]$
$\left[\begin{array}{cc}0+2 & 0-1 \\ 0-3 & 0+6\end{array}\right]=\left[\begin{array}{cc}2 & -6 \\ -3 & 6\end{array}\right]$
Question 8
Ans: $P=\left[\begin{array}{cc}1 & 2 \\ 2 & -1\end{array}\right]$ and $Q=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right]$
(i) $P^{2}-Q^{2}$
$P^{2}=\left[\begin{array}{rr}1 & 2 \\ 2 & -1\end{array}\right]\left[\begin{array}{rr}1 & 2 \\ 2 & -1\end{array}\right]$
$=\left[\begin{array}{ll}1 \times 1+2 \times 2 & 1 \times 2+2 \times(-1) \\ 2 \times 1+(-1) \times 2 & 2 \times 2+(-10 \times(-1)\end{array}\right]$
$=\left[\begin{array}{ll} & \end{array}\right]$
$=\left[\begin{array}{ll}1+4 & 2-2 \\ 2-2 & 4+1\end{array}\right]$
$=\left[\begin{array}{cc}5 & 0 \\ 0 & 5\end{array}\right]$
$Q^{2}=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right]\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right]$
$=\left[\begin{array}{ll}1 \times 1+0 \times 2 & 1\times 0+ 0 \times 1 \\ 2 \times 1+1 \times 2 & 2 \times 0+1 \times 1\end{array}\right]$
$=\left[\begin{array}{ll}1+0 & 0+0 \\ 2+2 & 0+1\end{array}\right]$
$=\left[\begin{array}{ll}1 & 0 \\ 4 & 1\end{array}\right]$
$\therefore P^{2}-Q^{2}$
$=\left[\begin{array}{cc}5 & 0 \\ 0 & 5\end{array}\right]-\left[\begin{array}{ll}1 & 0 \\ 4 & 1\end{array}\right]$
$=\left[\begin{array}{cc}5-1 & 0-0 \\ 0-4 & 5-1\end{array}\right]$
$=\left[\begin{array}{cc}4 & 0 \\ -4 & 4\end{array}\right]$
(ii) $(P+Q)(P- Q)$
$P+Q=\left[\begin{array}{cc}1 & 2 \\ 2 & -1\end{array}\right]+\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right]$
$=\left[\begin{array}{ll}1+1 & -2+0 \\ 2+2 & -1+1\end{array}\right]$
$=\left[\begin{array}{ll}2 & 2 \\ 4 & 0\end{array}\right]$
$\begin{aligned} P-Q &=\left[\begin{array}{cc}1 & 2 \\ 2 & -1\end{array}\right]-\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right] \\ &=\left[\begin{array}{cc}1-1 & 2-0 \\ 2-2 & -1-1\end{array}\right] \\ &=\left[\begin{array}{cc}0 & 2 \\ 0 & -2\end{array}\right] \end{aligned}$
$\therefore(P+Q)(P-Q)$
$\quad=\left[\begin{array}{ll}2 & 2 \\ 4 & 0\end{array}\right]\left[\begin{array}{cc}0 & 2 \\ 0 & -2\end{array}\right]$
$=\left[\begin{array}{cc}2 \times 0+2 \times 0 & 2 \times 2+2 \times(-2) \\ 4 \times 0+0 \times 0 & 4 \times 2+0 \times(-2)\end{array}\right]$
$=\left[\begin{array}{ll}0+0 & 4-4 \\ 0+0 & 8+0\end{array}\right]$
$=\left[\begin{array}{ll}0 & 0 \\ 0 & 8\end{array}\right]$
$\therefore(P+Q)(P-Q) \neq P^{2}-Q^{2}$ for Matrix.
Question 9
Ans: $\left[\begin{array}{ll}3 & 4 \\ 2 & 5\end{array}\right]=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$\left[\begin{array}{ll}3 & 4 \\ 2 & 5\end{array}\right]=$ $\left[\begin{array}{ll}a \times+b \times 0 & a \times0 + b \times 1 \\ c \times 1+d \times 0 & c \times 0 +d \times 1\end{array}\right]$
$\left[\begin{array}{ll}3 & 4 \\ 2 & 5\end{array}\right]=\left[\begin{array}{ll}a+0 & 0+b \\ c+0 & 0+d\end{array}\right]$
$\left[\begin{array}{ll}3 & 4 \\ 2 & 5\end{array}\right]=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$
On comparing their elements,
$a=3, b=4, c=2$ and $d=5$.
Question 10
Ans: $A=\left[\begin{array}{cc}2 & 0 \\ -3 & 1\end{array}\right]$ and $B=\left[\begin{array}{cc}0 & 1 \\ -2 & 3\end{array}\right]$
(i)$B A=\left[\begin{array}{cc}0 & 1 \\ -2 & 3\end{array}\right]\left[\begin{array}{cc}2 & 0 \\ -3 & 1\end{array}\right] .$
$=\left[\begin{array}{ll}0 \times 2+1 \times(-3) & 0 \times 0+1 \times 1 \\ -2 \times 2+3 \times(-3) & -2 \times 0+3 \times 1\end{array}\right]$
$=\left[\begin{array}{ccc}0 & -3 & 0+1 \\ -4 & -9 & 0+3\end{array}\right]$
$=\left[\begin{array}{cc}-3 & 1 \\ -13 & 3\end{array}\right]$
(ii) $A^{2}=A \times A$
$=\left[\begin{array}{cc}2 & 0 \\-3 & 1\end{array}\right]\left[\begin{array}{cc}2 & 0 \\-3 & 1\end{array}\right]$
$=\left[\begin{array}{cc}2 \times 2+0 \times(-3) & 2 \times 0+0 \times 1 \\ -3 \times 2+1 \times(-3) & -3 \times 0+1 \times 1\end{array}\right]$
$=\left[\begin{array}{cc}4+0 & 0+0 \\ -6-3 & 0+1\end{array}\right]$
$=\left[\begin{array}{cc}4 & 0 \\ -9 & 1\end{array}\right]$
Question 11
Ans: $A=\left[\begin{array}{ll}2 & x \\ 0 & 1\end{array}\right], B=\left[\begin{array}{ll}4 & 36 \\ 0 & 1\end{array}\right]$
$A^{2}=\left[\begin{array}{ll}2 & x \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}2 & x \\ 0 & 1\end{array}\right]$
$=\left[\begin{array}{ll}2 \times 2+x \times 0 & 2 \times x+x(1) \\ 0 \times 2+1 \times 0 & 0 \times x+1 \times 1\end{array}\right]$
$\therefore A^{2}=B$
$\quad\left[\begin{array}{ll}4 & 3 x \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}4 & 36 \\ 0 & 1\end{array}\right]$
On comparing their corresponding elements
$3 x=36$
$x=\frac{36}{3}$
$x=12$
Question 12
Ans: (i) $\left[\begin{array}{ll}p & q\end{array}\right]\left[\begin{array}{l}p \\ q\end{array}\right]=[25]$
$\left[p^{2}+q^{2}\right]=[25]$
$\therefore 25$ Sum of two squares which can be
$\therefore(\pm 0)^{2}=(\pm 5)^{2} ;(\pm 5)^{2}+(\pm 0)^{2} ;(\pm 3)^{2}+(\pm 4)^{2},(\pm 4)^{2}+(\pm 3)^{2}$
$\begin{aligned} \therefore \quad & p=0, \text { and } q=\pm 5 . \\ & p=\pm 5, q=0 \\ & p \neq \pm 3, q=\pm 4 . \\ p &=\pm 4, q= \pm 3 . \end{aligned}$
(ii) $\left[\begin{array}{ll}1 & 3 \\ 0 & 0\end{array}\right]\left[\begin{array}{c}2 \\ -1\end{array}\right]=\left[\begin{array}{l}p \\ q\end{array}\right]$
$\left[\begin{array}{l}1 \times 2+3 \times(-1) \\ 0 \times 2+0 \times(-1)\end{array}\right]=\left[\begin{array}{l}p \\ q\end{array}\right]$
$\left[\begin{array}{l}2+3 \\ 0+0\end{array}\right]=\left[\begin{array}{l}p \\ q\end{array}\right]$
$\left[\begin{array}{c}-1 \\ 0\end{array}\right]=\left[\begin{array}{l}p \\ q\end{array}\right]$
On comparing their corresponding elements,
p= -1
Question 13
Ans: A and B are any two 2 $\times $ 2 Matrix and AB = B
ஃ B is not a zero matrix
So, A matrix will be unit matrix or identify matrix $=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
Question 14
Ans: $A=\left[\begin{array}{cc}3 & 2 \\ 4 & -3\end{array}\right], B=\left[\begin{array}{cc}2 & -3 \\ -4 & 5\end{array}\right]$ and $C=\left[\begin{array}{cc}1 & -3 \\ -4 & 4\end{array}\right]$.
(i) $B C=\left[\begin{array}{cc}2 & -3 \\ -4 & 5\end{array}\right]\left[\begin{array}{cc}1 & -3 \\ -4 & 4\end{array}\right]$
$=\left[\begin{array}{ll}2 \times 1+(-3) \times(-4) & 2 \times(-3)+(-3) \times 4 \\ -4 \times 1+5 \times(-4) & -4 \times(-3)+5 \times 4\end{array}\right]$
$=\left[\begin{array}{ll}2+12 & -6-12 \\ -4-20 & -12+20\end{array}\right]$
$=\left[\begin{array}{cc}14 & -18 \\ -24 & 32\end{array}\right]$
(ii) $A^{2}=\left[\begin{array}{cc}3 & 2 \\ 4 & -3\end{array}\right]\left[\begin{array}{cc}3 & 2 \\ 4 & -3\end{array}\right]$
$=\left[\begin{array}{ll}3 \times 3+2 \times 4 & 3 \times 2+2 \times(-3) \\ 4 \times 3+(-3) \times 4 & 4 \times 2+(-3) \times(-3)\end{array}\right]$
$=\left[\begin{array}{cc}9+8 & 6-6 \\ 12 & -12 & 8+9\end{array}\right]$
$=\left[\begin{array}{cc}17 & 0 \\ 0 & 17\end{array}\right]$
$\begin{aligned} A^{2}+A &=\left[\begin{array}{cc}17 & 0 \\ 0 & 17\end{array}\right]+\left[\begin{array}{cc}3 & 2 \\ 4 & -3\end{array}\right] . \\ &=\left[\begin{array}{cc}17+3 & 0+2 \\ 0+4 & 17-3\end{array}\right] \\ &=\left[\begin{array}{cc}20 & 2 \\ 4 & 14\end{array}\right] . \end{aligned}$
Question 15
Ans: (i) $\left[\begin{array}{ll}a-b & b-4 \\ b+4 & a-2\end{array}\right]\left[\begin{array}{ll}2 & 0 \\ 0 & 2\end{array}\right]=\left[\begin{array}{cc}-2 & -2 \\ 14 & 0\end{array}\right]$
$\left[\begin{array}{ll}a-b) \times 2+(b-4) \times 0 & (a-b) \times 0+(b-y) \times 2 \\ (b+4) \times 2+(a-2) \times 0 & (a+4) \times 0+(a-2) \times 2\end{array}\right]=\left[\begin{array}{cc}-2 & -2 \\ 14 & 0\end{array}\right]$
$\left[\begin{array}{ll}2 a-2 b+0 & 0+2 b-8 \\ -2 b+8+0 & 0+2 a-4\end{array}\right]=\left[\begin{array}{cc}-2 & -2 \\ 14 & 0\end{array}\right]$
$\left[\begin{array}{cc}2 a-2 b & 2 b-8 \\ 2 b+8 & 2 a-4\end{array}\right]=\left[\begin{array}{cc}-2 & -2 \\ 14 & 0\end{array}\right]$
On comparing their corresponding elements,
$2 b-8=-2 \quad$ and $\quad 2 a-4=0$
$2, b=-2+8 \quad$ and $\quad 2 a=4$
$2 b=6 \quad$ and $\quad a=\frac{4}{2}$
$b=\frac{6}{2} \quad$ and $\quad a=2$
$b=3 \quad$ and $\quad a=2$.
Hence a = 2 and b = 3
(ii) $\left[\begin{array}{cc}8 & -2 \\ 1 & 4\end{array}\right] X=\left[\begin{array}{c}12 \\ 10\end{array}\right]$
(a) The order of matrix is of the order $2 \times 1$
(b) $\operatorname{Let} x=\left[\begin{array}{l}a \\ b\end{array}\right]$
$\therefore\left[\begin{array}{cc}8 & -2 \\ 1 & 4\end{array}\right]\left[\begin{array}{l}a \\ b\end{array}\right]=\left[\begin{array}{l}12 \\ 10\end{array}\right]$
$\left[\begin{array}{c}8 a-2 b \\ a+4 b\end{array}\right]=\left[\begin{array}{l}12 \\ 10\end{array}\right]$
$8 a-2 b=12$ ............(i)
$a+4 b=10$..........(ii)
Multiply by B in eq ( ii)
$\begin{aligned} 8 a-2 b &=12 \\8 a+32 b &=80 \end{aligned}$
$-34 b=-68$
$b=\frac{+68}{+34}$
b= 2
Put the value of b= 2 in eq (ii)
$a+4 \times 2=10$
$a+8=10$
$a=10-8$
$a=2 $
$\therefore X=\left[\begin{array}{l}2 \\ 2\end{array}\right]$
Question 16
Ans: Given,
$A=\left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right]$
$\begin{aligned} A^{2} &=A \cdot A \\ &=\left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right]\left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right] \end{aligned}$
$=\left[\begin{array}{cc}1 \times 1+1 \times(-2) & 1 \times(-1)+(-1) \times 3 \\ 2 \times 1+3 \times 2 & 2 \times(-1)+3 \times 3\end{array}\right]$
$=\left[\begin{array}{cc}1-2 & -1-3 \\ 2+6 & -2+9\end{array}\right]$
$=\left[\begin{array}{cc}-1 & -4 \\ 8 & 7\end{array}\right]$
$\begin{aligned} 4 A &=4\left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right] \\ &=\left[\begin{array}{cc}4 & -4 \\ 8 & 12\end{array}\right] \end{aligned}$
$\begin{aligned} 4 A &=4\left[\begin{array}{cc}-1 & -1 \\ 2 & 3\end{array}\right] \\ &=\left[\begin{array}{cc}4 & -4 \\ 8 & 12\end{array}\right] \\ 5 I &=5\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right] \\ &=\left[\begin{array}{ll}5 & 0 \\ 0 & 5\end{array}\right] \end{aligned}$
$\therefore A^{2}-4 A+5 I=0$
$\left[\begin{array}{cc}-1 & -4 \\ 8 & 7\end{array}\right]-\left[\begin{array}{cc}4 & -4 \\ 8 & 12\end{array}\right]+\left[\begin{array}{ll}5 & 0 \\ 0 & 5\end{array}\right]=0$
$\left[\begin{array}{cc}-1-4+5 & -4+4+0 \\ 8-8+0 & 7-12+5\end{array}\right]=0$
$\left[\begin{array}{cc}-5+5 & 0+0 \\ 0+0 & 7-7\end{array}\right]=0$
$\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]=0$, Hence proved.
Question 17
Ans: Given ,
$A=\left[\begin{array}{ll}1 & 4 \\ 1 & 0\end{array}\right], B=\left[\begin{array}{cc}2 & 1 \\ 3 & -1\end{array}\right]$ and $C=\left[\begin{array}{ll}2 & 3 \\ 0 & 5\end{array}\right]$.
$A B=\left[\begin{array}{ll}1 & 4 \\ 1 & 0\end{array}\right]\left[\begin{array}{rr}2 & 1 \\ 3 & -1\end{array}\right]$.
$=\left[\begin{array}{cc}1 \times 2+4 \times 3 & 1\times 1+4 \times(-1) \\ 1 \times 2+0 \times 3 & 1\times 1+0 \times(-1)\end{array}\right]$
$=\left[\begin{array}{cc}2+12 & 1-4 \\ 2+0 & 1+0\end{array}\right]$
$=\left[\begin{array}{cc}14 & -3 \\ 2 & 1\end{array}\right]$
$(A B) c=\left[\begin{array}{cc}14 & -3 \\ 2 & 1\end{array}\right]\left[\begin{array}{ll}2 & 3 \\ 0 & 5\end{array}\right]$
$=\left[\begin{array}{cc}14 \times 2+(-3) \times 0 & 14 \times 3+(-3) \times 5 \\ 2 \times 2+1 \times 0 & 2 \times 3+1 \times 5\end{array}\right]$
$=\left[\begin{array}{ll}2.8+0 & 42-15 \\ 4+0 & 6+5\end{array}\right]$
$=\left[\begin{array}{ll}28 & 27 \\ 4 & 11\end{array}\right]$
$C B=\left[\begin{array}{ll}2 & 3 \\ 0 & 5\end{array}\right]\left[\begin{array}{rr}2 & 1 \\ 3 & -1\end{array}\right]$
$=\left[\begin{array}{ll}2 \times 2+3 \times 3 & 2 \times 1+3 \times(-1) \\ 0 \times 2+5 \times 3 & 0 \times 1+5 \times(-1)\end{array}\right]$
$=\left[\begin{array}{ll}4+9 & 2-3 \\ 0+15 & 0-5\end{array}\right]$
$=\left[\begin{array}{ll}13 & -1 \\ 15 & -5\end{array}\right]$
$(C B) A=\left[\begin{array}{cc}13 & -1 \\ 15 & -5\end{array}\right]\left[\begin{array}{ll}1 & 4 \\ 1 & 0\end{array}\right]$
$=\left[\begin{array}{cc}1 3 \times 1+(-1) \times 1 & 13 \times 4+(-1) \times 0 \\ 15 \times 1+(-5) \times 1 & 15 \times 4+(-5) \times 0\end{array}\right]$
$=\left[\begin{array}{ll}13-1 & 52+0 \\ 15-5 & 60+0\end{array}\right]$
$=\left[\begin{array}{ll}12 & 54 \\ 10 & 60\end{array}\right]$
Hence , (AB) C $\neq(C B) A$
Question 18
Ans: (i)
$\left[\begin{array}{cc}6 & -2 \\1 & 2\end{array}\right]\left[\begin{array}{ll}5 & 3 \\2 & 4\end{array}\right]$
$=\left[\begin{array}{cc}6 \times 5+(-2) \times 2 & 6 \times 3+(-2) \times 4 \\ 1 \times 5+2 \times 2 & 1 \times 3+2 \times 4\end{array}\right]$
$=\left[\begin{array}{cc}30-4 & 18-8 \\ 5+4 & 3+8\end{array}\right]$
$=\left[\begin{array}{cc}26 & 10 \\ 9 & 11\end{array}\right]$
(ii) $\left[\begin{array}{cc}2 & 3 \\ -1 & 0\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}7 \\ -2\end{array}\right]$
$\left[\begin{array}{c}2 \times x+3 \times y \\ -1 times x+0 \times y\end{array}\right]=\left[\begin{array}{c}7 \\ -2\end{array}\right]$
On comparing their corresponding elements,
$-x+0=-2$
x = 2
$2 x+3 y=7$
Put the value of x = 2
$2 x 2+3 y=7$
$4+3 y=7$
$3 y=7-y$
$3 y=3$
$y=\frac{3}{3}$
y =1
Hence , x = 2 and y = 1
Question 19
Ans: Given ,
$A=\left[\begin{array}{ll}3 & 1 \\ 2 & 1\end{array}\right]$ and $B=\left[\begin{array}{cc}1 & -2 \\ 5 & 3\end{array}\right]$.
$\begin{aligned} A-B &=\left[\begin{array}{cc}3 & 1 \\ 2 & 1\end{array}\right]-\left[\begin{array}{cc}1 & -2 \\ 5 & 3\end{array}\right] \\ &=\left[\begin{array}{cc}3-1 & 1+2 \\ 2-5 & 1-3\end{array}\right] \\ &=\left[\begin{array}{cc}2 & 3 \\ -3 & -2\end{array}\right] \end{aligned}$
$(A-B)^{2}=\left[\begin{array}{cc}2 & 3 \\ -3 & -2\end{array}\right]\left[\begin{array}{cc}2 & 3 \\ -3 & -2\end{array}\right]$
$=\left[\begin{array}{cc}2 \times 2+3 \times(-3) & 2 \times 3+3 \times(-2) \\ -3 \times 2+(-2) \times(-3) & -3 \times 3+(-2) \times(-2)\end{array}\right]$
$=\left[\begin{array}{cc}4+9 & 6-6 \\ -6+6 & -9+4\end{array}\right]$
$=\left[\begin{array}{cc}-5 & 0 \\ 0 & -5\end{array}\right] .$
$A^{2}=\left[\begin{array}{ll}3 & 1 \\ 2 & 1\end{array}\right]\left[\begin{array}{ll}3 & 1 \\ 2 & 1\end{array}\right]$
$=\left[\begin{array}{ll}3 \times 3+1 \times 2 & 3 \times 1+1\times 1 \\ 2 \times 3+1 \times 2 & 2 \times 1+1\times 1\end{array}\right]$
$=\left[\begin{array}{ll}9+2 & 3+1 \\ 6+2 & 2+1\end{array}\right]$
$=\left[\begin{array}{ll}11 & 4 \\ 8 & 3\end{array}\right] .$
$2 A B=2\left[\begin{array}{ll}3 & 1 \\ 2 & 1\end{array}\right]\left[\begin{array}{cc}1 & -2 \\ 5 & 3\end{array}\right]$
$=2\left[\begin{array}{ll}3 \times 1+1 \times 5 & 3 \times(-2)+1+3 \\ 2 \times 1+1 \times 5 & 2 \times(-2)+1+3\end{array}\right]$
$=2\left[\begin{array}{cc}3+5 & -6+3 \\ 2+5 & -4+3\end{array}\right]$
$=2\left[\begin{array}{ll}8 & -3 \\ 7 & -1\end{array}\right]$
$=\left[\begin{array}{ll}16 & -6 \\ 14 & -2\end{array}\right]$
$B^{2}=\left[\begin{array}{cc}1 & -2 \\ 5 & 3\end{array}\right]\left[\begin{array}{cc}1 & -2 \\ 5 & 3\end{array}\right]$
$=\left[\begin{array}{ll}1 \times 1+(-2) \times 5 & 1 \times(-2)+(-2) \times 3 \\ 5 \times 1+3 \times 5 & 5 \times(-2)+3 \times 3\end{array}\right]$
$=\left[\begin{array}{cc}1-10 & -2-6 \\ 5+15 & -10+9\end{array}\right]$
$=\left[\begin{array}{cc}-9 & -8 \\ 20 & -1\end{array}\right] .$
$\therefore A^{2}-2 A B+B^{2}=\left[\begin{array}{ll}11 & 4 \\ 8 & 3\end{array}\right]-\left[\begin{array}{cc}16 & -6 \\ 14 & -2\end{array}\right]+\left[\begin{array}{cc}-9 & -8 \\ 20 & -1\end{array}\right]$
$=\left[\begin{array}{ll}11-16+9 & 4+6-8 \\ 8-14+20 & 3+2-1\end{array}\right]$
$=\left[\begin{array}{cc}-14 & 2 \\ 14 & 4\end{array}\right]$
$\therefore(A-B)^{2} \neq A^{2}-2 A B+B^{2}$
Question 20
Ans: Given,
$A=\left[\begin{array}{ll}3 & 0 \\ 0 & 4\end{array}\right]$ and $B=\left[\begin{array}{ll}a & b \\ 0 & c\end{array}\right]$.
$A B=\left[\begin{array}{ll}3 & 0 \\ 0 & 4\end{array}\right]\left[\begin{array}{ll}a & b \\ 0 & c\end{array}\right]$
$=\left[\begin{array}{ll}3 \times a+0 \times 0 & 3 \times 6+a \times c \\ 0 \times a+4 \times 0 & 0 \times b+4 \times c\end{array}\right]$
$=\left[\begin{array}{ll}3 a+0 & 3 b+0 \\ 0+0 & \text 0+ 4 c\end{array}\right]$
$=\left[\begin{array}{cc}3 a & 3 b \\ 0 & 4 c\end{array}\right]$
$\begin{aligned} A+B &=\left[\begin{array}{ll}3 & 0 \\ 0 & 4\end{array}\right]+\left[\begin{array}{ll}a & b \\ 0 & c\end{array}\right] . \\ &=\left[\begin{array}{ll}3+a & 0+b \\ 0+0 & 4+c\end{array}\right] \\ &=\left[\begin{array}{cc}3+a & b \\ 0 & 4+c\end{array}\right] . \\ \therefore A B &=A+B . \\ &\left[\begin{array}{cc}3 a & 3 b \\ 0 & 4 c\end{array}\right] \doteq\left[\begin{array}{ll}3+a & b \\ 0 & 4+c\end{array}\right] \end{aligned}$
On comparing their correspond elements,
$3 a=3+9,3 b=b$ and $4 c=4+c$ $3 a-a=3,3 b-b=0$ and $4 c-c=4$
$2 a=3, \quad 2 b=0$ and $\quad 3 c=4$
$a=\frac{2}{3}, b=0$ and $c=\frac{4}{3}$
Question 21
Ans: (i) $\left[\begin{array}{ll}2 & 1 \\ 5 & 0\end{array}\right]-3 x=\left[\begin{array}{ll}-7 & 4 \\ 2 & 6\end{array}\right]$
$\left[\begin{array}{ll}2 & 1 \\ 5 & 0\end{array}\right]-\left[\begin{array}{cc}-7 & 4 \\ 2 & 6\end{array}\right]=3 x$
$\left[\begin{array}{cc}2+7 & 1-4 \\ 5-2 & 0-6\end{array}\right]=3 x$
$\left[\begin{array}{cc}9 & -3 \\ 3 & -6\end{array}\right]=3 x$
$x=\frac{1}{3}\left[\begin{array}{cc}9 & -3 \\ 3 & -6\end{array}\right]$
$x=\left[\begin{array}{ll}\frac{1}{3} \times 9 & \frac{1}{3} \times(-3) \\ \frac{1}{3} \times 3 & \frac{1}{3} \times(-6)\end{array}\right]$
$x=\left[\begin{array}{cc}3 & -1 \\ 1 & -2\end{array}\right]$
(ii) Given,
$A=\left[\begin{array}{ll}1 & 2 \\2 & 1\end{array}\right] \text { and } B=\left[\begin{array}{ll}2 & 1 \\1 & 2\end{array}\right]$
$B A=\left[\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right]$
$=\left[\begin{array}{cc}2 \times 1+1 \times 2 & 2 \times 2+1 \times 1 \\ 1\times 1+2 \times 2 & 1 \times 2+2 \times 1\end{array}\right]$
$=\left[\begin{array}{cc}2+2 & 4+1 \\ 1+4 & 2+2\end{array}\right]$
$=\left[\begin{array}{ll}4 & 5 \\ 5 & 4\end{array}\right]$
$A(B A)=\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right]\left[\begin{array}{ll}4 & 5 \\ 5 & 4\end{array}\right] .$
$=\left[\begin{array}{ll}1 \times 4+2 \times 5 & 1 \times 5+2 \times 4 \\ 2 \times 4+1 \times 5 & 2 \times 5+1 \times 4\end{array}\right]$
$=\left[\begin{array}{cc}4+10 & 5+8 \\ 8+5 & 10+4\end{array}\right]$
$=\left[\begin{array}{cc}14 & 13 \\ 13 & 14\end{array}\right] .$
Question 22
Ans: Given,
$A=\left[\begin{array}{ll}9 & 1 \\ 5 & 3\end{array}\right]$ and $B=\left[\begin{array}{rr}1 & 5 \\ 7 & -11\end{array}\right]$
$3 A+5 B-2 x=0$
$3 A+5 B=2 x$
$2 x=3 A+5 B .$
$2 x=3\left[\begin{array}{ll}9 & 1 \\ 5 & 3\end{array}\right]+5\left[\begin{array}{cc}1 & 5 \\ 7 & -11\end{array}\right]$
$2 x=\left[\begin{array}{ll}27 & 3 \\ 15 & 9\end{array}\right]+\left[\begin{array}{c}5 \\ 35 \\ 35\end{array}-55\right]$
$2 x=\left[\begin{array}{cc}27+5 & 3125 \\ 15+35 & 9-55\end{array}\right]$
$2 x=\left[\begin{array}{cc}32 & 28 \\ 50 & -46\end{array}\right]$
$x=\frac{1}{2}\left[\begin{array}{cc}32 & 28 \\ 50 & -46\end{array}\right]$
$x=\left[\begin{array}{cc}16 & 14 \\ 25 & -23\end{array}\right] .$
Question 23
Ans: (1)
$\left[\begin{array}{ll}x & 3 x \\ y & 4 y\end{array}\right]\left[\begin{array}{l}2 \\ 1\end{array}\right]=\left[\begin{array}{c}5 \\ 12\end{array}\right] .$
$\left[\begin{array}{l}2 x+3 x \\ 2 y+4 y\end{array}\right]=\left[\begin{array}{c}5 \\ 12\end{array}\right]$
$\left[\begin{array}{c}5 x \\ 6 y\end{array}\right]=\left[\begin{array}{c}5 \\ 12\end{array}\right]$
On comparing their corresponding elements
$\begin{aligned} 5 x &=5 \quad \text { and } 6 y=12 \\ x &=\frac{5}{5} \quad \text { and } y=\frac{12}{6} \\ x &=1 \quad \text { and } y=2 . \end{aligned}$
Hence, $x=1$ and $y=2$
(ii) Given,
$A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right], B=\left[\begin{array}{ll}2 & 1 \\ 4 & 2\end{array}\right]$ and $C=\left[\begin{array}{ll}5 & 1 \\ 7 & 4\end{array}\right]$.
$\begin{aligned} B+C &=\left[\begin{array}{ll}2 & 1 \\ 4 & 2\end{array}\right]+\left[\begin{array}{ll}5 & 1 \\ 7 & 4\end{array}\right] \\ &=\left[\begin{array}{ll}2+5 & 1+1 \\ 4+7 & 2+4\end{array}\right] \\ &=\left[\begin{array}{cc}7 & 2 \\ 11 & 6\end{array}\right] . \end{aligned}$
$A(B+C)=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]\left[\begin{array}{ll}7 & 2 \\ 11 & 6\end{array}\right]$
$=\left[\begin{array}{cc}1 \times 7+2 \times 11 & 7 \times 2+2 \times 6 \\ 3 \times 7+4 \times 11 & 3 \times 2+4 \times 6\end{array}\right]$
$=\left[\begin{array}{cc}7+22 & 2+12 \\ 21+44 & 6+24\end{array}\right]$
$=\left[\begin{array}{cc}29 & 14 \\ 65 & 30\end{array}\right] .$
$(B+C) A=\left[\begin{array}{ll}7 & 2 \\ 11 & 6\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$
$=\left[\begin{array}{ll}7 \times 1+2 \times 3 & 7 \times 2+2 \times 4 \\ 11 \times 1+6 \times 3 & 11 \times 2+6 \times 4\end{array}\right]$
$=\left[\begin{array}{ll}7+6 & 17+8 \\ 11+18 & 22+24\end{array}\right]$
$=\left[\begin{array}{ll}13 & 22 \\ 29 & 46\end{array}\right] .$
Question 24
Ans (i) $\begin{aligned}\left[\begin{array}{cc}x-2 & 5 \\ 3 & 3\end{array}\right] &=\left[\begin{array}{ll}4 & 2 \\ y & 5\end{array}\right]+\left[\begin{array}{ll}-4 & 3 \\ -1 & -2\end{array}\right] . \\\left[\begin{array}{cc}x-2 & 5 \\ 3 & 3\end{array}\right] &=\left[\begin{array}{cc}4-4 & 2+3 \\ y-1 & 5-2\end{array}\right] \\\left[\begin{array}{cc}x-2 & 5 \\ 3 & 3\end{array}\right] &=\left[\begin{array}{ll}0 & 5 \\ y-1 & 3\end{array}\right] \end{aligned}$
On comparing their corresponding elements
$x-2=0$ and $y-1=3$.
$x=0+2$ and $y=3+1$
$x=2$. and $y=4$.
Hence, $x=2$ and $y=4$.
(ii) Given ,
$\begin{aligned} A &=\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right], \quad B=\left[\begin{array}{ll}2 & 1 \\ 3 & 2\end{array}\right], C=\left[\begin{array}{ll}1 & 3 \\ 3 & 1\end{array}\right] \\ B-A &=\left[\begin{array}{ll}2 & 1 \\ 3 & 2\end{array}\right]-\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right] . \\ &=\left[\begin{array}{cc}2-1 & 1-2 \\ 3-2 & 2-3\end{array}\right] . \\ &=\left[\begin{array}{cc}1 & -1 \\ 1 & -1\end{array}\right] . \end{aligned}$
$\therefore C(B-A)=\left[\begin{array}{ll}1 & 3 \\ 3 & 1\end{array}\right]\left[\begin{array}{ll}1 & -1 \\ 1 & -1\end{array}\right]$
$=\left[\begin{array}{cc} 1\times 1+3\times 1 & 1 \times (-1)+3 \times(-1) \\ 3 \times 1+ 1\times 1 & 3 \times(-1)+1 \times(-1)\end{array}\right]$
$=\left[\begin{array}{ll}1+3 & -1-3 \\ 3+1 & -3-1\end{array}\right] .$
$=\left[\begin{array}{ll}4 & -4 \\ 4 & -4\end{array}\right] .$
Question 25
Ans Given,
$A=\left[\begin{array}{ccc}1 & -2 & 1 \\ 2 & 1 & 3\end{array}\right]$ and $B=\left[\begin{array}{ll}2 & 1 \\ 3 & 2 \\ 1 & 1\end{array}\right]$
$A B=\left[\begin{array}{ccc}1 & -2 & 1 \\ 2 & 1 & 3\end{array}\right]\left[\begin{array}{ll}2 & 1 \\ 3 & 2 \\ 1 & 1\end{array}\right]$
$=\left[\begin{array}{cc} 1\times 2+(-2) \times 3+ 1 \times 1 & 1 \times 1+(-2) \times 2+1\times1 \\ 2 \times 2+1 \times 3+3 \times 1 & 2 \times 1+1 \times 2+3 \times1\end{array}\right]$
$=\left[\begin{array}{cc}-3 & -2 \\ 10 & 7\end{array}\right]$
$B A=\left[\begin{array}{ll}2 & 1 \\ 3 & 2 \\ 1 & 1\end{array}\right]\left[\begin{array}{ccc}1 & -2 & 1 \\ 2 & 1 & 3\end{array}\right] .$
$=\left[\begin{array}{ccc}2 \times 1+1 \times 2 & 2 \times(-2)+ 1 \times 1 & 2 \times 1+1 \times 3 \\ 3 \times 1+2 \times 2 & 3 \times(-2)+2 \times 1 & 3 \times 1+2 \times 3 \\ 1 \times 1+1 \times 2 & 1\times (-2)+1 \times 1 & 1 \times 1+1 \times 3\end{array}\right]$
$=\left[\begin{array}{ccc}2+2 & -9+1 & 2+3 \\ 3+4 & -6+2 & 3+6 \\ 1+2 & -2+1 & 1+3\end{array}\right]$
$=\left[\begin{array}{ccc}4 & -3 & 5 \\ 7 & -4 & 9 \\ 3 & -1 & 4\end{array}\right] .$
Hence it is a possible
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