Exercise 8 B
Question 1
$\Rightarrow A=\left[\begin{array}{c}-3 \\ 2\end{array}\right], B=\left[\begin{array}{c}-2 \\ -1\end{array}\right], C\left[\begin{array}{c}1 \\ -3\end{array}\right]$
Ans:
i)
$\begin{aligned}&\Rightarrow X=A+B+C=\left[\begin{array}{l}3 \\2\end{array}\right]+\left[\begin{array}{c}-2 \\-1\end{array}\right]+\left[\begin{array}{c}1 \\-3\end{array}\right] \\&=\left[\begin{array}{c}-3-2+1 \\2-1-3\end{array}\right]=\left[\begin{array}{c}2 \\-2\end{array}\right]\end{aligned}$
(ii)
$\Rightarrow X=A+B-C$
$=\left[\begin{array}{l}3 \\ 2\end{array}\right]+\left[\begin{array}{c}-2 \\ -1\end{array}\right]-\left[\begin{array}{c}1 \\ -3\end{array}\right]$
$=\left[\begin{array}{l}3-2-1 \\ 2-1+3\end{array}\right]=\left[\begin{array}{c}0 \\ 4\end{array}\right]$
(iii) $\Rightarrow X-A=B \Rightarrow X=A+B$
$X=\left[\begin{array}{l}3 \\ 2\end{array}\right]+\left[\begin{array}{c}-2 \\ -1\end{array}\right]=\left[\begin{array}{l}3-2 \\ 2-1\end{array}\right]=\left[\begin{array}{l}1 \\ 1\end{array}\right]$
(iv) $\Rightarrow A+X=B+C \Rightarrow X=B+C-A$
so, $X=\left[\begin{array}{c}-2 \\ -1\end{array}\right]+\left[\begin{array}{c}1 \\ -3\end{array}\right]-\left[\begin{array}{l}3 \\ 2\end{array}\right]$
$=\left[\begin{array}{c}-2+1-3 \\ -1-3-2\end{array}\right]=\left[\begin{array}{c}-4 \\ -6\end{array}\right]$
(v) $\Rightarrow 2 X=A-C$
$=\left[\begin{array}{c}3 \\ 2\end{array}\right]-\left[\begin{array}{c}1 \\ -3\end{array}\right]$
$=\left[\begin{array}{l}3-1 \\ 2+3\end{array}\right]$
$=\left[\begin{array}{c}2 \\ 5\end{array}\right]$
So $X=\frac{1}{2}\left[\begin{array}{c}2 \\ 5\end{array}\right]$
$=\left[\begin{array}{l}1 \\ \frac{5}{2}\end{array}\right]$
Question 2
Ans: $B=\left[\begin{array}{cc}2 & -3 \\ -4 & 5\end{array}\right], C=\left[\begin{array}{cc}1 & -3 \\ -4 & 4\end{array}\right]$
$B-C=\left[\begin{array}{cc}2 & -3 \\ -4 & 5\end{array}\right]-\left[\begin{array}{cc}1 & -3 \\ -4 & 4\end{array}\right]$
$=\left[\begin{array}{cc}2-1 & -3+3 \\ -4+4 & 5-4\end{array}\right]$
$=\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]$
Question 3
Ans: $\Rightarrow A=\left[\begin{array}{ll}2 & 0 \\ 3 & 1\end{array}\right], B=\left[\begin{array}{cc}0 & 1 \\ -2 & 3\end{array}\right]$
$2 A-3 B=2\left[\begin{array}{cc}2 & 0 \\ -3 & 1\end{array}\right]-3\left[\begin{array}{cc}0 & 1 \\ -2 & 3\end{array}\right]$
$=\left[\begin{array}{cc}4 & 0 \\ -6 & 2\end{array}\right]-\left[\begin{array}{cc}0 & 3 \\ -6 & 9\end{array}\right]$
$=\left[\begin{array}{ccc}4 & -0 & 0 & -3 \\ -6 & -(-6) & 2 & -9\end{array}\right]$
$=\left[\begin{array}{cc}4 & -3 \\ -6+6 & -7\end{array}\right]$
$=\left[\begin{array}{cc}4 & -3 \\ 0 & -7\end{array}\right]$
Question 4
$\Rightarrow A=\left[\begin{array}{ll}1 & 4 \\ 2 & 3\end{array}\right], B=\left[\begin{array}{cc}-4 & -1 \\ -3 & -2\end{array}\right]$
(i) $\Rightarrow 2 A+B=2\left[\begin{array}{cc}1 & 4 \\ 2 & 3\end{array}\right]+\left[\begin{array}{cc}-4 & -1 \\ -3 & -2\end{array}\right]$
$=\left[\begin{array}{ll}2 & 8 \\ 4 & 6\end{array}\right]+\left[\begin{array}{cc}-4 & -1 \\ -3 & -2\end{array}\right]$
$=\left[\begin{array}{ccc}2 & -4 & 8-1 \\ 4 & -3 & 6-2\end{array}\right]$
$2\left[\begin{array}{cc}-2 & 7 \\ 1 & 4\end{array}\right]$
(ii) $\Rightarrow C+B=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
=$\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]-B$
$\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]-\left[\begin{array}{ll}-4 & -1 \\ -3 & -1\end{array}\right]$
=$\left[\begin{array}{ll}4 & 1 \\ 3 & 2\end{array}\right]$
Question 5
Ans:
$\Rightarrow P=\left[\begin{array}{cc}4 & 6 \\ 2 & -8\end{array}\right], \quad Q=\left[\begin{array}{cc}2 & -3 \\ -1 & y\end{array}\right]$
So $ P+2Q=\left[\begin{array}{cc}4 & 6 \\ 2 & -8\end{array}\right]+2\left[\begin{array}{cc}2 & -3 \\ -1 & 1\end{array}\right]$
$=\left[\begin{array}{cc}4 & 6 \\ 2 & -8\end{array}\right]+\left[\begin{array}{cc}4 & -6 \\ -2 & 2\end{array}\right]$
$=\left[\begin{array}{cc}4 & +4 & 6-6 \\ 2 & -2 & -8+2\end{array}\right]$
$=\left[\begin{array}{cc}8 & 0 \\ 0 & -6\end{array}\right]$
Question 6
Ans:
$\Rightarrow 2\left[\begin{array}{ll}3 & 4 \\ 5 & u\end{array}\right]+\left[\begin{array}{ll}1 & y \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}7 & 0 \\ 10 & 5\end{array}\right]$
$\Rightarrow\left[\begin{array}{ll}6 & 8 \\ 10 & 2 x\end{array}\right]+\left[\begin{array}{ll}1 & y \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}7 & 0 \\ 10 & 5\end{array}\right]$
$\Rightarrow\left[\begin{array}{ll}6 & +1 & 8+y \\ 10 & 0 & 2 x+1\end{array}\right]=\left[\begin{array}{ll}7 & 0 \\ 10 & 5\end{array}\right]$
$\Rightarrow\left[\begin{array}{ll}7 & 8 & +y \\ 10 & 2 x+1\end{array}\right]$
$=\left[\begin{array}{ll}7 & 0 \\ 10 & 5\end{array}\right]$
Comparing the corresponding elements,
$\begin{aligned} 2 x+1 &=5 \Rightarrow 2 x=5-1=4 \\ \Rightarrow x=\frac{4}{2} &=2 \end{aligned}$
and $8+y=0 \Rightarrow y=-8$
So, $x=2, y=-8$
Question 7
Ans: $\Rightarrow\left[\begin{array}{cc}a & 3 \\ 4 & 2\end{array}\right]+\left[\begin{array}{cc}2 & b \\ 1 & -2\end{array}\right]-\left[\begin{array}{cc}1 & 1 \\ -2 & c\end{array}\right]$
$=\left[\begin{array}{ll}5 & 0 \\ 7 & 3\end{array}\right]$
$\Rightarrow\left[\begin{array}{ll}a+2-1 & 3+b-1 \\ 4+1+2 & 2-2-c\end{array}\right]$
$\Rightarrow\left[\begin{array}{ll}5 & 0 \\ 7 & 3\end{array}\right]$
Comparing the correspending elements
$\begin{aligned}&b+2=0 \Rightarrow b=-2 \\&-c=3 \Rightarrow c=-3 \\&So, a=4, b=-2, c=-3\end{aligned}$
Question 8
Ans: $\Rightarrow\left[\begin{array}{cc}2 & -1 \\ 2 & 0\end{array}\right]+2 A=\left[\begin{array}{cc}-3 & 5 \\ 4 & 3\end{array}\right]$
$\Rightarrow 2 A=\left[\begin{array}{cc}-3 & 5 \\ 4 & 3\end{array}\right]-\left[\begin{array}{cc}2 & -1 \\ 2 & 0\end{array}\right]$
$=\left[\begin{array}{ccc}-3-2 & 5 & +1 \\ 4 & -2 & 3 & -0\end{array}\right]=\left[\begin{array}{cc}-5 & 6 \\ 2 & 3\end{array}\right]$
$A=\frac{1}{2}\left[\begin{array}{cc}-5 & 6 \\ 2 & 3\end{array}\right]=\left[\begin{array}{cc}\frac{-5}{2} & 3 \\ 1 & \frac{3}{2}\end{array}\right]$
Question 9
Ans : $\Rightarrow 2\left[\begin{array}{cc}3 & x \\ 0 & 1\end{array}\right]+3\left[\begin{array}{ll}1 & 3 \\ 4 & 2\end{array}\right]$
$=\left[\begin{array}{cc}2 & -7 \\ 15 & 8\end{array}\right]$
$\Rightarrow\left[\begin{array}{cc}6 & 2 x \\ 0 & 2\end{array}\right]+\left[\begin{array}{ll}3 & 9 \\ 3 y & 6\end{array}\right]$
$=\left[\begin{array}{cc}2 & -7 \\ 15 & 8\end{array}\right]$
$\Rightarrow\left[\begin{array}{cc}6 & 2 x+9 \\ 0+3 y & 2+6\end{array}\right]$
$=\left[\begin{array}{cc}2 & -7 \\ 15 & 8\end{array}\right]$
$\Rightarrow\left[\begin{array}{cc}9 & 2 x+9 \\ 3 y & 8\end{array}\right]$
Comparing the corresponding elements $2 x+9=-7 \Rightarrow 2 x+-7-9 \Rightarrow 2 x=-16$ $\Rightarrow x=\frac{-16}{2}=-8$
$3 y=15 \Rightarrow y=\frac{15}{3}=5$
$z=9$
$\therefore x=-8, y=5, z=9$
Question 10
Ans : $\Rightarrow M=\left[\begin{array}{ll}2 & 0 \\ 1 & 2\end{array}\right]$ and $N=\left[\begin{array}{cc}2 & 0 \\ -1 & 2\end{array}\right]$
$M+2 N=\left[\begin{array}{ll}2 & 0 \\ 1 & 2\end{array}\right]+2\left[\begin{array}{cc}2 & 0 \\ -1 & 2\end{array}\right]$
$=\left[\begin{array}{ll}2 & 0 \\ 1 & 2\end{array}\right]+\left[\begin{array}{cc}4 & 0 \\ -2 & 4\end{array}\right]$
$=\left[\begin{array}{cc}2+4 & 0+0 \\ 1-2 & 2+4\end{array}\right]$
$=\left[\begin{array}{cc}6 & 0 \\ -1 & 6\end{array}\right]$
Question 11
(a)
$\Rightarrow A=\left[\begin{array}{ll}3 & 0 \\ 0 & 3\end{array}\right], B=\left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right], C=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
So $A-B+C=\left[\begin{array}{ll}3 & 0 \\ 0 & 3\end{array}\right]-\left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right]+\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$\Rightarrow\left[\begin{array}{cc}3 & -1+1 & 0-3+0 \\ 0 & -0+0 & 3-1+1\end{array}\right]$
$=\left[\begin{array}{cc}3 & -3 \\ 0 & 3\end{array}\right]$
(b)
= So , A and B are '2 $\times$ 2 Matrices
$\begin{aligned} & A-B=A+B \\ \Rightarrow &-B=B \end{aligned}$
It is possible if B is a zero matrix
Question 12
$\Rightarrow P=\left[\begin{array}{cc}-3 & -1 \\ 2 & 5\end{array}\right], Q=\left[\begin{array}{cc}1 & 6 \\ -4 & 0\end{array}\right], R=\left[\begin{array}{cc}4 & -1 \\ 2 & -3\end{array}\right]$
(i) $2 p+3 \varphi-R=2\left[\begin{array}{cc}-3 & -1 \\ 2 & 5\end{array}\right]+3\left[\begin{array}{cc}1 & 6 \\ -4 & 0\end{array}\right]$ - $\left[\begin{array}{cc}4 & -1 \\ 2 & 3\end{array}\right]$
$=\left[\begin{array}{cc}-6 & -2 \\ 4 & 10\end{array}\right]+\left[\begin{array}{cc}3 & 18 \\ -12 & 0\end{array}\right]-\left[\begin{array}{cc}4 & -1 \\ 2 & 3\end{array}\right]$
$=\left[\begin{array}{cc}-6 & -3-4 & -2+18+1 \\ 4 & -12-2 & 10+0-3\end{array}\right]$
$=\left[\begin{array}{cc}-7 & 17 \\ -10 & 7\end{array}\right]$
(ii) $\Rightarrow 4 P-2 Q+3 R=4\left[\begin{array}{cc}-3 & -1 \\ 2 & 5\end{array}\right]-2\left[\begin{array}{cc}1 & 6 \\ -4 & 0\end{array}\right]+3\left[\begin{array}{ll}4 & -1 \\ 2 & 3\end{array}\right]$
$=\left[\begin{array}{cc}-12 & -4 \\ 8 & 20\end{array}\right]-\left[\begin{array}{cc}2 & 12 \\ -8 & 0\end{array}\right]+\left[\begin{array}{cc}12 & -3 \\ 6 & 9\end{array}\right]$
$=\left[\begin{array}{cc}-12 & -2+12 & -4-12-3 \\ 8+8+6 & 20-0+9\end{array}\right]$
$=\left[\begin{array}{cc}-2 & -19 \\ 22 & 29\end{array}\right]$
Question 13
Ans: $\Rightarrow\left[\begin{array}{ll}\cos ^{2} \theta & 0 \\ \cot ^{2} \theta & 1\end{array}\right]+\left[\begin{array}{cc}\sin ^{2} \theta & 1 \\ -\operatorname{cosec}^{2} \theta & 0\end{array}\right]+\left[\begin{array}{cc}0 & -j \\ -1 & 0\end{array}\right]$
$=\left[\begin{array}{c}\cos ^{2} \theta+\sin ^{2} \theta+0 \\ \cot ^{2} \theta-\operatorname{cosec}^{2} \theta-1\end{array}\right]$
$=\left[\begin{array}{cc}1+0 & 0 \\ 1-1 & 1\end{array}\right] \quad\left\{\begin{array}{l}\because \sin ^{2} \theta+\cos ^{2} \theta=1 \\ \cot ^{2} \theta-\operatorname{cosec}^{2} \theta=1\end{array}\right\}$
$=\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]$
Question 14
Ans: $\Rightarrow A=\left[\begin{array}{ll}x & y \\ z & w\end{array}\right], B=\left[\begin{array}{cc}x & -y \\ -z & w\end{array}\right], C=\left[\begin{array}{ll}0 & y \\ 2z & 0\end{array}\right]$
Now L.H.S. $=(A+B)+C$
$=\left\{\left[\begin{array}{lll}x & y & \\ 2 & w & 1\end{array}\right]+\left[\begin{array}{cc}x & -y \\ -z & w\end{array}\right]\right\}+\left[\begin{array}{cc}0 & y \\ 2 z & 0\end{array}\right]$
$=\left[\begin{array}{ll}x+x & y-y \\ z-2 & w+w\end{array}\right]+\left[\begin{array}{cc}0 & y \\ 2 z & 0\end{array}\right]$
$=\left[\begin{array}{ll}2 x+0 & 0+y \\ 0 & +2 z & 2 w+0\end{array}\right]=\left[\begin{array}{l}2 x y \\ 2 z .2 w\end{array}\right]$
R.H.S $=A+(B+C)$
$=\left[\begin{array}{ll}x & y \\ z & \omega\end{array}\right]+\left\{\left[\begin{array}{cc}x & -y \\ -z & \omega\end{array}\right]+\left[\begin{array}{cc}0 & y \\ 2 z & 0\end{array}\right]\right\}$
$=\left[\begin{array}{cc}x & y \\ z & b\end{array}\right]+\left[\begin{array}{cc}x+0 & -y+y \\ -z+2 z & w+0\end{array}\right]$
$=\left[\begin{array}{ll}2x & y \\ 2z & 2w\end{array}\right]$ $=L \cdot H \cdot S=R \cdot H \cdot S$
Question 15
Ans: $\Rightarrow A=\left[\begin{array}{ll}8 & 1 \\ 5 & 2\end{array}\right], B=\left[\begin{array}{ll}-4 & 6 \\ -2 & 12\end{array}\right]$
Now 2A + 3B +4C is a null matrix
$2\left[\begin{array}{ll}8 & 1 \\ 5 & 2\end{array}\right]+3\left[\begin{array}{ll}-4 & 6 \\ -2 & 12\end{array}\right]+4 C=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
$\Rightarrow\left[\begin{array}{ll}16 & 2 \\ 10 & 4\end{array}\right]+\left[\begin{array}{cc}-12 & 18 \\ -6 & 36\end{array}\right]+4 C=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
$\Rightarrow\left[\begin{array}{lll}16 & -12 & 2 & +18 \\ 10 & -6 & 4+36\end{array}\right]+4 C=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
$\Rightarrow\left[\begin{array}{ll}4 & 20 \\ 4 & 40\end{array}\right]+4 C=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right] .$
$\Rightarrow 4 C=\left[\begin{array}{cc}0 & 0 \\ 0 & 0\end{array}\right]-\left[\begin{array}{ll}4 & 20 \\ 4 & 40\end{array}\right]$
$\Rightarrow 4 C=\left[\begin{array}{ccc}0 & -4 & 0 & -20 \\ 0 & -4 & 0 & -90\end{array}\right]=\left[\begin{array}{cc}-4 & -20 \\ -4 & -90\end{array}\right]$
$\Rightarrow C=\left[\begin{array}{cc}-1 & -5 \\ -1 & -10\end{array}\right]$
Question 16
$\Rightarrow P=\left[\begin{array}{ll}6 & -2 \\ 4 & -6\end{array}\right], Q=\left[\begin{array}{ll}5 & 3 \\ 2 & 0\end{array}\right]$
Now 3P - 2Q + 3X =O
So, 3X = - 3P+2Q = 2Q - 3p
$=3\left[\begin{array}{ll}5 & 3 \\ 2 & 0\end{array}\right]-3\left[\begin{array}{ll}6 & -2 \\ 4 & -6\end{array}\right]$
$=\left[\begin{array}{ll}10 & 6 \\ 4 & 0\end{array}\right]-\left[\begin{array}{cc}18 & -6 \\ 12 & -18\end{array}\right]$
$=\left[\begin{array}{ccc}10 & -18 & 6+6 \\ 4 & -12 & 0+18\end{array}\right]=\left[\begin{array}{cc}-8 & 12 \\ -8 & +18\end{array}\right]$
$So x=\frac{1}{3}\left[\begin{array}{cc}-8 & 12 \\ -8 & 18\end{array}\right]$
$=\left[\begin{array}{cc}\frac{-8}{3} & 4 \\ \frac{-8}{3} & 6\end{array}\right]$
Question 17
$\Rightarrow x+y=\left[\begin{array}{cc}9 & 7 \\ 2 & 11\end{array}\right]$
$x-y=\left[\begin{array}{cc}-5 & -6 \\ 4 & 3\end{array}\right]$
$=\left[\begin{array}{ccc}9 & -5 & 7 & -6 \\ 2 & +4 & 11 & -3\end{array}\right]=\left[\begin{array}{cc}4 & 1 \\ 6 & 14\end{array}\right]$
$\therefore X=\frac{1}{2}\left[\begin{array}{cc}4 & 1 \\ 6 & 14\end{array}\right]$
$\therefore\left[\begin{array}{ll}2 & \frac{1}{2} \\ 3 & 7\end{array}\right]$
Subtracting ,We get
$2 y=\left[\begin{array}{cc}9 & 7 \\ 2 & 11\end{array}\right]-\left[\begin{array}{cc}-5 & -6 \\ 4 & 3\end{array}\right]$
$=\left[\begin{array}{cc}9+5 & 7+6 \\ 2-4 & 11-3\end{array}\right]=\left[\begin{array}{cc}14 & 13 \\ -2 & 8\end{array}\right]$
$y=\frac{1}{2}\left[\begin{array}{cc}14 & 13 \\ -2 & 8\end{array}\right]=\left[\begin{array}{cc}7 & 13 \\ -1 & \frac{3}{4}\end{array}\right]$
Question 18
Ans: $\Rightarrow 2 A+B=\left[\begin{array}{cc}3 & -4 \\ 2 & 7\end{array}\right]$...........(i)
$A-2 B=\left[\begin{array}{ll}4 & 3 \\ 1 & 1\end{array}\right]$...........(ii)
$2 A-4 B=2\left[\begin{array}{ll}4 & 3 \\ 1 & 1\end{array}\right]=\left[\begin{array}{ll}8 & 6 \\ 2 & 2\end{array}\right] $..........(iii)
Multiplying (ii) by 2
Subtracting (iii) from (i)
$5 B=\left[\begin{array}{cc}3 & -4 \\ 2 & 7\end{array}\right]-\left[\begin{array}{ll}8 & 6 \\ 2 & 2\end{array}\right]$
$=\left[\begin{array}{cc}3-8 & 4-6 \\ 2-2 & 7-2\end{array}\right]=\left[\begin{array}{cc}-5 & -10 \\ 0 & 5\end{array}\right]$
$B=\frac{1}{5}\left[\begin{array}{cc}-5 & -10 \\ 0 & 5\end{array}\right]=\left[\begin{array}{cc}-1 & -2 \\ 0 & 1\end{array}\right]$
Now $2 A=\left[\begin{array}{cc}3 & -4 \\ 2 & 7\end{array}\right]-B$
$=\left[\begin{array}{cc}- & -4 \\ 2 & 7\end{array}\right]-\left[\begin{array}{cc}-1 & -2 \\ 0 & 1\end{array}\right]$
$=\left[\begin{array}{cc}3+1 & -4+2 \\ 2-0 & 7 & -1\end{array}\right]=\left[\begin{array}{cc}4 & -2 \\ 2 & 6\end{array}\right]$
$A=\frac{1}{2}\left[\begin{array}{cc}4 & -2 \\ 2 & 6\end{array}\right]=\left[\begin{array}{cc}2 & -1 \\ 1 & 3\end{array}\right]$
$\therefore A=\left[\begin{array}{cc}2 & -1 \\ 1 & 3\end{array}\right]$ and $B=\left[\begin{array}{cc}-1 & -2 \\ 0 & 1\end{array}\right]$
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