KC Sinha Mathematics Solution Class 12 Chapter 5 आव्यूह ( Matrices ) Exercise 5.2 (Q1-Q7)

Exercise 5.2


Question 1

(i) यदि (If) A=[2 3 5] तथा (and) $B=\begin{bmatrix}1\\2\\3\end{bmatrix}$ , find AB निकालें 
Sol :



(ii) यदि (if) A=$\begin{bmatrix}2&1&3\\4&1&0\end{bmatrix}$ तथा B=$\begin{bmatrix}1&-1\\0&2\\5&0\end{bmatrix}$ AB और BA निकालें । (Find AB and BA)
Sol :

AB=$\begin{bmatrix}2&1&3\\4&1&0\end{bmatrix}$×$\begin{bmatrix}1&-1\\0&2\\5&0\end{bmatrix}$

=$\begin{bmatrix}2+0+15&-2+2+0\\4+0+0&-4+2+0\end{bmatrix}$

=$\begin{bmatrix}17&0\\4&-2\end{bmatrix}$

BA=$\begin{bmatrix}1&-1\\0&2\\5&0\end{bmatrix}$×$\begin{bmatrix}2&1&3\\4&1&0\end{bmatrix}$

=$\begin{bmatrix}2-4&1-1&3-0\\0+8&0+2&0+0\\10+0&5+0&15+0\end{bmatrix}$

=$\begin{bmatrix}-2&0&3\\8&2&0\\10&5&15\end{bmatrix}$



Question 2
Evaluate the following :
(i) $\left[\begin{array}{ll}0 & 2 \\ 0 & 3\end{array}\right]\left[\begin{array}{ll}4 & 6 \\ 0 & 0\end{array}\right]$
Sol :

(ii) $\left[\begin{array}{ll}1 & 3 \\ 2 & 1\end{array}\right]\left[\begin{array}{r}4 \\ -1\end{array}\right]$
Sol :
$=\left[\begin{array}{rr}4 & -3 \\ 8 & -1\end{array}\right]$

$=\left[\begin{array}{l}1 \\ 7\end{array}\right]$

(iii) $\left[\begin{array}{l}2 \\ 4 \\ 6\end{array}\right][1~2~3]$
Sol :
$=\left[ \begin{array}{ccc}2 & 4 & 6 \\ 4 & 8 & 12 \\ 6 & 12 & 18\end{array}\right]$

(iv) [1~2~3]$\left[\begin{array}{l}2 \\ 4 \\ 6\end{array}\right]$
Sol :

(v) $\left[\begin{array}{rrr}1 & 2 & -3 \\ -2 & 1 & 7\end{array}\right]\left[\begin{array}{lll}2 & 3 & 1 \\ 5 & 4 & 2 \\ 1 & 6 & 3\end{array}\right]$

(vi) $\left[\begin{array}{rrr}1 & 4 & 2 \\ 5 & -2 & 3\end{array}\right]\left[\begin{array}{rr}2 & -4 \\ 1 & -3 \\ 4 & 0\end{array}\right]$

Question 3

यदि (if) A=$\begin{bmatrix}2&9\\4&3\end{bmatrix}$ तथा (and) B=$\begin{bmatrix}1&5\\7&2\end{bmatrix}$ AB-BA निकालें । (Find AB-BA)
Sol :

AB-BA=
$\begin{bmatrix}2&9\\4&3\end{bmatrix}\begin{bmatrix}1&5\\7&2\end{bmatrix}$-$\begin{bmatrix}1&5\\7&2\end{bmatrix}\begin{bmatrix}2&9\\4&3\end{bmatrix}$

=$\begin{bmatrix}2+63&10+18\\4+21&20+6\end{bmatrix}-\begin{bmatrix}2+28&9+15\\14+8&63+6\end{bmatrix}$

=$\begin{bmatrix}65&28\\25&26\end{bmatrix}-\begin{bmatrix}22&24\\22&69\end{bmatrix}$

=$\begin{bmatrix}43&4\\3&-43\end{bmatrix}$



Question 4
(i) यदि (If) A=$\begin{bmatrix}cos\theta&sin\theta&\\sin\theta&cos\theta\end{bmatrix}$ , B=$\begin{bmatrix}cos\phi&sin\phi&\\sin\phi&cos\phi\end{bmatrix}$तो साबित करें कि (then show that)  AB=BA
Sol :
L.H.S
AB=$\begin{bmatrix}cos\theta&sin\theta&\\sin\theta&cos\theta\end{bmatrix}\begin{bmatrix}cos\phi&sin\phi&\\sin\phi&cos\phi\end{bmatrix}$

=$\begin{bmatrix} cos\theta.cos\phi+sin\theta.sin\phi & cos\theta.sin\phi+sin\theta.cos\phi\\ sin\theta.cos\phi+cos\theta.sin\phi & sin\theta.sin\phi+cos\theta.cos\phi \end{bmatrix}$

=$\begin{bmatrix}cos(\theta-\phi)&sin(\theta+\phi)\\sin(\theta+\phi)&cos(\theta-\phi)\end{bmatrix}$


R.H.S
BA=$\begin{bmatrix}cos\phi&sin\phi&\\sin\phi&cos\phi\end{bmatrix}\begin{bmatrix}cos\theta&sin\theta&\\sin\theta&cos\theta\end{bmatrix}$

=$\begin{bmatrix}cos\theta.cos\phi+sin\theta.sin\phi&cos\phi.sin\theta+sin\phi.cos\theta\\sin\phi.cos\theta+cos\phi.sin\theta&sin\theta.sin\phi+cos\theta.cos\phi&\end{bmatrix}$

=$\begin{bmatrix}cos(\theta-\phi)&sin(\theta+\phi)\\sin(\theta+\phi)&cos(\theta-\phi)\end{bmatrix}$

∴AB=BA Proved


Question 5

(i) यदि (If) $A=\begin{bmatrix}1&2\\5&7\end{bmatrix}$ तथा (and) $B=\begin{bmatrix}2&0\\3&-4\end{bmatrix}$ साबित करे कि (show that) AB≠BA
Sol :


(ii) यदि (If) $A=\begin{bmatrix}1&3\\3&-4\\5&6\end{bmatrix}$ तथा (and) $B=\begin{bmatrix}4&5&6\\7&-8&2\end{bmatrix}$ क्या AB=BA है?
Sol :

(iii) यदि (If) $A=\begin{bmatrix}-1&2\\3&4\end{bmatrix}$ and $B=\begin{bmatrix}2&-3\\5&1\end{bmatrix}$ , दिखाएँ कि (show that) AB≠BA
Sol :


(iv) यदि (If) $A=\begin{bmatrix}1&2&3\\0&1&0\\1&1&0\end{bmatrix}$ तथा (and) $B=\begin{bmatrix}-1&1&0\\0&-1&1\\2&3&4\end{bmatrix}$ दिखाएँ कि (show that) AB≠BA
Sol :




Question 6
(i) निम्नलिखित ज्ञात करें (Evaluate the following):
$\left\{\begin{bmatrix}1&3\\-1&-4\end{bmatrix}+\begin{bmatrix}3&-2\\-1&1\end{bmatrix}\right\}\begin{bmatrix}1&3&5\\2&4&6\end{bmatrix}$
Sol :

=$\begin{bmatrix}4&1\\-2&-3\end{bmatrix}\begin{bmatrix}1&3&5\\2&4&6\end{bmatrix}$

=$\begin{bmatrix}4+2&112+4&20+6\\-2-6&-6-12&-10-18\end{bmatrix}$

=$\begin{bmatrix}6&16&26\\-8&-18&-28\end{bmatrix}$


(ii) निकाले (Find) $\begin{bmatrix}1&-1\\0&2\\2&3\end{bmatrix}\left(\begin{bmatrix}1&0&2\\2&0&1\end{bmatrix}-\begin{bmatrix}0&1&2\\1&0&2\end{bmatrix}\right)$
Sol :
=$\begin{bmatrix}1&-1\\0&2\\2&3\end{bmatrix} \begin{bmatrix}1&-1&0\\1&0&-1\end{bmatrix}$

=$\begin{bmatrix}1-1&-1-0&0+1\\0+2&-0+0&0-2\\2+3&-2-0&0-3\end{bmatrix}$

=$\begin{bmatrix}0&-1&1\\2&0&-2\\5&-2&-3\end{bmatrix}$


(iii) $\begin{bmatrix}1&1&1\end{bmatrix} \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix} \begin{bmatrix}4\\4\\4\end{bmatrix}$
Sol :
=$\begin{bmatrix}1&1&1\end{bmatrix}\begin{bmatrix}4+0+0\\0+4+0\\0+0+4\end{bmatrix}$

=$\begin{bmatrix}1&1&1\end{bmatrix}\begin{bmatrix}4\\4\\4\end{bmatrix}$

=$\begin{bmatrix}4+4+4\end{bmatrix}=12$


(iv) $[1~3~5]\begin{bmatrix}1&0&3\\2&0&1\\0&1&2\end{bmatrix}\begin{bmatrix}1&4&6\end{bmatrix}$
Sol :



(v) $\begin{bmatrix}1&-1\\0&2\\2&3\end{bmatrix}\left(\begin{bmatrix}1&0&2\\2&0&1\end{bmatrix}-\begin{bmatrix}0&1&3\\1&0&2\end{bmatrix}\right)$
Sol :




Question 7
(i) यदि (If) P(x)=$\begin{bmatrix}cosx&sinx&\\-sinx&cosx\end{bmatrix}$ , तो साबित करें कि (then show that) 
P(x).P(y)=P(y).P(x)
Sol :
P(x).P(y)=$\begin{bmatrix}cosx&sinx\\-sinx&cosx\end{bmatrix} \begin{bmatrix}cosy&siny\\-siny&cosy\end{bmatrix}$

=$\begin{bmatrix}cosx.cosy-sinx.siny&cosx.siny+sinx.cosy\\-sinx.cosy-cosx.siny&-sinx.siny+cosx.cosy\end{bmatrix}$

=$\begin{bmatrix}cos(x+y)&sin(x+y)\\-sin(x+y)&cos(x+y)\end{bmatrix}$

=P(x+y)

P(y).P(x)=$\begin{bmatrix}cosy&siny\\-siny&cosy\end{bmatrix} \begin{bmatrix}cosx&sinx\\-sinx&cosx\end{bmatrix}$

=$\begin{bmatrix}cosy.cosx-siny.sinx&cosy.sinx+siny.cosx\\-siny.cosx-cosy.sinx&-siny.sinx+cosy.cosx\end{bmatrix}$

=$\begin{bmatrix}cos(x+y)&sin(x+y)\\-sin(x+y)&cos(x+y)\end{bmatrix}$=P(x+y)

∴P(x).P(y)=P(x+y)=P(y).P(x)


(ii) यदि (If)$F(x)=\left[\begin{array}{ccc}\cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right]$ दिखाएँ कि (show that)
F(x).F(y)=F(x+y)
Sol :
L.H.S
F(x).F(y)=$\left[\begin{array}{ccc}\cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{ccc}cosy & -\sin y & 0 \\ sin y & \cos y & 0 \\ 0 & 0 & 1\end{array}\right]$

=$\left[\begin{array}{ccc}\cos x \cos y-\sin x.sin y+0 & -\cos x.siny-\sin x \cos y+0 & 0+0+0 \\ \sin x \cos y+\cos x \sin y+0 & -\sin x \sin y+\cos x & 0+0+0 \\ 0+0+0 & -0+0+0 & 0+0+1\end{array}\right]$

$=\left[\begin{array}{ccc}\cos (x+y) & -\sin (x+y) & 0 \\ \sin (x+y) & \cos (x+y) & 0 \\ 0 & 0 & 1\end{array}\right]$

=F(x+y) Proved


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