[mathjax]
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=$\begin{bmatrix}1&-1\\0&2\\5&0\end{bmatrix}$×$\begin{bmatrix}2&1&3\\4&1&0\end{bmatrix}$
=$\begin{bmatrix}2-4&1-1&3-0\\0+8&0+2&0+0\\10+0&5+0&15+0\end{bmatrix}$
=$\begin{bmatrix}-2&0&3\\8&2&0\\10&5&15\end{bmatrix}$
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) $\left[\begin{array}{rrr}1 & 4 & 2 \\ 5 & -2 & 3\end{array}\right]\left[\begin{array}{rr}2 & -4 \\ 1 & -3 \\ 4 & 0\end{array}\right]$
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=
$\begin{bmatrix}2&9\\4&3\end{bmatrix}\begin{bmatrix}1&5\\7&2\end{bmatrix}$-$\begin{bmatrix}1&5\\7&2\end{bmatrix}\begin{bmatrix}2&9\\4&3\end{bmatrix}$
=$\begin{bmatrix}2+63&10+18\\4+21&20+6\end{bmatrix}-\begin{bmatrix}2+28&9+15\\14+8&63+6\end{bmatrix}$
=$\begin{bmatrix}65&28\\25&26\end{bmatrix}-\begin{bmatrix}22&24\\22&69\end{bmatrix}$
=$\begin{bmatrix}43&4\\3&-43\end{bmatrix}$
(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
(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
(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?
(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
(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
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
Question 8
(i) यदि (If) $A=\left[\begin{array}{cc}2 & 3 \\ -1 & 5\end{array}\right], B=\left[\begin{array}{cc}3 & -1 \\ 4 & 7\end{array}\right]$ तथा (and) $c=\left[\begin{array}{cc}5 & -1 \\ 0 & 3\end{array}\right]$ दिखाएँ कि (show that)
A(B+C)=AB+AC
Sol :
L.H.S
A(B+C)=$=\left[\begin{array}{cc}2 & 3 \\ -1 & 5\end{array}\right]\left(\left[\begin{array}{cc}3 & -1 \\ 4 & 7\end{array}\right]+\left(\begin{array}{cc}5 & -1 \\ 0 & 3\end{array}\right]\right)$
$=\left[\begin{array}{cc}2 & 3 \\ -1 & 5\end{array}\right]\left[\begin{array}{cc}8 & -2 \\ 4 & 10\end{array}\right]$
$=\left[\begin{array}{cc}11+12 & -4+30 \\ -8+20 & 2+50\end{array}\right]$
$=\left[\begin{array}{ll}28 & 26 \\ 12 & 52\end{array}\right]$
R.H.S
AB+AC=$\left[\begin{array}{cc}2 & 3 \\ -1 & 5\end{array}\right]\left[\begin{array}{cc}3 & -1 \\ 4 & 7\end{array}\right]+\left[\begin{array}{cc}2 & 3 \\ -1 & 5\end{array}\right]\left[\begin{array}{cc}5 & -1 \\ 0 & 3\end{array}\right]$
$=\left[\begin{array}{cc}6+12 & -2+21 \\ -3+20 & 1+35\end{array}\right]+\left[\begin{array}{cc}10+0 & -2+9 \\ -5+0 & 1+15\end{array}\right]$
$=\left[\begin{array}{ll}18 & 19 \\ 17 & 31\end{array}\right]+\left[\begin{array}{cc}10 & 7 \\ -5 & 16\end{array}\right]$
$=\left[\begin{array}{cc}28 & 26 \\ 12 & 52\end{array}\right]$
∴A(B+C)=AB+AC
(ii) यदि (If) $A=\left[\begin{array}{cc}2 & -1 \\ -1 & 2\end{array}\right]$ तथा (and) $B=\left[\begin{array}{cc}1 & 4 \\ -1 & 1\end{array}\right]$ क्या (is)
(A+B)2=A2+2AB+B2
Sol :
L.H.S
(A+B)2$=\left(\left[\begin{array}{cc}2 & -1 \\ -1 & 2\end{array}\right]+\left[\begin{array}{cc}1 & 4 \\ -1 & 1\end{array}\right]\right)^{2}$
$=\left[\begin{array}{cc}3 & 3 \\ -2 & 3\end{array}\right]^{2}$
$=\left[\begin{array}{cc}3 & 3 \\ -2 & 3\end{array}\right]\left[\begin{array}{cc}3 & 3 \\ -2 & 3\end{array}\right]$
$=\left[\begin{array}{cc}9-6 & 9+7 \\ -6-6 & -6+9\end{array}\right]=\left[\begin{array}{cc}3 & 18 \\ -12 & 3\end{array}\right]$
R.H.S
A2+2AB+B2
$=\left[\begin{array}{cc}2 & -1 \\ -1 & 2\end{array}\right]\left[\begin{array}{cc}2 & -1 \\ -1 & 2\end{array}\right]+2\left[\begin{array}{cc}2 & -1 \\ -1 & 2\end{array}\right]\left[\begin{array}{cc}1 & 4 \\ -1 & 1\end{array}\right]+\begin{bmatrix}1&4\\-1&1\end{bmatrix}\begin{bmatrix}1&4\\-1&1\end{bmatrix}$
$=\left[\begin{array}{cc}4+1 & -2-2 \\ -2-2 & 1+4\end{array}\right]+2\left[\begin{array}{ccc}2+1 & 8-1 \\ -1 -2 & -4+2\end{array}\right]+\left[\begin{array}{cc}1-4 & 4+4 \\ -1-1 & -4+1\end{array}\right]$
$=\left[\begin{array}{cc}5 & -4 \\ -4 & 5\end{array}\right]+\left[\begin{array}{cc}6 & 14 \\ -6 & -4\end{array}\right]+\left[\begin{array}{cc}-3 & 8 \\ -2 & -3\end{array}\right]$
$=\left[\begin{array}{rr}8 & 18 \\ -12 & -2\end{array}\right]$
∴(A+B)2≠ A2+2AB+B2
Question 9
(i) यदि (If) $A=\left[\begin{array}{ll}2 & 3 \\ 4 & 5\end{array}\right], B=\left[\begin{array}{ll}3 & 4 \\ 7 & 2\end{array}\right], C=\left[\begin{array}{ll}1 & 0 \\ 0 & 7\end{array}\right]$ तो सत्यापित करें कि (verify that)
(AB)C=A(BC)
Sol :
L.H.S
(AB)C=$\left(\left[\begin{array}{ll}2 & 3 \\ 4 & 5\end{array}\right]\left[\begin{array}{ll}3 & 4 \\ 7 & 2\end{array}\right]\right)\left[\begin{array}{ll}1 & 0 \\ 0 & 7\end{array}\right]$
$=\left[\begin{array}{ll}6+21 & 8+6 \\ 12+35 & 16+10\end{array}\right]\left[\begin{array}{ll}1 & 0 \\ 0 & 7\end{array}\right]$
$=\left[\begin{array}{ll}27 & 14 \\ 4 7 & 26\end{array}\right]\left[\begin{array}{ll}1 & 0 \\ 0 & 7\end{array}\right]$
$=\left[\begin{array}{cc}27+0 & 0+91 \\ 47+0 & 0+182\end{array}\right]$
$=\left[\begin{array}{ll}2 7&98 \\ 4 7 & 182\end{array}\right]$
R.H.S
A(BC)=$=\left[\begin{array}{ll}2 & 3 \\ 4 & 5\end{array}\right]\left(\left[\begin{array}{ll}3 & 4 \\ 7 & 2\end{array}\right]\left[\begin{array}{ll}1 & 0 \\ 0 & 7\end{array}\right]\right)$
$=\left[\begin{array}{ll}2 & 3 \\ 4 & 5\end{array}\right]\left[\begin{array}{ll}3+0 & 0+28 \\ 7+0 & 0+14\end{array}\right]$
$=\left[\begin{array}{ll}2 & 3 \\ 4 & 5\end{array}\right]\left[\begin{array}{lll}3 & 2 8 \\ 7 & 14\end{array}\right]$
$=\left[\begin{array}{ll}6+21 & 56+42 \\ 12+35 & 112+70\end{array}\right]$
$=\left[\begin{array}{ll}2 7&98 \\ 4 7 & 182\end{array}\right]$
(AB)C=A(BC)
Question 10
(i) यदि (If) $A=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$ , साबित करें कि (show that)
$A^{2}=\left[\begin{array}{cc}\cos 2 \alpha & \sin 2 \alpha \\ -\sin 2 \alpha & \cos 2 \alpha\end{array}\right]$
Sol :
A2=A.A=$=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]\left[\begin{array}{cc}cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$
$=\left[\begin{array}{ll}\cos ^{2} \alpha-sin^{2}\alpha & \cos \alpha.sin\alpha+\sin \alpha \cos \alpha \\ -\sin \alpha \cos \alpha-\sin \alpha .cos\alpha & -\sin ^{2} \alpha+\cos ^{2} \alpha\end{array}\right]$
(ii) यदि (If) $A=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right] \cdot B=\left[\begin{array}{lr}0 & -i \\ i & 0\end{array}\right]$ तथा (and) $C=\begin{bmatrix}i&0\\0&-i\end{bmatrix}$ दिखाएँ कि (show that) $A^2=B^2=-C^2=I_2$ तथा (and) AB=-BA, AC=-CA तथा (and) BC=-CB
Sol :
A2=A.A$=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$
$= \left[\begin{array}{cc}0+1 & 0+0 \\ 0+0 & 1+0\end{array}\right]$
$=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=I_{2}$
B2=B.B$=\left[\begin{array}{cc}0 & -i \\ i & 0\end{array}\right]\left[\begin{array}{cc}0 & -i \\ i & 0\end{array}\right]$
$=\left[\begin{array}{cc}0-1^{2} & -0-0 \\ 0+0 & -i^{2}+0\end{array}\right]$
-C2=C.C$=-\left[\begin{array}{ll}i & 0 \\ 0 & -1\end{array}\right]\left[\begin{array}{ll}i & 0 \\ 0 & -i\end{array}\right]$
$=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=I_{2}$
$=-\left[\begin{array}{cc}i^{2}+0 & 0-0 \\ 0-0 & 0+i^{2}\end{array}\right]$
$=-\left[\begin{array}{rr}-1 & 0 \\ 0 & -1\end{array}\right]$
$=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=I_{2}$
$\therefore A^{2}=B^{2}=-C^{2}=I_2$
AB=$\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]\left[\begin{array}{ll}0 & -i \\ i & 0\end{array}\right]$
$=\left[\begin{array}{cc}0+i & -0+0 \\ 0+0 & -i+0\end{array}\right]$
$=\left[\begin{array}{cc}1 & 0 \\ 0 & -i\end{array}\right]$
-BA=$-\left[\begin{array}{cc}0 & -i \\ i & 0\end{array}\right]\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$
$=-\left[\begin{array}{cc}0-i & 0-0 \\ 0+0 & 1+0\end{array}\right]$
$=-\left[\begin{array}{cc}-i & 0\\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}i & 0 \\ 0 & -i\end{array}\right]$
$\therefore A B=-B A$
(iii) यदि (If) $A=\left[\begin{array}{lll}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right], B=\left[\begin{array}{lll}0 & 5 & 7 \\ 0 & 0 & 6 \\ 0 & 0 & 0\end{array}\right]$ तथा (and) $C=\left[\begin{array}{ccc}-1 & 3 & 5 \\ 1 & -3 & -5 \\ -1 & 3 & 5\end{array}\right]$ दिखाएँ कि (show that)
(i) $A^{2}=1$ (ii) $C^{2}=C$ (iii) $B^{4}=0$
Sol :
(i)
$A^{2}=A \cdot A=$ $\left[\begin{array}{lll}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right]\left[\begin{array}{lll}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right]$
$=\left[\begin{array}{ccc}0+0+1 & 0+0+0 & 0+0+0 \\ 0+0+0 & 0+1+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 1+0+0\end{array}\right]$
$=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=I$
(ii)
$c^{2}=c \cdot c$ $=\left[\begin{array}{ccc}-1 & 3 & 5 \\ 1 & -3 & -5 \\ -1 & 3 & 5\end{array}\right]\left[\begin{array}{ccc}-1 & 3 & 5 \\ 1 & -3 & -5 \\ -1 & 3 & 5\end{array}\right]$
$=\left[\begin{array}{ccc}1+3-5 & -3-9+15 & -5-15+25 \\ -1-3+5 & 3+9-15 & 5+15-25 \\ 1+3-5 & -3-9+15 & -5-15+25\end{array}\right]$
$=\left[\begin{array}{ccc}-1 & 3 & 5 \\ 1 & -3 & -5 \\ -1 & 3 & 5\end{array}\right]=C$
$\therefore c^{2}=c$
(iii) $B^{2}=B \cdot B$ $=\left[\begin{array}{ccc}0 & 5 & 7 \\ 0 & 0 & 6 \\ 0 & 0 & 0\end{array}\right]\left[\begin{array}{lll}0 & 5 & 7 \\ 0 & 0 & 6 \\ 0 & 0 & 0\end{array}\right]$
$=\left[\begin{array}{ccc}0+0+0 & 0+0+0 & 0+30+0 \\ 0+0+0 & 0+0+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+0\end{array}\right]$
$=\left[\begin{array}{lll}0 & 0 & 30 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$
$B^{4}=B^{2} \cdot B^{2}$
$=\left[\begin{array}{ccc}0 & 0 & 30 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]\left[\begin{array}{lll}0 & 0 & 30 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$
$=\left[\begin{array}{ccc}0+0+0 & 0+0+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+0\end{array}\right]$
$=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$
$B^{4}=0$
यदि (If) $\mathrm{A}=\left[\begin{array}{rrr}1 & 2 & 3 \\ 2 & 0 & -2\end{array}\right], \mathrm{B}=\left[\begin{array}{rrr}1 & 1 & -1 \\ 2 & 0 & 3 \\ 3 & -1 & 2\end{array}\right]$ तथा (and) $\mathrm{C}=\left[\begin{array}{rr}1 & 3 \\ 0 & 2 \\ -1 & 4\end{array}\right]$ तो A(BC) निकालें। इससे या किसी अन्य विधि से (AB)C को लिखें। (then find A(BC) . Hence or otherwise, write down (AB) C)
Question 15
साबित करे कि दो आव्यूह (Prove that the product of two matrices)
$\left[\begin{array}{cc}\cos ^{2} \theta & \cos \theta \cdot \sin \theta \\ \cos \theta \cdot \sin \theta & \sin ^{2} \theta\end{array}\right]$ तथा (and) $\left[\begin{array}{cc}\cos ^{2} \phi & \cos \phi \cdot \sin \phi \\ \cos \phi \cdot \sin \phi & \sin ^{2} \phi\end{array}\right]$ का गुणनफल एक शून्य आव्यूह है यदि θ और ɸ का अन्तप $\dfrac{\pi}{2}$ का विषम अपवर्त्य है । (is a zero matrix when θ and ɸ differ by an odd multiple of $\dfrac{\pi}{2}$)
Sol :
$\theta-\phi=(2 n-1) \frac{\pi}{2} \quad \ldots n \in 2$
$\left[\begin{array}{cc}\cos ^{2} \theta & \cos \theta sin\theta \\ \cos \theta \sin \theta & si n^{2} \theta\end{array}\right]\left[\begin{array}{cc}\cos ^{2} \phi & \cos \phi \sin \phi \\ \cos \phi \sin \phi & \sin ^{2} \phi\end{array}\right]$
$=\left[\begin{array}{ccc}\cos ^{2} \theta \cos ^{2} \phi+\cos \theta \sin \theta & \cos ^{2} \theta \cos \phi \sin \phi+\cos \theta sin \theta \sin ^{2} \phi \\ \cos \theta \sin \theta \cos ^{2} \phi+\sin ^{2} \theta \cos \phi+sin \phi & \cos \theta \sin \theta \cos \phi \sin \phi+\sin ^{2} \theta \sin ^{2} \phi\end{array}\right]$
$\left[\begin{array}{ll}\cos \theta \cos \phi(\cos \theta \cos \phi+\sin \theta \sin \phi) & \cos \theta \sin \phi(\cos \theta \cos \phi+\sin \theta \sin \phi) \\ \sin \theta \cos \phi(\cos \theta \cos \phi+\sin \theta sin \phi) & \sin \theta \sin \phi\left(\cos \theta \cos \phi+\sin \theta sin \phi\right)\end{array}\right]$
$=\left[\begin{array}{ll}\cos \theta \cos \phi \cos (\theta-\phi) & \cos \theta \sin \phi \cos (\theta-\phi) \\ \sin \theta \cos \phi \cos (\theta-\phi) & \sin \theta \sin \phi \cos (\theta-\phi)\end{array}\right]$
$=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]=0$
Question 19
यदि (If) $A=\begin{bmatrix}2&3&4\\1&2&3\\-1&1&2\end{bmatrix}$ , $B=\begin{bmatrix}1&3&0\\-1&2&1\\0&0&2\end{bmatrix}$ , AB तथा BA निकालें तथा दिखाएँ कि (find AB and BA and show that )AB≠BA
Sol :
Question 20
दिखाएँ कि (show that) $\left(\left[\begin{array}{ccc}1 & 0 & \omega^{2} \\ \omega & \omega^{2} & 1 \\ \omega^{2} & 1 & \omega\end{array}\right]+\left[\begin{array}{ccc}\omega & \omega^{2} & 1 \\ \omega^{2} & 1 & \omega \\ \omega & \omega^{2} & 1\end{array}\right]\right)\left[\begin{array}{c}1 \\ \omega \\ \omega^{2}\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$
Sol :
L.H.S
$=\left[\begin{array}{ccc}1+\omega & \omega+\omega^{2} & \omega^{2}+1 \\ \omega+\omega^{2} & \omega^{2}+1 & 1+\omega \\ \omega^{2}+\omega & 1+\omega^{2} & \omega+1\end{array}\right]\left[\begin{array}{c}1 \\ \omega \\ \omega^{2}\end{array}\right]$
$=\left[\begin{array}{ccc}-\omega^{2} & -1 & -\omega \\ -1 & -\omega & -\omega^{2} \\ -1 & -\omega & -\omega^{2}\end{array}\right]\left[\begin{array}{c}1 \\ \omega \\ \omega^{2}\end{array}\right]$
$=\left[\begin{array}{cc}-\omega^{2}-\omega-\omega^{3} \\ -1-\omega^{2}-\omega^{4} \\ -1-\omega^{2}-\omega^{4}\end{array}\right]=\left[\begin{array}{cc}-\omega^{2}-\omega-1 \\ -1 -\omega^{2}-\omega \\ -1-\omega^{2}-\omega\end{array}\right]$
$=\left[\begin{array}{c}-\left(\omega^{2}+\omega+1\right) \\ -\left(1+w^{2}+w\right) \\ -\left(1+w^{2}+w\right)\end{array}\right]$
$=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$
यदि (If) $A=\begin{bmatrix}1&0&-2\\3&-1&0\\-2&1&1\end{bmatrix}$ , $B=\begin{bmatrix}0&5&-4\\-2&1&3\\-1&0&2\end{bmatrix}$ तथा (and) $C=\begin{bmatrix}1&5&2\\-1&1&0\\0&-1&1\end{bmatrix}$ सत्यापित करें कि (verify that) A(B-C)=(AB-AC)
Sol :
Question 22
(i) x निकाले यदि (find x if) $\left[\begin{array}{ll}x & 1\end{array}\right]\left[\begin{array}{rr}1 & 0 \\ -2 & -3\end{array}\right]\left[\begin{array}{l}x \\ 3\end{array}\right]=0$
Sol :
(ii) $\left[\begin{array}{ll}2 x & 3\end{array}\right]\left[\begin{array}{rr}1 & 2 \\ -3 & 0\end{array}\right]\left[\begin{array}{l}x \\ 3\end{array}\right]=0$
Sol :
$\left[\begin{array}{cc}x & 1\end{array}\right]\left[\begin{array}{c}x+0 \\ -2 x-9\end{array}\right]=0$
$\left[x^{2}-2 x-9\right]=[0]$
$x^{2}-2 x-9=0$
a=1 , b=-2 , c=-9
$x=\frac{-(-2) \pm \sqrt{(-2)^{2}-4 \times1\times(-9)}}{2 \times 1}$
$x=\frac{2\pm \sqrt{4+36}}{2}$
$x=\frac{2 \pm 2 \sqrt{10}}{2}=$\frac{2(1 \pm \sqrt{10})}{2}$$
$x=1 \pm \sqrt{10}$
(iii) a और b का मान निकालें जिसके लिए निम्नलिखित सत्य है:
(Find the values of a and b for which the following holds)
$\begin{bmatrix}4&2\\3&-1\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}=\begin{bmatrix}-4\\2\end{bmatrix}$
Sol :
(iv) माना कि (Let) $A=\left[\begin{array}{rr}2 & -1 \\ 3 & 4\end{array}\right], B=\left[\begin{array}{ll}5 & 2 \\ 7 & 4\end{array}\right], C=\left[\begin{array}{ll}2 & 5 \\ 3 & 8\end{array}\right]$
एक आव्यूह D निकालें ताकि (find a matrix D such that) CD-AB=0
Sol :
C D-A B=D
$\left[\begin{array}{ll}2 & 5 \\ 3 & 8\end{array}\right] D-\left[\begin{array}{ll}2 & -1 \\ 3 & \phantom{-}4\end{array}\right]\left[\begin{array}{ll}5 & 2 \\ 7 & 4\end{array}\right]=0$
Let $D=\left[\begin{array}{ll}x & a \\ y & b\end{array}\right]$
$\left[\begin{array}{cc}2 & 5 \\ 3 & 8\end{array}\right]\left[\begin{array}{cc}x & a \\ y & b\end{array}\right]=\left[\begin{array}{cc}2 & -1 \\ 3 & 4\end{array}\right]\left[\begin{array}{cc}5 & 2 \\ 7 & 4\end{array}\right]$
$\left[\begin{array}{cc}2 x+5 y & 2 a+5 b \\ 3 x+8 y & 3 a+8 b\end{array}\right]=\left[\begin{array}{cc}10-7 & 4-4 \\ 15+28 & 6+16\end{array}\right]$
$\left[\begin{array}{cc}2 x+5 y & 2 a+5 b \\ 3 x+8 y & 3 a+8 b\end{array}\right]=\left[\begin{array}{cc}3 & 0 \\ 43 & 22\end{array}\right]$
$2 x+5 y=3..(i)\times 3$
$3 x+8 y=43..(ii)\times 2$
2a+5 b=0..(iii)
3a+8b=22...(iv)
$\begin{array}{rl}6 x+15 y^{2}=9 \\ -6 x+16 y=86 \\\hline y=77\end{array}$
Putting the value of y in equation (i)
2x+5(77)=3
2x+385=3
2x=-382
$x=\dfrac{-382}{2}=-191$
$\begin{aligned} 6 a +15 b &=0 \\6 a+16 b &=44 \\\hline b &=44 \end{aligned}$
Putting the value of b in equation (ii)
2a+5(44)=0
2a+220=0
$a=\dfrac{-220}{2}$
a=-110
$\therefore \quad D=\left[\begin{array}{cc}-191 & -110 \\ 77 & 44\end{array}\right]$
(v) आव्यूह X निकाले ताकि (Find the matrix X so that) $x\left[\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6\end{array}\right]=\left[\begin{array}{rrr}-7 & -8 & -9 \\ 2 & 4 & 6\end{array}\right]$
Sol :
Let $X=\left[\begin{array}{ll}x & a\\ y & b\end{array}\right]$
$\left[\begin{array}{ll}x & a \\ y & b\end{array}\right]\left[\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6\end{array}\right]=\left[\begin{array}{rrr}-7 & -8 & -9 \\ 2 & 4 & 6\end{array}\right]$
$\left[\begin{array}{ccc}x+4 a & 2 x+5 a & 3 x+64 \\ y+4 b & 2 y+5 b & 3 y+6 b\end{array}\right]=\left[\begin{array}{ccc}-7 & -8 & -9 \\ 2 & 4 & 6\end{array}\right]$
$x+4 a=-7..(i) \times 2$
$2x+5 a=-8..(ii) \times 1$
$y+4b=2..(iii)\times 2$
$2y+5b=4..(iv)\times 1$
Question 23
(i) x का मान निकाले ताकि (Find the value of x , such that)
$\left[\begin{array}{lll}1 & x & 1\end{array}\right]\left[\begin{array}{ccc}1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2\end{array}\right]\left[\begin{array}{l}1 \\ 2 \\ x\end{array}\right]=0$
Sol :
$\left[\begin{array}{lll}1 & x & 1\end{array}\right]\left[\begin{array}{c}1+6+2 x \\ 2+10+x \\ 15+6+2 x\end{array}\right]=0$
$\left[\begin{array}{lll}1 & x & 1\end{array}\right]\left[\begin{array}{c}7-2 x \\ 12+x \\ 21+2 x\end{array}\right]=0$
$\left[7+2 x+12 x+x^{2}+21+2 x\right]=[0]$
$2^{2}+16 x+28=0$
$x^{2}+14 x+2 x+28=0$
$x(x+14)+2(x+14)=0$
$(x+14)(x+2)=0$
⇒x=-14 , -2
(ii) x का मान निकालें यदि (Find the value of x , if)
$\begin{bmatrix}1&x&1\end{bmatrix}\begin{bmatrix}1&2&3\\4&5&6\\3&2&5\end{bmatrix}\begin{bmatrix}1\\-2\\3\end{bmatrix}=0$
Sol :
(iii) x निकालें यदि (Find x, if) $\left[\begin{array}{lll}x&-5 & -1\end{array}\right]\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]\left[\begin{array}{l}x \\ 4 \\ 1\end{array}\right]=0$
Sol :
(iv) x के किस मान के लिए
(For what value of x)
$\left[\begin{array}{lll}1 & 2 & 1\end{array}\right]\left[\begin{array}{lll}1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2\end{array}\right]\left[\begin{array}{l}0 \\ 2 \\ x\end{array}\right]=0$?
Sol :
(i) यदि (If) $\left[\begin{array}{lll}x & 4 & 1\end{array}\right]\left[\begin{array}{ccc}2 & 1 & 2 \\ 1 & 0 & 2 \\ 0 & 2 & -4\end{array}\right]\left[\begin{array}{r}x \\ 4 \\ -1\end{array}\right]=0$, x निकालें (find x)
Sol :
(ii) यदि (If) $\left[\begin{array}{lll}1 & 1 & x\end{array}\right]\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 1 & 0\end{array}\right]\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=0$, x निकाले (find x)
Sol :
एक 2×2 आवयूह B निकाले ताकि (Find the 2×2 matrix B such that)
$\left[\begin{array}{ll}6 & 5 \\ 5 & 6\end{array}\right] \mathrm{B}=\left[\begin{array}{ll}11 & 0 \\ 0 & 11\end{array}\right]$
Sol :
Let $B=\left[\begin{array}{ll}a & c \\ b & d\end{array}\right]$
$\left[\begin{array}{ll}6 & 5 \\ 5 & 6\end{array}\right]\left[\begin{array}{ll}a & c \\ b & d\end{array}\right]=\left[\begin{array}{ll}11 & 0 \\ 0 & 11\end{array}\right]$
$\left[\begin{array}{ll}6 a+5 b & 6 c+5 d \\ 5 a+6 b & 5 c+6 d\end{array}\right]=\left[\begin{array}{cc}11 & 0 \\ 0 & 1\end{array}\right]$
6a+5b=11..(i)
5a+6b=0..(ii)
6c+5d=0...(iii)
5c+6d=11...(iv)
$\left[\begin{array}{ll}5 & 4 \\ 1 & 1\end{array}\right] X=\left[\begin{array}{rr}1 & -2 \\ 1 & 3\end{array}\right]$
बिना आव्यूह के प्रतिलोम की अवधारणा का प्रयोग किए आव्यूह (Without using the concept of inverse of a matrix, find the matrix) $\left[\begin{array}{ll}x & y \\ z & u\end{array}\right]$ निकालें ताकि (such that) $\left[\begin{array}{rr}5 & -7 \\ -2 & 3\end{array}\right]\left[\begin{array}{ll}x & y \\ z & u\end{array}\right]=\left[\begin{array}{rr}-16 & -6 \\ 7 & 2\end{array}\right]$
Question 29
आव्यूह X निकालें ताकि
(Find the matrix X such that)
$X\left[\begin{array}{ll}2 & 3 \\ 4 & 5\end{array}\right]=\left[\begin{array}{rr}0 & -4 \\ 10 & 3\end{array}\right]$
Sol :
Question 30
यदि (if) $A=\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]$ , (find) $A^{2}-5 A-14 I$ , निकाले , जहाँ I एक इकाई आव्यूह हैं (where I is a unit matrix)
Sol :
$A^{2}-5 A-14 I$
$=\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]-5\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]-14\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]$
$=\left[\begin{array}{cc}9+20 & -15-10 \\ -12-8 & 20+4\end{array}\right]-\left[\begin{array}{cc}15 & -25 \\ -20 & 10\end{array}\right]-\left[\begin{array}{cc}14 & 0 \\ 0 & 14\end{array}\right]$
$=\left[\begin{array}{ccc}29 & -25 \\ -20 & \phantom{-} 24\end{array}\right]-\left[\begin{array}{cc}15 & -25 \\ -20 & 10\end{array}\right]-\left[\begin{array}{cc}14 & 0 \\ 0 & 14\end{array}\right]$
$=\left[\begin{array}{cc}14 & 0 \\ 0 & 14\end{array}\right]-\left[\begin{array}{cc}14 & 0 \\ 0 & 14\end{array}\right]$
$=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]=0$
Question 31
(i) यदि (If) $A=\left[\begin{array}{ll}2 & 3 \\ 4 & 5\end{array}\right]$, सत्यापित करें कि (verify that) $A^{2}-7 A-2 I=0$
Sol :
(ii) यदि आव्यूह (If the matrix) $A=\left[\begin{array}{ll}5 & 3 \\ 12 & 7\end{array}\right]$ , जहाँ I एक इकाई आव्यूह है , तो सत्यापित करें, कि (where I is a unit matrix , then verify that)
$A^{2}-12 A-I=O$
Sol :
Question 32
सत्यापित करे कि (Verify that) $A=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]$ समिकरण (satisfies the equation) $A^{3}-4 A^{2}+A=0$ को संतुष्ट करता है ।
Sol :
$A^{2}=A A=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]$
$=\left[\begin{array}{cc}4+3 & 6+6 \\ 2+2 & 3+4\end{array}\right]$
$=\left[\begin{array}{ll}7 & 12 \\ 4 & 7\end{array}\right]$
$A^{3}=A^{2} \cdot A=\left[\begin{array}{ll}7 & 12 \\ 4 & 7\end{array}\right]\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]$
$=\left[\begin{array}{cc}14+12 & 21+24 \\ 8+7 & 12+14\end{array}\right]$
$=\left[\begin{array}{cc}26 & 45 \\ 15 & 26\end{array}\right]$
L.H.S
$4^{3}-4 A^{2}+A$
$=\left[\begin{array}{cc}26 & 45 \\ 15 & 26\end{array}\right]-4\left[\begin{array}{cc}7 & 12 \\ 4 & 7\end{array}\right]+\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]$
$=\left[\begin{array}{cc}28 & 48 \\ 16 & 28\end{array}\right]-\left[\begin{array}{cc}28 & 48 \\ 16 & 28\end{array}\right]$
$=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]=0$
Question 33
(i) यदि (If) $A=\left[\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right]$ तथा (and) $\mathrm{I}_{2}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$, k ज्ञात करें ताकि (find k so that ) $A^{2}=8 A+k I_{2}$
Sol :
$\left[\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right]=8\left[\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right]+k\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]$
$\left[\begin{array}{cc}1-0 & 0+0 \\ -1-7 & -0+4 y\end{array}\right]=\left[\begin{array}{cc}8 & 0 \\ -8 & 56\end{array}\right]+\left[\begin{array}{cc}k & 0 \\ 0 & k\end{array}\right]$
$\left[\begin{array}{cc}1 & 0 \\ -8 & 4\end{array}\right]=\left[\begin{array}{cc}8+k & 0 \\ -8 & 56+k\end{array}\right]$
⇒1=8+k
⇒-7=k
(ii) यदि (If) $A=\left[\begin{array}{ll}3 & -2 \\ 4 & -2\end{array}\right]$, k ज्ञात करें ताकि find k such that $A^{2}=k A-2 I_{2}$
Sol :
(iii) यदि (If) $A=\left[\begin{array}{rr}1 & 0 \\ -1 & 7\end{array}\right]$ , k ज्ञात करें ताकि (find k such that)
$A^{2}-8 A+k I=O$
Sol :
Question 34
(i) यदि (If) $A=\left[\begin{array}{rr}3 & 1 \\ -1 & 2\end{array}\right], f(A)$ निकाले जहाँ (find $f(A)$ , where) $f(x)=x^{2}-5 x+7$
Sol :
$f(A)=A^{2}-5 A+7 I$
$=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]-5\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]+7\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]$
$=\left[\begin{array}{cc}9-1 & 3+2 \\ -3-2 & -1+4\end{array}\right]-\left[\begin{array}{cc}15 & 5 \\ -5 & 10\end{array}\right]+\left[\begin{array}{cc}7 & 0 \\ 0 & 7\end{array}\right]$
$=\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]-\left[\begin{array}{cc}15 & 5 \\ -5 & 10\end{array}\right]+\left[\begin{array}{cc}7 & 0 \\ 0 & 7\end{array}\right]$
$=\left[\begin{array}{cc}15 & 5 \\ -5 & 10\end{array}\right]-\left[\begin{array}{cc}15 & 5 \\ -5 & 10\end{array}\right]$
$=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]=0$
(ii) यदि (If) $A=\left[\begin{array}{rr}3 & 4 \\ -4 & -3\end{array}\right]$, f(A) निकालें जहाँ (find f(A), where) $f(x)=x^{2}-5 x+7$
Sol :
(iii) $A=\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right], f(x)=x^{2}-2 x-3$ दिखाएँ कि (show that) f(A)=0
Sol :
Question 35
(i) यदि (if) $A=\left[\begin{array}{ll}2 & -1 \\ 3 & 2\end{array}\right]$ तथा (and) $B=\left[\begin{array}{cc}0 & 4 \\ -1 & 7\end{array}\right]$ (find) $\left(3 A^{2}-2 B\right)$ निकाले ।
Sol :
$3 A^{2}-2 B=3\left[\begin{array}{cc}2 & -1 \\ 3 & 2\end{array}\right]\left[\begin{array}{cc}2 & -1 \\ 3 & 2\end{array}\right]-2\left[\begin{array}{cc}0 & 4 \\ -1 & 7\end{array}\right]$
$=3\left[\begin{array}{cc}4-3 & -2-2 \\ 6+6 & -3+4\end{array}\right]-\left[\begin{array}{cc}0 & 8 \\ -2 & 14\end{array}\right]$
$=3\left[\begin{array}{rr}1 & -4 \\ 12 & 1\end{array}\right]-\left[\begin{array}{cc}0 & 8 \\ -2 & 14\end{array}\right]$
$=\left[\begin{array}{rr}3 & -12 \\ 36 & 3\end{array}\right]-\left[\begin{array}{cc}0 & 8 \\ -2 & 14\end{array}\right]$
$=\left[\begin{array}{cc}3 & -20 \\ 38 & -11\end{array}\right]$
(ii) यदि (If)$A=\left[\begin{array}{cc}2 & -1 \\ 3 & 2\end{array}\right] $, $ B=\left[\begin{array}{cc}0 & 4 \\ -1 & 7\end{array}\right]$ ( find ) $3 A^{2}-2 B+I$ निकाले ।
Sol :
Question 36
(i) माना कि (Let) $f(x)=x^{2}-5 x+6$ (find) $f(A)$ निकाले यदि (if)
$A=\left[\begin{array}{lll}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0\end{array}\right]$
Sol :
$f(a)=A^{2}-5 A+6 I$
(ii) यदि (If) $A=\left[\begin{array}{rrr}1 & 2 & 3 \\ 3 & -2 & 1 \\ 4 & 2 & 1\end{array}\right]$
दिखाएँ कि (show that) $A^{3}-23 A-40 I=0$
Sol :
(iii) यदि (If) $A=\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]$, साबित करे कि (prove that) $A^{3}-6 A^{2}+7 A+2 I=0$
(iv) यदि (If) $A=\left[\begin{array}{rr}3 & 1 \\ -1 & 2\end{array}\right]$, दिखाएँ कि (show that)
Question 37
यदि (if) $A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]$ , साबित करे कि (prove that)
$A^{n}=\left[\begin{array}{ll}1 & n \\ 0 & 1\end{array}\right]$ सभी $n \in N$ के लिए (for all $n \in \mathbf{N}$)
Sol :
माना $P(n): A^{n}=\left[\begin{array}{ll}1 & n \\ 0 & 1\end{array}\right]$
$\begin{array}{rl}n=1 & A^{1}=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\end{array}$..(i)
P(1)=सत्य है ।
माना P(k) सत्य है
$P(k): A^{k}=\left[\begin{array}{ll}1 & k \\ 0 & 1\end{array}\right]$
तो साबित करना है कि P(k+1) भी सत्य होगा
$P(k+1): \quad A^{k+1}=\left[\begin{array}{cc}1 & k+1 \\ 0 & 1\end{array}\right]$
(i) मे दोनो तरफ A से गुना करने पर
$A^{k} \cdot A^{1}=\left[\begin{array}{ll}1 & k \\ 0 & 1\end{array}\right] \cdot A$
$A^{k+1}=\left[\begin{array}{ll}1 & k \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]$
$=\left[\begin{array}{cc}1+0 & 1+k \\ 0+0 & 0-1\end{array}\right]$
$=\left[\begin{array}{ll}1 & k+1 \\ 0 & 1\end{array}\right]$
$n \in N$ के लिए $P(n): A^{n}=\left[\begin{array}{ll}1 & n \\ 0 & 1\end{array}\right]$ सत्य है ।
Question 38
यदि (if)$A=\left[\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right]$ दिखाएँ कि (show that) $A^{n}=\left[\begin{array}{cc}1+2 n & -4 n \\ n & 1-2 n\end{array}\right]$
जहाँ n एक धन पूर्णाक है (where n is a positive integer)
Sol :
माना P(n): $A^{n}=\left[\begin{array}{cc}1+2 n & -4 n \\ n & 1-2 n\end{array}\right]$
n=1 , $A^{\prime}=\left[\begin{array}{ccc}1+2(1) & -4(1) \\ 1 & 1-2(1)\end{array}\right]$
$=\left[\begin{array}{rr}3 & -4 \\ 1 & -1\end{array}\right]$
P(1) सत्य हैं
माना P(k) सत्य हैं
P(k) : $A^{k}=\left[\begin{array}{cc}1+2 k & -4 k \\ k & 1-2 k\end{array}\right]$..(i)
तो , साबित करना है कि P(k+1) भी सत्य होगा ।
$A^{k+1}=\left[\begin{array}{cc}1+2(k-n) & -4(k+1] \\ k+1 & 1-2(k+1)\end{array}\right]$
(i) मे दोनो तरफ A से गुना करने पर ,
$A^{k} \cdot A=\left[\begin{array}{cc}1+2 k & -4 k \\ k & 1-2 k\end{array}\right]\left[\begin{array}{cc}3 & -4 \\ 1 & -1\end{array}\right]$
$A^{k+1}=\left[\begin{array}{cc}3+6 k-4 k & -4-8 k+4 k \\ 3 k+1-2 k & -4 k-1+2 k\end{array}\right]$
$A^{k-1}=\left[\begin{array}{cc}3+2 k & -4-4 k \\ k+1 & -2 k-1\end{array}\right]$
$A^{k+1}=\left[\begin{array}{cc}1+2 k+2 & -4 k-4 \\ k+1 & 1-2 k-2\end{array}\right]$
$=\left[\begin{array}{cc}1+2(k+1) & -4(k+1) \\ k+1 & 1-2(k+1)\end{array}\right]$
$\therefore n \in z^{+}$ के लिए $A^{n}=\left[\begin{array}{cc}1+2 n & -4 n \\ n & 1-2 n\end{array}\right]$ सत्य है ।
Question 39
यदि (if) A=diag.[a b c] दिखाएँ कि (show that) $\mathbf{A}^{a}=\operatorname{diag}\left[a^{n} \quad b^{n} \quad c^{n}\right]$ सभी (for all) $n \in \mathbf{N}$ के लिए
Sol :
माना $A^{n}=\left[\begin{array}{ccc}a^{n} & 0 & 0 \\ 0 & b^{n} & 0 \\ 0 & 0 & c^{n}\end{array}\right]$
n=1 , $A^{\prime}=\left[\begin{array}{lll}a^{1} & 0 & 0 \\ 0 & b^{1} & 0 \\ 0 & 0 & c^{1}\end{array}\right]$
$=\left[\begin{array}{lll}a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c\end{array}\right]$
P(1) सत्य हैं
माना P(k) सत्य हैं
P(k) : $A^{k}=\left[\begin{array}{lll}a^{k} & 0 & 0 \\ 0 & b^{k} & 0 \\ 0 & 0 & c^{k}\end{array}\right]$
तो , साबित करना है कि P(k+1) भी सत्य होगा ।
P(k+1): $A^{k+1}=\left[\begin{array}{ccc}a^{k+1} & 0 & 0 \\ 0 & b^{k+1} & 0 \\ 0 & 0 & c^{k+1}\end{array}\right]$
(i) मे दोनो तरफ A से गुना करने पर ,
$A^{k} \cdot A=\left[\begin{array}{lll}a^{k} & 0 & 0 \\ 0 & b^{k} & 0 \\ 0 & 0 & c^{k}\end{array}\right]\left[\begin{array}{lll}a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c\end{array}\right]$
$A^{k+1}=\left[\begin{array}{ccc}a^{k+1} & 0 & 0 \\ 0 & b^{k+1} & 0 \\ 0 & 0 & c^{k-1}\end{array}\right]$
$\therefore n \in z^{+}$ के लिए $A^{n}=\left[\begin{array}{lll}a^{n} & 0 & 0 \\ 0 & b^{n} & 0 \\ 0 & a & c^{1}\end{array}\right]$ सत्य है ।
Question 40
यदि (if) $A=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$ ,
तो गणितीय आगमन सिद्धान्त से साबित करे कि (prove by principle of mathematical induction that)
$A^{n}=\left[\begin{array}{cc}\cos n \alpha & \sin n \alpha \\ -\sin n \alpha & \cos n \alpha\end{array}\right]$
प्रत्येक प्राकृत संख्या n के लिए (for every natural number n)
Sol :
Let P(n): $A^{n}=\left[\begin{array}{cc}\cos n \alpha & \sin n \alpha \\ -\sin n \alpha & \cos n \alpha\end{array}\right]$
when n=1 ,
$A^{1}=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$
P(1) is true
Let P(k) be true
$P(k): A^{k}=\left[\begin{array}{cc}\cos k \alpha & \sin k \alpha \\ -\sin k \alpha & \cos k \alpha\end{array}\right]$
then prove that P(k+1) is true
P(k+1):
$A^{k+1}=\left[\begin{array}{cc}\cos (k+1) \alpha & \sin (k+1) \alpha \\ -\sin (k+1) \alpha & \cos (k+1) \alpha\end{array}\right]$
<to be added>
$A^{k} \cdot A=\left[\begin{array}{cc}\cos k \alpha & \sin k \alpha \\ -\sin k \alpha & \cos k \alpha\end{array}\right]\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$
$A^{k+1}=\left[\begin{array}{ll}\cos k \alpha+\cos \alpha-\sin \alpha \sin \alpha & cosk\alpha \sin \alpha+ \sin k \alpha \cos \alpha\\ -\cos \alpha \sin k \alpha-\cos k\alpha\sin \alpha & -\sin k \alpha\sin \alpha+\cos k\alpha \cos \alpha\end{array}\right]$
$=\left[\begin{array}{ll}\cos (k\alpha+\alpha) & \sin (k \alpha+\alpha) \\ -\sin (k\alpha+\alpha) & \cos (k+\alpha)\end{array}\right]$
$A^{k+1}=\left[\begin{array}{cc}\cos (k+1) \alpha & \sin (k+1)+\alpha \\ -\sin (k-1) \alpha & \cos (k+1) \alpha \end{array}\right]$
∴ nϵN के लिए , $A^{n}=\left[\begin{array}{cc}\cos n \alpha & \sin \alpha \\ -\sin n \alpha & \cos n \alpha\end{array}\right]$ सत्य है ।
यदि (If) $A=\left[\begin{array}{ll}\cos \theta & i \sin \theta \\ i \sin \theta & \cos \theta\end{array}\right]$ , तो गणितीय आगमन सिद्धान्त से साबित करें कि (then prove by principle f mathematical induction that) $A^{n}=\left[\begin{array}{cc}\cos n \theta & i \sin n \theta \\ i \sin n \theta & \cos n \theta\end{array}\right]$ , जहाँ (where) n∊N
Sol :
Question 42
यदि A और B समान कोटिवाले वर्ग आव्यूह इस प्रकार है कि AB=BA , तो गणितीय आगमन सिद्धान्त से साबित करे कि $AB^n=B^nA$ . इसके अतिरिक्त यह भी साबित करें कि $(\mathrm{AB})^{n}=\mathrm{A}^{n} \mathrm{~B}^{n}$ सभी n∈N के लिए।
[If A and B are square matrices of the smae order such that AB=BA then prove by principle of mathematical induction that $AB^n=B^nA$. Further, prove that $(\mathrm{AB})^{n}=\mathrm{A}^{n} \mathrm{~B}^{n}$ for all n∈N ]
Sol :
Let P(n): $A B^{n}=B^{n} A$
where n∈N
when n=1 ,
Question 43
किसी व्यापार संघ के पास 30000 रूपयों का कोष है जिसे दो भिन्न-भिन्न प्रकार के बांडों में निवेशित करना है। प्रथम बांड पर 5% वार्षिक तथा द्वितीय बांड पर 7% वार्षिक ब्याज प्राप्त होता है। आव्यूह गुणन के प्रयोग द्वारा यह निर्धारित कीजिए कि 30000 रुपयों के कोष को दों प्रकार के बांडों मे निवेश करने के लिए प्रकार बाँटे जिससे व्यापार संघ को प्राप्त कुल वार्षिक ब्याज
(a) 1800 रु हो।
(b) 2000 रु हो।
[A trust fund has 30000 that must be invested in two different types of bonds. The first pays 5% interest per year and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide 30000 among the two types of bonds. If the trust fund must obtain an annual total interest of :
(a) 1800
(b) 2000
]
Sol :
<to be added>
Question 44
एक दुकान के स्टाँक मे 20 दर्जन कमीज, 15 दर्जन पायजामा और 25 दर्जन मोजे है । यदि इनका विक्रय मूल्य 50 प्रति कमीज , 90 प्रति पायजामा तथा 12 प्रति जोड़ा मोजा हैं , तो दुकान के मालिक द्वारा सभी वस्तुओ को बेचने के बाद प्राप्त कुल राशि निकाले ।
[A store has in stock 20 dozen shirts, 15 dozen trousers and 25 dozen pair f socks. If the selling prices are 50 per shirt, 90 per trouser and 12 per pair of socks then find the total amount the store owner will get after selling all items in the stock]
Sol :
कुल धन राशि
$=\left[\begin{array}{lll}240 & 180 & 300\end{array}\right]\left[\begin{array}{l}50 \\ 90 \\ 12\end{array}\right]$
=12000+16200+3600
=31800
किसी विशेष विद्यालय के सहकारी दुकान में 10 दर्जन भौतिक विज्ञान, 8 दर्जन रसायन विज्ञान और 5 दर्जन गणित की पुस्तकें है । इनका विक्रय मूल्य क्रमशः 8.30 रु०, 3.45 रु० तथा 4.50 रु० प्रति पुस्तक है। सभी पुस्तकों का विक्रय करने पर दुकानदार द्वारा कुल प्राप्त धनराशि निकालें।
[Co-operative store of a particular school has 10 dozen physics books. 8 dozen chemistry books and 5 dozen mathematics books. Their selling prices are Rs. 8.30, Rs. 3.45 and Rs. 4.50 each book respectively. Find the total amount the store keeper will receive from selling all tile items.]
Sol :
किसी स्कूल की पुस्तकों की दुकान में 10 दर्जन रसायन विज्ञान, 8 दर्जन भौतिक विज्ञान तथा 10 दर्जन अर्थशास्त्र की पुस्तकें हैं। इन पुस्तकों का विक्रय मूल्य क्रमशः 80 रु०, 60 रु० तथा 40 रु० प्रति पुस्तक है। आव्यूह बीजगणित के प्रयोग द्वारा ज्ञात कीजिए कि सभी पुस्तकों को बेचने से दुकान को कुल कितनी धनराशि प्राप्त होगी ।
[The book-shop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling price are Rs. 80, Rs. 60 and Rs. 40 each respectively. Find the total amount the book-shop will receive from selling all the books using matrix algebra.]
Sol :
Question 47
एक निर्माता तीन प्रकार प्रकार की वस्तुएँ x, y तथा z का उत्पादन करता है जिन का वह दो बाजारों में विक्रय करता है। वस्तुओं की वार्षिक आय बिक्री नीचे सूचित (निर्देशित) है :
[A manufacturer produces three products x, y, z which he sells in two markets. Annual sale are indicated below.l
| बाजार (Market) | उत्पाद (Products) |
|---|---|
| I 10000 | 2000 18000 |
| II 6000 | 20000 8000 |
(a) यदि x, y तथा z के इकाई विक्रय मूल्य क्रमशः 2.50 रु० ,1.50 रु० तथा 1.00 रु० हो तो आव्यूह बीजगणित की सहायता से ज्ञात कीजिए कि प्रत्येक बाजार से कुल राजस्व कितना प्राप्त होगा।
[If unit sale prices of x, y and z are Rs. 2.50, Rs. 1.50 and Rs. 1.00 respectively. Find the total revenue in each market with the help of matrix algebra.]
(b) यदि उपरोक्त तीनों उत्पादों का इकाई लागत मूल्य क्रमशः 2.00 रु०, 1.00 रु० तथा 50 पैसे हो, तो कुल लाभ ज्ञात करें ।
[If the unit costs of the above three commodities are Rs. 2.00Rs. 1.00 and 50 paise respectively. Find the gross profit.]
Sol :
<to be added>
<to be added>
बाजार-I की वस्तुओ मे लागत
$=\left[\begin{array}{llll}10000 & 2000 &18000\end{array}\right]\begin{bmatrix}2\\1\\0.50\end{bmatrix}$
=20000+2000+9000
=31000
कुल लाभ =46000-31000
=15000
बाजार की वस्तुओ मे लागत$=\left[\begin{array}{llll}6 000 & 20000 &8000\end{array}\right]\begin{bmatrix}2\\1\\0.50\end{bmatrix}$
=12000+20000+4000
=36000
कुल लाभ =53000-36000
=17000