KC Sinha Mathematics Solution Class 12 Chapter 5 आव्यूह ( Matrices ) Exercise 5.2 (Q8-Q14)

Exercise 5.2

Ex-5.2

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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) $B^{2}=B \cdot B$ $=\left[\begin{array}{ccc}0 & 5 & 7 \\ 0 & 0 & 6 \\ 0 & 0 & 0\end{array}\right]\left[\begin{array}{lll}0 & 5 & 7 \\ 0 & 0 & 6 \\ 0 & 0 & 0\end{array}\right]$

$=\left[\begin{array}{ccc}0+0+0 & 0+0+0 & 0+30+0 \\ 0+0+0 & 0+0+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+0\end{array}\right]$

$=\left[\begin{array}{lll}0 & 0 & 30 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$

$B^{4}=B^{2} \cdot B^{2}$

$=\left[\begin{array}{ccc}0 & 0 & 30 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]\left[\begin{array}{lll}0 & 0 & 30 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$

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

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

$B^{4}=0$