Question 1
निम्नलिखित आव्यूहों के परिवर्त तथा सहखण्डज निकालें छ
[Find the transpose and adjoint of the following matrices]
(i) $\left[\begin{array}{ll}2 & 3 \\ 5 & 1\end{array}\right]$
Sol :
$A=\left[\begin{array}{ll}2 & 3 \\ 5 & 1\end{array}\right]$
$A^{\prime}orA^{T}=\left[\begin{array}{ll}2 & 5 \\ 3 & 1\end{array}\right]$
$\left|\begin{array}{ll}+ & – \\ – & +\end{array}\right|$
A11=1 , A12=-5
A21=-3 , A22=2
adjA $=\left[\begin{array}{cc}1 & -5 \\ -3 & 2\end{array}\right]^{\prime}$
$=\left[\begin{array}{cc}1 & -3 \\ -5 & 2\end{array}\right]$
(ii) $\left[\begin{array}{ll}\sec \theta & \operatorname{cosec} \theta \\ \sin \theta & \cos \theta\end{array}\right]$
Sol :
$A=\left[\begin{array}{cc}\sec \theta & cosec \theta \\ \sin \theta & \cos \theta\end{array}\right]$
$A^{\prime}=\left[\begin{array}{ll}sec \theta & \sin \theta \\ cosec \theta & \cos \theta\end{array}\right]$
A11=cosθ , A12=-sinθ
A21=-cosecθ , A22=secθ
adjA $=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ -\operatorname{cosec} \theta & \sec \theta\end{array}\right]^{\prime}$
$=\left[\begin{array}{cc}cos\theta & -\operatorname{cosec} \theta \\ -\sin \theta & \sec \theta\end{array}\right]$
(iii) $\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$
Sol :
(iv) $\left[\begin{array}{rr}2 & 3 \\ -4 & -6\end{array}\right]$
Sol :
(v) $\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$
Question 2
Find the adjoint of the following matrices:
(i) $\left[\begin{array}{lll}1 & 2 & 3 \\ 2 & 3 & 2 \\ 3 & 3 & 4\end{array}\right]$
Sol :
$A=\left[\begin{array}{lll}1 & 2 & 3 \\ 2 & 3 & 2 \\ 3 & 3 & 4\end{array}\right]$ $\left| \begin{array}{ccc}+ & -&+ \\ -&+ & – \\ + & -&+\end{array}\right|$
$A_{11}=\left|\begin{array}{cc}3 & 2 \\ 3 & 4\end{array}\right|$
=12-6
=6
$A_{12}=-\left|\begin{array}{cc}2 & 2 \\ 3 & 4\end{array}\right|$
=-(-8-6)
=-2
$A_{13}=\left|\begin{array}{ll}2 & 3 \\ 3 & 3\end{array}\right|$
=6-9
=-3
$A_{2 1}=-\left|\begin{array}{ll}2 & 3 \\ 3 & 4\end{array}\right|$
=-(8-9)
=1
$A_{22}=\left|\begin{array}{ll}1 & 3 \\ 3 & 4\end{array}\right|$
=4-9
=-5
$A_{23}=-\left|\begin{array}{cc}1 & 2 \\ 3 & 3\end{array}\right|$
=-(3-6)
=3
$A_{31}=\left|\begin{array}{cc}2 & 3 \\ 3 & 2\end{array}\right|$
=4-9
=-5
$A_{32}=-\left|\begin{array}{ll}1 & 3 \\ 2 & 2\end{array}\right|$
=-(2-6)
=4
$A_{33}=\left|\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right|$
=3-4
=-1
$\operatorname{adj} A=\left[\begin{array}{ccc}6 & -2 & -3 \\ 1 & -5 & 3 \\ -5 & 4 & -1\end{array}\right]’$
$=\left[\begin{array}{ccc}6 & 1 & -5 \\ -2 & -5 & 4 \\ -3 & 3 & -1\end{array}\right]$
(ii) $\left[\begin{array}{rrr}1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3\end{array}\right]$
Sol :
(iii) $\left[\begin{array}{rrr}1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1\end{array}\right]$
Sol :
Question 3
Find the inverse of the following matrices
(i) $\left[\begin{array}{rr}2 & 5 \\ -3 & 1\end{array}\right]$
Sol :
$A=\left[\begin{array}{ll}2 & 5 \\ -3 & 1\end{array}\right]$
$|A|=\left|\begin{array}{ll}2 & 5 \\ -3 & 1\end{array}\right|$
=2+15
=17≠0
<to be added>
$A_{11}=1$ ,$A_{12}=-(-3)=3$ ,$A_{21}=-5$
$A_{22}=2$
$a d{j} A-\left[\begin{array}{cc}1 & 3 \\ -5 & 2\end{array}\right]’$
$=\left[\begin{array}{cc}1 & -5 \\ 3 & 2\end{array}\right]$
$\therefore A^{-1}=\frac{1}{|A|} \cdot$ adj A
$=\frac{1}{17}\left[\begin{array}{ll}1 & -5 \\ 3 & 2\end{array}\right]$
(ii) $\left[\begin{array}{rr}2 & -2 \\ 1 & 3\end{array}\right]$
Sol :
(iii) $\left[\begin{array}{rr}a+i b & c+i d \\ -c+i d & a-i b\end{array}\right]$ where
$a^{2}+b^{2}+c^{2}+d^{2}=1$
Sol :
$A=\left[\begin{array}{rr}a+i b & c+i d \\ -c+i d & a-i b\end{array}\right]$
|A|=(a+ib)(a-ib)-(-c+id)(c+id)
$=a^{2}-1^{2} b^{2}+c^{2}-i^{2} d^{2}$
$=a^{2}+b^{2}+c^{2}+d^{2}$
=1≠0
<to be added>
$A_{11}=a-i b$ , $A_{12}=-(-c+i d)$=c-id
$A_{21}=-(c+i d)$
=-c-id
$A_{22}=a+ib$
$ad{j} A=\left[\begin{array}{cc}a-ib & c-i d \\ -c-i d & a+i b\end{array}\right]$
$=\left[\begin{array}{cc}a-i b & -c-i d \\ c-i d & a+i b\end{array}\right]$
$A^{-1}=\frac{1}{(A)} \cdot a d j A$
$=\frac{1}{1}\left[\begin{array}{cc}a-i b & -c-i d \\ c-i d & a+2 b\end{array}\right]$
$A^{-1}=\left[\begin{array}{cc}a-i b & -c-i d \\ c-i d & a+i b\end{array}\right]$
(iv) $\left[\begin{array}{cc}1 & \tan x \\ -\tan x & 1\end{array}\right]$
Sol :
(v) $\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$. जहाँ (where) ad-bc≠0
Sol :
Question 4
Find the inverse of the following matrices
(i) $\left[\begin{array}{lll}1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4\end{array}\right]$
Sol :
$A=\left[\begin{array}{lll}1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4\end{array}\right]$
$|A|=\left|\begin{array}{lll}1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4\end{array}\right|$
$\left(\begin{array}{l}R_{1} \rightarrow R_{1}-R_{2} \\ R_{2} \rightarrow R_{2}-R_{3}\end{array}\right)$
$=\left|\begin{array}{rrr}0 & -1 & 0 \\ 0 & 1 & -1 \\ 1 & 3 & 4\end{array}\right|$
$=1\left|\begin{array}{cc}-1 & 0 \\ 1 & -1\end{array}\right|$
=1-0
=1≠0
<to be added>
$A_{11}=\left|\begin{array}{rr}4 & 3 \\ 3 & 4\end{array}\right|$
=16-9
=7
$A_{12}=-\left|\begin{array}{ll}1 & 3 \\ 1 & 4\end{array}\right|-(4-3)$
=-1
$A_{13}=\left[\begin{array}{ll}1 & 4 \\ 1 & 3\end{array} \right]$
=3-4
=-1
$A_{21}=-\left|\begin{array}{ll}3 & 3 \\ 3 & 4\end{array}\right|$
=-(12-9)
=-3
$A_{22}=\left|\begin{array}{ll}1 & 3 \\ 1 & 4\end{array} \right|$
=4-3
=1
$A_{23}=-\left|\begin{array}{ll}1 & 3 \\ 1 & 3\end{array}\right|$
=-(3-3)
=0
$A_{31}=\left|\begin{array}{ll}3 & 3 \\ 4 & 3\end{array}\right|$
=9-12
=3
$A_{32}=-\left|\begin{array}{ll}1 & 3 \\ 1 & 3\end{array}\right|$
=-(3-3)
=0
$A_{33}=\left|\begin{array}{ll}1 & 3 \\ 1 & 4\end{array}\right|$
=4-3
=1
$a d j A=\left[\begin{array}{ccc}7 & -1 & -1 \\ -3 & 1 & 0 \\ -3 & 0 & 1\end{array}\right]^{\prime}$
$=\left[\begin{array}{rrr}7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1\end{array}\right]$
$\therefore A^{-1}=\frac{1}{| A|} \cdot a d j A$
$=\frac{1}{1}\left[\begin{array}{ccc}7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1\end{array}\right]$
$A^{-1}=\left[\begin{array}{ccc}7^{0} & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1\end{array}\right]$
(ii) $\left[\begin{array}{rrr}2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2\end{array}\right]$
Sol :
(iii) $\left[\begin{array}{rrr}-1 & 1 & 2 \\ 3 & -1 & 1 \\ -1 & 3 & 4\end{array}\right]$
(iv) $\left[\begin{array}{rrr}1 & 2 & 3 \\ -3 & 5 & 0 \\ 0 & 1 & 1\end{array}\right]$
Sol :
(v) $\left[\begin{array}{rrr}2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2\end{array}\right]$
(vi) $\left[\begin{array}{rrr}1 & 2 & 5 \\ 1 & -1 & -1 \\ 2 & 3 & -1\end{array}\right]$
(vii) $\left[\begin{array}{rrr}0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4\end{array}\right]$
Sol :
(viii) $\left[\begin{array}{rrr}2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2\end{array}\right]$
Sol :
(ix) diag[1 3 2]
Sol :
$=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 2\end{array}\right]$
(x) $\left[\begin{array}{rrr}1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha\end{array}\right]$
Sol :
(xi) $\left[\begin{array}{ccc}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1\end{array}\right]$
Sol :
(xii) $\left[\begin{array}{rrr}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]$
Sol :
(xiii) $\left[\begin{array}{rrr}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1\end{array}\right]$
Sol :
(xiv) $\left[\begin{array}{rrr}2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1\end{array}\right]$
Sol :
Question 5
आव्यूह A निकाले ताकि
[Find the matrix A such that]:
(i)$\left[\begin{array}{rr}5 & -7 \\ -2 & 3\end{array}\right] \mathbf{A}=\left[\begin{array}{rr}-16 & -6 \\ 7 & 2\end{array}\right]$
Sol :
$\left[\begin{array}{cc}5 & -7 \\ -2 & 3\end{array}\right] A=\left[\begin{array}{cc}-16 & -6 \\ 7 & 2\end{array}\right]$
माना $X=\left[\begin{array}{cc}5 & -7 \\ -2 & 3\end{array}\right]$ , $Y=\left[\begin{array}{cc}-16 & -6 \\ 7 & 2\end{array}\right]$
XA=Y
A=X-1Y
$|x|=\left|\begin{array}{cc}5 & -7 \\ -2 & 3\end{array}\right|$
=15-14=1≠0
C11=3 , C12=-(-2)=2
C21=-(-7)=7 ,C22=5
adj X$=\left[\begin{array}{ll}3 & 2 \\ 7 & 5\end{array}\right]^{\prime}$
$=\left[\begin{array}{ll}3 & 7 \\ 2 & 5\end{array}\right]$
$x^{-1}=\frac{1}{|x|} \cdot$ adj X
$=\frac{1}{1}\left[\begin{array}{ll}3 & 7 \\ 2 & 5\end{array}\right]$
$=\left[\begin{array}{ll}3 & 7 \\ 2 & 5\end{array}\right]$
$\therefore A=x^{-1} y=\left[\begin{array}{ll}3 & 7 \\ 2 & 5\end{array}\right]\left[\begin{array}{cc}-16 & -6 \\ 7 & 2\end{array}\right]$
$A=\left[\begin{array}{lr}-48+49 & -18+14 \\ -32+35 & -12+10\end{array}\right]$
$A=\left[\begin{array}{ll}1 & -4 \\ 3 & -2\end{array}\right]$
(ii)$A\left[\begin{array}{rr}3 & 2 \\ 1 & -1\end{array}\right]=\left[\begin{array}{ll}4 & 1 \\ 2 & 3\end{array}\right]$
Sol :
Question 6
(i) यदि(If) $A=\left[\begin{array}{rr}-1 & -1 \\ 2 & -2\end{array}\right]$ (दिखाएँ कि) show that
$A^{2}+3 A+4 I_{2}=0$ hence find $A^{-1}$
Sol :
L.H.S
$A^{2}+3 A+4 I_{2}$
$=\left[\begin{array}{cc}-1 & -1 \\ 2 & -2\end{array}\right]\left[\begin{array}{cc}-1 & -1 \\ 2 & -2\end{array}\right]+3\left[\begin{array}{cc}-1 & -1 \\ 2 & -2\end{array}\right]+4\left[\begin{array}{c}1&0 \\ 0&1\end{array}\right]$
$=\left[\begin{array}{ccc}1-2 & 1+2 \\ -2-4 & -2+4\end{array}\right]+\left[\begin{array}{cc}-3 & -3 \\ 6 & -6\end{array}\right]+\left[\begin{array}{cc}4 & 0 \\ 0 & 4\end{array}\right]$
$=\left[\begin{array}{cc}-1 & 3 \\ -6 & 2\end{array}\right]+\left[\begin{array}{cc}1 & -3 \\ 6 & -2\end{array}\right]$
$=\left[\begin{array}{cc}0 & 0 \\ 0 & 0\end{array}\right]=0$
$\because A^{2}+3 A+4 I_{2}=0$
<to be added>
$A^{2} \cdot A^{-1}+3 A \cdot A^{-1}+4 I_{2} \cdot A^{-1}=0 \cdot A^{-1}$
$A+3 I+4 A^{-1}=0$
$\left[\begin{array}{rr}-1 & -1 \\ 2 & -2\end{array}\right]+3\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]+4 A^{-1}=0$
$\left[\begin{array}{cc}-1 & -1 \\ 2 & -2\end{array}\right]+\left[\begin{array}{ll}3 & 0 \\ 0 & 3\end{array}\right]+4 A^{-1}=0$
$\left[\begin{array}{cc}2 & -1 \\ 2 & 1\end{array}\right]+4 A^{-1}=0$
$4 A^{-1}=\left[\begin{array}{cc}-2 & 1 \\ -2 & -1\end{array}\right]$
$A^{-1}=\frac{1}{4}\left[\begin{array}{cc}-2 & 1 \\ -2 & -1\end{array}\right]$
$A^{-1}=\left[\begin{array}{cc}-\frac{1}{2} & \frac{1}{4} \\ -\frac{1}{2} & -\frac{1}{4}\end{array}\right]$
(ii) यदि (If) $A=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]$ , दिखाएँ कि (show that) $\mathrm{A}^{2}-4 \mathrm{~A}+\mathrm{I}=\mathrm{O}$ तथा इससे (and hence find) $\mathrm{A}^{-1}$ निकालें।
Sol :
(iii) यदि (If) $\mathrm{A}=\left[\begin{array}{rrr}2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2\end{array}\right]$ तो सत्यापित करें कि (verify that) $\mathrm{A}^{3}-6 \mathrm{~A}^{2}+9 \mathrm{~A}-4 \mathrm{I}=\mathrm{O}$ तथा इससे (and hence find) $\mathrm{A}^{-1}$ निकालें।
Sol :
(iv) If $A=\left[\begin{array}{rrr}1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3\end{array}\right]$ show that
$A^{3}-6 A^{2}+5 A+11 I=O$
Hence find $\mathrm{A}^{-1}$
Sol :
$A^{2}=\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3\end{array}\right]\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3\end{array}\right]$
$=\left[\begin{array}{cccc}1+1+2 & 1+2-1 & 1-3+3 \\ 1+2-6 & 1+4+3 & 1-6-9 \\ 2-1+6 & 2-2-3 & 2+3+9\end{array}\right]$
$A^{3}=A^{2} \cdot A=\left[\begin{array}{ccc}4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14\end{array}\right]\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3\end{array}\right]$
$=\left[\begin{array}{ccc}4+2+2 & 4+4-1 & 4-6+3 \\ -3+8-28 & -3+16+14 & -3-24-42 \\ 7-3+28 & 7-6-14 & 7+9+42\end{array}\right]$
$A^{3}=\left[\begin{array}{ccc}8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58\end{array}\right]$
(iv) If $A=\left[\begin{array}{rrr}1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3\end{array}\right]$ show that
$A^{3}-6 A^{2}+5 A+11 I=O$
Hence find $\mathrm{A}^{-1}$
Sol :
$A^{2}=\left[\begin{array}{rrr}4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14\end{array}\right]$ , $A^{3}=\left[\begin{array}{ccc}8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58\end{array}\right]$
L.H.S
$A^{3}-6 A^{2}+5 A+11 I$
$=\left[\begin{array}{ccccc}8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58\end{array}\right]-6\left[\begin{array}{ccc}4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14\end{array}\right]+5\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3\end{array}\right]+11\left[\begin{array}{c}1&0&0 \\ 0&1&0 \\ 0&0&1\end{array}\right]$
$=\left[\begin{array}{ccc}8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58\end{array}\right]-\left[\begin{array}{ccc}24 & 12 & 6 \\ -18 & 48 & -84 \\ 42 & -18 & 84\end{array}\right]+\left[\begin{array}{ccc}5 & 5 & 5 \\ 5 & 10 & -15 \\ 20 & -5 & 15\end{array}\right]+\left[\begin{array}{ccc}11 & 0 & 0 \\ 0 & 11 & 0\\0&0&11 \end{array}\right]$
$=\left[\begin{array}{ccc}24 & 12 & 6 \\ -18 & 48 & -84 \\ 42 & -18 & 84\end{array}\right]-\left[\begin{array}{ccc}24 & 12 & 6 \\ -18 & 48 & -84 \\ 42 & -18 & 84\end{array}\right]$
$=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$
=0
(iv) If $A=\left[\begin{array}{rrr}1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3\end{array}\right]$ show that
$A^{3}-6 A^{2}+5 A+11 I=O$
Hence find $\mathrm{A}^{-1}$
Sol :
$A^{2}=\left[\begin{array}{ccc}4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14\end{array}\right]$
$A^{3}-6 A^{2}+5 A+11 I=0$
<to be added>
$A^{3} A^{-1}-6 A^{2} \cdot A^{-1}+5 \cdot A \cdot A^{-1}+11 I \cdot A^{-1}=0$
$A^{2}-6 A+5 I+11 A^{-1}=0$
$\left[\begin{array}{ccc}4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14\end{array}\right]-6\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3\end{array}\right]+5\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]+11 A^{-1}=0$
$\left[\begin{array}{ccc}4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14\end{array}\right]-\left[\begin{array}{ccc}6 & 6 & 6 \\ 6 & 12 & -18 \\ 12 & -6 & 18\end{array}\right]+\left[\begin{array}{ccc}5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5\end{array}\right]+11 A^{-1}=0$
$\left[\begin{array}{ccc}9 & 2 & 1 \\ -3 & 13 & -14 \\ 7 & -3 & 19\end{array}\right]-\left[\begin{array}{ccc}6 & 6 & 6 \\ 6 & 12 & -18 \\ 12 & -6 & 18\end{array}\right]+11 A^{-1}=0$
$\left[\begin{array}{ccc}3 & -4 & -5 \\ -9 & 1 & 4 \\ -5 & 3 & 1\end{array}\right]+11 A^{-1}=0$
$11 A^{-1}=\left[\begin{array}{rrr}-3 & 4 & 5 \\ 9 & -1 & -4 \\ 5 & -3 & -1\end{array}\right]$
$A^{-1}=\frac{1}{11}\left[\begin{array}{ccc}-3 & 4 & 5 \\ 9 & -1 & -4 \\ 5 & -3 & -1\end{array}\right]$
Question 7
(i) If $A=\left[\begin{array}{rr}2 & 3 \\ 5 & -2\end{array}\right]$ show that
$A^{-1}=\frac{1}{19} A$
Sol :
$|A|=\left|\begin{array}{cc}2 & 3 \\ 5 & -2\end{array}\right|$
=-4-15
=-19≠0
<to be added>
$A_{11}=-2, A_{12}=-5$
$A_{21}=-3, A_{22}=2$
$a d j A=\left[\begin{array}{cc}-2 & -5 \\ -3 & 2\end{array}\right]^{\prime}$
$=\left[\begin{array}{cc}-2 & -3 \\ -5 & 2\end{array}\right]$
$A^{-1}=\frac{1}{|A|} \cdot a d j A$
$=\frac{1}{-19}\left[\begin{array}{cc}-2 & -3 \\ -5 & 2\end{array}\right]$
$A^{-1}=\frac{1}{19}\left[\begin{array}{cc}2 & 3 \\ 5 & -2\end{array}\right]$
$\therefore A^{-1}=\frac{1}{19} \cdot A$
(ii) यदि (If) $A=\left[\begin{array}{rr}2 & 5 \\ 1 & -2\end{array}\right]$. दिखाएँ कि (show that) $A^{-1}=\frac{1}{9} A$
Sol :
(iii) If $A=\left[\begin{array}{rr}2 & -3 \\ -4 & 7\end{array}\right]$ show that $2 \mathrm{A}^{-1}=9 \mathrm{I}-\mathrm{A}$
Sol :
$|A|=\left|\begin{array}{cc}2 & -3 \\ -4 & 7\end{array}\right|$
=14-12
=2≠0
$A_{11}=7, A_{12}=-(-4)=4$
$A_{21}=-(-3)=3$ ,$A_{22}=2$
$a dj A=\left[\begin{array}{ll}7 & 4 \\ 3 & 2\end{array}\right]^{\prime}$
$=\left[\begin{array}{ll}7 & 3 \\ 4 & 2\end{array}\right]$
$A^{-1}=\frac{1}{|A|}$ adj $A$
$=\frac{1}{2}\left[\begin{array}{ll}7 & 3 \\ 4 & 2\end{array}\right]$
L.H.S
$2 A^{-1}=2 \times\frac{ 1}{2}\left[\begin{array}{ll}7 & 3 \\ 4 & 2\end{array}\right]$
$=\left[\begin{array}{ll}7 & 3 \\ 4 & 2\end{array}\right]$
R.H.S
$9I-A=9\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]-\left[\begin{array}{cc}2 & -3 \\ -4 & 7\end{array}\right]$
$=\left[\begin{array}{ll}9 & 0 \\ 0 & 9\end{array}\right]-\left[\begin{array}{cc}2 & -3 \\ -4 & 7\end{array}\right]$
$=\left[\begin{array}{lll} 7 & 3 \\ 4 & 2\end{array}\right]$
∴$2 A^{-1}=9 I-A$
Question 8
(i) यदि (If) $A=\left[\begin{array}{ll}3 & 1 \\ 4 & 0\end{array}\right], B=\left[\begin{array}{ll}4 & 0 \\ 2 & 5\end{array}\right]$, सत्यापित करें कि ( verify that)
$(A B)^{-1}=B^{-1} A^{-1}$
Sol :
$A B=\left[\begin{array}{ll}3 & 1 \\ 4 & 0\end{array}\right]\left[\begin{array}{ll}4 & 0 \\ 2 & 5\end{array}\right]$
$=\left[\begin{array}{ll}12+2 & 0+5 \\ 16+0 & 0+0\end{array}\right]$
$A B=\left[\begin{array}{lll}14& 5 \\ 16 & 0\end{array}\right]$
Let $C=A B=\left[\begin{array}{ll}14 & 5 \\ 16 & 0\end{array}\right]$
|C|=0-80
=-80≠0
$C_{11}=0, C_{12}=-16$
$C_{21}=-5, C_{22}=14$
$a d j C=\left[\begin{array}{cc}0 & -16 \\ -5 & 14\end{array}\right]^{\prime}$
$=\left[\begin{array}{cc}0 & -5 \\ -16 & 14\end{array}\right]$
$C^{-1}=(A B)^{-1}=\frac{1}{|c|} \cdot a d j C$
$=\frac{1}{-80}\left[\begin{array}{cc}0 & -5 \\ -16 & 14\end{array}\right]$
∵ $A=\left[\begin{array}{ll}3 & 1 \\ 4 & 0\end{array}\right]$
|A|=0-4=-4≠0
$a d j A=\left[\begin{array}{cc}0 & -4 \\ -1 & 3\end{array}\right]’$
$=\left[\begin{array}{cc}0 & -1 \\ -4 & 3\end{array}\right]$
$A^{-1}=\frac{1}{|a|} \cdot a d j A$
$=\frac{1}{-4}\left[\begin{array}{cc}0 & -1 \\ -4 & 3\end{array}\right]$
∵$B=\left[\begin{array}{ll}4 & 0 \\ 2 & 5\end{array}\right]$
|B|=20-0
=20≠0
$a d j B=\left[\begin{array}{cc}5 & -2 \\ 0 & 4\end{array}\right]^{\prime}$
$=\left[\begin{array}{cc}5 & 0 \\ -2 & 4\end{array}\right]$
$B^{-1}=\frac{1}{\left| B\right|} a d_{j} B$
$=\frac{1}{20}\left[\begin{array}{cc}5 & 0 \\ -2 & 4\end{array}\right]$
R.H.S
$B^{-1} A^{-1}=\frac{1}{20}\left[\begin{array}{cc}5 & 0 \\ -2 & 4\end{array}\right] \cdot \frac{1}{(-4)}\left[\begin{array}{cc}0 & -1 \\ -4 & 3\end{array}\right]$
$=\frac{1}{-80}\left[\begin{array}{cc}0-0 & -5+0 \\ -0-16 & 2+12\end{array}\right]$
$=-\frac{1}{80}\left[\begin{array}{cc}0 & -5 \\ -16 & 14\end{array}\right]$
$\therefore(A B)^{-1}=B^{-1} \cdot A^{-1}$
(ii) यदि (If) $\mathrm{A}=\left[\begin{array}{ll}3 & 2 \\ 7 & 5\end{array}\right], \mathrm{B}=\left[\begin{array}{ll}6 & 7 \\ 8 & 9\end{array}\right]$, सत्यापित करें कि (verify that)
$(\mathrm{AB})^{-1}=\mathrm{B}^{-1} \mathrm{~A}^{-1}$
Sol :
(iii) If $A=\left[\begin{array}{rr}2 & -3 \\ 4 & 6\end{array}\right]$ verify that
(a) $\operatorname{adj} A^{\prime}=(\operatorname{adj} A)^{\prime}$
Sol :
$A=\left[\begin{array}{cc}2 & -3 \\ 4 & 6\end{array}\right]$
$A^{\prime}=\left[\begin{array}{ll}2 & 4 \\ -3 & 6\end{array}\right]$
$C_{11}=6, C_{12}=-(-3)=3$
$C_{21}=-4, C_{22}=2$
$\operatorname{adj} A^{\prime}=\left[\begin{array}{cc}6 & 3 \\ -4 & 2\end{array}\right]^{\prime}$
$=\left[\begin{array}{cc}6 & -4 \\ 3 & 2\end{array}\right]$
L.H.S
$\because A=\left[\begin{array}{ll}2 & -3 \\ 4 & 6\end{array}\right]$
$a{d}jA=\left[\begin{array}{cc}6 & -4 \\ 3 & 2\end{array}\right]’$
$=\left[\begin{array}{cc}6 & 3 \\ -4 & 2\end{array}\right]$
R.H.S
$(a d j A)^{\prime}=\left[\begin{array}{cc}6 & -4 \\ 3 & 2\end{array}\right]$
$\therefore a dj A^{\prime}=(a d j A)^{\prime}$
(b) $(\operatorname{adj} A)^{-1}=\operatorname{adj}\left(A^{-1}\right)$
Sol :
$A=\left[\begin{array}{cc}2 & -3 \\ 4 & 6\end{array}\right]$
$a d j A=\left[\begin{array}{cc}6 & -4 \\ 3 & 2\end{array}\right]’$
$=\left[\begin{array}{cc}6 & 3 \\ -4 & 2\end{array}\right]$
|adj A|=12+12
=24≠0
$a d{j}(a d j A)=\left[\begin{array}{cc}2 & 4 \\ -3 & 6\end{array}\right]’$
$=\left[\begin{array}{cc}2 & -3 \\ 4 & 6\end{array}\right]$
$(a djA)^{-1}=\frac{1}{\left|a dj A\right|} \cdot a dj(a d j A)$
$=\frac{1}{24}\left[\begin{array}{cc}2 & -3 \\ 4 & 6\end{array}\right]$
∵ $A=\left[\begin{array}{ll}2 & -3 \\ 4 & 6\end{array}\right]$
|A|=12+12
=24≠0
$a d{j} A=\left[\begin{array}{cc}6 & -4 \\ 3 & 2\end{array}\right]’$
$=\left[\begin{array}{ll}6 & 3 \\ -4 & 2\end{array}\right]$
$A^{-1}=\frac{1}{24}\left[\begin{array}{cc}6 & 3 \\ -4 & 2\end{array}\right]$
$a d{j}\left(A^{-1}\right)=\frac{1}{24}\left[\begin{array}{cc}2 & 4 \\ -3 & 6\end{array}\right]’$
$=\frac{1}{24}\left[\begin{array}{cc}2 & -3 \\ 4 & 6\end{array}\right]$
$\therefore,(a d j A)^{-1}=a d{j}\left(A^{-1}\right)$
(iv) If $A=\left[\begin{array}{rr}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]$ verify that $\left(A^{\prime}\right)^{-1}=\left(A^{-1}\right)^{\prime}$
Sol :
$A=\left[\begin{array}{ll}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]$
$A^{\prime}=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right]$
$\left|A^{\prime}\right|=\cos ^{2} \theta+\sin ^{2} \theta=1 \neq \theta$
$\left|A^{\prime}\right|=\cos ^{2} \theta+\sin ^{2} \theta=1 \neq 0$
$a d{j} A^{\prime}=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right]’$
$\left[\begin{array}{cc}\cos \theta & -\tan \theta \\ \sin \theta & \cos \theta\end{array}\right]$
$\left(A^{\prime}\right)^{-1}=\frac{1}{\left|A^{\prime}\right|} \cdot a d{j} A$
$=\frac{1}{1}\left[\begin{array}{cc}\cos \theta& \sin \theta\\-\sin \theta& \cos \theta\end{array}\right]$
$\left(A^{\prime}\right)^{-1}=\left[\begin{array}{cc}\cos \theta& \sin \theta\\-\sin \theta& \cos \theta\end{array}\right]$
(v) If $A=\left[\begin{array}{rr}1 & -1 \\ 0 & 3\end{array}\right]$ and $B=\left[\begin{array}{rr}2 & 4 \\ -3 & 0\end{array}\right]$ verify that $\operatorname{adj}(A B)=(\operatorname{adj} B)(\operatorname{adj} A)$
Sol :
Question 9
(i) If $A=\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{array}\right]$ verify that
A(adj A)=|A|I
Sol :
$a d j A=\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{array}\right]’$
$=\left[\begin{array}{ccc}\cos \alpha&\sin \alpha&0 \\-\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{array}\right]$
$A(a dj A)=\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{ccc}\cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 0\end{array}\right]$
$=\left[\begin{array}{cccc}\cos^{2} \alpha +\sin ^{2} \alpha+0 & \cos \alpha\sin \alpha -\sin \alpha\cos \alpha+0 & 0+0+0 \\ \sin \alpha \cos \alpha-\sin \alpha \cos \alpha+0 & \sin ^{2} \alpha+\cos ^{2} \alpha+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+1\end{array}\right]$
$=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=I$
R.H.S
$|A|-I=\left|\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{array}\right| \left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$=\left(\cos ^{2} \alpha+\sin ^{2} \alpha\right)\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
=I
∴A(adj A)=|A|.I
(ii) If $f(x)=\left[\begin{array}{ccc}\cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right]$ verify that
$\{f(x)\}^{-1}=f(-x)$
Sol :
$|f(x)|=1\begin{vmatrix}\cos x&-\sin x\\\sin x& \cos x\end{vmatrix}$
$=\cos ^{2} x+\sin^{2} x=1 \neq 0$
$a d{j} f(x)=\left[\begin{array}{ccc}\cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right]$
$=\left[\begin{array}{ccc}\cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right]$
$\{f(x)\}^{-1}=\frac{1}{|f ( x) |} \cdot \operatorname{adj} f(x)$
$=\frac{1}{1}\left[\begin{array}{ccc}\cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right]$
$\{f(x)\}^{-1}=\left[\begin{array}{ccc}\cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1 \end{array}\right]$
$\{f(x)\}^{-1}=\left[\begin{array}{ccc}\cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right]$
$=\left[\begin{array}{ccc}\cos (-x) & -\sin (-x) & 0 \\ \sin (-x) & \cos (-x) & 0 \\ 0 & 0 & 1\end{array}\right]$
$\therefore\{f(x)\}^{-1}=f(-x)$
(iii) If $A=\left[\begin{array}{rrr}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{array}\right]$ verify that
$A^{3}=A^{-1}$
Sol :
(iv) If $A=\left[\begin{array}{rrr}1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0\end{array}\right]$ show that
$A^{-1}=A^{2}$
Question 10
यदि (If) $A^{-1}=\left[\begin{array}{rrr}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]$ and $B=\left[\begin{array}{rrr}1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1\end{array}\right]$
निकाले (Find) $(A B)^{-1}$
Sol :