KC Sinha Mathematics Solution Class 12 Chapter 7 वर्ग आव्यूह का सहखंडज और प्रतिलोम (adjoint and inverse of matrix) Exercise 7.1 (Q7-Q10)

Exercise 7.1




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 :


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