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
(Q1-Q3) (Q4-Q6) (Q7-Q10)
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} |A|=\left|\begin{array}{cc}2 & -3 \\ -4 & 7\end{array}\right| 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] 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] 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} 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] C=A B=\left[\begin{array}{ll}14 & 5 \\ 16 & 0\end{array}\right] 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 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] 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] 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}
(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} 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] \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] (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) 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] 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 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} 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)
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 thata 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 |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] (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) |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 thatA^{3}=A^{-1} (iv) If A=\left[\begin{array}{rrr}1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0\end{array}\right] show thatA^{-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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