Question
Find $A ^{-1}$ using adjoint method, where $A =\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]$

Answer

$|A|=\cos \theta(\cos \theta)-\sin \theta(-\sin \theta)$
$=\cos ^2 \theta+\sin ^2 \theta$
$=1 \neq 0$
$\therefore A^{-1} \text { exists. }$
$A_{11}=(-1)^{1+1} M_{11}=M_{11}=\cos \theta$
$A_{12}=(-1)^{1+2} M_{12}=-M_{12}=\sin \theta$
$A_{21}=(-1)^{2+1} M_{21}=-M_{21}=-\sin \theta$
$A_{22}=(-1)^{2+2} M_{22}=M_{22}=\cos \theta $
$ \therefore \operatorname{adj}(A)=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]^{ T }$
$\begin{array}{l}=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right] \end{array}$
$ \therefore A ^{-1}=\frac{1}{| A |} \operatorname{adj}( A ) $
$ =\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right]$

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