If $A=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ -\sin \theta & -\cos \theta\end{array}\right]$,…
If $A=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ -\sin \theta & -\cos \theta\end{array}\right]$, then $A^{-1}=$
- $\left[\begin{array}{cc}-\sin \theta & -\cos \theta \\ -\cos \theta & \sin \theta\end{array}\right]$
- $\left[\begin{array}{cc}\sin \theta & -\cos \theta \\ \cos \theta & -\sin \theta\end{array}\right]$
- $\left[\begin{array}{cc}-\cos \theta & \sin \theta \\ \sin \theta & \cos \theta\end{array}\right]$
- $\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ -\sin \theta & -\cos \theta\end{array}\right]$
Solution
We have $A=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ -\sin \theta & -\cos \theta\end{array}\right] \Rightarrow \operatorname{adj} A=\left[\begin{array}{cc}-\cos \theta & \sin \theta \\ \sin \theta & \cos \theta\end{array}\right]$
$|A|=\left|\begin{array}{cc}\cos \theta & -\sin \theta \\ -\sin \theta & -\cos \theta\end{array}\right|=-\cos ^{2} \theta-\sin ^{2} \theta=-\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=-1$
$\therefore A^{-1}=\frac{1}{(-1)}\left[\begin{array}{cc}-\cos \theta & \sin \theta \\ \sin \theta & \cos \theta\end{array}\right]=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ -\sin \theta & -\cos \theta\end{array}\right]$
Asked in: MHT CET 2020 (12 Oct Shift 2)
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