$\begin{aligned} & A(\alpha)=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right] \Rightarrow|\mathrm{A}(\alpha)|=\cos ^2 \alpha+\sin ^2 \alpha=1 \\ & \mathrm{~A}^2(\alpha)=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right] \\ & {\left[\begin{array}{cc}\cos ^2 \alpha-\sin ^2 \alpha & 2 \sin \alpha \cos \alpha \\ -\sin \alpha \cos \alpha & \cos ^2 \alpha-\sin ^2 \alpha\end{array}\right]=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]} \\ & \therefore \operatorname{adj}\left[\mathrm{A}^2(\alpha)\right]\left[\begin{array}{cc}\cos 2 \alpha & -\sin 2 \alpha \\ \sin 2 \alpha & \cos 2 \alpha\end{array}\right] \\ & \therefore\left[\mathrm{A}^2(\alpha)\right]^{-1}=\left[\begin{array}{cc}\cos 2 \alpha & -\sin 2 \alpha \\ \sin 2 \alpha & \cos 2 \alpha\end{array}\right] \quad \ldots\left[\because\left|\mathrm{A}^2\right|=|\mathrm{A}|^2=1\right] \\ & \therefore\left[\mathrm{A}^2(\alpha)\right]^{-1}=\left[\begin{array}{cc}\cos (-2 \alpha) & \sin (-2 \alpha) \\ -\sin (-2 \alpha) & \cos (-2 \alpha)\end{array}\right]=\mathrm{A}(-2 \alpha)\end{aligned}$