$A(\propto)\left[\begin{array}{cc}\cos \propto & \sin \propto \ -\sin \propto & \cos \propto\end{array} ight]$, then $\left[A^2(\propto) ight]^{-1}=$

Question

$A(\propto)\left[\begin{array}{cc}\cos \propto & \sin \propto \ -\sin \propto & \cos \propto\end{array}ight]$, then $\left[A^2(\propto)ight]^{-1}=$

Aicharya · Accepted Answer

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

$A(\propto)\left[\begin{array}{cc}\cos \propto & \sin \propto \\ -\sin \propto & \cos…

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