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If $A=\left[\begin{array}{cc}\sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha\end{array}\right]$ and $A+A^{-1}=I$, then $\alpha=$
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$\frac{\pi}{6}$
Given, $\begin{aligned} A & =\left[\begin{array}{cc}\sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha\end{array}\right] \\ \text { Adj } A & =\left[\begin{array}{cc}\sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha\end{array}\right] \\ |A| & =\sin ^2 \alpha+\cos ^2 \alpha=1 \\ A^{-1} & =\left[\begin{array}{cc}\sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha\end{array}\right] \\ A+A^{-1} & =\left[\begin{array}{cc}\sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha\end{array}\right]+\left[\begin{array}{cc}\sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha\end{array}\right] \\ & =\left[\begin{array}{cc}2 \sin \alpha & 0 \\ 0 & 2 \sin \alpha\end{array}\right] \\ A+A^{-1} & =I, \text { if } 2 \sin \alpha=1 \\ \sin \alpha & =\frac{1}{2} \Rightarrow \alpha=\frac{\pi}{6}\end{aligned}$
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