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(C) Given $A(\alpha) = \begin{bmatrix} \cos \alpha & \sin \alpha \ -\sin \alpha & \cos \alpha \end{bmatrix}$. First,calculate $A^2(\alpha) = A(\alpha

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$A(-2 \alpha)$

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(C) Given $A(\alpha) = \begin{bmatrix} \cos \alpha & \sin \alpha \ -\sin \alpha & \cos \alpha \end{bmatrix}$. First,calculate $A^2(\alpha) = A(\alpha) \cdot A(\alpha)$: $A^2(\alpha) = \begin{bmatrix} \cos \alpha & \sin \alpha \ -\sin \alpha & \cos \alpha \end{bmatrix} \begin{bmatrix} \cos \alpha & \sin \alpha \ -\sin \alpha & \cos \alpha \end{bmatrix} = \begin{bmatrix} \cos^2 \alpha - \sin^2 \alpha & \cos \alpha \sin \alpha + \sin \alpha \cos \alpha \ -\sin \alpha \cos \alpha - \cos \alpha \sin \alpha & -\sin^2 \alpha + \cos^2 \alpha \end{bmatrix}$ Using trigonometric identities $\cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha$ and $\sin 2\alpha = 2 \sin \alpha \cos \alpha$: $A^2(\alpha) = \begin{bmatrix} \cos 2\alpha & \sin 2\alpha \ -\sin 2\alpha & \cos 2\alpha \end{bmatrix} = A(2\alpha)$. Now,find the inverse $[A^2(\alpha)]^{-1} = [A(2\alpha)]^{-1}$. Since $|A(2\alpha)| = \cos^2 2\alpha + \sin^2 2\alpha = 1$,the inverse is given by the adjoint: $[A(2\alpha)]^{-1} = \begin{bmatrix} \cos 2\alpha & -\sin 2\alpha \ \sin 2\alpha & \cos 2\alpha \end{bmatrix}$. Using $\cos(-x) = \cos x$ and $\sin(-x) = -\sin x$: $[A^2(\alpha)]^{-1} = \begin{bmatrix} \cos(-2\alpha) & \sin(-2\alpha) \ -\sin(-2\alpha) & \cos(-2\alpha) \end{bmatrix} = A(-2\alpha)$.

If $A(\alpha) = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}$,then $[A^2(\alpha)]^{-1} = $

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