If A = begin{bmatrix} cos theta & -sin theta | vedclass

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(C) Given $A = \begin{bmatrix} \cos 	heta & -\sin 	heta \ \sin 	heta & \cos 	heta \end{bmatrix}$.By the property of rotation matrices,$A^n = \beg

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$\begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}$

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(C) Given $A = \begin{bmatrix} \cos 	heta & -\sin 	heta \ \sin 	heta & \cos 	heta \end{bmatrix}$.By the property of rotation matrices,$A^n = \begin{bmatrix} \cos(n	heta) & -\sin(n	heta) \ \sin(n	heta) & \cos(n	heta) \end{bmatrix}$ for any integer $n$.Therefore,$A^{-50} = \begin{bmatrix} \cos(-50	heta) & -\sin(-50	heta) \ \sin(-50	heta) & \cos(-50	heta) \end{bmatrix} = \begin{bmatrix} \cos(50	heta) & \sin(50	heta) \ -\sin(50	heta) & \cos(50	heta) \end{bmatrix}$.Given $	heta = \frac{\pi}{12}$,we have $50	heta = 50 	imes \frac{\pi}{12} = \frac{25\pi}{6} = 4\pi + \frac{\pi}{6}$.Since $\cos(4\pi + \frac{\pi}{6}) = \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$ and $\sin(4\pi + \frac{\pi}{6}) = \sin(\frac{\pi}{6}) = \frac{1}{2}$,$A^{-50} = \begin{bmatrix} \cos(\frac{\pi}{6}) & \sin(\frac{\pi}{6}) \ -\sin(\frac{\pi}{6}) & \cos(\frac{\pi}{6}) \end{bmatrix} = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}$.

If $A = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$,then the matrix $A^{-50}$ when $\theta = \frac{\pi}{12}$ is equal to:

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