If A = begin{bmatrix} cos frac{2 pi}{33} & si | vedclass

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(C) The given matrix $A$ is a rotation matrix of the form $R(	heta) = \begin{bmatrix} \cos 	heta & \sin 	heta \ -\sin 	heta & \cos 	heta \end{bm

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$A^4$

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(C) The given matrix $A$ is a rotation matrix of the form $R(	heta) = \begin{bmatrix} \cos 	heta & \sin 	heta \ -\sin 	heta & \cos 	heta \end{bmatrix}$ where $	heta = \frac{2 \pi}{33}$.By the property of rotation matrices,$A^n = R(n 	heta) = \begin{bmatrix} \cos(n 	heta) & \sin(n 	heta) \ -\sin(n 	heta) & \cos(n 	heta) \end{bmatrix}$.We need to find $A^{2017}$,so $n = 2017$.The angle becomes $n 	heta = 2017 	imes \frac{2 \pi}{33} = \frac{4034 \pi}{33}$.Dividing $4034$ by $33$: $4034 = 33 	imes 122 + 8$.So,$\frac{4034 \pi}{33} = 122 \pi + \frac{8 \pi}{33}$.Since $\cos(122 \pi + \alpha) = \cos \alpha$ and $\sin(122 \pi + \alpha) = \sin \alpha$,we have $A^{2017} = \begin{bmatrix} \cos \frac{8 \pi}{33} & \sin \frac{8 \pi}{33} \ -\sin \frac{8 \pi}{33} & \cos \frac{8 \pi}{33} \end{bmatrix}$.Note that $A^4 = R(4 	imes \frac{2 \pi}{33}) = R(\frac{8 \pi}{33})$.Therefore,$A^{2017} = A^4$.

If $A = \begin{bmatrix} \cos \frac{2 \pi}{33} & \sin \frac{2 \pi}{33} \\ -\sin \frac{2 \pi}{33} & \cos \frac{2 \pi}{33} \end{bmatrix}$,then $A^{2017} = $

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