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

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

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$\begin{bmatrix} 0 & 1 \ -1 & 0 \end{bmatrix}$

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(A) Given $A = \begin{bmatrix} \cos 	heta & -\sin 	heta \ \sin 	heta & \cos 	heta \end{bmatrix}$.Using the property of rotation matrices, $A^n = \begin{bmatrix} \cos(n	heta) & -\sin(n	heta) \ \sin(n	heta) & \cos(n	heta) \end{bmatrix}$.Thus, $A^{-3} = \begin{bmatrix} \cos(-3	heta) & -\sin(-3	heta) \ \sin(-3	heta) & \cos(-3	heta) \end{bmatrix} = \begin{bmatrix} \cos(3	heta) & \sin(3	heta) \ -\sin(3	heta) & \cos(3	heta) \end{bmatrix}$.Given $	heta = \pi/6$, then $3	heta = 3(\pi/6) = \pi/2$.Substituting the value, $A^{-3} = \begin{bmatrix} \cos(\pi/2) & \sin(\pi/2) \ -\sin(\pi/2) & \cos(\pi/2) \end{bmatrix} = \begin{bmatrix} 0 & 1 \ -1 & 0 \end{bmatrix}$.

If $A = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$, then the matrix $A^{-3}$ when $\theta = \pi/6$ is equal to...

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