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

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

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$\begin{bmatrix} \cos 2 	heta & -\sin 2 	heta \ \sin 2 	heta & \cos 2 	heta \end{bmatrix}$

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(A) 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 $n = 100$,we have $A^{100} = \begin{bmatrix} \cos 100 	heta & -\sin 100 	heta \ \sin 100 	heta & \cos 100 	heta \end{bmatrix}$.Given $	heta = \frac{2 \pi}{7}$,so $100 	heta = \frac{200 \pi}{7}$.We can write $\frac{200 \pi}{7} = \frac{196 \pi + 4 \pi}{7} = 28 \pi + \frac{4 \pi}{7}$.Since $\cos(28 \pi + \alpha) = \cos \alpha$ and $\sin(28 \pi + \alpha) = \sin \alpha$,we get $A^{100} = \begin{bmatrix} \cos \frac{4 \pi}{7} & -\sin \frac{4 \pi}{7} \ \sin \frac{4 \pi}{7} & \cos \frac{4 \pi}{7} \end{bmatrix} = A^2$.Comparing this with the options,none of the provided options match $A^2$ exactly. However,evaluating the expression,$A^{100} = A^2$ is the correct result.

If $A = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$ and $\theta = \frac{2 \pi}{7}$,then $A^{100} = A \times A \times \dots \times A$ ($100$ times) is equal to:

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