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(B) 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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$A^3$

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(B) Given $A = \begin{bmatrix} \cos 	heta & \sin 	heta \ -\sin 	heta & \cos 	heta \end{bmatrix}$.By the property of rotation matrices,$A^n = \begin{bmatrix} \cos(n	heta) & \sin(n	heta) \ -\sin(n	heta) & \cos(n	heta) \end{bmatrix}$.Here,$n = 2017$ and $	heta = \frac{2\pi}{19}$.So,$n	heta = 2017 	imes \frac{2\pi}{19} = \frac{4034\pi}{19}$.Dividing $4034$ by $19$: $4034 = 19 	imes 212 + 6$.Thus,$n	heta = (212 	imes 19 + 6) 	imes \frac{2\pi}{19} = 212(2\pi) + \frac{12\pi}{19} = 424\pi + \frac{12\pi}{19}$.Since $\cos(2k\pi + \alpha) = \cos \alpha$ and $\sin(2k\pi + \alpha) = \sin \alpha$,we have $A^{2017} = \begin{bmatrix} \cos(\frac{12\pi}{19}) & \sin(\frac{12\pi}{19}) \ -\sin(\frac{12\pi}{19}) & \cos(\frac{12\pi}{19}) \end{bmatrix}$.Note that $A^5 = \begin{bmatrix} \cos(5	heta) & \sin(5	heta) \ -\sin(5	heta) & \cos(5	heta) \end{bmatrix} = \begin{bmatrix} \cos(\frac{10\pi}{19}) & \sin(\frac{10\pi}{19}) \ -\sin(\frac{10\pi}{19}) & \cos(\frac{10\pi}{19}) \end{bmatrix}$.Wait,checking the options again,$A^{2017} = A^{19 	imes 106 + 3} = (A^{19})^{106} 	imes A^3$. Since $A^{19} = I$,$A^{2017} = I^{106} 	imes A^3 = A^3$.

If $A = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$ where $\theta = \frac{2 \pi}{19}$,then $A^{2017} = $

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