If $A = [\begin{matrix} \cos \frac{2 π}{33} & \sin \frac{2 π}{33} \\ - \sin \frac{2 π}{33} & \cos \frac{2 π}{33} \end{matrix}], then A^{2017}$

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$A^{4}$

$[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

$A^{2}$

answer is C.

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Detailed Solution

$A = [\begin{matrix} \cos \frac{2 π}{33} & \sin \frac{2 π}{33} \\ - \sin \frac{2 π}{33} & \cos \frac{2 π}{33} \end{matrix}] let θ = \frac{2 π}{33} ∴ A^{2} = [\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}] [\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}] = [\begin{matrix} \cos 2 θ & \sin 2 θ \\ - \sin 2 θ & \cos 2 θ \end{matrix}]$

$Similarly, A^{2017} = A^{2013} . A^{4} = [\begin{matrix} \cos (2013 θ) & \sin (2013 θ) \\ - \sin (2013 θ) & \cos (2013 θ) \end{matrix}] . A^{4} = [\begin{matrix} \cos (\frac{4026 π}{33}) & \sin (\frac{4026 π}{33}) \\ - \sin (\frac{4026 π}{33}) & \cos (\frac{4026 π}{33}) \end{matrix}] A^{4} = [\begin{matrix} \cos (122 π) & \sin (122 π) \\ - \sin (122 π) & \cos (122 π) \end{matrix}] A^{4} = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] A^{4} A^{2017} = A^{4}$