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Let $\theta = 2\pi/67$. Now consider the matrix $$ A = \begin{pmatrix} \phantom{-}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}. $$ Then the matrix $A^{2010}$ is

\begin{align*} &\text{(A)}\; \begin{pmatrix} \phantom{-}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}, & &\text{(B)}\; \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \\ &\text{(C)}\; \begin{pmatrix} \phantom{-}\cos^{30} \theta & \sin^{30} \theta \\ -\sin^{30} \theta & \cos^{30} \theta \end{pmatrix}, & &\text{(D)}\; \begin{pmatrix} \phantom{-}0 & 1 \\ -1 & 0 \end{pmatrix}. \end{align*}

I think answer is B. But I am not sure.

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marked as duplicate by Arnaud D., Lee David Chung Lin, Alex Provost, Lord Shark the Unknown, Jyrki Lahtonen Apr 10 at 5:36

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Note that $A$ is a rotation matrix that takes a vector and rotates it by $1/67$th of a revolution. Now notice that $2010 = 67 \times 30$, so the answer is indeed (B).

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  • $\begingroup$ The signs on the $\sin \theta$ terms are reversed from what is usually considered the standard rotation matrices, so it takes a small bit of rewriting to express that this is a clockwise (instead of the usual counter-clockwise) rotation. $\endgroup$ – Aaron Dec 31 '15 at 7:44
  • $\begingroup$ @Aaron Quite right, didn't notice that. I've reworded. $\endgroup$ – user296602 Dec 31 '15 at 7:46
  • $\begingroup$ @user can you explain line 2 of answer $\endgroup$ – Taylor Ted Dec 31 '15 at 7:52
  • $\begingroup$ @TaylorTed If you rotate $1/67$th of a circle $67 x 30$ times, where do you end up? $\endgroup$ – user296602 Dec 31 '15 at 7:55
  • $\begingroup$ @user as far as i have understood it, your saying that this matrix rotates given point by angle $2\pi/67$.And we are to apply this transformation 2010 times so final rotated angle will be $2pi/67 * 2010=60\pi$ $\endgroup$ – Taylor Ted Dec 31 '15 at 7:58
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Note that $$ A=PDP^{-1} $$ where \begin{align*} P &= \begin{bmatrix} i & -i \\ 1 & 1 \end{bmatrix} & D &= \begin{bmatrix} \cos\theta+i\sin\theta & 0 \\ 0 & \cos\theta-i\sin\theta \end{bmatrix} \end{align*} De Moivre's formula then implies \begin{align*} A^{2010} &= PD^{2010} P^{-1} \\ &= P \begin{bmatrix} \cos(2010\,\theta)+i\sin(2010\,\theta) & 0 \\ 0 & \cos(2010\,\theta)-i\sin(2010\,\theta) \end{bmatrix}P^{-1}\\ &= P \begin{bmatrix} \cos(60\,\pi) & \sin(60\,\pi) \\ -\sin(60\,\pi) & \cos(60\,\pi) \end{bmatrix} P^{-1} \\ &= P \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} P^{-1} \\ &= PIP^{-1} \\ &= PP^{-1} \\ &= I \end{align*}

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$A$ can be thought as $e^{-i\theta}$. So that for the natural homeomorphism $\phi:\mathbb{C}\rightarrow \mathbb{R}^2:x+iy \mapsto (x,y)$, $(\phi^{-1}\circ A\circ \phi )(z)=e^{-i\theta } z$. Hence, $(\phi^{-1} \circ A^{2010} \circ \phi)(z)= e^{- 2010 i \theta} z = e^{-60\pi i}z=z$. Hence, the answer is (B).

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