Given matrix A. To find $A^{2010}$ 
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.

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).
