If f(\theta)=\begin{bmatrix}\cos~\theta&\sin~\theta\\ -\sin~\theta&\cos~\theta\end{bmatrix} then what is \{f(\pi)\}^{2} equal to?

  1. A. \begin{bmatrix}-1&0\\ 0&-1\end{bmatrix}
  2. B. \begin{bmatrix}1&1\\ 1&1\end{bmatrix}
  3. C. \begin{bmatrix}-1&0\\ 0&1\end{bmatrix}
  4. D. \begin{bmatrix}1&0\\ 0&1\end{bmatrix}

Correct Answer: D. \begin{bmatrix}1&0\\ 0&1\end{bmatrix}

Explanation

Substitute \theta = \pi to get f(\pi) = \begin{bmatrix}-1&0\\ 0&-1\end{bmatrix} = -I. Squaring this matrix gives (-I)^2 = I = \begin{bmatrix}1&0\\ 0&1\end{bmatrix}.

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