Question

If \(A = \begin{bmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{bmatrix}\), then the matrix \(A^{-50}\) when \(\theta = \frac{\pi}{12}\), is equal to :

• A. \(\begin{bmatrix}\frac{\sqrt3}{2} & -\frac12\\ \frac12 & \frac{\sqrt3}{2}\end{bmatrix}\)
• B. \(\begin{bmatrix}\frac12 & -\frac{\sqrt3}{2}\\ \frac{\sqrt3}{2} & \frac{-1}{2}\end{bmatrix}\)
• C. \(\begin{bmatrix}\frac{\sqrt3}{2} & \frac12\\ -\frac12 & \frac{\sqrt3}{2}\end{bmatrix}\)
• D. \(\begin{bmatrix}\frac12 & \frac{\sqrt3}{2}\\ -\frac{\sqrt3}{2} & \frac12\end{bmatrix}\)

Hint:

\(A\) is a rotation matrix: \(A^n\) is the rotation by \(n\theta\) for every integer \(n\), including negative.

Answer 1

The Correct Option is C

By induction \(A^n = \begin{bmatrix}\cos n\theta & -\sin n\theta\\ \sin n\theta & \cos n\theta\end{bmatrix}\) for all integers \(n\) (\(A^{-1}\) is the rotation by \(-\theta\)).
For \(n = -50\), \(\theta = \frac\pi{12}\): \(n\theta = -\frac{50\pi}{12} = -\frac{25\pi}{6}\).
\(\cos\left(-\frac{25\pi}{6}\right) = \cos\left(4\pi + \frac\pi6\right) = \frac{\sqrt3}{2}\), \(\sin\left(-\frac{25\pi}{6}\right) = -\sin\frac\pi6 = -\frac12\).
\(A^{-50} = \begin{bmatrix}\frac{\sqrt3}{2} & \frac12\\ -\frac12 & \frac{\sqrt3}{2}\end{bmatrix}\).
Answer: (3) \(\begin{bmatrix}\frac{\sqrt3}{2} & \frac12\\ -\frac12 & \frac{\sqrt3}{2}\end{bmatrix}\)