Question:

<p><strong>Step 1:</strong> We have $$A = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}$$</p><p><strong>Step 2:</strong> Calculate the determinant: $$|A| = \cos^2 \theta + \sin^2 \theta = 1$$</p><p><strong>Step 3:</strong> The adjugate matrix is: $$\text{adj}(A) = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}$$</p><p><strong>Step 4:</strong> Since $|A| = 1$, we have $$A^{-1} = \text{adj}(A) = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}$$</p><p><strong>Step 5:</strong> This is a rotation matrix by angle $-\theta$. Thus $A^{-50}$ corresponds to rotation by angle $-50\theta$.</p><p><strong>Step 6:</strong> When $\theta = \frac{\pi}{12}$, we have $-50\theta = -\frac{50\pi}{12} = -\frac{25\pi}{6} = -4\pi - \frac{\pi}{6}$, which is equivalent to $-\frac{\pi}{6}$.</p><p><strong>Step 7:</strong> Therefore: $$A^{-50} = \begin{pmatrix} \cos(-\frac{\pi}{6}) & \sin(-\frac{\pi}{6}) \\ -\sin(-\frac{\pi}{6}) & \cos(-\frac{\pi}{6}) \end{pmatrix} = \begin{pmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{pmatrix}$$</p><p>∴ Answer is (c).</p>