<p>Let \(A = \begin{pmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{pmatrix}\), \((\alpha \in \mathbb{R})\) such that \(A^{32} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\). Then a value of \(\alpha\) is:</p>
<p>\(\dfrac{\pi}{32}\)</p>
<p>\(0\)</p>
<p>\(\dfrac{\pi}{64}\)</p>
<p>\(\dfrac{\pi}{16}\)</p>
Step-by-Step Solution
Key Concept: Matrix A is a rotation matrix. Computing A^32 means applying the rotation 32 times, so 32α must equal the angle corresponding to the target matrix (which is a 90° counterclockwise rotation), modulo 2π.
<p><strong>Step 1:</strong> Recognize that A is a rotation matrix by angle α. For rotation matrices: A^n rotates by angle nα.</p><p><strong>Step 2:</strong> The target matrix ⎣0 -1⎦ is a 90° counterclockwise rotation, corresponding to angle π/2.
⎡1 0⎤</p><p><strong>Step 3:</strong> Therefore A^32 represents a rotation by 32α, which must equal a rotation by π/2 (plus any multiple of 2π):</p><p>32α = π/2 + 2πk, where k ∈ ℤ</p><p><strong>Step 4:</strong> Solve for α:</p><p>α = π/64 + πk/16</p><p><strong>Step 5:</strong> For the principal values in [0, 2π), we get α = π/64, 9π/64, 17π/64, 25π/64, etc. The most common answer is <strong>α = π/64</strong> (corresponding to k = 0).</p><p>∴ Answer: C</p>
Correct Answer: C