Matrices & Determinants
Powers of a Matrix
Grade 12
Question:
<p>Given <br>\(A = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}\)<br>and \(A^{32} = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\), then \(\alpha =\)</p>
<p>\(\dfrac{\pi}{32}\)</p>
<p>\(\dfrac{\pi}{64}\)</p>
<p>\(\dfrac{\pi}{16}\)</p>
<p>\(\dfrac{\pi}{128}\)</p>
Step-by-Step Solution
Key Concept: Matrix A is a rotation matrix that rotates by angle α. When raised to power 32, it rotates by 32α. Since A³² equals a 90° rotation matrix, we need 32α ≡ 90° (mod 360°).
<p><strong>Step 1:</strong> Recognize that matrix A is a rotation matrix by angle α. Therefore:</p><p>A^n = ⎡cos(nα) -sin(nα)⎤</p><p> ⎣sin(nα) cos(nα)⎦</p><p><strong>Step 2:</strong> Given A³² = ⎡0 -1⎤, this is a rotation by 90°.</p><p> ⎣1 0⎦</p><p>Thus: cos(32α) = 0 and sin(32α) = 1</p><p>This means 32α = 90° + 360°k for integer k</p><p><strong>Step 3:</strong> Solve for α:</p><p>32α = 90° + 360°k</p><p>α = (90° + 360°k)/32 = (90 + 360k)°/32</p><p><strong>Step 4:</strong> For k = 0: α = 90°/32 = 45°/16</p><p>For k = 1: α = 450°/32 = 225°/16</p><p>For k = 7: α = 2610°/32 = 1305°/16 = 81.5625° (simplifying to principal value)</p><p>The most common answer is α = 45°/16 or equivalently α = 2.8125° (or in radians: π/64)</p><p>∴ Answer: <strong>B</strong> (typically α = π/64 rad or 45°/16)</p>
Correct Answer: B