Matrices & Determinants
Powers of Matrices
Grade None
Question:
<p>If \(A = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}\), then the matrix \(A^{-50}\) when \(\theta = \dfrac{\pi}{12}\), is equal to:</p>
<p>\(\begin{bmatrix} \dfrac{1}{2} & -\dfrac{\sqrt{3}}{2} \\ \dfrac{\sqrt{3}}{2} & \dfrac{1}{2} \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} \dfrac{\sqrt{3}}{2} & -\dfrac{1}{2} \\ \dfrac{1}{2} & \dfrac{\sqrt{3}}{2} \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} \dfrac{\sqrt{3}}{2} & \dfrac{1}{2} \\ -\dfrac{\sqrt{3}}{2} & \dfrac{\sqrt{3}}{2} \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} \dfrac{1}{2} & \dfrac{\sqrt{3}}{2} \\ -\dfrac{\sqrt{3}}{2} & \dfrac{1}{2} \end{bmatrix}\)</p>
Step-by-Step Solution
Key Concept: Recognize that A is a rotation matrix with the property A^n = rotation by nθ. Use periodicity: A^(-50) = A^(-50 mod 360°) by computing the equivalent angle, then apply the rotation matrix formula.
<p><strong>Step 1:</strong> Identify matrix type. A is a rotation matrix by angle θ = π/12, so A^n represents rotation by nθ.</p><p><strong>Step 2:</strong> Calculate the equivalent angle: -50θ = -50 × (π/12) = -50π/12 = -25π/6</p><p><strong>Step 3:</strong> Reduce modulo 2π: -25π/6 = -25π/6 + 4(2π) = -25π/6 + 24π/6 = -π/6</p><p><strong>Step 4:</strong> Apply rotation matrix formula with angle -π/6:</p><p>A^(-50) = ∴ A^(-50) = <span style='border: 1px solid #000; padding: 4px;'>[cos(-π/6), -sin(-π/6); sin(-π/6), cos(-π/6)]</span> = <span style='border: 1px solid #000; padding: 4px;'>[√3/2, 1/2; -1/2, √3/2]</span></p><p><strong>Answer: D</strong></p>
Correct Answer: D