Matrices & Determinants
Inverse of a Matrix
Grade 12
Question:
<p>Given \(A = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}\). If \(A^{-50}\) is evaluated at \(\theta = \dfrac{\pi}{12}\), find \(A^{-50}\).</p>
<p>\(\begin{bmatrix} \dfrac{\sqrt{3}}{2} & \dfrac{1}{2} \\ -\dfrac{1}{2} & \dfrac{\sqrt{3}}{2} \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} \dfrac{1}{2} & \dfrac{\sqrt{3}}{2} \\ -\dfrac{\sqrt{3}}{2} & \dfrac{1}{2} \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} \dfrac{\sqrt{3}}{2} & -\dfrac{1}{2} \\ \dfrac{1}{2} & \dfrac{\sqrt{3}}{2} \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} \dfrac{1}{2} & -\dfrac{\sqrt{3}}{2} \\ \dfrac{\sqrt{3}}{2} & \dfrac{1}{2} \end{bmatrix}\)</p>
Step-by-Step Solution
Key Concept: Matrix A is a rotation matrix, so A^n represents rotation by nθ. For negative powers, A^(-50) = (A^(-1))^50, and since A^(-1) rotates by -θ, we get A^(-50) rotates by -50θ. Thus A^(-50) = rotation matrix with angle -50θ.
<p><strong>Step 1:</strong> Recognize that A is a 2D rotation matrix rotating by angle θ. For rotation matrices, A^n rotates by angle nθ.</p><p><strong>Step 2:</strong> For A^(-50), we rotate by angle (-50θ) = -50 · (π/12) = -50π/12 = -25π/6</p><p><strong>Step 3:</strong> Simplify the angle: -25π/6 = -24π/6 - π/6 = -4π - π/6 ≡ -π/6 (mod 2π)</p><p><strong>Step 4:</strong> Therefore A^(-50) is the rotation matrix with angle -π/6:</p><p>A^(-50) = \begin{bmatrix} cos(-π/6) & -sin(-π/6) \\ sin(-π/6) & cos(-π/6) \end{bmatrix} = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}</p><p>∴ Answer: A</p>
Correct Answer: A