Matrices & Determinants
Matrix Powers and Orthogonal Matrices
Grade 12
Question:
<p>Let \(P = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}\), \(Q = PAP^T\) and \(X = P^T Q^{2005} P\). Find \(X\), given \(A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\).</p>
<p>\(\begin{bmatrix} 1 & 2005 \\ 0 & 1 \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} 1 & 0 \\ 2005 & 1 \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} 2005 & 0 \\ 0 & 2005 \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} 0 & 2005 \\ 2005 & 0 \end{bmatrix}\)</p>
Step-by-Step Solution
Key Concept: Matrix P is a rotation matrix (orthogonal with P^T P = I), so Q = PAP^T represents a similarity transformation that simplifies A. Computing powers of Q becomes tractable by recognizing the block structure or eigenvalues of the transformed matrix.
<p><strong>Step 1:</strong> Verify P is orthogonal. Since cos(30°) = √3/2, sin(30°) = 1/2, matrix P is a rotation by 30°. Check: P^T P = I, so P^(-1) = P^T.</p><p><strong>Step 2:</strong> Use similarity: Q = PAP^T implies Q^n = PA^nP^T. Therefore, X = P^T Q^2005 P = P^T(PA^2005 P^T)P = (P^T P)A^2005(P^T P) = A^2005.</p><p><strong>Step 3:</strong> Compute A^2005 where A = [1,1; 0,1]. By induction, A^n = [1,n; 0,1]. For upper triangular unipotent matrices, A^n = I + n(A-I).</p><p><strong>Step 4:</strong> Thus A^2005 = [1, 2005; 0, 1].</p><p>∴ Answer: X = <strong>A</strong> = [1, 2005; 0, 1]</p>
Correct Answer: A