Matrices & Determinants
Matrix powers
Grade 12
Question:
<p>If \( P = \begin{bmatrix} \dfrac{\sqrt{3}}{2} & \dfrac{1}{2} \\ -\dfrac{1}{2} & \dfrac{\sqrt{3}}{2} \end{bmatrix} \), \( A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \) and \( Q = PAP^T \), then \( P^T Q^{2015} P \) is</p>
<p>\(\begin{bmatrix} 0 & 2015 \\ 0 & 0 \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} 2015 & 0 \\ 1 & 2015 \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} 1 & 2015 \\ 0 & 1 \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} 2015 & 1 \\ 0 & 2015 \end{bmatrix}\)</p>
Step-by-Step Solution
Key Concept: Recognize that P is a rotation matrix (cos 60° and sin 60° entries), so P^T = P^(-1). This means Q = PAP^T transforms A into a similar matrix, and Q^n follows a predictable pattern based on A's structure.
<p><strong>Step 1:</strong> Identify that P is a rotation matrix by 60° counterclockwise. Since orthogonal matrices satisfy P^T P = I, we have P^(-1) = P^T.</p><p><strong>Step 2:</strong> Given Q = PAP^T, we have Q = PAP^(-1), meaning Q is similar to A with the same eigenvalues and structural properties.</p><p><strong>Step 3:</strong> Compute A^2: A^2 = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}</p><p><strong>Step 4:</strong> Observe the pattern: A^n = \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}. Therefore A^2015 = \begin{bmatrix} 1 & 2015 \\ 0 & 1 \end{bmatrix}</p><p><strong>Step 5:</strong> Since Q^2015 = (PAP^T)^2015 = PA^2015P^T (by the property of similarity), we have:</p><p>P^T Q^2015 P = P^T(PA^2015P^T)P = (P^T P)A^2015(P^T P) = A^2015 = \begin{bmatrix} 1 & 2015 \\ 0 & 1 \end{bmatrix}</p><p>∴ Answer: C</p>
Correct Answer: C