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 \( Q = PAP^T \) and \( X = P^T Q^{2005} P \), then \( X \) is equal to</p>
<p>\(\begin{bmatrix} 1 & 2005 \\ 0 & 1 \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} 4 + 2005\sqrt{3} & 6015 \\ 2005 & 4 - 2005\sqrt{3} \end{bmatrix}\)</p>
<p>\(\dfrac{1}{4}\begin{bmatrix} 2+\sqrt{3} & 1 \\ -1 & 2-\sqrt{3} \end{bmatrix}\)</p>
<p>\(\dfrac{1}{4}\begin{bmatrix} 2005 & 2-\sqrt{3} \\ 2+\sqrt{3} & 2005 \end{bmatrix}\)</p>
Step-by-Step Solution
Key Concept: Recognize that P is a rotation matrix (rotation by 60°), so P^T = P^(-1). This means Q = PAP^T is a similarity transformation, and Q^n has the same eigenvalue structure as A^n, allowing us to use the recurrence relation for powers of A.
<p><strong>Step 1:</strong> Verify that P is a rotation matrix. Check that det(P) = (√3/2)(√3/2) + (1/2)(1/2) = 3/4 + 1/4 = 1, and P·P^T = I, confirming P^(-1) = P^T.</p><p><strong>Step 2:</strong> Since Q = PAP^T, we have Q = P·A·P^(-1), which means Q and A are similar matrices. Therefore Q^n = P·A^n·P^(-1).</p><p><strong>Step 3:</strong> Express X in terms of A^2005:<br/>X = P^T·Q^2005·P = P^T·(PAP^T)^2005·P = P^T·P·A^2005·(P^T)^2005·P<br/>Since P^T = P^(-1), we get: X = I·A^2005·(P^(-1))^2005·P = A^2005·P^(-2005)·P<br/>More directly: X = P^T·Q^2005·P = (P^T P)·A^2005·(P^T P) = A^2005</p><p><strong>Step 4:</strong> Compute A^2005. Notice that A = I + N where N = [0 1; 0 0] is nilpotent with N^2 = 0.<br/>Therefore A^2005 = (I + N)^2005 = I + 2005N = [1 2005; 0 1]</p><p>∴ Answer: A</p>
Correct Answer: A