<p>Let <br>\(A = \begin{bmatrix} \tan\frac{\pi}{3} & \sec\frac{2\pi}{3} \\ \cot\left(2013\frac{\pi}{2}\right) & \cos(2012\pi) \end{bmatrix}\)<br> and \(P\) be a \(2 \times 2\) matrix such that \(PP^T = I\), where \(I\) is an identity matrix of order 2. If \(Q = PAP^T\) and \(R = [r_{ij}]_{2\times 2} = P^T Q^8 P\), then find \(r_{11}\).</p>
Step-by-Step Solution
Key Concept: First simplify matrix A using trigonometric identities (tan(π/3)=√3, sec(2π/3)=-2, cot(1006.5π)=0, cos(2012π)=1), then recognize that P is orthogonal (PP^T=I), making Q similar to A with the same eigenvalues. The crucial insight is that R=P^T Q^8 P = P^T P A^8 P^T P = A^8, so we only need to compute A^8 and extract r₁₁.
<p><strong>Step 1: Simplify trigonometric values</strong></p><p>tan(π/3) = √3</p><p>sec(2π/3) = 1/cos(2π/3) = 1/(-1/2) = -2</p><p>cot(2013π/2): Since 2013 = 4(503) + 1, we have cot(2013π/2) = cot(π/2) = 0</p><p>cos(2012π) = cos(0) = 1</p><p>Therefore, A = [√3, -2; 0, 1]</p><p><strong>Step 2: Recognize similarity invariance</strong></p><p>Since PP^T = I (P is orthogonal), Q = PAP^T is similar to A.</p><p>Thus Q^8 is similar to A^8, and R = P^T Q^8 P = A^8</p><p><strong>Step 3: Compute A^8 by finding eigenvalues</strong></p><p>For upper triangular A, eigenvalues are λ₁ = √3 and λ₂ = 1</p><p>Diagonalize: A = PDP^(-1) where D = diag(√3, 1)</p><p>Then A^8 = PD^8P^(-1) = P·diag((√3)^8, 1)·P^(-1)</p><p><strong>Step 4: Calculate (√3)^8</strong></p><p>(√3)^8 = ((√3)^2)^4 = (3)^4 = 81</p><p><strong>Step 5: Compute A^8 directly</strong></p><p>A^2 = [√3, -2; 0, 1]·[√3, -2; 0, 1] = [3, -2√3-2; 0, 1]</p><p>A^4 = [9, -8√3-8; 0, 1]</p><p>A^8 = [81, r₁₂; 0, 1]</p><p>∴ r₁₁ = <strong>81</strong></p>
Correct Answer: 81