<p>If \(A = \begin{bmatrix}i & -i\\ -i & i\end{bmatrix}\) and \(B = \begin{bmatrix}1 & -1\\ -1 & 1\end{bmatrix}\), then \(A^8\) equals</p>
Step-by-Step Solution
Key Concept: Recognize that A = iB, so A⁸ = i⁸B⁸. Since i⁴ = 1, we have i⁸ = 1. Then compute B⁸ by finding the pattern in powers of B, noting that B² = 2B (idempotent-like behavior).
<p><strong>Step 1:</strong> Observe that A = iB, where B = [[1,-1],[-1,1]]</p><p><strong>Step 2:</strong> Therefore A⁸ = i⁸B⁸. Since i⁴ = 1, we have i⁸ = (i⁴)² = 1</p><p><strong>Step 3:</strong> Find the pattern for powers of B:</p><p>B² = [[1,-1],[-1,1]][[1,-1],[-1,1]] = [[1+1, -1-1],[-1-1, 1+1]] = [[2,-2],[-2,2]] = 2B</p><p><strong>Step 4:</strong> By induction: Bⁿ = 2ⁿ⁻¹B for n ≥ 1</p><p><strong>Step 5:</strong> Therefore B⁸ = 2⁷B = 128[[1,-1],[-1,1]] = [[128,-128],[-128,128]]</p><p><strong>Step 6:</strong> Thus A⁸ = 1 · B⁸ = [[128,-128],[-128,128]] = 2⁷B</p><p>∴ Answer: B</p>
Correct Answer: B