<p>The number of diagonal matrices <em>A</em> of order \(n\) for which \(A^3 = A\) is</p>
Step-by-Step Solution
Key Concept: A diagonal matrix A satisfies A³ = A if and only if each diagonal element d satisfies d³ = d, which means d ∈ {-1, 0, 1}. Since there are n independent diagonal positions, each with 3 choices, the total count is 3ⁿ.
<p><strong>Step 1:</strong> A diagonal matrix has the form A = diag(d₁, d₂, ..., dₙ) where dᵢ are the diagonal entries.</p><p><strong>Step 2:</strong> For a diagonal matrix, A³ = diag(d₁³, d₂³, ..., dₙ³).</p><p><strong>Step 3:</strong> The condition A³ = A requires: diag(d₁³, d₂³, ..., dₙ³) = diag(d₁, d₂, ..., dₙ).</p><p><strong>Step 4:</strong> This means dᵢ³ = dᵢ for each i, so dᵢ(dᵢ² - 1) = 0, giving dᵢ(dᵢ - 1)(dᵢ + 1) = 0.</p><p><strong>Step 5:</strong> Therefore, each diagonal element must satisfy dᵢ ∈ {-1, 0, 1}.</p><p><strong>Step 6:</strong> Since there are n diagonal positions and each can independently take any of 3 values, the total number of such matrices is 3ⁿ.</p><p>∴ Answer: <strong>3ⁿ</strong></p>
Correct Answer: D