<p>How many different diagonal matrices of order <em>n</em> can be formed which are idempotent?</p>
Step-by-Step Solution
Key Concept: An idempotent matrix satisfies A² = A. For a diagonal matrix with entries d₁, d₂, ..., dₙ on the diagonal, this means dᵢ² = dᵢ for each i, so each diagonal entry must be either 0 or 1.
<p><strong>Step 1:</strong> A diagonal matrix D has the form diag(d₁, d₂, ..., dₙ)</p><p><strong>Step 2:</strong> For idempotency, we require D² = D, which gives diag(d₁², d₂², ..., dₙ²) = diag(d₁, d₂, ..., dₙ)</p><p><strong>Step 3:</strong> This means dᵢ² = dᵢ for each i = 1, 2, ..., n</p><p><strong>Step 4:</strong> Solving dᵢ² = dᵢ gives dᵢ(dᵢ - 1) = 0, so dᵢ ∈ {0, 1}</p><p><strong>Step 5:</strong> Each of the n diagonal positions has 2 independent choices (0 or 1)</p><p><strong>Step 6:</strong> Total number of such matrices = 2 × 2 × ... × 2 (n times)</p><p>∴ Answer: <strong>2ⁿ</strong></p>
Correct Answer: 2^n