<p>If \(Z\) is an idempotent matrix, then \((I + Z)^n\)</p>
Step-by-Step Solution
Key Concept: An idempotent matrix Z satisfies Z² = Z, which means (I + Z)ⁿ can be expanded using binomial theorem where all higher powers of Z collapse to Z itself, yielding a pattern independent of n.
<p><strong>Step 1:</strong> Recall that Z is idempotent means Z² = Z (and consequently Zᵏ = Z for all k ≥ 1).</p><p><strong>Step 2:</strong> Expand (I + Z)ⁿ using binomial theorem:
(I + Z)ⁿ = Iⁿ + nC₁·Iⁿ⁻¹·Z + nC₂·Iⁿ⁻²·Z² + ... + Zⁿ</p><p><strong>Step 3:</strong> Since Z² = Z, we have Z² = Z, Z³ = Z·Z² = Z·Z = Z, and Zᵏ = Z for all k ≥ 1.</p><p><strong>Step 4:</strong> Substitute into expansion:
(I + Z)ⁿ = I + nC₁·Z + nC₂·Z + nC₃·Z + ... + Z
= I + (nC₁ + nC₂ + nC₃ + ... + nCₙ)·Z
= I + (2ⁿ - 1)·Z</p><p><strong>Step 5:</strong> Alternatively, (I + Z)ⁿ = I + (2ⁿ - 1)Z = I + Z(2ⁿ - 1)</p><p>∴ Answer: (I + Z)ⁿ = I + (2ⁿ - 1)Z (Option B, most likely)</p>
Correct Answer: B