<p>Let \(\omega\) be a complex cube root of unity with \(\omega \neq 1\) and \(P = [p_{ij}]\) be a \(n \times n\) matrix with \(p_{ij} = \omega^{i+j}\). Then \(P^2 \neq O\), when \(n =\)</p>
Step-by-Step Solution
Key Concept: For P² = O, the matrix P must be nilpotent; analyzing the (i,j)-th entry of P² = Σₖ ω^(i+k) · ω^(k+j) = ω^(i+j) Σₖ ω^(2k) reveals that P² = O only when Σₖ ω^(2k) = 0, which requires n ≡ 0 (mod 3) since ω²ᵏ cycles with period 3.
<p><strong>Step 1:</strong> Write the matrix P where p_{ij} = ω^(i+j) for i,j ∈ {1,2,...,n}.</p><p><strong>Step 2:</strong> Compute (P²)_{ij} = Σₖ₌₁ⁿ p_{ik}·p_{kj} = Σₖ₌₁ⁿ ω^(i+k)·ω^(k+j) = ω^(i+j) Σₖ₌₁ⁿ ω^(2k).</p><p><strong>Step 3:</strong> Factor out ω^(i+j) from each entry. Thus P² = ω^(i+j)·S where S = Σₖ₌₁ⁿ ω^(2k) is independent of i and j.</p><p><strong>Step 4:</strong> For P² = O, we need S = 0. Since ω is a primitive cube root of unity, ω² ≠ 1. The sum S = ω² + ω⁴ + ω⁶ + ... + ω^(2n) = ω²(1 + ω² + ω⁴ + ... + ω^(2(n-1))).</p><p><strong>Step 5:</strong> This is a geometric series with ratio ω². When n ≡ 0 (mod 3), we get ω^(2n) = 1 and S = ω²·(1-1)/(1-ω²) = 0. When n ≢ 0 (mod 3), S ≠ 0.</p><p><strong>Step 6:</strong> Therefore P² ≠ O precisely when n ≢ 0 (mod 3), i.e., n = 1, 2, 4, 5, 7, 8, ...</p><p>∴ Answer: ACD (selecting options where n is NOT divisible by 3)</p>
Correct Answer: ACD