<p>If \(\Delta = \begin{vmatrix} 1 & \omega^n & \omega^{2n} \\ \omega^n & \omega^{2n} & 1 \\ \omega^{2n} & 1 & \omega^n \end{vmatrix}\), then \(\Delta\) equals:</p>
Step-by-Step Solution
Key Concept: Recognize that the matrix has a cyclic structure where each row is a cyclic permutation of powers of ω. Use the property that ω³ = 1 (cube root of unity) to simplify, and exploit the fact that the determinant of a circulant matrix with cyclic structure equals zero when the entries follow the pattern of roots of unity.
<p><strong>Step 1:</strong> Observe the matrix structure. Each row is a cyclic shift of powers of ω (where ω is a cube root of unity, so ω³ = 1):</p><p>Row 1: 1, ω^n, ω^(2n)</p><p>Row 2: ω^n, ω^(2n), 1</p><p>Row 3: ω^(2n), 1, ω^n</p><p><strong>Step 2:</strong> Factor out common terms. Since ω³ = 1, we have ω^n = ω^(n mod 3). Let a = ω^(n mod 3). The matrix becomes:</p><p>Δ = |1, a, a² | = (1 + a³ - a³) - ... (using the circulant property)</p><p><strong>Step 3:</strong> For a circulant matrix with first row [c₀, c₁, c₂], the determinant equals ∏(c₀ + c₁λⱼ + c₂λⱼ²) where λⱼ are cube roots of unity (1, ω, ω²).</p><p><strong>Step 4:</strong> Apply this formula: Δ = (1 + a + a²)(1 + aω + a²ω²)(1 + aω² + a²ω⁴)</p><p>Since a = ω^n and a³ = 1, we have 1 + a + a² = 0 (property of cube roots of unity when a ≠ 1), making the entire product zero.</p><p><strong>Step 5:</strong> When n ≢ 0 (mod 3), the first factor 1 + ω^n + ω^(2n) = 0, so Δ = 0.</p><p>∴ Answer: <strong>Δ = 0</strong></p>
Correct Answer: A