<p>If 1, \(\omega\), \(\omega^2\) are the cube roots of unity, then \[\Delta = \begin{vmatrix} 1 & \omega^2 & \omega^{2n} \\ \omega^n & \omega^{2n} & 1 \\ \omega^{2n} & 1 & \omega^n \end{vmatrix}\] is equal to</p>
Step-by-Step Solution
Key Concept: Use the properties of cube roots of unity (1 + ω + ω² = 0 and ω³ = 1) to reduce powers of ω modulo 3, then recognize that the determinant has a cyclic structure that simplifies based on the value of n mod 3.
<p><strong>Step 1:</strong> Use ω³ = 1 to reduce powers. Let n = 3q + r where r ∈ {0, 1, 2}. Then ω^n = ω^r and ω^(2n) = ω^(2r).</p><p><strong>Step 2:</strong> The determinant becomes:</p><p>Δ = |1 ω² ω^(2r)|</p><p> |ω^r ω^(2r) 1|</p><p> |ω^(2r) 1 ω^r|</p><p><strong>Step 3:</strong> For n ≡ 0 (mod 3): r = 0, so ω^r = 1 and ω^(2r) = 1. The matrix becomes all 1's in this configuration, giving Δ = 0.</p><p><strong>Step 4:</strong> For n ≡ 1 (mod 3): r = 1, so the matrix is |1 ω² ω| / |ω ω² 1| / |ω² 1 ω|. Expanding: Δ = 1(ω³ - 1) - ω²(ω² - ω²) + ω(ω - ω⁴) = 0 - 0 + ω(ω - ω) = 0.</p><p><strong>Step 5:</strong> For n ≡ 2 (mod 3): r = 2, giving the same cyclic structure. Computing the determinant yields Δ = 0 in all cases due to linear dependence of rows/columns.</p><p><strong>Alternatively:</strong> Note that each row is a cyclic permutation, and with cube roots of unity symmetry, the determinant equals <strong>0</strong>.</p><p>∴ Answer: <strong>A</strong> (Δ = 0)
Correct Answer: A