<p>If 1, <span class="math">\omega</span> and <span class="math">\omega^2</span> are the cube roots of unity, then <span class="math">\begin{vmatrix} 1 & \omega^n & \omega^{2n} \\ \omega^n & \omega^{2n} & 1 \\ \omega^{2n} & 1 & \omega^n \end{vmatrix}</span> is equal to</p>
Step-by-Step Solution
Key Concept: The determinant exhibits a cyclic pattern in its rows and columns. Since 1, ω, ω² are cube roots of unity, we have ω³ = 1 and 1 + ω + ω² = 0. The matrix has a special structure where each row is a cyclic shift of the previous one, making it a circulant-type matrix whose determinant vanishes.
<p><strong>Step 1:</strong> Recognize the matrix structure. The determinant is:</p><p>$$\Delta = \begin{vmatrix} 1 & \omega^n & \omega^{2n} \\ \omega^n & \omega^{2n} & 1 \\ \omega^{2n} & 1 & \omega^n \end{vmatrix}$$</p><p>Notice that each row is a cyclic permutation of the previous row (with powers reducing modulo 3).</p><p><strong>Step 2:</strong> Add all rows together. Form (R₁ + R₂ + R₃):</p><p>$$R_1 + R_2 + R_3 = (1 + \omega^n + \omega^{2n}, \omega^n + \omega^{2n} + 1, \omega^{2n} + 1 + \omega^n)$$</p><p>Each entry equals: $(1 + \omega^n + \omega^{2n})$</p><p><strong>Step 3:</strong> Apply the cube roots of unity property. Since ω³ = 1, the exponents repeat with period 3. We have three cases:</p><p><strong>Case 1:</strong> If n ≡ 0 (mod 3): $1 + \omega^n + \omega^{2n} = 1 + 1 + 1 = 3$</p><p><strong>Case 2:</strong> If n ≡ 1 (mod 3): $1 + \omega + \omega^2 = 0$</p><p><strong>Case 3:</strong> If n ≡ 2 (mod 3): $1 + \omega^2 + \omega^4 = 1 + \omega^2 + \omega = 0$</p><p><strong>Step 4:</strong> For Cases 2 and 3, when we add rows, we get (R₁ + R₂ + R₃) = (0, 0, 0), making the rows linearly dependent. The determinant is 0.</p><p><strong>Step 5:</strong> For Case 1 (n ≡ 0 mod 3), factor out 3 from (R₁ + R₂ + R₃) = (3, 3, 3) = 3(1, 1, 1). After row reduction, columns become identical, confirming linear dependence and determinant = 0.</p><p><strong>Verification by direct expansion:</strong> Computing the determinant directly using Sarrus' rule or cofactor expansion and applying 1 + ω + ω² = 0 repeatedly confirms all terms cancel.</p><p><strong>∴ Answer:</strong> A</p>
Correct Answer: A