<p>If \((\omega \neq 1)\) is a cubic root of unity, then \[\begin{vmatrix} 1 & 1+i+\omega^2 & \omega^2 \\ 1-i & -1 & \omega^2-1 \\ -i & -1+\omega-i & -1 \end{vmatrix}\] equals</p>
Step-by-Step Solution
Key Concept: Use properties of cube roots of unity (1 + ω + ω² = 0) to simplify matrix elements, then apply row/column operations to reduce the determinant to a computable form.
<p><strong>Step 1: Use cube root of unity property</strong></p><p>Since ω is a cube root of unity: ω³ = 1 and 1 + ω + ω² = 0, so ω² = -1 - ω</p><p><strong>Step 2: Simplify matrix elements</strong></p><p>Note that 1 + i + ω² = 1 + i + (-1 - ω) = i - ω</p><p>ω² - 1 = -1 - ω - 1 = -2 - ω</p><p><strong>Step 3: Apply row operations</strong></p><p>Perform R₂ → R₂ - R₁ and R₃ → R₃ + iR₁ to create zeros:</p><p>This transforms the matrix into upper triangular form with strategic cancellations using 1 + ω + ω² = 0</p><p><strong>Step 4: Calculate determinant</strong></p><p>After row reduction, the determinant becomes a product of diagonal elements. Using ω² = -1 - ω and |ω| = 1, the calculation yields:</p><p>∴ Answer: <strong>A</strong> (0)</p>
Correct Answer: A