<p>Let \(\omega\) be a complex number such that \(2\omega + 1 = z\) where \(z = \sqrt{-3}\). If \(\begin{vmatrix} 1 & 1 & 1 \\ 1 & -\omega^2 - 1 & \omega^2 \\ 1 & \omega^2 & \omega^7 \end{vmatrix} = 3k\), then \(k\) is equal to</p>
Step-by-Step Solution
Key Concept: Recognize that ω is a primitive cube root of unity (ω³ = 1, 1 + ω + ω² = 0), which allows simplification of powers like ω⁷ = ω and the symmetric structure of the determinant matrix.
<p><strong>Step 1:</strong> From 2ω + 1 = √(-3) = i√3, we get 2ω = -1 + i√3, so ω = (-1 + i√3)/2. This is a primitive cube root of unity: ω³ = 1 and 1 + ω + ω² = 0.</p><p><strong>Step 2:</strong> Since ω³ = 1, we have ω⁷ = ω⁶·ω = (ω³)²·ω = ω.</p><p><strong>Step 3:</strong> From 1 + ω + ω² = 0, we get ω² + 1 = -ω, so -ω² - 1 = ω.</p><p><strong>Step 4:</strong> The determinant becomes: <br/>$$\begin{vmatrix} 1 & 1 & 1 \\ 1 & ω & ω^2 \\ 1 & ω^2 & ω \end{vmatrix}$$</p><p><strong>Step 5:</strong> Using R₂ → R₂ - R₁ and R₃ → R₃ - R₁:<br/>$$\begin{vmatrix} 1 & 1 & 1 \\ 0 & ω-1 & ω^2-1 \\ 0 & ω^2-1 & ω-1 \end{vmatrix} = (ω-1)² - (ω²-1)²$$</p><p><strong>Step 6:</strong> Computing: (ω-1)² - (ω²-1)² = (ω-1)² - (ω-1)²(ω+1)² = (ω-1)²[1 - (ω+1)²] = (ω-1)²(-ω² - 2ω) = (ω-1)²·ω(1-ω) = -ω(ω-1)³ = 3√3i = 3k, so k = √3i.</p><p>∴ Answer: B</p>
Correct Answer: B