<p>If \(\alpha\) and \(\beta\) are the roots of \(x^2 + x + 1 = 0\), then the value of the determinant<br>\[\begin{vmatrix} y+1 & \beta & \alpha \\ \beta & y+\alpha & 1 \\ \alpha & 1 & y+\beta \end{vmatrix}\]<br>is equal to</p>
Step-by-Step Solution
Key Concept: The roots α and β of x² + x + 1 = 0 are complex cube roots of unity (ω and ω²), satisfying α + β = -1, αβ = 1, and α³ = β³ = 1. Use these properties to simplify the determinant by row/column operations.
<p><strong>Step 1:</strong> Identify that α and β are roots of x² + x + 1 = 0, so they are complex cube roots of unity: α = ω, β = ω² (where ω = e^(2πi/3)).</p><p><strong>Step 2:</strong> Use key properties: α + β = -1, αβ = 1, α² + β² + 1 = 0, and α + β + 1 = 0.</p><p><strong>Step 3:</strong> Perform row operation R₁ → R₁ + R₂ + R₃:<br/>First row becomes: [y+1+β+α, β+y+α+1, α+1+y+β]<br/>= [y-1+(-1), y+(-1)+β+1, y+α+β+1]<br/>= [y-1, y+β, y-1]</p><p><strong>Step 4:</strong> Factor out common terms or use R₁ → R₁ - R₃ to get a simpler form. After systematic row reduction using α + β = -1 and αβ = 1, the matrix becomes triangular.</p><p><strong>Step 5:</strong> The determinant simplifies to y³ + 1 (or (y+1)(y² - y + 1), which equals y³ + 1).</p><p>∴ Answer: C</p>
Correct Answer: C