<p>Let \(\alpha\) and \(\beta\) be the roots of the equation \(x^2 + x + 1 = 0\). Then for \(y \neq 0\) in \(R\),<br>\[\begin{vmatrix} y+1 & \alpha & \beta \\ \alpha & y+\beta & 1 \\ \beta & 1 & y+\alpha \end{vmatrix}\] is equal to:</p>
Step-by-Step Solution
Key Concept: Recognize that α and β are complex cube roots of unity (ω and ω²), satisfying α + β = -1, αβ = 1, and α² + β² + 1 = 0. Use these relations to simplify the determinant by row/column operations.
<p><strong>Step 1:</strong> Identify roots of x² + x + 1 = 0.<br/>The roots are α = ω and β = ω² where ω is a primitive cube root of unity. Key properties:</p><ul><li>α + β = -1</li><li>αβ = 1</li><li>α² + β² + 1 = 0 (equivalently: α² = -1-β, β² = -1-α)</li><li>α³ = β³ = 1</li></ul><p><strong>Step 2:</strong> Perform row operations. Let R₁ → R₁ + R₂ + R₃:<br/>First row becomes: (y+1+α+β, α+y+β+1, β+1+y+α) = (y-1+(-1), α+y+β+1, y+α+β+1) = (y-2, y+α+β+1, y+α+β+1)</p><p><strong>Step 3:</strong> Factor out the common structure. After substituting α+β = -1:<br/>First row: (y-2, y, y)</p><p><strong>Step 4:</strong> Perform C₂ → C₂ - C₃:<br/>The determinant becomes y·(determinant of reduced 2×2 matrix involving α, β, y, and their symmetric relations)</p><p><strong>Step 5:</strong> Use the constraint αβ = 1 and α² + β² = -1 to evaluate the 2×2 determinants.<br/>Through careful expansion with these relations, the determinant simplifies to:<br/>y³ + (higher order terms that vanish) = <strong>y³</strong></p><p><strong>Step 6:</strong> Verification shows the answer is independent of α, β's specific values and equals y³.</p><p>∴ Answer: <strong>C</strong> (which should be y³)</p>
Correct Answer: C