<p>If \(f(x) = a + bx + cx^2\) and \(\alpha, \beta\) and \(\gamma\) are the roots of the equation \(x^3 = 1\), then \(\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}\) is equal to</p>
<p>(a) \(f(\alpha) + f(\beta) + f(\gamma)\)</p>
<p>(b) \(f(\alpha)f(\beta) + f(\beta)f(\gamma) + f(\gamma)f(\alpha)\)</p>
<p>(c) \(f(\alpha)f(\beta)f(\gamma)\)</p>
<p>(d) \(-f(\alpha)f(\beta)f(\gamma)\)</p>
Step-by-Step Solution
Key Concept: The determinant with cyclic structure can be evaluated using properties of cube roots of unity. Recognize that α, β, γ satisfy α³ = β³ = γ³ = 1, and use the circulant matrix determinant formula involving f(α), f(β), f(γ).
<p><strong>Step 1:</strong> Identify the roots of x³ = 1. These are α = 1, β = ω, γ = ω², where ω = e^(2πi/3) is a primitive cube root of unity. These satisfy: 1 + ω + ω² = 0 and ω³ = 1.</p><p><strong>Step 2:</strong> Recognize the matrix structure. The matrix is circulant with first row [a, b, c]:</p><p>$$\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}$$</p><p><strong>Step 3:</strong> For a circulant matrix with first row [a, b, c], the determinant formula is:</p><p>$$\det = f(1) \cdot f(ω) \cdot f(ω²)$$</p><p>where f(x) = a + bx + cx² and ω is a primitive cube root of unity.</p><p><strong>Step 4:</strong> Apply to our notation. Since α = 1, β = ω, γ = ω² are the three cube roots of unity:</p><p>$$\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = f(1) \cdot f(ω) \cdot f(ω²) = f(α) \cdot f(β) \cdot f(γ)$$</p><p><strong>Step 5:</strong> Verify using circulant matrix property. The determinant of a circulant matrix with entries c₀, c₁, c₂ equals ∏(c₀ + c₁ωᵏ + c₂ω²ᵏ) for k = 0, 1, 2, which is exactly f(1)·f(ω)·f(ω²).</p><p><strong>∴ Answer:</strong> C</p>
Correct Answer: C