<p>If \(f(x) = a + bx + cx^2\) and \(\alpha, \beta, \gamma\) are the roots of the equation \(x^2 = 1\), then \(\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}\) is equal to</p>
<p>\(f(\alpha) + f(\beta) + f(\gamma)\)</p>
<p>\(f(\alpha)f(\beta) + f(\beta)f(\gamma) + f(\gamma)f(\alpha)\)</p>
<p>\(f(\alpha)f(\beta)f(\gamma)\)</p>
<p>\(-f(\alpha)f(\beta)f(\gamma)\)</p>
Step-by-Step Solution
Key Concept: The determinant of a circulant matrix equals f(α)f(β)f(γ) where α, β, γ are cube roots of unity (not roots of x²=1). The roots of x²=1 are ±1, but the circulant structure suggests we should use ω, ω², 1 where ω is a primitive cube root of unity.
<p><strong>Step 1:</strong> Recognize the matrix structure. The matrix is circulant with entries [a, b, c] in the first row:</p><p>$$\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}$$</p><p><strong>Step 2:</strong> For a circulant matrix with first row [a, b, c], the determinant equals:</p><p>$$\det = f(1) \cdot f(\omega) \cdot f(\omega^2)$$</p><p>where f(x) = a + bx + cx², and ω, ω² are primitive cube roots of unity (1, ω, ω² where ω³=1, ω≠1)</p><p><strong>Step 3:</strong> Calculate each factor:</p><p>- f(1) = a + b + c</p><p>- f(ω) = a + bω + cω²</p><p>- f(ω²) = a + bω² + cω⁴ = a + bω² + cω</p><p><strong>Step 4:</strong> The determinant is:</p><p>$$(a + b + c)(a + bω + cω²)(a + bω² + cω) = a³ + b³ + c³ - 3abc$$</p><p>∴ Answer: D</p>
Correct Answer: D