<p>If \(\alpha, \beta, \gamma\) are the roots of \(px^3 + qx^2 + r = 0\), then the value of the determinant \(\begin{vmatrix} \alpha\beta & \beta\gamma & \gamma\alpha \\ \beta\gamma & \gamma\alpha & \alpha\beta \\ \gamma\alpha & \alpha\beta & \beta\gamma \end{vmatrix}\) is</p>
Step-by-Step Solution
Key Concept: Recognize that the determinant has a cyclic pattern where each row is a cyclic permutation of products αβ, βγ, γα. Use Vieta's formulas (αβ + βγ + γα = r/p) and the fact that this circulant matrix has eigenvalues that can be computed using the sum of elements.
<p><strong>Step 1:</strong> Observe the matrix structure. Each row contains the cyclic products {αβ, βγ, γα} in different orders. The rows are cyclic permutations of each other.</p><p><strong>Step 2:</strong> Notice that Row 1 = (αβ, βγ, γα), Row 2 = (βγ, γα, αβ), Row 3 = (γα, αβ, βγ). This is a circulant matrix.</p><p><strong>Step 3:</strong> For a 3×3 circulant matrix with first row (a, b, c), the determinant equals (a + b + c)³ - 3(a + b + c)(ab + bc + ca) + 3abc when using the circulant property, but more directly: add all rows together.</p><p><strong>Step 4:</strong> Sum of each row = αβ + βγ + γα (from Vieta's formulas = r/p). Adding all rows gives a common factor (αβ + βγ + γα) in each element position.</p><p><strong>Step 5:</strong> Factor out: The matrix becomes (αβ + βγ + γα) times a matrix with all entries equal to 1 in certain positions, leading to linear dependence.</p><p><strong>Step 6:</strong> Since all rows become proportional when factored, or applying the circulant determinant formula directly: The determinant = 0 (as the three rows are related through cyclic symmetry, making them linearly dependent).</p><p>∴ Answer: <strong>C (which is 0)</strong></p>
Correct Answer: C