<p>The determinant \(\Delta = \begin{vmatrix} a^2 + x & ab & ac \\ ab & b^2 + x & bc \\ ac & bc & c^2 + x \end{vmatrix}\) is divisible by</p>
Step-by-Step Solution
Key Concept: Factor out x from the determinant by recognizing the structure: the matrix equals x·I + (outer product of vector [a,b,c]). Use the matrix determinant lemma: det(A + uv^T) = det(A)(1 + v^T·A^(-1)·u) when A is invertible, or recognize this as a rank-1 perturbation of xI.
<p><strong>Step 1:</strong> Rewrite the matrix as M = xI + vv<sup>T</sup> where v = [a, b, c]<sup>T</sup></p><p>The given matrix = x·I + matrix with all (i,j) entries = a·b·c contributions</p><p><strong>Step 2:</strong> For rank-1 perturbation, use: det(xI + vv<sup>T</sup>) = x<sup>n-1</sup>[x + v·v]</p><p>Here, v·v = a² + b² + c², and we have a 3×3 matrix</p><p><strong>Step 3:</strong> Therefore Δ = x²(x + a² + b² + c²)</p><p><strong>Step 4:</strong> The determinant is divisible by:</p><p>• <strong>x</strong> (appears with power 2)</p><p>• <strong>x + a² + b² + c²</strong> (the second factor)</p><p>∴ Answer: A, B</p>
Correct Answer: A,B