<p>If \(\begin{vmatrix} x-4 & 2x & 2x \\ 2x & x-4 & 2x \\ 2x & 2x & x-4 \end{vmatrix} = (A + Bx)(x - A)^2\), then the ordered pair \((A, B)\) is equal to</p>
Step-by-Step Solution
Key Concept: Factor out common terms from rows/columns, then recognize the determinant as a product of (sum of eigenvalues) and squared deviations. The matrix has the form (x-4)I + 2x(J-I) where J is the all-ones matrix, making it a circulant-like structure with eigenvalues (x-4)+2x·3 and (x-4)-2x (with multiplicity 2).
<p><strong>Step 1:</strong> Rewrite the matrix as M = (x-4)I + 2xE, where E has all entries as 1 and I is identity. Notice the symmetric structure.</p><p><strong>Step 2:</strong> Use the property that for this symmetric circulant-type matrix, subtract row 1 from rows 2 and 3 to factor:</p><p>R₂ - R₁ and R₃ - R₁ give rows with pattern differences. This yields a factor from the first row operations.</p><p><strong>Step 3:</strong> Apply C₁ + C₂ + C₃ (add all columns to first column):</p><p>Column 1 becomes: (x-4+2x+2x, 2x+x-4+2x, 2x+2x+x-4) = (5x-4)(1,1,1)ᵀ</p><p>So determinant = (5x-4) × [minor determinant]</p><p><strong>Step 4:</strong> The minor (after removing common factor 5x-4) factors as (x-4-2x)² = (-x-4)² = (x+4)²</p><p>Therefore: det = (5x-4)(x+4)²</p><p><strong>Step 5:</strong> Rewrite as (5x-4)(x+4)² = (-4+5x)(x-(-4))² = (5x-4)(x-(-4))²</p><p>Comparing with (A+Bx)(x-A)²: A = -4, B = 5</p><p>∴ Answer: (A,B) = (-4, 5) → <strong>B</strong></p>
Correct Answer: B