<p>By using properties of determinants, show that \(\begin{vmatrix} x+4 & 2x & 2x \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{vmatrix} = (5x+4)(4-x)^2\)</p>
Step-by-Step Solution
Key Concept: Add all rows to the first row to create a common factor, then use row operations to extract (5x+4), leaving a 2×2 determinant that factors as (4-x)².
<p><strong>Step 1:</strong> Observe the cyclic symmetry. Apply R₁ → R₁ + R₂ + R₃</p><p>First row becomes: (x+4+2x+2x, 2x+x+4+2x, 2x+2x+x+4) = (5x+4, 5x+4, 5x+4)</p><p>The determinant is now: ∣(5x+4, 5x+4, 5x+4), (2x, x+4, 2x), (2x, 2x, x+4)∣</p><p><strong>Step 2:</strong> Factor out (5x+4) from Row 1</p><p>Determinant = (5x+4) × ∣(1, 1, 1), (2x, x+4, 2x), (2x, 2x, x+4)∣</p><p><strong>Step 3:</strong> Apply column operations: C₂ → C₂ - C₁ and C₃ → C₃ - C₁</p><p>∣(1, 0, 0), (2x, x+4-2x, 2x-2x), (2x, 2x-2x, x+4-2x)∣ = ∣(1, 0, 0), (2x, x+4-2x, 0), (2x, 0, x+4-2x)∣</p><p>= ∣(1, 0, 0), (2x, 4-x, 0), (2x, 0, 4-x)∣</p><p><strong>Step 4:</strong> Expand along Row 1 (C₁)</p><p>Determinant = 1 × ∣(4-x, 0), (0, 4-x)∣ = (4-x)(4-x) = (4-x)²</p><p><strong>Step 5:</strong> Combine results</p><p>Original determinant = (5x+4) × (4-x)²</p><p>∴ Answer: <strong>(5x+4)(4-x)²</strong></p>
Correct Answer: (5x+4)(4-x)^2