<p>Given <br/> \[\begin{vmatrix} a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix}\] Find the value of \(x\) such that the determinant equals \((x+a+b+c)(a+b+c)^2\).</p>
Step-by-Step Solution
Key Concept: Add all rows to the first row to factor out (a+b+c), then perform column operations to reveal the squared term. The determinant has the special form where each diagonal element equals -(sum of other two variables), allowing systematic factorization.
<p><strong>Step 1: Add all rows to R₁</strong></p><p>R₁ → R₁ + R₂ + R₃</p><p>First row becomes: [(a-b-c+2b+2c), (2a+b-c-a+2c), (2a+2b+c-a-b)]</p><p>= [(a+b+c), (a+b+c), (a+b+c)]</p><p><strong>Step 2: Factor out (a+b+c) from R₁</strong></p><p>Det = (a+b+c) · ∣1, 1, 1 ∣</p><p> ∣2b, b-c-a, 2b ∣</p><p> ∣2c, 2c, c-a-b ∣</p><p><strong>Step 3: Perform column operations (C₂ → C₂ - C₁, C₃ → C₃ - C₁)</strong></p><p>Det = (a+b+c) · ∣1, 0, 0 ∣</p><p> ∣2b, -c-a-b, 2b-2b ∣</p><p> ∣2c, 2c-2c, c-a-b-2c ∣</p><p>= (a+b+c) · ∣1, 0, 0 ∣</p><p> ∣2b, -(a+b+c), 0 ∣</p><p> ∣2c, 0, -(a+b+c) ∣</p><p><strong>Step 4: Expand along R₁</strong></p><p>Det = (a+b+c) · 1 · [-(a+b+c) · (-(a+b+c))]</p><p>= (a+b+c) · (a+b+c)²</p><p>= (a+b+c)³</p><p><strong>Step 5: Match with (x+a+b+c)(a+b+c)²</strong></p><p>We have (a+b+c)³ = (x+a+b+c)(a+b+c)²</p><p>Dividing both sides by (a+b+c)²:</p><p>a+b+c = x+a+b+c</p><p>∴ <strong>x = 0</strong></p>
Correct Answer: B