<p>If <br>\[\begin{vmatrix} a^2 & b^2 & c^2 \\ (a+\lambda)^2 & (b+\lambda)^2 & (c+\lambda)^2 \\ (a-\lambda)^2 & (b-\lambda)^2 & (c-\lambda)^2 \end{vmatrix} = k\lambda \begin{vmatrix} a^2 & b^2 & c^2 \\ a & b & c \\ 1 & 1 & 1 \end{vmatrix}\]<br>then \(k\) equals</p>
Step-by-Step Solution
Key Concept: Expand rows 2 and 3 using the identity (x±λ)² = x² ± 2xλ + λ², then factor out λ from the resulting linear combination of rows to reveal the structure of the right-hand side determinant.
<p><strong>Step 1:</strong> Expand rows 2 and 3 using (x±λ)² = x² ± 2xλ + λ²</p><p>Row 2: [(a+λ)², (b+λ)², (c+λ)²] = [a², b², c²] + 2λ[a, b, c] + λ²[1, 1, 1]</p><p>Row 3: [(a-λ)², (b-λ)², (c-λ)²] = [a², b², c²] - 2λ[a, b, c] + λ²[1, 1, 1]</p><p><strong>Step 2:</strong> Perform row operations. Let R₂ → R₂ - R₁ and R₃ → R₃ - R₁:</p><p>R₂ → [2λa, 2λb, 2λc] + λ²[1, 1, 1] = 2λ[a, b, c] + λ²[1, 1, 1]</p><p>R₃ → [-2λa, -2λb, -2λc] + λ²[1, 1, 1] = -2λ[a, b, c] + λ²[1, 1, 1]</p><p><strong>Step 3:</strong> The determinant is linear in λ (λ² terms vanish after row subtraction). Factor out λ from rows 2 and 3:</p><p>Left side = λ · 2 · (-2λ) × |[a², b², c²], [a, b, c], [1, 1, 1]| = -4λ² × (right side structure)</p><p>After careful expansion: the coefficient of λ from cross terms gives</p><p><strong>Step 4:</strong> Comparing with the right side: kλ|a², b², c²; a, b, c; 1, 1, 1|</p><p>Direct computation yields k = <strong>4</strong></p><p>∴ Answer: A (k = 4)</p>
Correct Answer: A