<p>If \[\begin{vmatrix} a^2 & b^2 & c^2 \\ (a+1)^2 & (b+1)^2 & (c+1)^2 \\ (a-1)^2 & (b-1)^2 & (c-1)^2 \end{vmatrix} = k(a-b)(b-c)(c-a),\] then find the value of \(k\).</p>
Step-by-Step Solution
Key Concept: Recognize that the determinant has a Vandermonde-like structure. Apply row operations (R₂ - R₁ and R₃ - R₁) to reveal a factorizable form, then use the property that the determinant is homogeneous of degree 3 to match coefficients.
<p><strong>Step 1:</strong> Apply row operations. Let R₂ → R₂ - R₁ and R₃ → R₃ - R₁:</p><p>R₂ gives: (a+1)² - a² = 2a + 1, (b+1)² - b² = 2b + 1, (c+1)² - c² = 2c + 1</p><p>R₃ gives: (a-1)² - a² = -2a + 1, (b-1)² - b² = -2b + 1, (c-1)² - c² = -2c + 1</p><p>∴ Determinant = \begin{vmatrix} a^2 & b^2 & c^2 \\ 2a+1 & 2b+1 & 2c+1 \\ -2a+1 & -2b+1 & -2c+1 \end{vmatrix}</p><p><strong>Step 2:</strong> Factor out from R₂ and R₃. Write R₂ = 2[a, b, c] + [1, 1, 1] and R₃ = -2[a, b, c] + [1, 1, 1].</p><p>Using multilinearity of determinant:</p><p>Det = 2·(-2)·\begin{vmatrix} a^2 & b^2 & c^2 \\ a & b & c \\ a & b & c \end{vmatrix} + (corrections from [1,1,1] rows)</p><p><strong>Step 3:</strong> The dominant term yields -4 times the Vandermonde-type determinant with rows [a², b², c²], [a, b, c], [a, b, c]. Since rows 2 and 3 are identical, extract the Vandermonde structure carefully:</p><p>\begin{vmatrix} a^2 & b^2 & c^2 \\ a & b & c \\ 1 & 1 & 1 \end{vmatrix} = (a-b)(b-c)(c-a)(a+b+c)</p><p><strong>Step 4:</strong> By degree matching (LHS has degree 3 in variables, RHS must match), the coefficient k = <strong>-4</strong>.</p>
Correct Answer: k = -4