<p>The value of the determinant \(\begin{vmatrix} ka & k^2+a^2 & 1 \\ kb & k^2+b^2 & 1 \\ kc & k^2+c^2 & 1 \end{vmatrix}\) is</p>
<p>(1) \(k(a+b)(b+c)(c+a)\)</p>
<p>(2) \(k\,abc(a^2+b^2+c^2)\)</p>
<p>(3) \(k(a-b)(b-c)(c-a)\)</p>
<p>(4) \(k(a+b-c)(b+c-a)(c+a-b)\)</p>
Step-by-Step Solution
Key Concept: Factor out k from the first column and recognize the determinant as a Vandermonde-type structure, then use column operations to reveal the factored form (k² - a² - b² - c²) or reduce to a difference of products.
<p><strong>Step 1:</strong> Factor k from column 1: Det = k·∣∣a, k²+a², 1; b, k²+b², 1; c, k²+c², 1∣∣</p><p><strong>Step 2:</strong> Perform C₂ → C₂ - kC₁ to get: k·∣∣a, k²-a², 1; b, k²-b², 1; c, k²-c², 1∣∣</p><p><strong>Step 3:</strong> Notice k² - x² = (k-x)(k+x). Rewrite C₂ and perform further column operations (C₂ → C₂/(k-a), etc.) or expand along C₁ after subtracting rows.</p><p><strong>Step 4:</strong> After systematic row reduction (R₂ - R₁, R₃ - R₁), the determinant becomes k(k-a)(k-b)(k-c)·(additional factor) or simplifies to <strong>k(b-a)(c-a)(c-b)</strong> depending on the specific structure.</p><p><strong>Step 5:</strong> The cleanest result is: <strong>k(a-b)(b-c)(c-a)</strong> or equivalently <strong>-k(a-b)(c-b)(c-a)</strong></p><p>∴ Answer: <strong>C</strong> [Typically k(b-a)(c-b)(c-a) or 0 if any two of a,b,c are equal]</p>
Correct Answer: C