<p>\(\Delta = \begin{vmatrix} 1 & 1+ac & 1+bc \\ 1 & 1+ad & 1+bd \\ 1 & 1+ae & 1+be \end{vmatrix}\) is independent of</p>
Step-by-Step Solution
Key Concept: Factor out common terms from columns by recognizing that column operations preserve determinant value. Rewrite each column as a sum and use linearity of determinants in columns to isolate terms independent of variables.
<p><strong>Step 1:</strong> Apply column operations. Subtract Column 1 from Columns 2 and 3:</p><p>C₂ → C₂ - C₁ and C₃ → C₃ - C₁</p><p>Δ = |1 ac bc|</p><p> |1 ad bd|</p><p> |1 ae be|</p><p><strong>Step 2:</strong> Factor out c from Column 2 and b from Column 3:</p><p>Δ = c·b |1 a 1|</p><p> |1 d 1|</p><p> |1 e 1|</p><p><strong>Step 3:</strong> Subtract Column 1 from Column 3 again:</p><p>Δ = bc |1 a 0|</p><p> |1 d 0|</p><p> |1 e 0|</p><p><strong>Step 4:</strong> Expand along Column 3 (which gives 0). Instead expand along Column 1 or recognize the pattern. Actually, doing R₂-R₁ and R₃-R₁:</p><p>Δ = bc |1 a 0|</p><p> |0 d-a 0|</p><p> |0 e-a 0|</p><p>The determinant equals 0 since Column 3 is all zeros, OR factor Column 2: Δ = abc|(d-a)(e-a)| structure shows Δ is independent of <strong>b and c</strong> (when properly simplified, the determinant vanishes or depends only on a, d, e).</p><p>∴ Answer: <strong>b, c</strong> (or equivalent based on which variables are shown to not appear in the final reduced form)</p>
Correct Answer: A,B