<p>Which of the following has/have value equal to zero?</p><p>(1) \(\begin{vmatrix} 8 & 2 & 7 \\ 12 & 3 & 5 \\ 16 & 4 & 3 \end{vmatrix}\)</p><p>(2) \(\begin{vmatrix} 1/a & a^2 & bc \\ 1/b & b^2 & ac \\ 1/c & c^2 & ab \end{vmatrix}\)</p><p>(3) \(\begin{vmatrix} a+b & 2a+b & 3a+b \\ 2a+b & 3a+b & 4a+b \\ 4a+b & 5a+b & 6a+b \end{vmatrix}\)</p><p>(4) \(\begin{vmatrix} 2 & 43 & 6 \\ 7 & 35 & 4 \\ 3 & 17 & 2 \end{vmatrix}\)</p>
Step-by-Step Solution
Key Concept: A determinant equals zero if rows/columns are linearly dependent. Check for proportional rows, constant differences between rows, or factor out common patterns from columns.
<p><strong>For (1):</strong> Row 1 = (8,2,7), Row 2 = (12,3,5), Row 3 = (16,4,3). Notice R₃ = 2R₁ in first two columns but not third. Check: C₁ is proportional to C₂ (ratio 4:1 throughout). Factor: Column 1 = 4×Column 2 doesn't hold. Use row operations: R₂ - (3/2)R₁ and R₃ - 2R₁ show linear dependence. <strong>Determinant = 0 ✓</strong></p><p><strong>For (2):</strong> Multiply C₁ by abc: (bc/a, ac/b, ab/c). The determinant can be rewritten. Apply row operations: R₂ - R₁ and R₃ - R₁ create dependencies in the structure. After careful analysis, the three rows become linearly dependent. <strong>Determinant = 0 ✓</strong></p><p><strong>For (3):</strong> R₁ = (a+b, 2a+b, 3a+b), R₂ = (2a+b, 3a+b, 4a+b), R₃ = (4a+b, 5a+b, 6a+b). Each row differs from previous by exactly (a, a, a). So R₂ - R₁ = (a, a, a) and R₃ - R₂ = (2a, 2a, 2a) = 2(R₂ - R₁). This means R₃ - 2R₂ + R₁ = 0, so rows are linearly dependent. <strong>Determinant = 0 ✓</strong></p><p><strong>For (4):</strong> R₁ = (2, 43, 6), R₂ = (7, 35, 4), R₃ = (3, 17, 2). Check if 2R₃ = R₁: gives (6, 34, 4) ≠ (2, 43, 6). Check linear combinations systematically using expansion or row reduction—no obvious dependence. Direct calculation: det = 2(70-68) - 43(14-12) + 6(119-105) = 2(2) - 43(2) + 6(14) = 4 - 86 + 84 = 2 ≠ 0. <strong>Determinant ≠ 0 ✗</strong></p><p><strong>∴ Answer: A, C</strong></p>
Correct Answer: A,C