<p><strong>For Problems 22–24</strong><br>Consider the system of equations<br>\(x + y + z = 6\)<br>\(x + 2y + 3z = 10\)<br>\(x + 2y + \lambda z = \mu\)<br>The system has no solution if</p>
<p>\(\lambda \neq 3\)</p>
<p>\(\lambda = 3,\ \mu = 10\)</p>
<p>\(\lambda = 3,\ \mu \neq 10\)</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: A system of linear equations has no solution when the coefficient matrix and augmented matrix have different ranks. For the third equation to create inconsistency, we need the coefficient rows to be proportional but the augmented columns to be non-proportional.
<p><strong>Step 1:</strong> Write the augmented matrix and perform row operations.</p><p>R₂ - R₁: (0, 1, 2, 4)</p><p>R₃ - R₁: (0, 1, λ-1, μ-6)</p><p><strong>Step 2:</strong> For no solution, rows 2 and 3 must be proportional in coefficients but inconsistent in constants.</p><p>Rows 2 and 3 have coefficients (1, 2) and (1, λ-1) respectively. For these to be proportional: λ - 1 = 2, so <strong>λ = 3</strong></p><p><strong>Step 3:</strong> With λ = 3, row 3 becomes (0, 1, 2, μ-6). For inconsistency with row 2: (0, 1, 2, 4), we need μ - 6 ≠ 4, so <strong>μ ≠ 10</strong></p><p><strong>Step 4:</strong> The system has no solution when <strong>λ = 3 and μ ≠ 10</strong></p><p>∴ Answer: C</p>
Correct Answer: C