<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 infinite solutions 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 has infinite solutions when the coefficient matrix and augmented matrix have the same rank, and this rank is less than the number of variables. The third equation must be a linear combination of the first two equations.
<p><strong>Step 1:</strong> Write the augmented matrix:</p><p>$$\begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 1 & 2 & 3 & | & 10 \\ 1 & 2 & \lambda & | & \mu \end{bmatrix}$$</p><p><strong>Step 2:</strong> For infinite solutions, rank(A) = rank(A|B) < 3. Perform row operations:</p><p>R₂ → R₂ - R₁: [0, 1, 2 | 4]</p><p>R₃ → R₃ - R₁: [0, 1, λ-1 | μ-6]</p><p><strong>Step 3:</strong> For the system to be consistent with dependent equations, R₃ must be a multiple of R₂:</p><p>The row [0, 1, λ-1 | μ-6] must equal k times [0, 1, 2 | 4]</p><p><strong>Step 4:</strong> Comparing coefficients:</p><p>From the second element: 1 = k(1) ⟹ k = 1</p><p>From the third element: λ - 1 = 1(2) ⟹ <strong>λ = 3</strong></p><p>From the fourth element: μ - 6 = 1(4) ⟹ <strong>μ = 10</strong></p><p><strong>Step 5:</strong> Verification: When λ = 3 and μ = 10, the third equation becomes x + 2y + 3z = 10, which is identical to the second equation. Rank = 2 < 3 variables, confirming infinite solutions.</p><p>∴ Answer: <strong>B</strong> (λ = 3, μ = 10)</p>
Correct Answer: B