<p>The system of linear equations<br>\(x + y + z = 5\)<br>\(x + 2y + 2z = 6\)<br>\(x + 3y + \lambda z = \mu\)<br>has infinitely many solutions. Find \(\lambda + \mu\).</p>
Step-by-Step Solution
Key Concept: For infinitely many solutions, the system must be consistent and dependent—the third equation must be a linear combination of the first two. This occurs when the coefficient matrix and augmented matrix have the same rank < 3.
<p><strong>Step 1:</strong> For infinitely many solutions, rank of coefficient matrix = rank of augmented matrix < 3 (system is consistent and dependent).</p><p><strong>Step 2:</strong> From equations (1) and (2):<br/>Eq(1): x + y + z = 5<br/>Eq(2): x + 2y + 2z = 6<br/>Subtracting: y + z = 1, so y = 1 - z</p><p><strong>Step 3:</strong> The third equation must be satisfiable by the same solution set. Substituting y = 1 - z into the general solution from (1): x = 5 - y - z = 5 - (1 - z) - z = 4<br/>So solutions are of form: (4, 1-z, z) for any z ∈ ℝ</p><p><strong>Step 4:</strong> For the third equation x + 3y + λz = μ to be satisfied by all these solutions:<br/>4 + 3(1 - z) + λz = μ<br/>4 + 3 - 3z + λz = μ<br/>7 + (λ - 3)z = μ</p><p><strong>Step 5:</strong> For this to hold for all values of z, we need:<br/>Coefficient of z: λ - 3 = 0 ⟹ λ = 3<br/>Constant term: μ = 7</p><p><strong>Step 6:</strong> Verify: Third equation becomes x + 3y + 3z = 7, which when combined with first two gives dependent system with infinitely many solutions.</p><p>∴ λ + μ = 3 + 7 = <strong>10</strong></p>
Correct Answer: 10