Question

The system of equations: \[ x + 2y + 2z = 5, \quad x + 2y + 3z = 6, \quad x + 2y + \lambda z = \mu \] has infinitely many solutions if:

• A. \( \lambda \ne 2 \)
• B. \( \lambda \ne 2, \mu \ne 5 \)
• C. \( \lambda = 2, \mu = 5 \)
• D. \( \mu \ne 5 \)

Hint:

For infinitely many solutions, the system must reduce to dependent equations—use substitution and compare reduced equations.

Answer 1

The Correct Option is C

Step 1: Subtract equation 1 from equation 2: \[ (x + 2y + 3z) - (x + 2y + 2z) = 6 - 5 \Rightarrow z = 1 \] Step 2: Substitute \( z = 1 \) into both equations. Substitute into equation 1: \[ x + 2y + 2 = 5 \Rightarrow x + 2y = 3 \quad \cdots (A) \] Substitute into equation 3: \[ x + 2y + \lambda(1) = \mu \Rightarrow x + 2y = \mu - \lambda \quad \cdots (B) \] For infinitely many solutions, equations (A) and (B) must be identical: \[ 3 = \mu - \lambda \Rightarrow \mu = \lambda + 3 \] Step 3: Use \( z = 1 \) from earlier. Substituting back into third equation gives: \[ \mu = \lambda + 3 \Rightarrow {Pick } \lambda = 2 \Rightarrow \mu = 5 \]