Question

If the system of linear equations : \[ x+y+z=4,\quad x+2y+3z=6,\quad 4x+5y+\lambda z=\mu \] has more than one solution, then the value of \( \lambda+\mu \) is equal to:

• A. \(18\)
• B. \(9\)
• C. \(12\)
• D. \(24\)

Hint:

For infinitely many solutions, one equation must be a linear combination of the others, including the constant terms.

Answer 1

The Correct Option is D

Concept: A system of linear equations has more than one solution if:
The equations are consistent, and
The rank of the coefficient matrix is less than the number of variables. This happens when one equation is a linear combination of the others, including the constant terms.
Step 1: Express the third equation as a combination of the first two Assume: \[ a(x+y+z) + b(x+2y+3z) = 4x+5y+\lambda z \] Comparing coefficients: \[ a+b=4 \quad (1) \] \[ a+2b=5 \quad (2) \] \[ a+3b=\lambda \quad (3) \]
Step 2: Solve for \(a\) and \(b\) Subtract (1) from (2): \[ b=1 \] Substitute in (1): \[ a+1=4 \Rightarrow a=3 \]
Step 3: Find \( \lambda \) From (3): \[ \lambda=a+3b=3+3=6 \]
Step 4: Find \( \mu \) The constants must satisfy the same linear combination: \[ \mu = 4a + 6b = 4(3)+6(1)=12+6=18 \]
Step 5: Required value \[ \lambda+\mu = 6+18 = 24 \]