Question

The values of \(\lambda\) and \(\mu\) such that the system of equations \(x + y + z = 6\), \(3x + 5y + 5z = 26\), \(x + 2y + \lambda z = \mu\) has no solution, are :

• A. \(\lambda = 3,\ \mu = 5\)
• B. \(\lambda = 3,\ \mu \ne 10\)
• C. \(\lambda \ne 2,\ \mu = 10\)
• D. \(\lambda = 2,\ \mu \ne 10\)

Hint:

Eliminate x from the equations; the system is inconsistent when the reduced equations contradict each other.

Answer 1

The Correct Option is D

Subtract 3 times equation (1) from equation (2): \(2y + 2z = 8 \Rightarrow y + z = 4\).
Subtract equation (1) from equation (3): \(y + (\lambda - 1)z = \mu - 6\).
Subtract \(y + z = 4\) from this: \((\lambda - 2)z = \mu - 10\).
No solution requires \(\lambda - 2 = 0\) and \(\mu - 10 \ne 0\), i.e. \(\lambda = 2\), \(\mu \ne 10\).
Answer: (4) \(\lambda = 2,\ \mu \ne 10\)