JEE Mains · Maths · STD 12 - 3 and 4 . metrices and determinant
The value of \(\lambda\) and \(\mu\) such that the system of equations \(x+y+z=6,3 x+5 y+5 z=26, x+2 y+\lambda z=\mu\) has no solution, are :
- A \(\lambda=3, \mu \neq 10\)
- B \(\lambda \neq 2, \mu=10\)
- C \(\lambda=3, \mu=5\)
- D \(\lambda=2, \mu \neq 10\)
Answer & Solution
Correct Answer
(D) \(\lambda=2, \mu \neq 10\)
Step-by-step Solution
Detailed explanation
\(D=\left|\begin{array}{lll}1 & 1 & 1 \\ 3 & 5 & 5 \\ 1 & 2 & \lambda\end{array}\right|=0\) \((5 \lambda-10)-1(3 \lambda-5)+(6-5)=0\) \(2 \lambda-10+5+1=0\) \(\lambda=2\) only one option satisfy Atleast one If \(\mathrm{D}_{1}, \mathrm{D}_{2}, \mathrm{D}_{3} \neq 0\)
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