TS EAMCET · Maths · Determinants
If the system of equations \(x+y+z=5, x+2 y+2 z=6\) and \(x+3 y+\lambda z=\mu(\lambda, \mu \in \mathbb{R})\) is solvable by Matrix Inversion Method, then
- A \(\lambda \neq 3, \mu \in \mathbb{R}\)
- B \(\lambda=3, \mu=0\)
- C \(\lambda \neq 3, \mu \neq 5\)
- D \(\lambda=3, \mu \in \mathbb{R}\)
Answer & Solution
Correct Answer
(A) \(\lambda \neq 3, \mu \in \mathbb{R}\)
Step-by-step Solution
Detailed explanation
\(x+y+z=5\) \(\begin{aligned} & x+2 y+2 z=6 \\ & x+3 y+\lambda z=\mu\end{aligned}\) is solvable \(\Rightarrow|A| \neq 0\) \(\left|\begin{array}{lll}1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 3 & \lambda\end{array}\right| \neq 0\) clearly \(\lambda \neq 3\) and at \(\mu \in R\), solution…
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