Q47 / 160 · TS EAMCET 2023 (14 May Shift 2) · Mathematics

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

  • Correct answer
    A
    $\lambda \neq 3, \mu \in \mathbb{R}$
  • B
    $\lambda=3, \mu=0$
  • C
    $\lambda \neq 3, \mu \neq 5$
  • D
    $\lambda=3, \mu \in \mathbb{R}$

Solution

$x+y+z=5$ $\begin{aligned} & x+2 y+2 z=6 \\ & x+3 y+\lambda z=\mu\end{aligned}$ 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 exist $\therefore \lambda \neq 3, \mu \in \mathrm{R}$

Hence, the correct answer is option: 'A'.

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