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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
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Verified Answer
The correct answer is:
$\lambda \neq 3, \mu \in \mathbb{R}$
$x+y+z=5$
$\begin{aligned} & x+2 y+2 z=6 \\ & x+3 y+\lambda z=\mu\end{aligned}$
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
$\therefore \lambda \neq 3, \mu \in \mathrm{R}$
$\begin{aligned} & x+2 y+2 z=6 \\ & x+3 y+\lambda z=\mu\end{aligned}$
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
$\therefore \lambda \neq 3, \mu \in \mathrm{R}$
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