Question:
If the system of equations
$x+y+z=2$
$2 x+4 y-z=6$
$3 x+2 y+\lambda z=\mu$
has infinitely many solutions, then :
Correct Option: , 3
Solution:
For infinite solutions
$\Delta=\Delta_{x}=\Delta_{y}=\Delta_{z}=0$
Now $\Delta=0 \Rightarrow\left|\begin{array}{ccc}1 & 1 & 1 \\ 2 & 4 & -1 \\ 3 & 2 & \lambda\end{array}\right|=0$
$\Rightarrow \lambda=\frac{9}{2}$
$\Delta_{x=0} \Rightarrow\left|\begin{array}{ccc}2 & 1 & 1 \\ 6 & 4 & -1 \\ \mu & 2 & -\frac{9}{2}\end{array}\right|=0$
$\Rightarrow \mu=5$
For $\lambda=\frac{9}{2} \& \mu=5, \Delta_{\mathrm{y}}=\Delta_{\mathrm{z}}=0$
Now check option $2 \lambda+\mu=14$
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