If the system of equations

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 :

  1. $\lambda-2 \mu=-5$

  2. $2 \lambda-\mu=5$

  3. $2 \lambda+\mu=14$

  4. $\lambda+2 \mu=14$


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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