Question:
The values of $\lambda$ and $\mu$ for which the system of linear equations
$x+y+z=2$
$x+2 y+3 z=5$
$x+3 y+\lambda z=\mu$
has infinitely many solutions are, respectively
Correct Option:
Solution:
For infinite many solutions
$\mathrm{D}=\mathrm{D}_{1}=\mathrm{D}_{2}=\mathrm{D}_{3}=0$
Now $\mathrm{D}=\left|\begin{array}{lll}1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & \lambda\end{array}\right|=0$
1. $(2 \lambda-9)-1 .(\lambda-3)+1 .(3-2)=0$
$\therefore \lambda=5$
Now $D_{1}=\left|\begin{array}{lll}2 & 1 & 1 \\ 5 & 2 & 3 \\ \mu & 3 & 5\end{array}\right|=0$
$2(10-9)-1(25-3 \mu)+1(15-2 \mu)=0$
$\mu=8$
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