Question:
If the system of linear equations
$x+y+z=5$
$x+2 y+2 z=6$
$x+3 y+\lambda z=\mu,(\lambda, \mu \in R)$, has infinitely many
solutions, then the value of $\lambda+\mu$ is :
Correct Option: , 2
Solution:
$x+3 y+\lambda z-\mu=p(x+y+z-5)+$
$q(x+2 y+2 z-6)$
on comparing the coefficient;
$p+q=1$ and $p+2 q=3$
$\Rightarrow(\mathrm{p}, \mathrm{q})=(-1,2)$
Hence $x+3 y+\lambda z-\mu=x+3 y+3 z-7$
$\Rightarrow \lambda=3, \mu=7$