Question

If the system of linear equations
$x+y+z=5$
$x+2y+2z=6$
$x+3y+\lambda z=\mu ,(\lambda ,\mu \in R)$, has infinitely many solutions, then the value of $\lambda +\mu$ is:

• A. $12$
• B. $10$
• C. $9$
• D. $7$

Answer 1

The correct option is B $10$

$x+3y+\lambda z-\mu =p(x+y+z-5)+q(x+2y+2z-6)$
on comparing the coefficient;
$p+q=1$ and $p+2q=3$
$\Rightarrow (p,q)=(-1,2)$
Hence $x+3y+\lambda x-\mu =x+3y+3z-7$
$\Rightarrow \lambda =3,\mu =7$