If the system of equation
$\begin{aligned}
& 2 x+\lambda y+3 z=5 \\ & 3 x+2 y-z=7 \\ & 4 x+5 y+\mu z=9
\end{aligned}$
has infinitely many solutions, then $\left(\lambda^2+\mu^2\right)$ is equal to :

Question

If the system of equation $\begin{aligned} & 2 x+\lambda y+3 z=5 \ & 3 x+2 y-z=7 \ & 4 x+5 y+\mu z=9 \end{aligned}$ has infinitely many solutions, then $\left(\lambda^2+\mu^2ight)$ is equal to :

Aicharya · Accepted Answer

$\Delta=0 \Rightarrow\left|\begin{array}{ccc}2 & \lambda & 3 \\ 3 & 2 & -1 \\ 4 & 5 & \mu\end{array}\right|=0$
$\Rightarrow 2(2 \mu+5)+\lambda(-4-3 \mu)+3(7)=0$
$\Rightarrow 4 \mu-3 \lambda \mu-4 \lambda+31=0$ ...(1)
$\begin{aligned}
& \Delta_3=0 \Rightarrow\left|\begin{array}{lll}
2 & \lambda & 5 \\ 3 & 2 & 7 \\ 4 & 5 & 9
\end{array}\right|=0 \\ & \Rightarrow 2(-17)+\lambda(1)+5(7)=0 \\ & \Rightarrow \lambda=-1
\end{aligned}$
from equation (1)
$4 \mu+3 \mu+4+31=0 \Rightarrow \mu=-5$
$\therefore \lambda^2+\mu^2=26$

If the system of equation $\begin{aligned} & 2 x+\lambda y+3 z=5 \\ & 3 x+2 y-z=7 \\ & 4 x+5 y+\mu z=9…

Solution