If the system of equations $x+4 y-z=\lambda, 7 x+9 y+\mu z=-3,5 x+y+2 z=-1$ has infinitely many solutions, then $(2 \mu+3 \lambda)$ is equal to :

Question

Aicharya · Accepted Answer

$\begin{aligned} & \Delta=\left|\begin{array}{ccc}1 & 4 & -1 \ 7 & 9 & \mu \ 5 & 1 & 2\end{array}ight|=0 \ & \Rightarrow(18-\mu)-4(14-5 \mu)-(7-45)=0 \Rightarrow \mu=0 \ & \Delta=\Delta_{\mathrm{x}}=\Delta_{\mathrm{y}}=\Delta_{\mathrm{z}}=0 	ext { (For infinite solution) } \ & \Delta_{\mathrm{x}}=\left|\begin{array}{ccc}\lambda & 4 & -1 \ -3 & 9 & \mu \ -1 & 1 & 2\end{array}ight|=0 \ & \lambda(18-\mu)-4(-6+\mu)-1(-3+9)=0 \ & 18 \lambda+24-6=0 \Rightarrow \lambda=-1\end{aligned}$

If the system of equations $x+4 y-z=\lambda, 7 x+9 y+\mu z=-3,5 x+y+2 z=-1$ has infinitely many solutions,…

Solution