If the system of equations [ aligned & (-1) x+(-4) y+ z=5 & x+(-1) y+(-4) z=7 & (+1) x+(+2…

Question

If the system of equations

$$
\begin{aligned}
& (\lambda-1) x+(\lambda-4) y+\lambda z=5 \
& \lambda x+(\lambda-1) y+(\lambda-4) z=7 \
& (\lambda+1) x+(\lambda+2) y-(\lambda+2) z=9
\end{aligned}$$

has infinitely many solutions, then $\lambda^2+\lambda$ is equal to

Accepted Answer

$$\begin{aligned}
&\begin{aligned}
& (\lambda-1) x+(\lambda-4) y+\lambda z=5 \
& \lambda x+(\lambda-1) y+(\lambda-4) z=7 \
& (\lambda+1) x+(\lambda+2) y-(\lambda+2) z=9
\end{aligned}\
\&\text { For infinitely many solutions }\
\&\begin{aligned}
& \mathrm{D}=\left|\[\begin{array}{ccc} \lambda-1 & \lambda-4 & \lambda \ \lambda & \lambda-1 & \lambda-4 \ \lambda+1 & \lambda+2 & -(\lambda+2) \end{array}\end{aligned}\end{aligned}$$\right|=0 \
\& (\lambda-3)(2 \lambda+1)=0 \
\& \mathrm{D}\_{\mathrm{x}}=\left|$$\begin{array}{ccc} 5 & \lambda-4 & \lambda \ 7 & \lambda-1 & \lambda-4 \ 9 & \lambda+2 & -(\lambda+2) \end{array}$$\right|=0 \
\& 2(3-\lambda)(23-2 \lambda)=0 \
\& \lambda=3 \
\& \therefore \lambda\^2+\lambda=9+3=12

\]

Solutions · 1