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

Question

Aicharya · Accepted Answer

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

If the system of equations $\begin{aligned} & (\lambda-1) x+(\lambda-4) y+\lambda z=5 \\ & \lambda…

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