Q37 / 75 · JEE Main 2025 (23 Jan Shift 1) · Mathematics
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
- A$6$
- B$10$
- C$20$
- Correct answerD$12$
Solution
$\begin{aligned}
& (\lambda-1) x+(\lambda-4) y+\lambda z=5 \\
& \lambda x+(\lambda-1) y+(\lambda-4) z=7 \\
& (\lambda+1) x+(\lambda+2) y-(\lambda+2) z=9
\end{aligned}$
For infinitely many solutions
$\begin{aligned}
& \mathrm{D}=\left|\begin{array}{ccc}
\lambda-1 & \lambda-4 & \lambda \\
\lambda & \lambda-1 & \lambda-4 \\
\lambda+1 & \lambda+2 & -(\lambda+2)
\end{array}\right|=0 \\
& (\lambda-3)(2 \lambda+1)=0 \\
& \mathrm{D}_{\mathrm{x}}=\left|\begin{array}{ccc}
5 & \lambda-4 & \lambda \\
7 & \lambda-1 & \lambda-4 \\
9 & \lambda+2 & -(\lambda+2)
\end{array}\right|=0 \\
& 2(3-\lambda)(23-2 \lambda)=0 \\
& \lambda=3 \\
& \therefore \lambda^2+\lambda=9+3=12
\end{aligned}$
Hence, the correct answer is option: 'D'.