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3 and 4 .Determinants and Matrices
medium
If the system of equations $(\lambda-1) x+(\lambda-4) y+\lambda z=5$, $\lambda x+(\lambda-1) y+(\lambda-4) z=7$, $(\lambda+1) x+(\lambda+2) y-(\lambda+2) z=9$ has infinitely many solutions, then $\lambda^2+\lambda$ is equal to
A$10$
B$12$
C$6$
D$20$
(JEE MAIN-2025)
Solution
$(\lambda-1) x+(\lambda-4) y+\lambda z=5$
$\lambda x+(\lambda-1) y+(\lambda-4) z=7$
$(\lambda+1) x+(\lambda+2) y-(\lambda+2) z=9$
For infinitely many solutions
$D =\left|\begin{array}{ccc}\lambda-1 & \lambda-4 & \lambda \\ \lambda & \lambda-1 & \lambda-4 \\ \lambda+1 & \lambda+2 & -(\lambda+2)\end{array}\right|=0$
$(\lambda-3)(2 \lambda+1)=0$
$D _{ x }=\left|\begin{array}{ccc}5 & \lambda-4 & \lambda \\ 7 & \lambda-1 & \lambda-4 \\ 9 & \lambda+2 & -(\lambda+2)\end{array}\right|=0$
$2(3-\lambda)(23-2 \lambda)=0$
$\lambda=3$
$\therefore \lambda^2+\lambda=9+3=12$
$\lambda x+(\lambda-1) y+(\lambda-4) z=7$
$(\lambda+1) x+(\lambda+2) y-(\lambda+2) z=9$
For infinitely many solutions
$D =\left|\begin{array}{ccc}\lambda-1 & \lambda-4 & \lambda \\ \lambda & \lambda-1 & \lambda-4 \\ \lambda+1 & \lambda+2 & -(\lambda+2)\end{array}\right|=0$
$(\lambda-3)(2 \lambda+1)=0$
$D _{ x }=\left|\begin{array}{ccc}5 & \lambda-4 & \lambda \\ 7 & \lambda-1 & \lambda-4 \\ 9 & \lambda+2 & -(\lambda+2)\end{array}\right|=0$
$2(3-\lambda)(23-2 \lambda)=0$
$\lambda=3$
$\therefore \lambda^2+\lambda=9+3=12$
Standard 12
Mathematics