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

Mathematics · Matrices

JEE Main 2025 — 23 January, Morning Shift — Question 19

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

(\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 = \begin{vmatrix} \lambda - 1 & \lambda - 4 & \lambda \\ \lambda & \lambda - 1 & \lambda - 4 \\ \lambda + 1 & \lambda + 2 & -(\lambda + 2) \end{vmatrix} = 0

(\lambda - 3)(2\lambda + 1) = 0

D_x = \begin{vmatrix} 5 & \lambda - 4 & \lambda \\ 7 & \lambda - 1 & \lambda - 4 \\ 9 & \lambda + 2 & -(\lambda + 2) \end{vmatrix} = 0

2(3 - \lambda)(23 - 2\lambda) = 0

\lambda = 3

\therefore \lambda^2 + \lambda = 9 + 3 = 12

\boxed{\lambda^2 + \lambda = 12}

Answer key and solution verified before publishing.

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