If the system of equations $(\lambda-1) x+(\lambda-4) y+\lambda z=5$ $\lambda \mathrm{x}+(\lambda-1) \mathrm{y}+(\lambda-4) \mathrm{z}=7$ $(\lambda+1) \mathrm{x}+(\lambda+2) \mathrm{y}-(\lambda+2) \mathrm{z}=9$ has infinitely many solutions, then $\lambda^{2}+\lambda$ is equal to

Matrices — JEE Main · 2025 · 23 Jan 2025 · Shift 1 Question with Solution

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

Answer: (B)

For infinitely many solutions, the coefficient determinant and every numerator determinant of Cramer's rule must vanish.

D = λ - 1 λ λ + 1 λ - 4 λ - 1 λ + 2 λ λ - 4 - (λ + 2)

Expanding along the first row and simplifying gives

D = - 4 λ^{2} + 10 λ + 6 = - 2 (2 λ^{2} - 5 λ - 3) = - 2 (2 λ + 1) (λ - 3)

D = 0

λ = 3

λ = - \frac{1}{2}

.

Replacing the first column by the constants

(5, 7, 9)

gives

D_{x} = 579 λ - 4 λ - 1 λ + 2 λ λ - 4 - (λ + 2) = 4 λ^{2} - 58 λ + 138 = 2 (2 λ^{2} - 29 λ + 69)

2 λ^{2} - 29 λ + 69 = 0 ⟹ λ = \frac{29 \pm 841 - 552}{4} = \frac{29 \pm 17}{4}

λ = 3

λ = \frac{23}{2}

.

The value common to both

D = 0

and

D_{x} = 0

λ = 3

λ^{2} + λ = 9 + 3 = 12

JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Single correct option