Let the system of equations \(x+2 y+3 z=5\), \(2 x+3 y+z=9,4 x+3 y+\lambda z=\mu\) have infinite number of solutions. Then \(\lambda+2 \mu\) is equal to :

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Matrices

Let the system of equations

x + 2 y + 3 z = 5

2 x + 3 y + z = 9, 4 x + 3 y + λ z = μ

have infinite number of solutions. Then

λ + 2 μ

is equal to :

Answer: (B)

For the system

x + 2 y + 3 z = 5, 2 x + 3 y + z = 9, 4 x + 3 y + λ z = μ

to have infinitely many solutions, the coefficient determinant and all the numerator determinants (Cramer's rule) must vanish.

Δ = 124233 31 λ = 1 (3 λ - 3) - 2 (2 λ - 4) + 3 (6 - 12) = - λ - 13.

Setting

Δ = 0

λ = - 13

Δ_{1} = 59 μ 233 31 λ = 5 (3 λ - 3) - 2 (9 λ - μ) + 3 (27 - 3 μ) = - 3 λ + 66 - 7 μ .

Setting

Δ_{1} = 0

with

λ = - 13

- 3 (- 13) + 66 - 7 μ = 0 ⟹ 39 + 66 = 7 μ ⟹ μ = 15.

(This choice of

λ, μ

also makes

Δ_{2} = Δ_{3} = 0

, confirming infinitely many solutions.)

λ + 2 μ = - 13 + 30 = 17.

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Single correct option