If the system of equations $x+2 y-3 z=2$ $2 \mathrm{x}+\lambda \mathrm{y}+5 \mathrm{z}=5$ $14 \mathrm{x}+3 \mathrm{y}+\mu \mathrm{z}=33$ has infinitely many solutions, then $\lambda+\mu$ is equal to:

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Determinants

If the system of equations

x + 2 y - 3 z = 2

2 x + λ y + 5 z = 5

14 x + 3 y + μ z = 33

has infinitely many solutions, then

λ + μ

is equal to:

Answer: (D)

For infinitely many solutions, the third equation must be a linear combination of the first two. Find

m, n

such that

14 x + 3 y + μ z = m (x + 2 y - 3 z) + n (2 x + λ y + 5 z), 33 = 2 m + 5 n .

Matching the

x

-coefficient and the constant/RHS term gives two equations in

m, n

m + 2 n = 14, 2 m + 5 n = 33.

From the first,

m = 14 - 2 n

; substituting into the second:

2 (14 - 2 n) + 5 n = 33 \Rightarrow 28 + n = 33 \Rightarrow n = 5

, so

m = 4

.

Matching the

y

-coefficient:

3 = 2 m + λn = 8 + 5 λ \Rightarrow λ = - 1

.

Matching the

z

-coefficient:

μ = - 3 m + 5 n = - 12 + 25 = 13

.

So

λ + μ = - 1 + 13 = 12.

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Single correct option