If the system of equations $2 \mathrm{x}-\mathrm{y}+\mathrm{z}=4$ $5 \mathrm{x}+\lambda \mathrm{y}+3 \mathrm{z}=12$ $100 \mathrm{x}-47 \mathrm{y}+\mu \mathrm{z}=212$, has infinitely many solutions, then $\mu-2 \lambda$ is equal to

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Determinants

If the system of equations

2 x - y + z = 4

5 x + λ y + 3 z = 12

100 x - 47 y + μ z = 212

,
has infinitely many solutions, then

μ - 2 λ

is equal to

Answer: (D)

For the system to have infinitely many solutions, the third equation must be a linear combination of the first two: row

_{3}

a \cdot

row

_{1}

b \cdot

row

_{2}

for scalars

a, b

.

Matching the

x

-coefficient and the constant term:

100 = 2 a + 5 b, 212 = 4 a + 12 b .

Doubling the first equation:

200 = 4 a + 10 b

. Subtracting from the second:

12 = 2 b \Rightarrow b = 6

, then

100 = 2 a + 30 \Rightarrow a = 35

.

Matching the

y

-coefficient:

- 47 = - a + λb = - 35 + 6 λ \Rightarrow 6 λ = - 12 \Rightarrow λ = - 2

.

Matching the

z

-coefficient:

μ = a + 3 b = 35 + 18 = 53

μ - 2 λ = 53 - 2 (- 2) = 57.

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Single correct option