Let the system of equations \(\mathrm{x}+5 \mathrm{y}-\mathrm{z}=1\) \(4 \mathrm{x}+3 \mathrm{y}-3 \mathrm{z}=7\) \(24 \mathrm{x}+\mathrm{y}+\lambda \mathrm{z}=\mu\) \(\lambda, \mu \in \mathrm{R}\), have infinitely many solutions. Then the number of the solutions of this system, If \(\mathrm{x}, \mathrm{y}, \mathrm{z}\) are integers and satisfy \(7 \leq \mathrm{x}+\mathrm{y}+\mathrm{z} \leq 77\), is

JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Matrices

Let the system of equations

x + 5 y - z = 1

4 x + 3 y - 3 z = 7

24 x + y + λ z = μ

λ, μ \in R

, have infinitely many solutions. Then the number of the solutions of this system,
If

x, y, z

are integers and satisfy

7 \leq x + y + z \leq 77

, is

Answer: (A)

For infinitely many solutions the third equation must be a consequence of the first two. Solving

x + 5 y - z = 1

and

4 x + 3 y - 3 z = 7

with

z = t

gives

x = \frac{32 + 12 t}{17}

and

y = \frac{t - 3}{17}

, whence

24 x + y = 45 + 17 t

. Consistency of

24 x + y + λ z = μ

then forces

λ = - 17

and

μ = 45

. Integer solutions require

t \equiv 3 (mod 17)

; writing

t = 3 + 17 k

gives

x = 4 + 12 k, y = k, z = 3 + 17 k

, so

x + y + z = 7 + 30 k

. The bound

7 \leq 7 + 30 k \leq 77

gives

k = 0, 1, 2

, which is

3

solutions.

JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Single correct option