If the system of linear equations : $x+y+2 z=6$, $2 \mathrm{x}+3 \mathrm{y}+\mathrm{az}=\mathrm{a}+1$, $-\mathrm{x}-3 \mathrm{y}+\mathrm{bz}=2 \mathrm{~b}$, where $a, b \in \mathbf{R}$, has infinitely many solutions, then $7 \mathrm{a}+3 \mathrm{~b}$ is equal to :

Matrices — JEE Main · 2025 · 22 Jan 2025 · Shift 2 Question with Solution

If the system of linear equations :

x + y + 2 z = 6

2 x + 3 y + az = a + 1

- x - 3 y + bz = 2 b

,
where

a, b \in R

, has infinitely many solutions, then

7 a + 3 b

is equal to :

Answer: (C)

The coefficient determinant must vanish for infinitely many solutions:

Δ = 12 - 1 13 - 3 2 a b = 3 b + 3 a - (2 b + a) - 6 = 2 a + b - 6 = 0. (i)

Eliminating using

2 \times Eq (1) - Eq (2)

and

Eq (3) + Eq (1)

y + (a - 4) z = a - 11, - 2 y + (b + 2) z = 2 b + 6.

Substituting

y

from the first into the second:

- 2 a + 22 + (2 a + b - 6) z = 2 b + 6.

Since

2 a + b - 6 = 0

from (i), the z-term vanishes and consistency (infinitely many solutions) requires

- 2 a + 22 = 2 b + 6 \Rightarrow a + b = 8. (ii)

Solving (i) and (ii): from (ii),

b = 8 - a

; substituting into (i),

2 a + 8 - a - 6 = 0 \Rightarrow a = - 2

, so

b = 10

7 a + 3 b = 7 (- 2) + 3 (10) = - 14 + 30 = 16.

JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Single correct option