If the system of linear equations $2 x+2 y+3 z=a$ $3 x-y+5 z=b$ $x-3 y+2 z=c$ Where $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are non zero real numbers, has more than one solution, then:

(A) $\mathrm{b}-\mathrm{c}+\mathrm{a}=0$

JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Matrices

If the system of linear equations

2 x + 2 y + 3 z = a

3 x - y + 5 z = b

x - 3 y + 2 z = c

Where

a, b, c

are non zero real numbers, has more than one solution, then:

(A) $b - c + a = 0$
(B) $b - c - a = 0$ ✓ Correct
(C) $a + b + c = 0$
(D) $b + c - a = 0$

Answer: (B)

Solution

More than one solution requires the coefficient determinant to vanish, which it does, and the three equations to be linearly dependent in a way the constants respect. Searching for the combination that kills the left sides,

R_{1} - R_{2} + R_{3}

gives

(2 - 3 + 1) x + (2 + 1 - 3) y + (3 - 5 + 2) z = 0

, so the same combination must annihilate the right side:

a - b + c = 0

. Rearranged, this is

b - a - c = 0

, that is

b - c - a = 0

JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Single correct option

Practise more: all JEE Main PYQs on Matrices →

JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Matrices

Solution

More JEE Main questions on Matrices