If the system of linear equations $2x+2ay+az=0, 2x+3by+bz=0, 2x+4cy+cz=0$ where $a, b, c \in \mathbb{R}$ are non-zero and distinct; has a non-zero solution, then:

(C) $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P.

JEE Main · 2020 · 07 Jan 2020 · Shift 1 · Determinants

If the system of linear equations

2 x + 2 a y + a z = 0, 2 x + 3 b y + b z = 0, 2 x + 4 cy + cz = 0

where

a, b, c \in R

are non-zero and distinct; has a non-zero solution, then:

Answer: (C)

A non-zero solution needs a vanishing coefficient determinant,

222 2 a 3 b 4 c a b c = 0

. Taking the

2

out of the first column and expanding gives

- b c + 2 a c - ab = 0

, i.e.

2 a c = ab + b c

. Dividing by

ab c

yields

\frac{2}{b} = \frac{1}{a} + \frac{1}{c}

, which says

\frac{1}{a}, \frac{1}{b}, \frac{1}{c}

are in arithmetic progression.

JEE Main · 2020 · 07 Jan 2020 · Shift 1 · Single correct option