Consider the following system of equations $\alpha \mathrm{x}+2 \mathrm{y}+\mathrm{z}=1$ $2 \alpha \mathrm{x}+3 \mathrm{y}+\mathrm{z}=1$ $3 \mathrm{x}+\alpha \mathrm{y}+2 \mathrm{z}=\beta$ For some $\alpha, \beta \in \mathbb{R}$. Then which of the following is NOT correct.

(B) It has no solution for $\alpha=-1$ and for all $\beta \in \mathbb{R}$

(A) It has no solution if $\alpha=-1$ and $\beta \neq 2$

(C) It has no solution for $\alpha=3$ and for all $\beta \neq 2$

(D) It has a solution for all $\alpha \neq-1$ and $\beta=2$

Matrices — JEE Main · 2023 · 29 Jan 2023 · Shift 1 Question with Solution

Consider the following system of equations

α x + 2 y + z = 1

2 α x + 3 y + z = 1

3 x + α y + 2 z = β

For some

α, β \in R

. Then which of the following is NOT correct.

Answer: (B)

The coefficient determinant equals

(α - 3) (α + 1)

, so it vanishes only at

α = - 1

and

α = 3

. For

α = - 1

the system is consistent when

β = 2

(then it has infinitely many solutions) and inconsistent when

β \neq = 2

; for

α = 3

it is inconsistent for every

β \neq = 2

; and for

α \neq = - 1

with

β = 2

a solution always exists. The assertion that the system has no solution when

α = - 1

for every value of

β

overlooks the consistent case

β = 2

, and that is the statement which does not hold.

JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Single correct option