Consider the system of linear equations $-x+y+2 z=0$ $3 x-a y+5 z=1$ $2 x-2 y-a z=7$ Let $S_{1}$ be the set of all $a \in R$ for which the system is inconsistent and $S_{2}$ be the set of all $\mathrm{a} \in \mathrm{R}$ for which the system has infinitely many solutions. If $\mathrm{n}\left(\mathrm{S}_{1}\right)$ and $\mathrm{n}\left(\mathrm{S}_{2}\right)$ denote the number of elements in $\mathrm{S}_{1}$ and $\mathrm{S}_{2}$ respectively, then

(C) $\mathrm{n}\left(\mathrm{S}_{1}\right)=2, \mathrm{n}\left(\mathrm{S}_{2}\right)=0$

(A) $\mathrm{n}\left(\mathrm{S}_{1}\right)=1, \mathrm{n}\left(\mathrm{S}_{2}\right)=0$

(B) $\mathrm{n}\left(\mathrm{S}_{1}\right)=0, \mathrm{n}\left(\mathrm{S}_{2}\right)=2$

(D) $\mathrm{n}\left(\mathrm{S}_{1}\right)=2, \mathrm{n}\left(\mathrm{S}_{2}\right)=2$

Matrices — JEE Main · 2021 · 01 Sep 2021 · Shift 2 Question with Solution

Consider the system of linear equations

- x + y + 2 z = 0

3 x - a y + 5 z = 1

2 x - 2 y - a z = 7

Let

S_{1}

be the set of all

a \in R

for which the system is inconsistent and

S_{2}

be the set of all

a \in R

for which the system has infinitely many solutions. If

n (S_{1})

and

n (S_{2})

denote the number of elements in

S_{1}

and

S_{2}

respectively, then

Answer: (C)

The determinant of the coefficient matrix is

Δ = - 1 32 1 - a - 2 25 - a = - a^{2} + 7 a - 12 = - (a - 3) (a - 4),

Δ = 0

only at

a = 3

and

a = 4

. Using

y = x - 2 z

from the first equation: for

a = 3

the second equation gives

11 z = 1

while the third gives

z = 7

, a contradiction; for

a = 4

the third equation reduces to

0 = 7

. Both values make the system inconsistent, and neither admits infinitely many solutions, so

n (S_{1}) = 2

and

n (S_{2}) = 0

JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Single correct option