If the following system of linear equations
Question:

If the following system of linear equations

$2 x+y+z=5$

$x-y+z=3$

$x+y+a z=b$

has no solution, then :

1. $\mathrm{a}=-\frac{1}{3}, \mathrm{~b} \neq \frac{7}{3}$

2. $\mathrm{a} \neq \frac{1}{3}, \mathrm{~b}=\frac{7}{3}$

3. $\mathrm{a} \neq-\frac{1}{3}, \mathrm{~b}=\frac{7}{3}$

4. $\mathrm{a}=\frac{1}{3}, \mathrm{~b} \neq \frac{7}{3}$

Correct Option: , 4

Solution:

Here $D=\left|\begin{array}{ccc}2 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & a\end{array}\right|=2(-a-1)-1(a-1)+1+1$

$D_{3}=\left|\begin{array}{ccc}2 & 1 & 5 \\ 1 & -1 & 3 \\ 1 & 1 & b\end{array}\right|=2(-b-3)-1(b-3)+5(1+1)$

for $a=\frac{1}{3}, b \neq \frac{7}{3}$, system has no solutions