If $S$ is the set of distinct values of ${}^{\prime }{b}^{\prime }$ for which the following system of linear equations
$x+y+z=1$
$x+ay+z=1$
$ax+by+z=0$
has no solution, then $S$ is

The correct option is A a singleton set
The given set of equation can be written in matrix form as:
$\left[\begin{array}{ccc}1& 1& 1\\ 1& a& 1\\ a& b& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right]$
Since this is a non-homogeneous equation, the determinant of the co-efficient matrix should be $0$ for no solution to exist.

$\therefore \left|\begin{array}{ccc}1& 1& 1\\ 1& a& 1\\ a& b& 1\end{array}\right|=0$

$\Rightarrow 1(a-b)-1(1-a)+1(b-a^2)=0$

$\Rightarrow a-b-1+a+b-a^2=0$

$\Rightarrow 2a-1-a^2=0$

$\Rightarrow -(a-1{)}^2=0$

$\Rightarrow a=1$

For $a=1,$ the equation become:

$x+y+z=1$

$x+y+z=1$

$x+by+z=0$

From the above three equations, we can see that if $b=1,$ the system will be inconsistent and hence will produce no solution. For $b\ne 1$, the system will produce infinite solutions.

Hence, for no solution, $S$ has to be a singleton set