For which sets of parameters will this system of equations have no solution? 
If $S$ is the set of distinct values of '$b$' for which the following system of linear equations
\begin{align} 
 x + y + z = 1\\
 x + ay + z = 1\\ 
 ax + by + z = 0
 \end{align}
  has no solution, then $S$ is:
  
  
*
  
*a singleton 
  
*an empty set
  
*an infinite set 
  
*a finite set containing two or more elements
  

There's no information given about value of $a$.
You will be getting two cases:


*

*For $a=1, b$ has one value $b=1$

*For $a$ not equal to $1 , b$ has no value.

 A: Suppose $a=1$. Then, the first two equations are the same.
The third equation is $x+by+z = 0$. We know that $x+y+z=1$, so subtracting, we get $y(1-b) = 1$, so that if $b\neq 1$, we have a solution for $y$, and trivially for $x,z$. Hence, $b=1$ is the only value that doesn't give solutions.
Suppose $a \neq 1$. Then, subtracting second equation from first, $(a-1)y = 0$, so that $y=0$.
Now the third equation changes to $ax + z =0$, and we know $x+z = 1$. These equations can be solved, with $x = \frac 1{1-a}$ and $z = 1-x$. These solutions exist regardless of the value of $b$, since they don't depend on $b$.
Hence, solutions will exist whenever $a \neq 1$.
This means that $b$ is a singleton (and you are correct, good job!)
A: When $a=1$ and $b=1$ The first two equations are coincident planes ($x+y+z=1$)and the last one ($x+y+z=0$) is a plane parallel to them . 
Therefore for $b=1$, these equations have no solutions at all.(Since parallel planes will intersect no where.)
Hence, your set $S=\{1 \}$
A: $$\begin{bmatrix} 1 & 1 & 1 \\ 1 & a & 1 \\ a & b & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 0\end{bmatrix}$$
$$\begin{bmatrix} 1 & 1 & 1 \\ 0 & a-1 & 0 \\ 0 & b-a & 1-a \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ -a\end{bmatrix}$$
The determinant of the matrix is $-(a-1)^2$, to have no solution, the determinant has to be $0$, hence $a=1$.
$$\begin{bmatrix} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & b-1 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ -1\end{bmatrix}$$
To have no solution $b$ has to be $1$.
Hence $S=\{1 \}$.
Remark:
I interpret the question as $S=\{b: \exists a \text{ such that the system has no solution.} \}$
If we interpret $S=\{b: \forall a \text{ such that the system has no solution.} \}$, then $S = \emptyset$.
