For what values of $a$ does the system have infinite solutions? Find the solutions. The system is 
$$\left\{
\begin{array}{rcr}
x+ay+z & = & 1 \\
ax+y+z & = & 1+a \\
x-y+z  & = & 2+a
\end{array}
\right.$$
After row reducing I got
$$\left\{
\begin{array}{rcr}
x+ay+z & = & 1 \\
-(1+a)y+0 & = & 1+a \\
(1-a)z& = & 2-a + (1-a)(1+a)
\end{array}
\right.$$
In order to have an infinite set of solutions, I want to have $0=0$ in the last row, meaning that $a=1\Rightarrow 0=1,$ which is not what I want. Have I made any arithmetical mistake? Did this computation times now.

The system reduces to 
$\left[\begin{matrix}
1 & 0 & 1 & 1+a\\
0 & -1 & 0 & 1 \\
0 & 0 & 1-a & 1+(1+a)^2
\end{matrix}\right]$ which means $z=\frac{1+(1+a)^2}{1-a}$
It implies for $a$ any value other than $1$ the system can have infinite solutions.
