<s>[INST] 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.
 [/INST] The only way for this system not to have a unique solution is if the matrix
$$
\left[
\begin{array}{ccc}
1 & a &1 \\
a & 1 & 1 \\
1 & -1 &1 
\end{array}
\right]
$$
is singular.  Its determinant is $1 - a^2$.  So it's singular if $a =1 $ or $a = -1$.  If $a = 1$, the first two rows are inconsistent equations.  If $a = -1$, the first and the third equations are identical, and the solutions are of the form $(x,y,z) = (y+1/2,y,1/2)$, where $y$ is a slack variable. 
</s>