Solving systems of equation with unknown How do i solve this system? 
$$
\left\{ 
\begin{array}{c}
ax+y+z=1 \\ 
x+ay+z=a \\ 
x+y+az=a^2
\end{array}
\right. 
$$
Ive reduced to this form. How should i continue to get infinitely many solutions, no solution and unique?
$$ \left[
    \begin{array}{ccc|c}
      1&a&1&a\\
      0&(a^2)-1&a-1&(a^2)-1\\
      0&(a-1)(a-2)&0&a-1
    \end{array}
\right] $$

If $A$ is the matrix of the coefficients, then for Cramer's theorem the system has one unique solution if $\det A\ne 0$
$\det A=\left|
\begin{array}{lll}
 a & 1 & 1 \\
 1 & a & 1 \\
 1 & 1 & a \\
\end{array}
\right|=a^3-3 a+2$
$\det A=0 \to (a-1)^2 (a+2)=0\to a_1=1;\;a_2=-2$
Then for $a\ne 1;\;a\ne 1$ the system has one and only one solution
$$x=-\frac{a+1}{a+2},\;y=\frac{1}{a+2},\;z=\frac{(a+1)^2}{a+2}$$
$$.$$
If $a=1$ the completed matrix $A|B$ of the system becomes
$
A|B=\left(
\begin{array}{lll|l}
 1 & 1 & 1 & 1 \\
 1 & 1 & 1 & 1 \\
 1 & 1 & 1 & 1 \\
\end{array}
\right)
$
As $\text{rank } A= \text{rank } A|B=1$
the system has $\infty^{3-1}=\infty^2$ solutions given by the equation
$x+y+z=1$ whose solutions are $(t,u,1-t-u)$ 
they have two parameters which is linked to the $\infty^2$ solutions
$$
.
$$
If $a=-2$
$
A|B=\left(
\begin{array}{rrr|r}
 -2 & 1 & 1 & 1 \\
 1 & -2 & 1 & -2 \\
 1 & 1 & -2 & 4 \\
\end{array}
\right)$
$\text{rank } A|B=3$ while $\text{rank }A=2$
they are different so the system is impossible, has no solutions.
Hope this helps
