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?
$$ \left[
    \begin{array}{ccc|c}
      1&a&1&a\\
      0&(a^2)-1&a-1&(a^2)-1\\
      0&(a-1)(a-2)&0&a-1
    \end{array}
\right] $$
 A: hint:$$\left\{ 
\begin{array}{c}
ax+y+z=1 \\ 
x+ay+z=a \\ 
x+y+az=a^2
\end{array}
\right.$$sum of the equations :
$$x(a+2)+y(a+2)+z(a+2)=1+a+a2\\x+y+z=\frac{1+a+a^2}{a+2}\\$$now $$ax+y+z=1-(x+y+z=\frac{1+a+a^2}{a+2})\\(a-1)x=1-\frac{1+a+a^2}{a+2}\\(a-1)=\frac{a+2-1-a-a^2}{a+2}\\x=\frac{1-a^2}{(a+2)(a-1)}$$
A: 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
A: We have to obtain the reduced row echelon form of the augmented matri, starting from
$$\begin{bmatrix}\!\!\begin{array}{@{}ccc|c}
1&a&1&a\\0&a^2-1&a-1&a^2-1\\0&(a-1)(a-2)&0&a-1
\end{array}\!\!\end{bmatrix}$$


*

*If $a=1$, this matrix becomes
$\;\smash[b]{\begin{bmatrix}\!\!\begin{array}{@{}ccc|c}
1&1&1&1\\0&0&0&0\\0&0&0&0
\end{array}\!\!\end{bmatrix}}$, which corresponds to the plane with equation:
$$\color{red}{x+y+z=1}.$$

*If $a\ne 1$, we can factor out $x-1$, and thus obtaining the matrix
$$\begin{bmatrix}\!\!\begin{array}{@{}ccc|c}
1&a&1&a\\0&a+1&1&a+1\\0&a-2&0&1
\end{array}\!\!\end{bmatrix},\enspace\text{and swapping columns $2$ and $3$}:\quad \begin{bmatrix}\!\!\begin{array}{@{}ccc|c}
1&1&a&a\\0&1&a+1&a+1\\0&0&a-2&1
\end{array}\!\!\end{bmatrix}$$
(Swapping these columns amounts to exchange unknowns $y$ and $z$).

*If $a=2$, the linear system is inconsistent.

*If $a\ne2$, we can proceed with row reduction:
\begin{align}
\begin{bmatrix}\!\!\begin{array}{@{}ccc|c}
1&1&a&a\\0&1&a+1&a+1\\0&0&a-2&1
\end{array}\!\!\end{bmatrix}&\rightsquigarrow\begin{bmatrix}\!\!\begin{array}{@{}ccc|c}
1&1&a&a\\0&1&a+1&a+1\\0&0&1&\frac1{a-2}
\end{array}\!\!\end{bmatrix}\rightsquigarrow\begin{bmatrix}\!\!\begin{array}{@{}ccc|c}
1&1&0&\frac{a(a-3)}{a-2}\\0&1&0&\frac{(a+1)(a-3)}{a-2}\\0&0&1&\frac1{a-2}
\end{array}\!\!\end{bmatrix}\\
&\rightsquigarrow 
\begin{bmatrix}\!\!\begin{array}{@{}ccc|c}
1&0&0&-\frac{(a-3)}{a-2}\\0&1&0&\\0&0&1&\frac1{a-2}
\end{array}\!\!\end{bmatrix} 
\end{align}
So the solutions are
$$\color{red}{x=-\frac{(a-3)}{a-2}},\quad \color{red}{y=\frac1{a-2}},\quad \color{red}{z=\frac{(a+1)(a-3)}{a-2}}.$$

