Linear system in $\mathbb{Z_3}$ Given the linear system in $\mathbb{Z_3}$:
$$
\left\{ 
\begin{array}{c}
a+b+c+d=1 \\ 
b+c+e=2 \\ 
a+2e=0
\end{array}
\right. 
$$
I used the row reduction with matrices and I got:
$$
\left\{ 
\begin{array}{c}
a+b+c+d=1 \\ 
b+c+e=2 \\ 
d=2
\end{array}
\right. 
$$
But now I don't know how to find the solutions.
 A: What you probably did is add the last two equations to get 
$$
2=a+b+c,
$$
since $3e=0$. Now the first equation gives you $d=1-2=2$. Going back to the system (and again using $2=-1$), 
$$
\left\{ 
\begin{array}{c}
a+b+c=2 \\ 
b+c+e=2 \\ 
a-e=0
\end{array}
\right.
$$
and we already know $d=2$, $a=e$. Now theh first two equations are $a+b+c=2$, and so we are free to prescribe two of them. So, say, if you prescribe $b=t$, $c=s$, you have 
$$
a=2+2t+2s,\ \ d=2,\ \ e=a. 
$$
A: I'd use the symbols $0,1,-1$ for the elements of $\mathbf Z/3\mathbf Z$. Here's how to put the augmented matrix in reduced row echelon form:
\begin{align}
&\begin{bmatrix}
1&1&1&1&0&\hspace{-0.6em}|\phantom{-}~1\\
0&1&1&0&1&\hspace{-0.6em}|\:{-}1 \\1&0&0&0&-1&\hspace{-0.6em}|\phantom{-}~0
\end{bmatrix}\xrightarrow{R_1\leftarrow R_1-R_2}
\begin{bmatrix}
1&0&0&1&-1&\hspace{-0.6em}|\:{-}1\\
0&1&1&0&1&\hspace{-0.6em}|\,{-}1 \\1&0&0&0&-1&\hspace{-0.6em}|\phantom{-}~0
\end{bmatrix}\\[1ex]
\xrightarrow[R_3\leftarrow R_1-R_3]{}
&\begin{bmatrix}
1&0&0&1&-1&\hspace{-0.6em}|\:{-}1\\
0&1&1&0&1&\hspace{-0.6em}|\,{-}1 \\0&0&0&1&0&\hspace{-0.6em}|\:{-}1
\end{bmatrix}
\xrightarrow{R_1\leftarrow R_1-R_3}
\begin{bmatrix}
\color{red}1&\color{red}0&0&\color{red}0&-1&\hspace{-0.6em}|\:\phantom{-}0\\
\color{red}0&\color{red}1&1&\color{red}0&1&\hspace{-0.6em}|\,{-}1 \\\color{red}0&\color{red}0&0&\color{red}1&0&\hspace{-0.6em}|\:{-}1
\end{bmatrix}
\end{align}
Thus the solution are:
$$\begin{cases}
a=e\\b=-1-c-e\\d=-1
\end{cases}\enspace\text{or, in vector form:}\quad
\begin{bmatrix}a\\b\\c\\d\\e\end{bmatrix}=\begin{bmatrix}\phantom{-}0\\-1\\\phantom{-}0\\-1\\\phantom{-}0
\end{bmatrix}-c\begin{bmatrix}\phantom{-}0\\\phantom{-}1\\-1\\\phantom{-}0\\\phantom{-}0\end{bmatrix} +e\begin{bmatrix}\phantom{-}1\\-1\\\phantom{-}0\\\phantom{-}0\\\phantom{-}1
\end{bmatrix}.$$
