The system is in $Z_3$
$$ \left\{ \begin{array}{c} 2a+b+2c=1 \\ 2d+e=1 \\ a+e=2 \end{array} \right. $$
My attempt:
$$ \begin{matrix} 2 & 1 & 2 & 0 & 0 & 1 \\ 0 & 0 & 0 & 2 & 1 & 1 \\ 1 & 0 & 0 & 0 & 1 & 2 \\ \end{matrix} $$
$R_3 <-> R_2$
$$ \begin{matrix} 2 & 1 & 2 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 2 & 1 & 1 \\ \end{matrix} $$
$R_2 -> 3R_2$
$$ \begin{matrix} 2 & 1 & 2 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 & 1 & 1 \\ \end{matrix} $$
$R_1 -> 2R_1$
$$ \begin{matrix} 1 & 2 & 1 & 0 & 0 & 2 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 & 1 & 1 \\ \end{matrix} $$
At this point I don't know what to do.
$R$ stands for "line" of the matrix
multiplying the first equation by $-2$ and adding to the last we get $$-2b-4c+e=-1$$ and also is $$2d+e=1$$ can you finish?