For $z, a, b, c \in \mathbb{R}$ solve this system of linear equations For $z, a, b, c  \in\mathbb{R}$ solve this system of linear equations using matrix methods,
$$
\left[
\begin{array}{ccc|c}
1 & z & 1 & a \\
z & 2 & 2 & b \\
2 & 1 & z & c \\
\end{array}
\right]
$$
I just added first and second row to the third row and got:
$$
\left[
\begin{array}{ccc|c}
1 & z & 1 & a \\
z & 2 & 2 & b \\
z + 3 & z + 3 & z + 3 & a + b + c \\
\end{array}
\right]
$$
But I don't know what should I do in next steps to solve this system for $z, a, b, c \in \mathbb{R}$

The next steps could be as follows:
$$
\left(
  \begin{array}{ccc|c}
    1 & z & 1 & a \\
    z & 2 & 2 & b\\
    z+3 & z+3 & z+3 & a+b+c \\
  \end{array}
\right)\sim
\left(
  \begin{array}{ccc|c}
    1 & z & 1 & a \\
    z & 2 & 2 & b\\
    1 & 1 & 1 & \Delta \\
  \end{array}
\right),\
\Delta=\frac{a+b+c}{z+3}
$$
$$
\left(
  \begin{array}{ccc|c}
    0 & z-1 & 0 & a-\Delta \\
    z-2 & 0 & 0 & b-2\Delta\\
    1 & 1 & 1 & \Delta \\
  \end{array}
\right).
$$
It follows that if $z\neq1,2,-3$, then
$$
x_1=\frac{b-2\Delta}{z-2},\ 
x_2=\frac{a-\Delta}{z-1},\ 
x_3=\Delta-\frac{b-2\Delta}{z-2}-\frac{a-\Delta}{z-1}.
$$
Now each of the three cases must be examined separately: $z=1,2,-3$.
