Let's walk through elimination:
\begin{align}
\left[
\begin{array}{cccc|c}
1 & -1 & 2 & 1 & 2\\
2 & -3 & 2 & 0 & 3\\
-1 & 1 & 2 & 3 & 6\\
-3 & 2 & 0 & 3 & 9
\end{array}\right]
&\to
\left[
\begin{array}{cccc|c}
1 & -1 & 2 & 1 & 2 \\
0 & -1 & -2 & -2 & -1 \\
0 & 0 & 4 & 4 & 8 \\
0 & -1 & 6 & 6 & 15
\end{array}\right]
&&\begin{aligned} R_2&\gets R_2-2R_1\\R_3&\gets R_3+R_1\\R_4&\gets R_4+3R_1\end{aligned}
\tag{1}\\[6px]&\to
\left[
\begin{array}{cccc|c}
1 & -1 & 2 & 1 & 2 \\
0 & 1 & 2 & 2 & 1 \\
0 & 0 & 4 & 4 & 8 \\
0 & 1 & 8 & 8 & 16
\end{array}\right]
&&\begin{aligned} R_2&\gets -R_2\\R_4&\gets R_4+R_2\end{aligned}
\tag{2}\\[6px]&\to
\left[
\begin{array}{cccc|c}
1 & -1 & 2 & 1 & 2 \\
0 & 1 & 2 & 2 & 1 \\
0 & 0 & 1 & 1 & 2 \\
0 & 0 & 0 & 0 & 0
\end{array}\right]
&&\begin{aligned} R_3&\gets \tfrac{1}{4}R_3\\R_4&\gets R_4-8R_3\end{aligned}
\tag{3}\\[6px]&\to
\left[
\begin{array}{cccc|c}
1 & -1 & 0 & -1 & -2 \\
0 & 1 & 0 & 0 & -3 \\
0 & 0 & 1 & 1 & 2 \\
0 & 0 & 0 & 0 & 0
\end{array}\right]
&&\begin{aligned}R_2&\gets R_2-2R_3\\R_1&\gets R_1-2R_3\end{aligned}
\tag{4}\\[6px]&\to
\left[
\begin{array}{cccc|c}
1 & 0 & 0 & -1 & -5 \\
0 & 1 & 0 & 0 & -3 \\
0 & 0 & 1 & 1 & 2 \\
0 & 0 & 0 & 0 & 0
\end{array}\right]
&&R_1\gets R_1+R_2
\tag{5}
\end{align}
Already after step $(3)$ you can conclude that the columns of the matrix $A$ are linearly dependent, because you get a null row in a row echelon form (just consider the first four columns); the fact that you get a full null row in the augmented matrix tells you that the linear system is solvable, because the last column cannot contain a leading $1$.
Steps $(4)$ and $(5)$ are backwards elimination, that lead to the reduced row echelon form, very useful to easily express all solutions.
Indeed, if we denote by $x_1$, $x_2$, $x_3$ and $x_4$ the unknowns, we see that $x_4$ is free and the equations read
$$
\begin{cases}
x_1-x_4=-5 \\[4px]
x_2 = -3 \\[4px]
x_3+x_4=2
\end{cases}
$$
so the generic solution is
$$
\begin{bmatrix}
-5+h\\
-3\\
2-h\\
h
\end{bmatrix}
$$
The advantage to go into this order for elimination is that it makes it possible to write the $LU$ decomposition of the matrix $A$. I use to denote the elementary operations we have performed up to step $(3)$ as
$$
E_{21}(-2),\quad
E_{31}(1),\quad
E_{41}(3),\quad
E_2(-1),\quad
E_{42}(1),\quad
E_3(\tfrac{1}{4}),\quad
E_{43}(-8)
$$
and this allows to write
$$
L=E_{21}(2)E_{31}(-1)E_{41}(-3)E_2(-1)E_{42}(-1)E_3(4)E_{43}(8)
=\begin{bmatrix}
1 & 0 & 0 & 0 \\
2 & -1 & 0 & 0 \\
-1 & 0 & 4 & 0 \\
-3 & -1 & 8 & 1
\end{bmatrix}
$$
and
$$
U=\begin{bmatrix}
1 & -1 & 2 & 1 \\
0 & 1 & 2 & 2 \\
0 & 0 & 1 & 1 \\
0 & 0 & 0 & 0
\end{bmatrix}
$$
You can check that $A=LU$.
