The following example, part of
Finding a sequence of elementary matrices, compliments the insight of @Xoque55

The target matrix is
$$
\left[
\begin{array}{cc|cc}
2 & 4 \\
1 & 1 \\
\end{array}
\right]
$$

Use elementary row operations for Gaussian elimination. $\color{blue}{Blue}$ coloring denotes changed entries in the output matrix.

**Row echelon form**
Form the augmented matrix

$$
\left[
\begin{array}{c|c}
\mathbf{A} & b
\end{array}
\right]
=
\left[
\begin{array}{cc|c}
2 & 4 & b_{1} \\
1 & 1 & b_{2} \\
\end{array}
\right]
$$

Normalize row 1:
$$
\left[
\begin{array}{cc}
\frac{1}{2} & 0 \\
0 & 1 \\
\end{array}
\right]
%
\left[
\begin{array}{cc|c}
2 & 4 & b_{1} \\
1 & 1 & b_{2} \\
\end{array}
\right]
=
\left[
\begin{array}{cc|c}
\color{blue}{1} & \color{blue}{2} & \frac{1}{2}b_{1} \\
1 & 1 & b_{2}\\
\end{array}
\right]
$$

Clear column 1
$$
\left[
\begin{array}{rc}
1 & 0 \\
-1 & 1 \\
\end{array}
\right]
%
\left[
\begin{array}{cc|c}
\color{blue}{1} & \color{blue}{2} & \frac{1}{2}b_{1} \\
1 & 1 & b_{2}\\
\end{array}
\right]
=
\left[
\begin{array}{cr|c}
1 & 2 & \frac{1}{2}b_{1} \\
\color{blue}{0} & \color{blue}{-1} & b_{2} - \frac{1}{2}b_{1} \\
\end{array}
\right]
$$

The system can be solved via back substitution.

**Reduced row echelon form**

The reduction process is
$$
%
\left[
\begin{array}{c|c}
\mathbf{A} & \mathbf{I}
\end{array}
\right]
%
\qquad \Rightarrow \qquad
%
\left[
\begin{array}{c|c}
\mathbf{E_{A}} & \mathbf{R}
\end{array}
\right]
$$

Form the augmented matrix

$$
\left[
\begin{array}{c|c}
\mathbf{A} & \mathbf{I}
\end{array}
\right]
=
\left[
\begin{array}{cc|cc}
2 & 4 & 1 & 0 \\
1 & 1 & 0 & 1 \\
\end{array}
\right]
$$

Normalize row 1:
$$
\left[
\begin{array}{cc}
\frac{1}{2} & 0 \\
0 & 1 \\
\end{array}
\right]
%
\left[
\begin{array}{cc|cc}
2 & 4 & 1 & 0 \\
1 & 1 & 0 & 1 \\
\end{array}
\right]
=
\left[
\begin{array}{cc|cc}
\color{blue}{1} & \color{blue}{2} & \frac{1}{2} & 0 \\
1 & 1 & 0 & 1 \\
\end{array}
\right]
$$

Clear column 1
$$
\left[
\begin{array}{rc}
1 & 0 \\
-1 & 1 \\
\end{array}
\right]
%
\left[
\begin{array}{cc|cc}
1 & 2 & \frac{1}{2} & 0 \\
1 & 1 & 0 & 1 \\
\end{array}
\right]
=
\left[
\begin{array}{cr|rc}
1 & 2 & \frac{1}{2} & 0 \\
\color{blue}{0} & \color{blue}{-1} & -\frac{1}{2} & 1 \\
\end{array}
\right]
$$

Normalize row 2
$$
\left[
\begin{array}{cr}
1 & 0 \\
0 & -1 \\
\end{array}
\right]
%
\left[
\begin{array}{cc|cr}
1 & 2 & \frac{1}{2} & 0 \\
0 & 1 & \frac{1}{2} & -1 \\
\end{array}
\right]
=
\left[
\begin{array}{cc|cr}
1 & 2 & \frac{1}{2} & 0 \\
\color{blue}{0} & \color{blue}{1} & \frac{1}{2} & -1 \\
\end{array}
\right]
$$

Clear column 2
$$
\left[
\begin{array}{cr}
1 & -2 \\
0 & 1 \\
\end{array}
\right]
%
\left[
\begin{array}{cc|cr}
1 & 2 & \frac{1}{2} & 0 \\
0 & 1 & \frac{1}{2} & -1 \\
\end{array}
\right]
=
\left[
\begin{array}{cc|rr}
\color{blue}{1} & \color{blue}{0} & -\frac{1}{2} & 2 \\
0 & 1 & \frac{1}{2} & -1 \\
\end{array}
\right]
$$
Result
$$
\left[
\begin{array}{c|c}
\mathbf{E_{A}} & \mathbf{R}
\end{array}
\right]
=
\left[
\begin{array}{cc|rr}
1 & 0 & -\frac{1}{2} & 2 \\
0 & 1 & \frac{1}{2} & -1 \\
\end{array}
\right]
$$

The solution is
$$
\mathbf{A} x = b
\quad \Rightarrow \quad
x = \mathbf{A}^{-1} b
\quad \Rightarrow \quad
x = \frac{1}{2}\left[
\begin{array}{rr}
-1 & 4 \\
1 & -2 \\
\end{array}
\right]
%
\left[
\begin{array}{c}
b_{1} \\
b_{2} \\
\end{array}
\right]
\quad \Rightarrow \quad
x =
\left[
\begin{array}{l}
-\frac{1}{2} b_{1} + 2b_{2} \\
\phantom{-}\frac{1}{2} b_{1} - b_{2} \\
\end{array}
\right]
$$

**Product of reduction matrices**

The product of the sequence of reduction matrices is the inverse:
$$
% four
\left[
\begin{array}{cr}
1 & -2 \\
0 & 1 \\
\end{array}
\right]
% third
\left[
\begin{array}{cr}
1 & 0 \\
0 & -1 \\
\end{array}
\right]
% second
\left[
\begin{array}{rc}
1 & 0 \\
-1 & 1 \\
\end{array}
\right]
% first
\left[
\begin{array}{cc}
\frac{1}{2} & 0 \\
0 & 1 \\
\end{array}
\right]
=
\left[
\begin{array}{rr}
-\frac{1}{2} & 2 \\
\frac{1}{2} & -1 \\
\end{array}
\right]
=
\mathbf{A}^{-1}
$$