Given is a linear system of the form
\begin{pmatrix}A & B & C & \bigl | &M \\
R & D & E & \bigl | &N \\
S & F & G & \bigl | &L
\end{pmatrix} with the unique solution
\begin{pmatrix} {\color{red}X} \\
{\color{blue}Y} \\ Z \end{pmatrix}.
Determine the entries
\color{red} X
=
X
\color{blue} Y
=
Y
Z
=
Z
In the 1sr step towards the row echelon form, we must get \fbox{\text{a zero}} in
\begin{pmatrix}A & B & C & \bigl | &M \\
\boxed{R} & D & E & \bigl | &N \\
\boxed{S} & F & G & \bigl | &L
\end{pmatrix} , respectively.
This is done by replacing the {2nd row} with
negParens(R) \; \cdot {1. row} - A \; \cdot {2. row}
and replacing the {3rd row} with
negParens(S) \; \cdot {1. row} - A \; \cdot {3. row}.
In this way we get
\begin{pmatrix}A & B & C & \bigl | &M \\
\boxed{0} & B*R-D*A & C*R-E*A & \bigl | &M*R-A*N \\
\boxed{0} & B*S-F*A & C*S-G*A & \bigl | &M*S-A*L
\end{pmatrix} .
For the row echelon form, we must further get \fbox{\text{a zero}} in
\begin{pmatrix}A & B & C & \bigl | &M \\
0 & B*R-D*A & C*R-E*A & \bigl | &M*R-A*N \\
0 & \boxed{FF} & GG & \bigl | & LL
\end{pmatrix}
.
For this we replace the {3rd row} with
negParens(FF) \; \cdot {2. row} - DD \; \cdot {3. row}.
This leads to
\begin{pmatrix}A & B & C & \bigl | &M \\
0 & DD & EE & \bigl | &NN \\
0 & \boxed{0} & FF*EE-DD*GG & \bigl | &FF*NN-DD*LL
\end{pmatrix}
and implies Z = Z.
The second row
DD Y + EE Z = NN
rearranganges to
DD Y = NN - EE Z , so making use of Z = Z, we find {\color{blue}Y = Y}.
Using Z = Z and {\color{blue} Y =
Y}
in the first row
A X + B Y + C Z = M
leads to
A {\color{red}X} = M - B {\color{blue}Y} - C Z , from which we find that
{\color{red}X = X}.