<s>[INST] Triangular System Forward substitution column oriented I have been reading a way to improve forward substitution when solving a triangular matrix, it says that once $x_1$ is resolved it can be removed from other equations $2$ through $n$ ( I understand that) and then proceed with the reduced system 
$L(2:n, 2:n)x(2:n)=b(2:n) - x(1)L(2:n,1) $, then compute $x_2$ and remove it from equations $3$ through $n$, sofor example:
$ \left( \begin{array}{ccc}
2 & 0 & 0 \\
1 & 5 & 0 \\
7 & 9 & 8 \end{array} \right)      \left( \begin{array}{ccc}
x_1  \\
x_2  \\
x_3  \end{array} \right) = \left( \begin{array}{ccc}
6  \\
2  \\
5  \end{array} \right)$
they find $x_1=3$ so there is no problem there, they did $x_1=6/2=3$, so then a $2 by 2 system$:
$ \left( \begin{array}{ccc}
5 & 0  \\
9 & 8  \end{array} \right)      \left( \begin{array}{ccc}
x_2  \\
x_3  \end{array} \right) = \left( \begin{array}{ccc}
2  \\
5  \end{array} \right) -  3 
\left( \begin{array}{ccc}
1  \\
7  \end{array} \right)=
\left( \begin{array}{ccc}
-1  \\
-16 \end{array} \right)
$
Could anyone explain a bit more the numbers?
 [/INST] The last two equations of the system can be written as
$$
  x_1\begin{pmatrix}1\\7\end{pmatrix}
 +x_2\begin{pmatrix}5\\9\end{pmatrix}
 +x_3\begin{pmatrix}0\\8\end{pmatrix}=
  \begin{pmatrix}2\\5\end{pmatrix}
$$
Now substitute $x_1=3$, bring the corresponding term to the other side, and re-factor the remaining left-hand-side as a matrix product.
</s>