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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?

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up vote 1 down vote accepted

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.

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