Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.