Backwards substitution for solving triangular linear systems.
Consider a triangular system of the form 
, where the vector 
is given, and 
is upper-triangular. Let us first consider the case when 
, and is invertible. Thus, has the form
![\[ R = \begin{pmatrix} r_{11} & r_{12} & \ldots & r_{1n} \\ 0 & r_{22} & & r_{2n} \\ \vdots & & \ddots & \vdots \\ 0 & & 0 & r_{nn} \end{pmatrix} \]](https://pressbooks.pub/app/uploads/quicklatex/quicklatex.com-d1115d963f8acf5aae842b9885fb5740_l3.png "Rendered by QuickLaTeX.com")
with each 
, 
, non-zero.
The backwards substitution first solves for the last component of 
using the last equation:
![\[ x_n = \frac{1}{r_{nn}} y_n, \]](https://pressbooks.pub/app/uploads/quicklatex/quicklatex.com-1604fecb63469d477e1c705436a89dac_l3.png "Rendered by QuickLaTeX.com")
and then proceeds with the following recursion, for 
:
![\[ x_j = \frac{1}{r_{jj}} \left( y_j - \sum_{k=j+1}^n r_{jk} x_k \right). \]](https://pressbooks.pub/app/uploads/quicklatex/quicklatex.com-3e82c9228abb8b0991f9b9eff8b03a98_l3.png "Rendered by QuickLaTeX.com")
Example: Solving a 
triangular system by backward substitution
