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Backward Substitution

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An upper triangular system

$\displaystyle \left( \begin{array}{lcl} r_{1,1} & \cdots & r_{1,n} \\ & \ddo... ...ght)= \left( \begin{array}{l} b_1 \\ \vdots \\ b_n \end{array} \right) $

with $ \det R = r_{1,1} \cdots r_{n,n} \neq 0$ can be solved successively for $ x_n, \ldots , x_1$:

$\displaystyle r_{n,n} x_n = b_n \rightarrow x_n = b_n/r_{n,n}, $

and, for $ \ell= n-1, \ldots 1$,

$\displaystyle r_{\ell,\ell} x_\ell + \cdots + r_{\ell,n} x_n = b_\ell, \righta... ...l -r_{\ell,\ell+1}x_{\ell+1}- \cdots - r_{\ell,n}x_n \right)/r_{\ell,\ell}, $

using the previously computed values for $ x_{\ell+1}, \ldots, x_n$.

Similarly, a lower triangular system can be solved successively for $ x_1 , \ldots , x_n$. In both cases, multiple right-hand sides can be processed simultaneously.

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  automatically generated 3/ 8/2007