effiziente Technik zur Lösung eines linearen Gleichungssystems Lx = c mit unterer Dreiecksmatrix \(L{\rm{\hspace{0.17em}}}\in {\rm{\hspace{0.17em}}}{{\mathbb{R}}}^{n\times n}\) und rechter Seite \(c\in {\rm{\hspace{0.17em}}}{{\mathbb{R}}}^{n}\).

Beim Vorwärtseinsetzen löst man das Gleichungssystem

\begin{eqnarray}\left[\begin{array}{cccc}{\ell}_{11} & & & \\ {\ell}_{21} & {\ell}_{22} & & \\ \vdots & \vdots & \ddots & \\ {\ell}_{n1} & {\ell}_{n2} & \cdots & {\ell}_{nn}\end{array}\right]\left[\begin{array}{l}{x}_{1}\\ {x}_{2}\\ \vdots \\ {x}_{n}\end{array}\right]=\left[\begin{array}{l}{c}_{1}\\ {c}_{2}\\ \vdots \\ {c}_{n}\end{array}\right]\end{eqnarray}

\begin{eqnarray}\begin{array}{ll}{x}_{1} & =\displaystyle \frac{{c}_{1}}{{\ell}_{11}},\\ {x}_{i} & =\displaystyle \frac{1}{{\ell}_{ii}}\left({c}_{i}-\displaystyle \sum _{k=1}^{i=1}{\ell}_{ik}{x}_{k}\right)\quad{\text{f}}{\rm{\ddot {u}}}{\text{r}}\quad {i}=2,3,\ldots,n.\end{array}\end{eqnarray}