Lineares GS A x = b[br][br][math]\left(\begin{array}{r}a_{11} \; x_{1} + a_{12} \; x_{2} + a_{13} \; x_{3} + a_{14} \; x_{4}\\a_{21} \; x_{1} + a_{22} \; x_{2} + a_{23} \; x_{3} + a_{24} \; x_{4}\\a_{31} \; x_{1} + a_{32} \; x_{2} + a_{33} \; x_{3} + a_{34} \; x_{4}\\a_{41} \; x_{1} + a_{42} \; x_{2} + a_{43} \; x_{3} + a_{44} \; x_{4}\\\end{array}\right)=\vec{b}[br][/math][br][br]Vorwärts Substitution 1..2..3 Spalte[br][math]A \, := \, \left(\begin{array}{rrrr}a_{11}&a_{12}&a_{13}&a_{14}\\a_{21}&a_{22}&a_{23}&a_{24}\\a_{31}&a_{32}&a_{33}&a_{34}\\a_{41}&a_{42}&a_{43}&a_{44}\\\end{array}\right)[/math][br][math]A1 \, := \, \textcolor{red}{\left(\begin{array}{rrrr}1&0&0&0\\\frac{-a_{21}}{a_{11}}&1&0&0\\\frac{-a_{31}}{a_{11}}&0&1&0\\\frac{-a_{41}}{a_{11}}&0&0&1\\\end{array}\right) }\, \left(\begin{array}{rrrr}a_{11}&a_{12}&a_{13}&a_{14}\\a_{21}&a_{22}&a_{23}&a_{24}\\a_{31}&a_{32}&a_{33}&a_{34}\\a_{41}&a_{42}&a_{43}&a_{44}\\\end{array}\right)= \, \left(\begin{array}{rrrr}a_{11}&a_{12}&a_{13}&a_{14}\\0&c_{22}&c_{23}&c_{24}\\0&c_{32}&c_{33}&c_{34}\\0&c_{42}&c_{43}&c_{44}\\\end{array}\right)[/math] 1.Spalte (a[sub]i1[/sub])/a[sub]11, i=1..4[/sub][br][math] A2 \, := \, \textcolor{red}{\left(\begin{array}{rrrr}1&0&0&0\\0&1&0&0\\0&\frac{-c_{32}}{c_{22}}&1&0\\0&\frac{-c_{42}}{c_{22}}&0&1\\\end{array}\right)} \, \left(\begin{array}{rrrr}a_{11}&a_{12}&a_{13}&a_{14}\\0&c_{22}&c_{23}&c_{24}\\0&c_{32}&c_{33}&c_{34}\\0&c_{42}&c_{43}&c_{44}\\\end{array}\right) = \left(\begin{array}{rrrr}a_{11}&a_{12}&a_{13}&a_{14}\\0&c_{22}&c_{23}&c_{24}\\0&0&e_{33}&e_{34}\\0&0&e_{43}&e_{44}\\\end{array}\right)[/math] 2. Spalte (c[sub]i2[/sub])/c[sub]22, i=2..4[/sub][br][math]A3 \;:=\; \textcolor{red}{\left(\begin{array}{rrrr}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&\frac{-e_{43}}{e_{33}}&1\\\end{array}\right)} \;\left(\begin{array}{rrrr}a_{11}&a_{12}&a_{13}&a_{14}\\0&c_{22}&c_{23}&c_{24}\\0&0&e_{33}&e_{34}\\0&0&e_{43}&e_{44}\\\end{array}\right)=\left(\begin{array}{rrrr}a_{11}&a_{12}&a_{13}&a_{14}\\0&c_{22}&c_{23}&c_{24}\\0&0&e_{33}&e_{34}\\0&0&0&n_{44}\\\end{array}\right)[/math] 3. Spalte (e[sub]i3[/sub])/e[sub]33 i=3..4 [/sub][br][br]Diagonale[br][math]A4:=\textcolor{red}{\left(\begin{array}{rrrr}\frac{1}{a_{11}}&0&0&0\\0&\frac{1}{c_{22}}&0&0\\0&0&\frac{1}{e_{33}}&0\\0&0&0&\frac{1}{n_{44}}\\\end{array}\right) } \; \left(\begin{array}{rrrr}a_{11}&a_{12}&a_{13}&a_{14}\\0&c_{22}&c_{23}&c_{24}\\0&0&e_{33}&e_{34}\\0&0&0&n_{44}\\\end{array}\right)= \left(\begin{array}{rrrr}1&o_{12}&o_{13}&o_{14}\\0&1&o_{23}&o_{24}\\0&0&1&o_{34}\\0&0&0&1\\\end{array}\right)[/math] diag = 1[br][br]Rückwärts Substitution 4..3..2 Spalte[br][math]A5:=\textcolor{red}{\left(\begin{array}{rrrr}1&0&0&-o_{14}\\0&1&0&-o_{24}\\0&0&1&-o_{34}\\0&0&0&1\\\end{array}\right) }\; \left(\begin{array}{rrrr}1&o_{12}&o_{13}&o_{14}\\0&1&o_{23}&o_{24}\\0&0&1&o_{34}\\0&0&0&1\\\end{array}\right)=\left(\begin{array}{rrrr}1&o'_{12}&o'_{13}&0\\0&1&o'_{23}&0\\0&0&1&0\\0&0&0&1\\\end{array}\right)[/math] [br][math]A6:=\textcolor{red}{\left(\begin{array}{rrrr}1&0&-o'_{13}&0\\0&1&-o'_{23}&0\\0&0&1&0\\0&0&0&1\\\end{array}\right)} \; \left(\begin{array}{rrrr}1&o'_{12}&o'_{13}&0\\0&1&o'_{23}&0\\0&0&1&0\\0&0&0&1\\\end{array}\right)=\left(\begin{array}{rrrr}1&o''_{12}&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\\\end{array}\right)[/math][br][math]A7:=\textcolor{red}{\left(\begin{array}{rrrr}1&-o''_{12}&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\\\end{array}\right) }\; \left(\begin{array}{rrrr}1&o''_{12}&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\\\end{array}\right)=\left(\begin{array}{rrrr}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\\\end{array}\right)[/math][br][code][/code][br]Gleicher Algorithmus für[br]A wird um eine Spalte für b erweitert ===> A7 Spalte für b steht die Lösung des LGS[br]A wird um Einheitsmatrix erweitert ===> A7 enthält an der Stelle der Einheitsmatrix die Inverse A[sup]-1[/sup]