Beispiel Nr: 06
$\text{Gegeben:} \\ \text{Lineares Gleichungssytem} \\ a1 \cdot x_1 + b1\cdot x_2 + c1\cdot x_3 ....=d1 \\ a2\cdot x_1 + b2\cdot x_2 + c2\cdot x_3 .....=d2\\ a3\cdot x_1 + b3\cdot x_2 + c3\cdot x_3....=d3\\ ..... \\ \text{Gesucht: }x_1,x_2,x_3.... \\ \\$
$\textbf{Aufgabe:}$ f
$\textbf{Rechnung:}\\ \small \begin{array}{l} 3x_1+x_2 -x_3=0 \\ x_1+2x_2+4x_3=1 \\ 2x_1-2x_2+x_3=1 \\ \\ \end{array} \qquad \small \begin{array}{ccc|cc } x_1 & x_2 & x_3 & & \\ \hline3 & 1 & -1 & 0 \\ 1 & 2 & 4 & 1 \\ 2 & -2 & 1 & 1 \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}2=\text{Zeile}2\text{-Zeile}1\cdot \frac{1}{3}\\z2s1=1-3\cdot \frac{1}{3}=0 \\ z2s2=2-1\cdot \frac{1}{3}=1\frac{2}{3} \\ z2s3=4-(-1)\cdot \frac{1}{3}=4\frac{1}{3} \\ z2s4=1-0\cdot \frac{1}{3}=1 \\ \end{array}\qquad \small \begin{array}{ccc|cc } x_1 & x_2 & x_3 & & \\ \hline3 & 1 & -1 & 0 \\ 0 & 1\frac{2}{3} & 4\frac{1}{3} & 1 \\ 2 & -2 & 1 & 1 \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}3=\text{Zeile}3\text{-Zeile}1\cdot \frac{2}{3}\\z3s1=2-3\cdot \frac{2}{3}=0 \\ z3s2=-2-1\cdot \frac{2}{3}=-2\frac{2}{3} \\ z3s3=1-(-1)\cdot \frac{2}{3}=1\frac{2}{3} \\ z3s4=1-0\cdot \frac{2}{3}=1 \\ \end{array}\qquad \small \begin{array}{ccc|cc } x_1 & x_2 & x_3 & & \\ \hline3 & 1 & -1 & 0 \\ 0 & 1\frac{2}{3} & 4\frac{1}{3} & 1 \\ 0 & -2\frac{2}{3} & 1\frac{2}{3} & 1 \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}1=\text{Zeile}1\text{-Zeile}2\cdot \frac{1}{1\frac{2}{3}}\\z1s2=1-1\frac{2}{3}\cdot \frac{1}{1\frac{2}{3}}=0 \\ z1s3=-1-4\frac{1}{3}\cdot \frac{1}{1\frac{2}{3}}=-3\frac{3}{5} \\ z1s4=0-1\cdot \frac{1}{1\frac{2}{3}}=-\frac{3}{5} \\ \end{array}\qquad \small \begin{array}{ccc|cc } x_1 & x_2 & x_3 & & \\ \hline3 & 0 & -3\frac{3}{5} & -\frac{3}{5} \\ 0 & 1\frac{2}{3} & 4\frac{1}{3} & 1 \\ 0 & -2\frac{2}{3} & 1\frac{2}{3} & 1 \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}3=\text{Zeile}3\text{-Zeile}2\cdot \frac{-2\frac{2}{3}}{1\frac{2}{3}}\\z3s2=-2\frac{2}{3}-1\frac{2}{3}\cdot \frac{-2\frac{2}{3}}{1\frac{2}{3}}=0 \\ z3s3=1\frac{2}{3}-4\frac{1}{3}\cdot \frac{-2\frac{2}{3}}{1\frac{2}{3}}=8\frac{3}{5} \\ z3s4=1-1\cdot \frac{-2\frac{2}{3}}{1\frac{2}{3}}=2\frac{3}{5} \\ \end{array}\qquad \small \begin{array}{ccc|cc } x_1 & x_2 & x_3 & & \\ \hline3 & 0 & -3\frac{3}{5} & -\frac{3}{5} \\ 0 & 1\frac{2}{3} & 4\frac{1}{3} & 1 \\ 0 & 0 & 8\frac{3}{5} & 2\frac{3}{5} \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}1=\text{Zeile}1\text{-Zeile}3\cdot \frac{-3\frac{3}{5}}{8\frac{3}{5}}\\z1s3=-3\frac{3}{5}-8\frac{3}{5}\cdot \frac{-3\frac{3}{5}}{8\frac{3}{5}}=0 \\ z1s4=-\frac{3}{5}-2\frac{3}{5}\cdot \frac{-3\frac{3}{5}}{8\frac{3}{5}}=\frac{21}{43} \\ \end{array}\qquad \small \begin{array}{ccc|cc } x_1 & x_2 & x_3 & & \\ \hline3 & 0 & 0 & \frac{21}{43} \\ 0 & 1\frac{2}{3} & 4\frac{1}{3} & 1 \\ 0 & 0 & 8\frac{3}{5} & 2\frac{3}{5} \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}2=\text{Zeile}2\text{-Zeile}3\cdot \frac{4\frac{1}{3}}{8\frac{3}{5}}\\z2s3=4\frac{1}{3}-8\frac{3}{5}\cdot \frac{4\frac{1}{3}}{8\frac{3}{5}}=0 \\ z2s4=1-2\frac{3}{5}\cdot \frac{4\frac{1}{3}}{8\frac{3}{5}}=-0,31 \\ \end{array}\qquad \small \begin{array}{ccc|cc } x_1 & x_2 & x_3 & & \\ \hline3 & 0 & 0 & \frac{21}{43} \\ 0 & 1\frac{2}{3} & 0 & -0,31 \\ 0 & 0 & 8\frac{3}{5} & 2\frac{3}{5} \\ \end{array} \\ \\ x_1=\frac{\frac{21}{43}}{3}=\frac{7}{43}\\x_2=\frac{-0,31}{1\frac{2}{3}}=-\frac{8}{43}\\x_3=\frac{2\frac{3}{5}}{8\frac{3}{5}}=\frac{13}{43}\\L=\{\frac{7}{43}/-\frac{8}{43}/\frac{13}{43}\}$