Algebra-Lineare Algebra-Lineare Gleichungssysteme und Gauß-Algorithmus

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$n-Gleichungen$
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Beispiel Nr: 09
$\begin{array}{l} \text{Gegeben:} \\ a1 \cdot x + b1\cdot y + c1\cdot z=d1\\ a2\cdot x + b2\cdot y + c2\cdot z=d2\\ a3\cdot x + b3\cdot y + c3\cdot z=d3\\ \\ \text{Gesucht:} \\\text{x,y,z} \\ \\ \textbf{Gegeben:} \\ 4 x -3 + 2 z=10\\ 5 x +6 y + -7 z=4\\ 10 x -2 y + -3 z=7\\ \\ \\ \textbf{Rechnung:} \\\small \begin{array}{l} 4x-3y+2z=10 \\ 5x+6y-7z=4 \\ 10x-2y-3z=7 \\ \\ \end{array} \qquad \small \begin{array}{ccc|cc } x & y & z & & \\ \hline4 & -3 & 2 & 10 \\ 5 & 6 & -7 & 4 \\ 10 & -2 & -3 & 7 \\ \end{array} \\ \\ \begin{array}{l}\text{Zeile}2=\text{Zeile}2\text{-Zeile}1\cdot \frac{5}{4}\\z2s1=5-4\cdot \frac{5}{4}=0 \\ z2s2=6-(-3)\cdot \frac{5}{4}=9\frac{3}{4} \\ z2s3=-7-2\cdot \frac{5}{4}=-9\frac{1}{2} \\ z2s4=4-10\cdot \frac{5}{4}=-8\frac{1}{2} \\ \end{array}\qquad \small \begin{array}{ccc|cc } x & y & z & & \\ \hline4 & -3 & 2 & 10 \\ 0 & 9\frac{3}{4} & -9\frac{1}{2} & -8\frac{1}{2} \\ 10 & -2 & -3 & 7 \\ \end{array} \\ \\ \begin{array}{l}\text{Zeile}3=\text{Zeile}3\text{-Zeile}1\cdot \frac{10}{4}\\z3s1=10-4\cdot \frac{10}{4}=0 \\ z3s2=-2-(-3)\cdot \frac{10}{4}=5\frac{1}{2} \\ z3s3=-3-2\cdot \frac{10}{4}=-8 \\ z3s4=7-10\cdot \frac{10}{4}=-18 \\ \end{array}\qquad \small \begin{array}{ccc|cc } x & y & z & & \\ \hline4 & -3 & 2 & 10 \\ 0 & 9\frac{3}{4} & -9\frac{1}{2} & -8\frac{1}{2} \\ 0 & 5\frac{1}{2} & -8 & -18 \\ \end{array} \\ \\ \begin{array}{l}\text{Zeile}1=\text{Zeile}1\text{-Zeile}2\cdot \frac{-3}{9\frac{3}{4}}\\z1s2=-3-9\frac{3}{4}\cdot \frac{-3}{9\frac{3}{4}}=0 \\ z1s3=2-(-9\frac{1}{2})\cdot \frac{-3}{9\frac{3}{4}}=-\frac{12}{13} \\ z1s4=10-(-8\frac{1}{2})\cdot \frac{-3}{9\frac{3}{4}}=7\frac{5}{13} \\ \end{array}\qquad \small \begin{array}{ccc|cc } x & y & z & & \\ \hline4 & 0 & -\frac{12}{13} & 7\frac{5}{13} \\ 0 & 9\frac{3}{4} & -9\frac{1}{2} & -8\frac{1}{2} \\ 0 & 5\frac{1}{2} & -8 & -18 \\ \end{array} \\ \\ \begin{array}{l}\text{Zeile}3=\text{Zeile}3\text{-Zeile}2\cdot \frac{5\frac{1}{2}}{9\frac{3}{4}}\\z3s2=5\frac{1}{2}-9\frac{3}{4}\cdot \frac{5\frac{1}{2}}{9\frac{3}{4}}=0 \\ z3s3=-8-(-9\frac{1}{2})\cdot \frac{5\frac{1}{2}}{9\frac{3}{4}}=-2\frac{25}{39} \\ z3s4=-18-(-8\frac{1}{2})\cdot \frac{5\frac{1}{2}}{9\frac{3}{4}}=-13\frac{8}{39} \\ \end{array}\qquad \small \begin{array}{ccc|cc } x & y & z & & \\ \hline4 & 0 & -\frac{12}{13} & 7\frac{5}{13} \\ 0 & 9\frac{3}{4} & -9\frac{1}{2} & -8\frac{1}{2} \\ 0 & 0 & -2\frac{25}{39} & -13\frac{8}{39} \\ \end{array} \\ \\ \begin{array}{l}\text{Zeile}1=\text{Zeile}1\text{-Zeile}3\cdot \frac{-\frac{12}{13}}{-2\frac{25}{39}}\\z1s3=-\frac{12}{13}-(-2\frac{25}{39})\cdot \frac{-\frac{12}{13}}{-2\frac{25}{39}}=0 \\ z1s4=7\frac{5}{13}-(-13\frac{8}{39})\cdot \frac{-\frac{12}{13}}{-2\frac{25}{39}}=12 \\ \end{array}\qquad \small \begin{array}{ccc|cc } x & y & z & & \\ \hline4 & 0 & 0 & 12 \\ 0 & 9\frac{3}{4} & -9\frac{1}{2} & -8\frac{1}{2} \\ 0 & 0 & -2\frac{25}{39} & -13\frac{8}{39} \\ \end{array} \\ \\ \begin{array}{l}\text{Zeile}2=\text{Zeile}2\text{-Zeile}3\cdot \frac{-9\frac{1}{2}}{-2\frac{25}{39}}\\z2s3=-9\frac{1}{2}-(-2\frac{25}{39})\cdot \frac{-9\frac{1}{2}}{-2\frac{25}{39}}=0 \\ z2s4=-8\frac{1}{2}-(-13\frac{8}{39})\cdot \frac{-9\frac{1}{2}}{-2\frac{25}{39}}=39 \\ \end{array}\qquad \small \begin{array}{ccc|cc } x & y & z & & \\ \hline4 & 0 & 0 & 12 \\ 0 & 9\frac{3}{4} & 0 & 39 \\ 0 & 0 & -2\frac{25}{39} & -13\frac{8}{39} \\ \end{array} \\ \\ x=\frac{12}{4}=3\\y=\frac{39}{9\frac{3}{4}}=4\\z=\frac{-13\frac{8}{39}}{-2\frac{25}{39}}=5\\L=\{3/4/5\} \end{array}$