Algebra-Lineare Algebra-Lineare Gleichungssysteme und Gauß-Algorithmus

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$n-Gleichungen$
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Beispiel Nr: 15
$\begin{array}{l} \text{Gegeben:} \\ a1 \cdot x + b1\cdot y + c1\cdot z=d1\\ a2\cdot x + b2\cdot y + c2\cdot z=d2\\ a3\cdot x + b3\cdot y + c3\cdot z=d3\\ \\ \text{Gesucht:} \\\text{x,y,z} \\ \\ \textbf{Gegeben:} \\ 1 x -2 + 3 z=9\\ 3 x +8 y + 9 z=5\\ 2 x +3 y + 6 z=7\\ \\ \\ \textbf{Rechnung:} \\\small \begin{array}{l} x-2y+3z=9 \\ 3x+8y+9z=5 \\ 2x+3y+6z=7 \\ \\ \end{array} \qquad \small \begin{array}{ccc|cc } x & y & z & & \\ \hline1 & -2 & 3 & 9 \\ 3 & 8 & 9 & 5 \\ 2 & 3 & 6 & 7 \\ \end{array} \\ \\ \begin{array}{l}\text{Zeile}2=\text{Zeile}2\text{-Zeile}1\cdot \frac{3}{1}\\z2s1=3-1\cdot \frac{3}{1}=0 \\ z2s2=8-(-2)\cdot \frac{3}{1}=14 \\ z2s3=9-3\cdot \frac{3}{1}=0 \\ z2s4=5-9\cdot \frac{3}{1}=-22 \\ \end{array}\qquad \small \begin{array}{ccc|cc } x & y & z & & \\ \hline1 & -2 & 3 & 9 \\ 0 & 14 & 0 & -22 \\ 2 & 3 & 6 & 7 \\ \end{array} \\ \\ \begin{array}{l}\text{Zeile}3=\text{Zeile}3\text{-Zeile}1\cdot \frac{2}{1}\\z3s1=2-1\cdot \frac{2}{1}=0 \\ z3s2=3-(-2)\cdot \frac{2}{1}=7 \\ z3s3=6-3\cdot \frac{2}{1}=0 \\ z3s4=7-9\cdot \frac{2}{1}=-11 \\ \end{array}\qquad \small \begin{array}{ccc|cc } x & y & z & & \\ \hline1 & -2 & 3 & 9 \\ 0 & 14 & 0 & -22 \\ 0 & 7 & 0 & -11 \\ \end{array} \\ \\ \begin{array}{l}\text{Zeile}1=\text{Zeile}1\text{-Zeile}2\cdot \frac{-2}{14}\\z1s2=-2-14\cdot \frac{-2}{14}=0 \\ z1s3=3-0\cdot \frac{-2}{14}=3 \\ z1s4=9-(-22)\cdot \frac{-2}{14}=5\frac{6}{7} \\ \end{array}\qquad \small \begin{array}{ccc|cc } x & y & z & & \\ \hline1 & 0 & 3 & 5\frac{6}{7} \\ 0 & 14 & 0 & -22 \\ 0 & 7 & 0 & -11 \\ \end{array} \\ \\ \begin{array}{l}\text{Zeile}3=\text{Zeile}3\text{-Zeile}2\cdot \frac{7}{14}\\z3s2=7-14\cdot \frac{7}{14}=0 \\ z3s3=0-0\cdot \frac{7}{14}=0 \\ z3s4=-11-(-22)\cdot \frac{7}{14}=0 \\ \end{array}\qquad \small \begin{array}{ccc|cc } x & y & z & & \\ \hline1 & 0 & 3 & 5\frac{6}{7} \\ 0 & 14 & 0 & -22 \\ 0 & 0 & 0 & 0 \\ \end{array} \\ \\ \\ L= unendlich \end{array}$