Algebra-Lineare Algebra-Lineare Gleichungssysteme und Gauß-Algorithmus

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$n-Gleichungen$
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Beispiel Nr: 10
$\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:} \\ 9 x +5 + 4 z=13\\ 6 x +3 y + -5 z=17\\ 3 x -10 y + 6 z=23\\ \\ \\ \textbf{Rechnung:} \\\small \begin{array}{l} 9x+5y+4z=13 \\ 6x+3y-5z=17 \\ 3x-10y+6z=23 \\ \\ \end{array} \qquad \small \begin{array}{ccc|cc } x & y & z & & \\ \hline9 & 5 & 4 & 13 \\ 6 & 3 & -5 & 17 \\ 3 & -10 & 6 & 23 \\ \end{array} \\ \\ \begin{array}{l}\text{Zeile}2=\text{Zeile}2\text{-Zeile}1\cdot \frac{6}{9}\\z2s1=6-9\cdot \frac{6}{9}=0 \\ z2s2=3-5\cdot \frac{6}{9}=-\frac{1}{3} \\ z2s3=-5-4\cdot \frac{6}{9}=-7\frac{2}{3} \\ z2s4=17-13\cdot \frac{6}{9}=8\frac{1}{3} \\ \end{array}\qquad \small \begin{array}{ccc|cc } x & y & z & & \\ \hline9 & 5 & 4 & 13 \\ 0 & -\frac{1}{3} & -7\frac{2}{3} & 8\frac{1}{3} \\ 3 & -10 & 6 & 23 \\ \end{array} \\ \\ \begin{array}{l}\text{Zeile}3=\text{Zeile}3\text{-Zeile}1\cdot \frac{3}{9}\\z3s1=3-9\cdot \frac{3}{9}=0 \\ z3s2=-10-5\cdot \frac{3}{9}=-11\frac{2}{3} \\ z3s3=6-4\cdot \frac{3}{9}=4\frac{2}{3} \\ z3s4=23-13\cdot \frac{3}{9}=18\frac{2}{3} \\ \end{array}\qquad \small \begin{array}{ccc|cc } x & y & z & & \\ \hline9 & 5 & 4 & 13 \\ 0 & -\frac{1}{3} & -7\frac{2}{3} & 8\frac{1}{3} \\ 0 & -11\frac{2}{3} & 4\frac{2}{3} & 18\frac{2}{3} \\ \end{array} \\ \\ \begin{array}{l}\text{Zeile}1=\text{Zeile}1\text{-Zeile}2\cdot \frac{5}{-\frac{1}{3}}\\z1s2=5-(-\frac{1}{3})\cdot \frac{5}{-\frac{1}{3}}=0 \\ z1s3=4-(-7\frac{2}{3})\cdot \frac{5}{-\frac{1}{3}}=-111 \\ z1s4=13-8\frac{1}{3}\cdot \frac{5}{-\frac{1}{3}}=138 \\ \end{array}\qquad \small \begin{array}{ccc|cc } x & y & z & & \\ \hline9 & 0 & -111 & 138 \\ 0 & -\frac{1}{3} & -7\frac{2}{3} & 8\frac{1}{3} \\ 0 & -11\frac{2}{3} & 4\frac{2}{3} & 18\frac{2}{3} \\ \end{array} \\ \\ \begin{array}{l}\text{Zeile}3=\text{Zeile}3\text{-Zeile}2\cdot \frac{-11\frac{2}{3}}{-\frac{1}{3}}\\z3s2=-11\frac{2}{3}-(-\frac{1}{3})\cdot \frac{-11\frac{2}{3}}{-\frac{1}{3}}=0 \\ z3s3=4\frac{2}{3}-(-7\frac{2}{3})\cdot \frac{-11\frac{2}{3}}{-\frac{1}{3}}=273 \\ z3s4=18\frac{2}{3}-8\frac{1}{3}\cdot \frac{-11\frac{2}{3}}{-\frac{1}{3}}=-273 \\ \end{array}\qquad \small \begin{array}{ccc|cc } x & y & z & & \\ \hline9 & 0 & -111 & 138 \\ 0 & -\frac{1}{3} & -7\frac{2}{3} & 8\frac{1}{3} \\ 0 & 0 & 273 & -273 \\ \end{array} \\ \\ \begin{array}{l}\text{Zeile}1=\text{Zeile}1\text{-Zeile}3\cdot \frac{-111}{273}\\z1s3=-111-273\cdot \frac{-111}{273}=0 \\ z1s4=138-(-273)\cdot \frac{-111}{273}=27 \\ \end{array}\qquad \small \begin{array}{ccc|cc } x & y & z & & \\ \hline9 & 0 & 0 & 27 \\ 0 & -\frac{1}{3} & -7\frac{2}{3} & 8\frac{1}{3} \\ 0 & 0 & 273 & -273 \\ \end{array} \\ \\ \begin{array}{l}\text{Zeile}2=\text{Zeile}2\text{-Zeile}3\cdot \frac{-7\frac{2}{3}}{273}\\z2s3=-7\frac{2}{3}-273\cdot \frac{-7\frac{2}{3}}{273}=0 \\ z2s4=8\frac{1}{3}-(-273)\cdot \frac{-7\frac{2}{3}}{273}=\frac{2}{3} \\ \end{array}\qquad \small \begin{array}{ccc|cc } x & y & z & & \\ \hline9 & 0 & 0 & 27 \\ 0 & -\frac{1}{3} & 0 & \frac{2}{3} \\ 0 & 0 & 273 & -273 \\ \end{array} \\ \\ x=\frac{27}{9}=3\\y=\frac{\frac{2}{3}}{-\frac{1}{3}}=-2\\z=\frac{-273}{273}=-1\\L=\{3/-2/-1\} \end{array}$