Algebra-Lineare Algebra-Lineare Gleichungssysteme und Gauß-Algorithmus

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$n-Gleichungen$
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Beispiel Nr: 25
$\begin{array}{l} \\ \begin{array} \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.... \\ \\ \end{array} \\ \textbf{Aufgabe:}\\ y\\ \textbf{Rechnung:}\\ \small \begin{array}{l} 4x_1+x_2+8x_3=2 \\ 8x_1 -x_2+x_3=1 \\ 3x_1-6x_2+2x_3=10 \\ \\ \end{array} \qquad \small \begin{array}{ccc|cc } x_1 & x_2 & x_3 & & \\ \hline4 & 1 & 8 & 2 \\ 8 & -1 & 1 & 1 \\ 3 & -6 & 2 & 10 \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}2=\text{Zeile}2\text{-Zeile}1\cdot \frac{8}{4}\\z2s1=8-4\cdot \frac{8}{4}=0 \\ z2s2=-1-1\cdot \frac{8}{4}=-3 \\ z2s3=1-8\cdot \frac{8}{4}=-15 \\ z2s4=1-2\cdot \frac{8}{4}=-3 \\ \end{array}\qquad \small \begin{array}{ccc|cc } x_1 & x_2 & x_3 & & \\ \hline4 & 1 & 8 & 2 \\ 0 & -3 & -15 & -3 \\ 3 & -6 & 2 & 10 \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}3=\text{Zeile}3\text{-Zeile}1\cdot \frac{3}{4}\\z3s1=3-4\cdot \frac{3}{4}=0 \\ z3s2=-6-1\cdot \frac{3}{4}=-6\frac{3}{4} \\ z3s3=2-8\cdot \frac{3}{4}=-4 \\ z3s4=10-2\cdot \frac{3}{4}=8\frac{1}{2} \\ \end{array}\qquad \small \begin{array}{ccc|cc } x_1 & x_2 & x_3 & & \\ \hline4 & 1 & 8 & 2 \\ 0 & -3 & -15 & -3 \\ 0 & -6\frac{3}{4} & -4 & 8\frac{1}{2} \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}1=\text{Zeile}1\text{-Zeile}2\cdot \frac{1}{-3}\\z1s2=1-(-3)\cdot \frac{1}{-3}=0 \\ z1s3=8-(-15)\cdot \frac{1}{-3}=3 \\ z1s4=2-(-3)\cdot \frac{1}{-3}=1 \\ \end{array}\qquad \small \begin{array}{ccc|cc } x_1 & x_2 & x_3 & & \\ \hline4 & 0 & 3 & 1 \\ 0 & -3 & -15 & -3 \\ 0 & -6\frac{3}{4} & -4 & 8\frac{1}{2} \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}3=\text{Zeile}3\text{-Zeile}2\cdot \frac{-6\frac{3}{4}}{-3}\\z3s2=-6\frac{3}{4}-(-3)\cdot \frac{-6\frac{3}{4}}{-3}=0 \\ z3s3=-4-(-15)\cdot \frac{-6\frac{3}{4}}{-3}=29\frac{3}{4} \\ z3s4=8\frac{1}{2}-(-3)\cdot \frac{-6\frac{3}{4}}{-3}=15\frac{1}{4} \\ \end{array}\qquad \small \begin{array}{ccc|cc } x_1 & x_2 & x_3 & & \\ \hline4 & 0 & 3 & 1 \\ 0 & -3 & -15 & -3 \\ 0 & 0 & 29\frac{3}{4} & 15\frac{1}{4} \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}1=\text{Zeile}1\text{-Zeile}3\cdot \frac{3}{29\frac{3}{4}}\\z1s3=3-29\frac{3}{4}\cdot \frac{3}{29\frac{3}{4}}=0 \\ z1s4=1-15\frac{1}{4}\cdot \frac{3}{29\frac{3}{4}}=-\frac{64}{119} \\ \end{array}\qquad \small \begin{array}{ccc|cc } x_1 & x_2 & x_3 & & \\ \hline4 & 0 & 0 & -\frac{64}{119} \\ 0 & -3 & -15 & -3 \\ 0 & 0 & 29\frac{3}{4} & 15\frac{1}{4} \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}2=\text{Zeile}2\text{-Zeile}3\cdot \frac{-15}{29\frac{3}{4}}\\z2s3=-15-29\frac{3}{4}\cdot \frac{-15}{29\frac{3}{4}}=0 \\ z2s4=-3-15\frac{1}{4}\cdot \frac{-15}{29\frac{3}{4}}=4\frac{82}{119} \\ \end{array}\qquad \small \begin{array}{ccc|cc } x_1 & x_2 & x_3 & & \\ \hline4 & 0 & 0 & -\frac{64}{119} \\ 0 & -3 & 0 & 4\frac{82}{119} \\ 0 & 0 & 29\frac{3}{4} & 15\frac{1}{4} \\ \end{array} \\ \\ x_1=\frac{-\frac{64}{119}}{4}=-\frac{16}{119}\\x_2=\frac{4\frac{82}{119}}{-3}=-1\frac{67}{119}\\x_3=\frac{15\frac{1}{4}}{29\frac{3}{4}}=\frac{61}{119}\\L=\{-\frac{16}{119}/-1\frac{67}{119}/\frac{61}{119}\} \end{array}$