«K12/K13» Gauß-Algorithmus «PDF», «POD»

0.0.1 ↑ Der Gauß-Algorithmus

\begin{array}{rrr|l} {} 10 & 1 & -2 & 2 \\ {} 1 & 2 & 2 & 3 \\ {} 4 & 4 & 3 & 5 \\ \\ {} 1 & 2 & 2 & 3 \\ {} 10 & 1 & -2 & 2 \\ {} 4 & 4 & 3 & 5 \\ \\ {} 1 & 2 & 2 & 3 \\ {} 0 & -19 & -22 & -28 \\ {} 4 & 4 & 3 & 5 \\ \\ {} 1 & 2 & 2 & 3 \\ {} 0 & -19 & -22 & -28 \\ {} 0 & -4 & -5 & -7 \\ \\ {} 1 & 2 & 2 & 3 \\ {} 0 & -4 & -5 & -7 \\ {} 0 & -19 & -22 & -28 \\ \\ {} 1 & 2 & 2 & 3 \\ {} 0 & 1 & \frac{5}{4} & \frac{7}{4} \\ {} 0 & -19 & -22 & -28 \\ \\ {} 1 & 2 & 2 & 3 \\ {} 0 & 1 & \frac{5}{4} & \frac{7}{4} \\ {} 0 & 0 & \frac{7}{4} & \frac{21}{4} \\ \\ {} 1 & 2 & 2 & 3 \\ {} 0 & 1 & \frac{5}{4} & \frac{7}{4} \\ {} 0 & 0 & 1 & 3 \\ \end{array}$\begin{array}{cccc}\hfill 10& \hfill 1& \hfill -2& 2\hfill \\ \hfill 1& \hfill 2& \hfill 2& 3\hfill \\ \hfill 4& \hfill 4& \hfill 3& 5\hfill \\ \hfill \\ \hfill 1& \hfill 2& \hfill 2& 3\hfill \\ \hfill 10& \hfill 1& \hfill -2& 2\hfill \\ \hfill 4& \hfill 4& \hfill 3& 5\hfill \\ \hfill \\ \hfill 1& \hfill 2& \hfill 2& 3\hfill \\ \hfill 0& \hfill -19& \hfill -22& -28\hfill \\ \hfill 4& \hfill 4& \hfill 3& 5\hfill \\ \hfill \\ \hfill 1& \hfill 2& \hfill 2& 3\hfill \\ \hfill 0& \hfill -19& \hfill -22& -28\hfill \\ \hfill 0& \hfill -4& \hfill -5& -7\hfill \\ \hfill \\ \hfill 1& \hfill 2& \hfill 2& 3\hfill \\ \hfill 0& \hfill -4& \hfill -5& -7\hfill \\ \hfill 0& \hfill -19& \hfill -22& -28\hfill \\ \hfill \\ \hfill 1& \hfill 2& \hfill 2& 3\hfill \\ \hfill 0& \hfill 1& \hfill \frac{5}{4}& \frac{7}{4}\hfill \\ \hfill 0& \hfill -19& \hfill -22& -28\hfill \\ \hfill \\ \hfill 1& \hfill 2& \hfill 2& 3\hfill \\ \hfill 0& \hfill 1& \hfill \frac{5}{4}& \frac{7}{4}\hfill \\ \hfill 0& \hfill 0& \hfill \frac{7}{4}& \frac{21}{4}\hfill \\ \hfill \\ \hfill 1& \hfill 2& \hfill 2& 3\hfill \\ \hfill 0& \hfill 1& \hfill \frac{5}{4}& \frac{7}{4}\hfill \\ \hfill 0& \hfill 0& \hfill 1& 3\hfill \\ \hfill \end{array}$

x_1 = 3 - 2x_2 - 2x_3 = 1;$x_1=3-2x_2-2x_3=1;$

x_2 = \frac{7}{4} - \frac{5}{4}x_3 = -2;$x_2=\frac{7}{4}-\frac{5}{4}{x}_3=-2;$

x_3 = 3;$x_3=3;$