Algebra-Lineares Gleichungssystem-Determinantenverfahren (2)
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Beispiel Nr: 14
$\begin{array}{l}
D_h=\begin{array}{|cc|}a1\ & b1 \\ a2&b2 \\ \end{array}=
a1 \cdot b2 -b1 \cdot a2 \\
D_x=\begin{array}{|cc|}c1\ & b1 \\ c2&b2 \\ \end{array}=
c1 \cdot b2 -b1 \cdot c2 \\
D_y=\begin{array}{|cc|}a1\ & c1 \\ a2&c2 \\ \end{array}=
a1 \cdot c2 -c1 \cdot a2\\
x=\frac{D_x}{D_h} \\
y=\frac{D_y}{D_h}
\\ \\ \textbf{Gegeben:} \\
\\
2 x +3 y =4\\
\frac{1}{3} x -\frac{1}{5} y = 12 \\
\\
\\ \\ \textbf{Rechnung:} \\
D_h=\begin{array}{|cc|}2\ & 3 \\ \frac{1}{3}&-\frac{1}{5} \\ \end{array}=
2 \cdot \left(-\frac{1}{5}\right) -3 \cdot \frac{1}{3}=-1\frac{2}{5} \\
D_x=\begin{array}{|cc|}4\ & 3 \\ 12&-\frac{1}{5} \\ \end{array}=
4 \cdot \left(-\frac{1}{5}\right) -3 \cdot 12=-36\frac{4}{5} \\
D_y=\begin{array}{|cc|}2\ & 4 \\ \frac{1}{3}&12 \\ \end{array}=
2 \cdot 12 -4 \cdot \frac{1}{3}=22\frac{2}{3} \\
\
x=\frac{-36\frac{4}{5}}{-1\frac{2}{5}} \\
x=26\frac{2}{7} \\
y=\frac{22\frac{2}{3}}{-1\frac{2}{5}} \\
y=-16\frac{4}{21} \\
L=\{26\frac{2}{7}/-16\frac{4}{21}\}\\
\,
\end{array}$