Algebra-Lineares Gleichungssystem-Determinantenverfahren (2)

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Beispiel Nr: 10
$\begin{array}{l} D_h=\begin{array}{|cc|}a1\ & b1 \\ a2&b2 \\ \end{array}= a1 \cdot b2 -b1 \cdot a2 \\ D_x=\begin{array}{|cc|}c1\ & b1 \\ c2&b2 \\ \end{array}= c1 \cdot b2 -b1 \cdot c2 \\ D_y=\begin{array}{|cc|}a1\ & c1 \\ a2&c2 \\ \end{array}= a1 \cdot c2 -c1 \cdot a2\\ x=\frac{D_x}{D_h} \\ y=\frac{D_y}{D_h} \\ \\ \textbf{Gegeben:} \\ \\ -1 x +1 y =3\\ \frac{1}{2} x -4 y = 5 \\ \\ \\ \\ \textbf{Rechnung:} \\ D_h=\begin{array}{|cc|}-1\ & 1 \\ \frac{1}{2}&-4 \\ \end{array}= -1 \cdot \left(-4\right) -1 \cdot \frac{1}{2}=3\frac{1}{2} \\ D_x=\begin{array}{|cc|}3\ & 1 \\ 5&-4 \\ \end{array}= 3 \cdot \left(-4\right) -1 \cdot 5=-17 \\ D_y=\begin{array}{|cc|}-1\ & 3 \\ \frac{1}{2}&5 \\ \end{array}= -1 \cdot 5 -3 \cdot \frac{1}{2}=-6\frac{1}{2} \\ \ x=\frac{-17}{3\frac{1}{2}} \\ x=-4\frac{6}{7} \\ y=\frac{-6\frac{1}{2}}{3\frac{1}{2}} \\ y=-1\frac{6}{7} \\ L=\{-4\frac{6}{7}/-1\frac{6}{7}\}\\ \, \end{array}$