Algebra-Lineares Gleichungssystem-Determinantenverfahren (2)

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Beispiel Nr: 20
$\begin{array}{l} D_h=\begin{array}{|cc|}a1\ & b1 \\ a2&b2 \\ \end{array}= a1 \cdot b2 -b1 \cdot a2 \\ D_x=\begin{array}{|cc|}c1\ & b1 \\ c2&b2 \\ \end{array}= c1 \cdot b2 -b1 \cdot c2 \\ D_y=\begin{array}{|cc|}a1\ & c1 \\ a2&c2 \\ \end{array}= a1 \cdot c2 -c1 \cdot a2\\ x=\frac{D_x}{D_h} \\ y=\frac{D_y}{D_h} \\ \\ \textbf{Gegeben:} \\ \\ -1\frac{4}{5} x +1\frac{1}{3} y =-1\\ -\frac{2}{3} x +\frac{1}{9} y = 9 \\ \\ \\ \\ \textbf{Rechnung:} \\ D_h=\begin{array}{|cc|}-1\frac{4}{5}\ & 1\frac{1}{3} \\ -\frac{2}{3}&\frac{1}{9} \\ \end{array}= -1\frac{4}{5} \cdot \frac{1}{9} -1\frac{1}{3} \cdot \left(-\frac{2}{3}\right)=\frac{31}{45} \\ D_x=\begin{array}{|cc|}-1\ & 1\frac{1}{3} \\ 9&\frac{1}{9} \\ \end{array}= -1 \cdot \frac{1}{9} -1\frac{1}{3} \cdot 9=-12\frac{1}{9} \\ D_y=\begin{array}{|cc|}-1\frac{4}{5}\ & -1 \\ -\frac{2}{3}&9 \\ \end{array}= -1\frac{4}{5} \cdot 9 -\left(-1\right) \cdot \left(-\frac{2}{3}\right)=-16\frac{13}{15} \\ \ x=\frac{-12\frac{1}{9}}{\frac{31}{45}} \\ x=-17\frac{18}{31} \\ y=\frac{-16\frac{13}{15}}{\frac{31}{45}} \\ y=-24\frac{15}{31} \\ L=\{-17\frac{18}{31}/-24\frac{15}{31}\}\\ \, \end{array}$