$\begin{array}{l} \\ \text{ Gegeben:} \\ \begin{array}{c} Matrix A \\ \left[ \begin{array}{cccc} a_{11} & a_{12} & \ldots & a_{1n}\\ a_{21} & a_{22} & \ldots & a_{2n}\\ \vdots & \vdots &\vdots & \vdots \\ a_{m1} & a_{m2} & \ldots & a_{mn}\\ \end{array} \right] \\ Matrix B \\ \left[ \begin{array}{cccc} b_{11} & b_{12} & \ldots & b_{1n}\\ b_{21} & b_{22} & \ldots & b_{2n}\\ \vdots & \vdots &\vdots & \vdots \\ b_{m1} & b_{m2} & \ldots & b_{mn}\\ \end{array} \right] \end{array} \\ \textbf{Aufgabe:}\\Invertieren\\ \textbf{Rechnung:}\\ A= \small \left[ \begin{array}{cc} 3 & 5 \\ 6 & 7 \\ \end{array} \right] \\det(A)=(-9)\Rightarrow \text{Matrix ist invertierbar} \\\\\text{Matrix invertieren}\\ \small \left[ \begin{array}{cc} 3 & 5 \\ 6 & 7 \\ \end{array} \right] \small \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right] \\\tiny \begin{array}{l} \text{Zeile}2=\text{Zeile}2\text{ - Zeile}1\cdot \frac{6}{3}\\a21=6-3\cdot \frac{6}{3}=0\\a22=7-5\cdot \frac{6}{3}=-3\\b21=0-1\cdot \frac{6}{3}=0\\b22=1-0\cdot \frac{6}{3}=1\\ \end{array}\qquad \\ \small \left[ \begin{array}{cc} 3 & 5 \\ 0 & -3 \\ \end{array} \right] \small \left[ \begin{array}{cc} 1 & 0 \\ -2 & 1 \\ \end{array} \right] \\\tiny \begin{array}{l} \text{Zeile}1=\text{Zeile}1\text{ - Zeile}2\cdot \frac{5}{-3}\\a12=5-(-3)\cdot \frac{5}{-3}=0\\b11=1-(-2)\cdot \frac{5}{-3}=1\\b12=0-1\cdot \frac{5}{-3}=0\\ \end{array}\qquad \\ \small \left[ \begin{array}{cc} 3 & 0 \\ 0 & -3 \\ \end{array} \right] \small \left[ \begin{array}{cc} -2\frac{1}{3} & 1\frac{2}{3} \\ -2 & 1 \\ \end{array} \right] \\\tiny \begin{array}{l}\text{Zeile}1=\text{ Zeile}1 : 3\\\text{Zeile}2=\text{ Zeile}2 : -3\\ \end{array} \\A^{-1}= \small \left[ \begin{array}{cc} -\frac{7}{9} & \frac{5}{9} \\ \frac{2}{3} & -\frac{1}{3} \\ \end{array} \right] \end{array}$