Invertierbare Matrix/Finden der inversen Matrix/37/45/Aufgabe/Lösung

${}{\begin{pmatrix}3&7\\4&5\end{pmatrix}}$	${}{\begin{pmatrix}1&0\\0&1\end{pmatrix}}$
${}{\begin{pmatrix}3&7\\0&-{\frac {13}{3}}\end{pmatrix}}$	${}{\begin{pmatrix}1&0\\-{\frac {4}{3}}&1\end{pmatrix}}$
${}{\begin{pmatrix}3&0\\0&-{\frac {13}{3}}\end{pmatrix}}$	${}{\begin{pmatrix}-{\frac {15}{13}}&{\frac {21}{13}}\\-{\frac {4}{3}}&1\end{pmatrix}}$
${}{\begin{pmatrix}1&0\\0&1\end{pmatrix}}$	${}{\begin{pmatrix}-{\frac {5}{13}}&{\frac {7}{13}}\\{\frac {4}{13}}&-{\frac {3}{13}}\end{pmatrix}}$

Die inverse Matrix ist also ${}{\begin{pmatrix}-{\frac {5}{13}}&{\frac {7}{13}}\\{\frac {4}{13}}&-{\frac {3}{13}}\end{pmatrix}}$.

Zur gelösten Aufgabe