Invertierbare Matrix/Finden der inversen Matrix/41/53/Aufgabe/Lösung

${}{\begin{pmatrix}4&1\\5&3\end{pmatrix}}$	${}{\begin{pmatrix}1&0\\0&1\end{pmatrix}}$
${}{\begin{pmatrix}4&1\\0&{\frac {7}{4}}\end{pmatrix}}$	${}{\begin{pmatrix}1&0\\-{\frac {5}{4}}&1\end{pmatrix}}$
${}{\begin{pmatrix}4&0\\0&{\frac {7}{4}}\end{pmatrix}}$	${}{\begin{pmatrix}{\frac {12}{7}}&-{\frac {4}{7}}\\-{\frac {5}{4}}&1\end{pmatrix}}$
${}{\begin{pmatrix}1&0\\0&1\end{pmatrix}}$	${}{\begin{pmatrix}{\frac {3}{7}}&-{\frac {1}{7}}\\-{\frac {5}{7}}&{\frac {4}{7}}\end{pmatrix}}$

Die inverse Matrix ist also ${}{\begin{pmatrix}{\frac {3}{7}}&-{\frac {1}{7}}\\-{\frac {5}{7}}&{\frac {4}{7}}\end{pmatrix}}$.

Zur gelösten Aufgabe