R^2/Dualbasis zu (1,3),(2,-5)/Standarddualbasis/Aufgabe/Lösung

Wir invertieren die Matrix ${}{\begin{pmatrix}1&2\\3&-5\end{pmatrix}}$.

${}{\begin{pmatrix}1&2\\3&-5\end{pmatrix}}$	${}{\begin{pmatrix}1&0\\0&1\end{pmatrix}}$
${}{\begin{pmatrix}1&2\\0&-11\end{pmatrix}}$	${}{\begin{pmatrix}1&0\\-3&1\end{pmatrix}}$
${}{\begin{pmatrix}1&2\\0&1\end{pmatrix}}$	${}{\begin{pmatrix}1&0\\{\frac {3}{11}}&-{\frac {1}{11}}\end{pmatrix}}$
${}{\begin{pmatrix}1&0\\0&1\end{pmatrix}}$	${}{\begin{pmatrix}{\frac {5}{11}}&{\frac {2}{11}}\\{\frac {3}{11}}&-{\frac {1}{11}}\end{pmatrix}}$

Daher ist

${}u_{1}^{*}={\frac {5}{11}}e_{1}^{*}+{\frac {2}{11}}e_{2}^{*}\,$

und

${}u_{2}^{*}={\frac {3}{11}}e_{1}^{*}-{\frac {1}{11}}e_{2}^{*}\,.$

Zur gelösten Aufgabe