# Basiswechsel/Übergangsmatrix/Standardbasis und 237/1-34/569/Aufgabe/Lösung

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Es ist

${\displaystyle {}M_{\mathfrak {u}}^{\mathfrak {v}}={\begin{pmatrix}2&1&5\\3&-3&6\\7&4&9\end{pmatrix}}\,.}$

Für die umgekehrte Übergangsmatrix müssen wir diese Matrix invertieren. Es ist

${\displaystyle {}{\begin{pmatrix}2&1&5\\3&-3&6\\7&4&9\end{pmatrix}}}$ ${\displaystyle {}{\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}}}$
${\displaystyle {}{\begin{pmatrix}2&1&5\\0&-{\frac {9}{2}}&-{\frac {3}{2}}\\0&{\frac {1}{2}}&-{\frac {17}{2}}\end{pmatrix}}}$ ${\displaystyle {}{\begin{pmatrix}1&0&0\\-{\frac {3}{2}}&1&0\\-{\frac {7}{2}}&0&1\end{pmatrix}}}$
${\displaystyle {}{\begin{pmatrix}2&1&5\\0&-{\frac {9}{2}}&-{\frac {3}{2}}\\0&0&-{\frac {26}{3}}\end{pmatrix}}}$ ${\displaystyle {}{\begin{pmatrix}1&0&0\\-{\frac {3}{2}}&1&0\\-{\frac {11}{3}}&{\frac {1}{9}}&1\end{pmatrix}}}$
${\displaystyle {}{\begin{pmatrix}1&{\frac {1}{2}}&{\frac {5}{2}}\\0&1&{\frac {1}{3}}\\0&0&1\end{pmatrix}}}$ ${\displaystyle {}{\begin{pmatrix}{\frac {1}{2}}&0&0\\{\frac {1}{3}}&-{\frac {2}{9}}&0\\{\frac {11}{26}}&-{\frac {1}{78}}&-{\frac {3}{26}}\end{pmatrix}}}$
${\displaystyle {}{\begin{pmatrix}1&0&{\frac {7}{3}}\\0&1&{\frac {1}{3}}\\0&0&1\end{pmatrix}}}$ ${\displaystyle {}{\begin{pmatrix}{\frac {1}{3}}&{\frac {1}{9}}&0\\{\frac {1}{3}}&-{\frac {2}{9}}&0\\{\frac {11}{26}}&-{\frac {1}{78}}&-{\frac {3}{26}}\end{pmatrix}}}$
${\displaystyle {}{\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}}}$ ${\displaystyle {}{\begin{pmatrix}-{\frac {17}{26}}&{\frac {11}{78}}&{\frac {7}{26}}\\{\frac {5}{26}}&-{\frac {17}{78}}&{\frac {1}{26}}\\{\frac {11}{26}}&-{\frac {1}{78}}&-{\frac {3}{26}}\end{pmatrix}}}$

Es ist also

${\displaystyle {}M_{\mathfrak {v}}^{\mathfrak {u}}={\begin{pmatrix}-{\frac {17}{26}}&{\frac {11}{78}}&{\frac {7}{26}}\\{\frac {5}{26}}&-{\frac {17}{78}}&{\frac {1}{26}}\\{\frac {11}{26}}&-{\frac {1}{78}}&-{\frac {3}{26}}\end{pmatrix}}\,.}$