Determinante/Multiplikation/6/Aufgabe/Lösung

Es ist

${}{\begin{aligned}\det {\begin{pmatrix}{\frac {2}{3}}&2&-1\\0&0&-1\\1&{\frac {1}{2}}&1\end{pmatrix}}&={\frac {2}{3}}\cdot {\frac {1}{2}}-2\\&=-{\frac {5}{3}}\end{aligned}}$

und

${}{\begin{aligned}\det {\begin{pmatrix}2&4&-5\\1&1&0\\0&2&-{\frac {1}{3}}\end{pmatrix}}&=-2\cdot {\frac {1}{3}}-1{\left(-4\cdot {\frac {1}{3}}+10\right)}\\&={\frac {-2-26}{3}}\\&=-{\frac {28}{3}}.\end{aligned}}$

Es ist

${}{\begin{pmatrix}{\frac {2}{3}}&2&-1\\0&0&-1\\1&{\frac {1}{2}}&1\end{pmatrix}}\cdot {\begin{pmatrix}2&4&-5\\1&1&0\\0&2&-{\frac {1}{3}}\end{pmatrix}}={\begin{pmatrix}{\frac {10}{3}}&{\frac {8}{3}}&-3\\0&-2&{\frac {1}{3}}\\{\frac {5}{2}}&{\frac {13}{2}}&-{\frac {16}{3}}\end{pmatrix}}\,.$

Die Determinante davon ist

${}{\begin{aligned}\det {\begin{pmatrix}{\frac {10}{3}}&{\frac {8}{3}}&-3\\0&-2&{\frac {1}{3}}\\{\frac {5}{2}}&{\frac {13}{2}}&-{\frac {16}{3}}\end{pmatrix}}&={\frac {10}{3}}{\left({\frac {32}{3}}-{\frac {13}{6}}\right)}+{\frac {5}{2}}{\left({\frac {8}{9}}-6\right)}\\&={\frac {10}{3}}\cdot {\frac {51}{6}}-{\frac {5}{2}}\cdot {\frac {46}{9}}\\&={\frac {5}{3}}\cdot 17-5\cdot {\frac {23}{9}}\\&={\frac {255-115}{9}}\\&={\frac {140}{9}},\end{aligned}}$

was mit dem Produkt der beiden Determinanten übereinstimmt.

Zur gelösten Aufgabe