Inverse Matrix/3/Explizit/Aufgabe/Lösung

Es ist

{}{\begin{aligned}{\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}}{\begin{pmatrix}ei-fh&ch-bi&bf-ce\\fg-di&ai-cg&cd-af\\dh-eg&bg-ah&ae-bd\end{pmatrix}}&={\begin{pmatrix}a(ei-fh)+b(fg-di)+c(dh-eg)&a(ch-bi)+b(ai-cg)+c(bg-ah)&a(bf-ce)+b(cd-af)+c(ae-bd)\\d(ei-fh)+e(fg-di)+f(dh-eg)&d(ch-bi)+e(ai-cg)+f(bg-ah)&d(bf-ce)+e(cd-af)+f(ae-bd)\\g(ei-fh)+h(fg-di)+i(dh-eg)&g(ch-bi)+h(ai-cg)+i(bg-ah)&g(bf-ce)+h(cd-af)+i(ae-bd)\end{pmatrix}}\\&={\begin{pmatrix}aei-afh+bfg-bdi+cdh-ceg&ach-abi+bai-bcg+cbg-cah&abf-ace+bcd-baf+cae-cbd\\dei-dfh+efg-edi+fdh-feg&dch-dbi+eai-ecg+fbg-fah&dbf-dce+ecd-eaf+fae-fbd\\gei-gfh+hfg-hdi+idh-ieg&gch-gbi+hai-hcg+ibg-iah&gbf-gce+hcd-haf+iae-ibd\end{pmatrix}}\\&={\begin{pmatrix}aei-afh+bfg-bdi+cdh-ceg&0&0\\0&dch-dbi+eai-ecg+fbg-fah&0\\0&0&gbf-gce+hcd-haf+iae-ibd\end{pmatrix}}.\,\end{aligned}}

In der Diagonalen steht immer der gleiche Eintrag, nämlich

{}aei-afh+bfg-bdi+cdh-ceg=a(ei-fh)-b(di-fg)+c(dh-eg)=\det M\,.

Mit dem Vorfaktor

{}{\frac {1}{\det M}}

ergibt sich also bei Multiplikation die Einheitsmatrix.

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