Algebra-Lineare Algebra-Determinante
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$Determinante$
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Beispiel Nr: 01
$\begin{array}{l} \\
\begin{array}
\text{Gegeben:} \\
\text{Determinante von der quadratischen Matrix:} \\
\left| \begin{array}{cccc}
a_{11} & a_{12} & \ldots & a_{1n}\\
a_{21} & a_{22} & \ldots & a_{2n}\\
\vdots & \vdots &\vdots & \vdots \\
a_{m1} & a_{m2} & \ldots & a_{mn}\\
\end{array} \right|
\end{array}
\\ \textbf{Aufgabe:}\\a\\ \textbf{Rechnung:}\\
\small D_4=\left| \begin{array}{cccc}
1&-2&6&-1\\2&0&3&2\\0&3&2&0\\5&4&1&1\end{array} \right|=1\cdot\left| \begin{array}{ccc}
0&3&2\\3&2&0\\4&1&1\end{array} \right|-2\cdot\left| \begin{array}{ccc}
-2&6&-1\\3&2&0\\4&1&1\end{array} \right|-5\cdot\left| \begin{array}{ccc}
-2&6&-1\\0&3&2\\3&2&0\end{array} \right|=-250\\
\small D_3=\left| \begin{array}{ccc}
-2&6&-1\\0&3&2\\3&2&0\end{array} \right|=(-2)\cdot\left| \begin{array}{cc}
3&2\\2&0\end{array} \right|+3\cdot\left| \begin{array}{cc}
6&-1\\3&2\end{array} \right|=53\\
\small D_2=\left| \begin{array}{cc}
6&-1\\3&2\end{array} \right|=6\cdot2-3\cdot(-1)=15\\
\small D_2=\left| \begin{array}{cc}
3&2\\2&0\end{array} \right|=3\cdot0-2\cdot2=-4\\
\small D_3=\left| \begin{array}{ccc}
-2&6&-1\\3&2&0\\4&1&1\end{array} \right|=(-2)\cdot\left| \begin{array}{cc}
2&0\\1&1\end{array} \right|-3\cdot\left| \begin{array}{cc}
6&-1\\1&1\end{array} \right|+4\cdot\left| \begin{array}{cc}
6&-1\\2&0\end{array} \right|=-17\\
\small D_2=\left| \begin{array}{cc}
6&-1\\2&0\end{array} \right|=6\cdot0-2\cdot(-1)=2\\
\small D_2=\left| \begin{array}{cc}
6&-1\\1&1\end{array} \right|=6\cdot1-1\cdot(-1)=7\\
\small D_2=\left| \begin{array}{cc}
2&0\\1&1\end{array} \right|=2\cdot1-1\cdot0=2\\
\small D_3=\left| \begin{array}{ccc}
0&3&2\\3&2&0\\4&1&1\end{array} \right|=-3\cdot\left| \begin{array}{cc}
3&2\\1&1\end{array} \right|+4\cdot\left| \begin{array}{cc}
3&2\\2&0\end{array} \right|=-19\\
\small D_2=\left| \begin{array}{cc}
3&2\\2&0\end{array} \right|=3\cdot0-2\cdot2=-4\\
\small D_2=\left| \begin{array}{cc}
3&2\\1&1\end{array} \right|=3\cdot1-1\cdot2=1\\det(D)=(-250)\\ \end{array}$