Algebra-Lineare Algebra-Determinante
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$Determinante$
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Beispiel Nr: 02
$\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:}\\ b\\ \textbf{Rechnung:}\\
\small D_4=\left| \begin{array}{cccc}
-4&-2&5&1\\3&3&3&2\\12&-2&3&4\\5&4&-4&1\end{array} \right|=(-4)\cdot\left| \begin{array}{ccc}
3&3&2\\-2&3&4\\4&-4&1\end{array} \right|-3\cdot\left| \begin{array}{ccc}
-2&5&1\\-2&3&4\\4&-4&1\end{array} \right|+12\cdot\left| \begin{array}{ccc}
-2&5&1\\3&3&2\\4&-4&1\end{array} \right|-5\cdot\left| \begin{array}{ccc}
-2&5&1\\3&3&2\\-2&3&4\end{array} \right|=-423\\
\small D_3=\left| \begin{array}{ccc}
-2&5&1\\3&3&2\\-2&3&4\end{array} \right|=(-2)\cdot\left| \begin{array}{cc}
3&2\\3&4\end{array} \right|-3\cdot\left| \begin{array}{cc}
5&1\\3&4\end{array} \right|+(-2)\cdot\left| \begin{array}{cc}
5&1\\3&2\end{array} \right|=-77\\
\small D_2=\left| \begin{array}{cc}
5&1\\3&2\end{array} \right|=5\cdot2-3\cdot1=7\\
\small D_2=\left| \begin{array}{cc}
5&1\\3&4\end{array} \right|=5\cdot4-3\cdot1=17\\
\small D_2=\left| \begin{array}{cc}
3&2\\3&4\end{array} \right|=3\cdot4-3\cdot2=6\\
\small D_3=\left| \begin{array}{ccc}
-2&5&1\\3&3&2\\4&-4&1\end{array} \right|=(-2)\cdot\left| \begin{array}{cc}
3&2\\-4&1\end{array} \right|-3\cdot\left| \begin{array}{cc}
5&1\\-4&1\end{array} \right|+4\cdot\left| \begin{array}{cc}
5&1\\3&2\end{array} \right|=-21\\
\small D_2=\left| \begin{array}{cc}
5&1\\3&2\end{array} \right|=5\cdot2-3\cdot1=7\\
\small D_2=\left| \begin{array}{cc}
5&1\\-4&1\end{array} \right|=5\cdot1-(-4)\cdot1=9\\
\small D_2=\left| \begin{array}{cc}
3&2\\-4&1\end{array} \right|=3\cdot1-(-4)\cdot2=11\\
\small D_3=\left| \begin{array}{ccc}
-2&5&1\\-2&3&4\\4&-4&1\end{array} \right|=(-2)\cdot\left| \begin{array}{cc}
3&4\\-4&1\end{array} \right|-(-2)\cdot\left| \begin{array}{cc}
5&1\\-4&1\end{array} \right|+4\cdot\left| \begin{array}{cc}
5&1\\3&4\end{array} \right|=48\\
\small D_2=\left| \begin{array}{cc}
5&1\\3&4\end{array} \right|=5\cdot4-3\cdot1=17\\
\small D_2=\left| \begin{array}{cc}
5&1\\-4&1\end{array} \right|=5\cdot1-(-4)\cdot1=9\\
\small D_2=\left| \begin{array}{cc}
3&4\\-4&1\end{array} \right|=3\cdot1-(-4)\cdot4=19\\
\small D_3=\left| \begin{array}{ccc}
3&3&2\\-2&3&4\\4&-4&1\end{array} \right|=3\cdot\left| \begin{array}{cc}
3&4\\-4&1\end{array} \right|-(-2)\cdot\left| \begin{array}{cc}
3&2\\-4&1\end{array} \right|+4\cdot\left| \begin{array}{cc}
3&2\\3&4\end{array} \right|=103\\
\small D_2=\left| \begin{array}{cc}
3&2\\3&4\end{array} \right|=3\cdot4-3\cdot2=6\\
\small D_2=\left| \begin{array}{cc}
3&2\\-4&1\end{array} \right|=3\cdot1-(-4)\cdot2=11\\
\small D_2=\left| \begin{array}{cc}
3&4\\-4&1\end{array} \right|=3\cdot1-(-4)\cdot4=19\\det(D)=(-423)\\ \end{array}$