Algebra-Lineare Algebra-Determinante

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
$Determinante$
1 2
Beispiel Nr: 01
$\begin{array}{l} \\ \begin{array} \text{Gegeben:} \\ \text{Determinante von der quadratischen Matrix:} \\ \left| \begin{array}{cccc} a_{11} & a_{12} & \ldots & a_{1n}\\ a_{21} & a_{22} & \ldots & a_{2n}\\ \vdots & \vdots &\vdots & \vdots \\ a_{m1} & a_{m2} & \ldots & a_{mn}\\ \end{array} \right| \end{array} \\ \textbf{Aufgabe:}\\a\\ \textbf{Rechnung:}\\ \small D_4=\left| \begin{array}{cccc} 1&-2&6&-1\\2&0&3&2\\0&3&2&0\\5&4&1&1\end{array} \right|=1\cdot\left| \begin{array}{ccc} 0&3&2\\3&2&0\\4&1&1\end{array} \right|-2\cdot\left| \begin{array}{ccc} -2&6&-1\\3&2&0\\4&1&1\end{array} \right|-5\cdot\left| \begin{array}{ccc} -2&6&-1\\0&3&2\\3&2&0\end{array} \right|=-250\\ \small D_3=\left| \begin{array}{ccc} -2&6&-1\\0&3&2\\3&2&0\end{array} \right|=(-2)\cdot\left| \begin{array}{cc} 3&2\\2&0\end{array} \right|+3\cdot\left| \begin{array}{cc} 6&-1\\3&2\end{array} \right|=53\\ \small D_2=\left| \begin{array}{cc} 6&-1\\3&2\end{array} \right|=6\cdot2-3\cdot(-1)=15\\ \small D_2=\left| \begin{array}{cc} 3&2\\2&0\end{array} \right|=3\cdot0-2\cdot2=-4\\ \small D_3=\left| \begin{array}{ccc} -2&6&-1\\3&2&0\\4&1&1\end{array} \right|=(-2)\cdot\left| \begin{array}{cc} 2&0\\1&1\end{array} \right|-3\cdot\left| \begin{array}{cc} 6&-1\\1&1\end{array} \right|+4\cdot\left| \begin{array}{cc} 6&-1\\2&0\end{array} \right|=-17\\ \small D_2=\left| \begin{array}{cc} 6&-1\\2&0\end{array} \right|=6\cdot0-2\cdot(-1)=2\\ \small D_2=\left| \begin{array}{cc} 6&-1\\1&1\end{array} \right|=6\cdot1-1\cdot(-1)=7\\ \small D_2=\left| \begin{array}{cc} 2&0\\1&1\end{array} \right|=2\cdot1-1\cdot0=2\\ \small D_3=\left| \begin{array}{ccc} 0&3&2\\3&2&0\\4&1&1\end{array} \right|=-3\cdot\left| \begin{array}{cc} 3&2\\1&1\end{array} \right|+4\cdot\left| \begin{array}{cc} 3&2\\2&0\end{array} \right|=-19\\ \small D_2=\left| \begin{array}{cc} 3&2\\2&0\end{array} \right|=3\cdot0-2\cdot2=-4\\ \small D_2=\left| \begin{array}{cc} 3&2\\1&1\end{array} \right|=3\cdot1-1\cdot2=1\\det(D)=(-250)\\ \end{array}$