Algebra-Lineare Algebra-Determinante



Beispiel Nr: 02
$ \\ \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)\\ $