Algebra-Lineare Algebra-Matrix
$Matrix$
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Beispiel Nr: 03
$\begin{array}{l} \\
\text{ Gegeben:} \\
\begin{array}{c}
Matrix A \\
\left[ \begin{array}{cccc}
a_{11} & a_{12} & \ldots & a_{1n}\\
a_{21} & a_{22} & \ldots & a_{2n}\\
\vdots & \vdots &\vdots & \vdots \\
a_{m1} & a_{m2} & \ldots & a_{mn}\\
\end{array} \right] \\
Matrix B \\
\left[ \begin{array}{cccc}
b_{11} & b_{12} & \ldots & b_{1n}\\
b_{21} & b_{22} & \ldots & b_{2n}\\
\vdots & \vdots &\vdots & \vdots \\
b_{m1} & b_{m2} & \ldots & b_{mn}\\
\end{array} \right]
\end{array}
\\ \textbf{Aufgabe:}\\ Mutliplizieren\\ \textbf{Rechnung:}\\ A= \small \left[ \begin{array}{cc}
3 & 5 \\
6 & 7 \\
\end{array} \right]
\\det(A)=(-9)\Rightarrow \text{Matrix ist invertierbar} \\\\\text{Matrix invertieren}\\ \small \left[ \begin{array}{cc}
3 & 5 \\
6 & 7 \\
\end{array} \right]
\small \left[ \begin{array}{cc}
1 & 0 \\
0 & 1 \\
\end{array} \right]
\\\tiny \begin{array}{l} \text{Zeile}2=\text{Zeile}2\text{ - Zeile}1\cdot \frac{6}{3}\\a21=6-3\cdot \frac{6}{3}=0\\a22=7-5\cdot \frac{6}{3}=-3\\b21=0-1\cdot \frac{6}{3}=0\\b22=1-0\cdot \frac{6}{3}=1\\ \end{array}\qquad \\ \small \left[ \begin{array}{cc}
3 & 5 \\
0 & -3 \\
\end{array} \right]
\small \left[ \begin{array}{cc}
1 & 0 \\
-2 & 1 \\
\end{array} \right]
\\\tiny \begin{array}{l} \text{Zeile}1=\text{Zeile}1\text{ - Zeile}2\cdot \frac{5}{-3}\\a12=5-(-3)\cdot \frac{5}{-3}=0\\b11=1-(-2)\cdot \frac{5}{-3}=1\\b12=0-1\cdot \frac{5}{-3}=0\\ \end{array}\qquad \\ \small \left[ \begin{array}{cc}
3 & 0 \\
0 & -3 \\
\end{array} \right]
\small \left[ \begin{array}{cc}
-2\frac{1}{3} & 1\frac{2}{3} \\
-2 & 1 \\
\end{array} \right]
\\\tiny \begin{array}{l}\text{Zeile}1=\text{ Zeile}1 : 3\\\text{Zeile}2=\text{ Zeile}2 : -3\\ \end{array} \\A^{-1}= \small \left[ \begin{array}{cc}
-\frac{7}{9} & \frac{5}{9} \\
\frac{2}{3} & -\frac{1}{3} \\
\end{array} \right]
\end{array}$