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$\vert A\vert=2$. $\vert_{11}\vert=+\left\vert\begin{array}{cc} 3 & 4\\ 5 & 7 \end{array}\right\vert=1$, $\vert_{12}\vert=-\left\vert\begin{array}{cc} 2 & 4\\ 1 & 7 \end{array}\right\vert=-10$, $\vert_{13}\vert=+\left\vert\begin{array}{cc} 2 & 3\\ 1 & 5 \end{array}\right\vert=7$, $\vert_{21}\vert=-\left\vert\begin{array}{cc} 2 & 3\\ 5 & 7 \end{array}\right\vert=1$, $\vert_{22}\vert=+\left\vert\begin{array}{cc} 1 & 3\\ 1 & 7 \end{array}\right\vert=4$, $\vert_{23}\vert=-\left\vert\begin{array}{cc} 1 & 2\\ 1 & 5 \end{array}\right\vert=-3$, $\vert_{31}\vert=+\left\vert\begin{array}{cc} 2 & 3\\ 3 & 4 \end{array}\right\vert=-1$, $\vert_{32}\vert=-\left\vert\begin{array}{cc} 1 & 3\\ 2 & 4 \end{array}\right\vert=2$, $\vert_{33}\vert=+\left\vert\begin{array}{cc} 1 & 2\\ 2 & 3 \end{array}\right\vert=-1$. ,

\begin{displaymath}A^{-1}=\frac{1}{2}\left(\begin{array}{ccc}
1 & 1 & -1\\
-1...
...
\frac{7}{2} & -\frac{3}{2} & -\frac{1}{2}
\end{array}\right).\end{displaymath}



Vassilis Metaftsis
1999-09-15