Λύση

Εύκολα βρίσκουμε ότι $\vert A\vert=2$. Από την άλλη έχουμε $\vertΑ_{11}\vert=+\left\vert\begin{array}{cc} 3 & 4\\ 5 & 7 \end{array}\right\vert=1$, $\vertΑ_{12}\vert=-\left\vert\begin{array}{cc} 2 & 4\\ 1 & 7 \end{array}\right\vert=-10$, $\vertΑ_{13}\vert=+\left\vert\begin{array}{cc} 2 & 3\\ 1 & 5 \end{array}\right\vert=7$, $\vertΑ_{21}\vert=-\left\vert\begin{array}{cc} 2 & 3\\ 5 & 7 \end{array}\right\vert=1$, $\vertΑ_{22}\vert=+\left\vert\begin{array}{cc} 1 & 3\\ 1 & 7 \end{array}\right\vert=4$, $\vertΑ_{23}\vert=-\left\vert\begin{array}{cc} 1 & 2\\ 1 & 5 \end{array}\right\vert=-3$, $\vertΑ_{31}\vert=+\left\vert\begin{array}{cc} 2 & 3\\ 3 & 4 \end{array}\right\vert=-1$, $\vertΑ_{32}\vert=-\left\vert\begin{array}{cc} 1 & 3\\ 2 & 4 \end{array}\right\vert=2$, $\vertΑ_{33}\vert=+\left\vert\begin{array}{cc} 1 & 2\\ 2 & 3 \end{array}\right\vert=-1$. Άρα,

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