Λύση

Εύκολα βρίσκουμε ότι $\vert B\vert=-2$. Από την άλλη έχουμε $\vert B_{11}\vert=+\left\vert\begin{array}{cc} 3 & 4\\ 8 & 9 \end{array}\right\vert=-5$, $\vert B_{12}\vert=-\left\vert\begin{array}{cc} 2 & 4\\ 5 & 9 \end{array}\right\vert=2$, $\vert B_{13}\vert=+\left\vert\begin{array}{cc} 2 & 3\\ 5 & 8 \end{array}\right\vert=1$, $\vert B_{21}\vert=-\left\vert\begin{array}{cc} 1 & 1\\ 8 & 9 \end{array}\right\vert=-1$, $\vert B_{22}\vert=+\left\vert\begin{array}{cc} 1 & 1\\ 5 & 9 \end{array}\right\vert=4$, $\vert B_{23}\vert=-\left\vert\begin{array}{cc} 1 & 1\\ 5 & 8 \end{array}\right\vert=-3$, $\vert B_{31}\vert=+\left\vert\begin{array}{cc} 1 & 1\\ 3 & 4 \end{array}\right\vert=1$, $\vert B_{32}\vert=-\left\vert\begin{array}{cc} 1 & 1\\ 2 & 4 \end{array}\right\vert=-2$, $\vert B_{33}\vert=+\left\vert\begin{array}{cc} 1 & 1\\ 2 & 3 \end{array}\right\vert=1$. Άρα,

$$B^{-1}=\frac{1}{-22}\left(\begin{array}{ccc} -5 & -1 & 1\\ ... ... -\frac{1}{2} & \frac{3}{2} & -\frac{1}{2} \end{array}\right).$$