Lösung.

Mit

$ \mbox{$\displaystyle \left( \begin{array}{rrrrr} 1 & 5 & -3 & 2 & -1 \\ 0... ...\ -1 & 2 & 0 & 0 & 0 \\ 0 & 2 & 0 & 1 & 0 \\ \end{array}\right)^{-1} $}$
erhalten wir
$ \mbox{$\displaystyle \begin{array}{rcl} \exp\left( t\,\left( \begin{array}{rr... ... 2t & 4t & -2t & 0 & -2t + 2 \\ \end{array}\right) \;\; . \\ \end{array}$}$