Für eine invertierbare ${\displaystyle {}2\times 2}$-Matrix ${\displaystyle {}{\begin{pmatrix}a&b\\c&d\end{pmatrix}}}$ ist die inverse Matrix generell gleich

${\displaystyle {\frac {1}{ad-bc}}{\begin{pmatrix}d&-b\\-c&a\end{pmatrix}}.}$

Im vorliegenden Fall ist die Determinante gleich

${\displaystyle {}{\begin{aligned}{\frac {t}{t^{2}-1}}\cdot {\frac {t-1}{t+1}}-{\frac {1}{t^{3}}}\cdot {\frac {t^{2}-4}{t}}&={\frac {t}{(t+1)^{2}}}-{\frac {t^{2}-4}{t^{4}}}\\&={\frac {t^{5}-(t+1)^{2}(t^{2}-4)}{(t+1)^{2}t^{4}}}\\&={\frac {t^{5}-t^{4}-2t^{3}+3t^{2}+8t+4}{t^{6}+2t^{5}+t^{4}}}.\end{aligned}}}$

Somit ist die inverse Matrix gleich

${\displaystyle {\frac {t^{6}+2t^{5}+t^{4}}{t^{5}-t^{4}-2t^{3}+3t^{2}+8t+4}}{\begin{pmatrix}{\frac {t-1}{t+1}}&-{\frac {1}{t^{3}}}\\-{\frac {t^{2}-4}{t}}&{\frac {t}{t^{2}-1}}\end{pmatrix}}.}$