Impossible hill cipher question
Two independent ideas:

In the Hill cipher, elements of matrices are in the ring of integers modulo the number of characters in the alphabet used, usually $n=26$. Therefore, as long as the denominator of a fraction is coprime with $n$ (that is, for $n=26$, if the denominator is divisible neither by $2$ not by $13$), that fraction reduces to an integer modulo $n$.E.g. $\displaystyle\frac23\equiv18\pmod{26}$ since $2\equiv3\cdot18\pmod{26}$.
The Hill cipher is sometime instantiated with a $3\times3$ or larger matrix.