To be more specific than the title:
Let $A$ (an $n\times n$ matrix) be over $\mathbb Q$, so of course $A$ is over $\mathbb R$ too. Now, how can I prove that if $A$ is invertible over $\mathbb R$, it's also invertible over $\mathbb Q$?
I'm not sure that what I've tried is enough to prove it:
If $A$ is invertible over $\mathbb R$ so the $\det(A)\neq 0$ so its also not equal to zero over $\mathbb Q$.
The entries of $A^{-1}$ are obtained from the entries of $A$ by a lot of multiplications and additions of these entries (which is "harmless", i.e., will never take us out of $\Bbb Q$), and then dividing by $\det A$. We already know that $\det A\ne 0$, so this division is possible. But also $\det A$ is obtained from the entries of $A$ by harmless additions and multiplications, so $\det A\in\Bbb Q$ in our case. Of course, division by a (non-zero) element of $\Bbb Q$ still cannot take us out of $\Bbb Q$. So indeed all entries of $A^{-1}$ are rational.
The argument generalizes: if all entries of an invertible $A$ are in some ring (e.g., $\Bbb Z$), then all entries of $\det(A)\cdot A^{-1}$ are also in that ring.