Suppose we have an invertible matrix (Over $F$) $A$ with entries over $F$, say, where $F$ is any field of characteristic $k \neq 0$. The first question is:
Is it $A$ always invertible over $\mathbb{Q}$? If so...
If the inverse of $A$ over $\mathbb{Q}$ is $A^{-1}_{\mathbb{Q}}$, and $A^{-1}_{\mathbb{Q}}$ is in the form $A^{-1}_{\mathbb{Q}} = (a_{ij})$,
Is it true that $A_{F}^{-1} = (a_{ij} \ \ mod \ k) $?
Example as it seems what I am asking is unclear: Let $$A= \begin{pmatrix} 1 & 0 & 2 \\ 2 & 2 & 0\\ 1 & 0 & 0 \end{pmatrix}, $$ with entries over $F_5$, the finite field of characteristic $5$. Now this matrix is invertible over $F_{5}$ and $$A_{F_5}^{-1}= \begin{pmatrix} 0 & 0 & 1 \\ 0 & 3 & 4\\ 3 & 0 & 2 \end{pmatrix}.$$ And if we consider the same matrix but we inverse it over $\mathbb{Q}$, we get $$A_{\mathbb{Q}}^{-1}= \begin{pmatrix} 0 & 0 & 1 \\ 0 & \frac{1}{2} & -1 \\ \frac{1}{2} & 0 & -\frac{1}{2} \end{pmatrix}. $$
But considering $\frac{1}{2}_{F_5}=(2)^{-1}_{F_5}=3$ and $-1=4$, it is clear that the result holds, i.e. $$A_{F_5}^{-1} = A_{\mathbb{Q}}^{-1} \ mod \ 5 . $$
Is this always true for invertible matrices over the finite field?
Look at the adjugate matrix.
$M_n(R)$ is the ring of $n \times n$ matrices with coefficients in a commutative ring $R$.
If $A \in M_n(\mathbb{Z})$ and $A_p \in M_n(\mathbb{F}_p),\ A_p \equiv A \bmod p$ then $$A\ \text{adj}(A) = \det(A) I \qquad \text{in } M_n(\mathbb{Z})$$ Whose reduction $\bmod p$ of each term is $$A_p\ \text{adj}(A_p) = \det(A_p) I \qquad \text{in } M_n(\mathbb{F}_p)$$
If $\det(A_p) \not\equiv 0 \bmod p$ then $\det(A) \ne 0$ and $$A^{-1} = \frac{1}{\det(A)}\text{adj}(A)\qquad \text{in } M_n(\mathbb{Q})$$ whose reduction $\bmod p$ of each term is $$ \qquad A_p^{-1} =\frac{1}{\det(A_p)}\text{adj}(A_p) \qquad \text{in } M_n(\mathbb{F}_p)$$