How to prove or disprove matrix $A$ is invertible $\iff$ $\det{A}\ne 0$ if it's defined over field $F_p$
What's special about $A$ if it's defined over field $F_p$?
For normal procedure, we have $A$ is invertible $\iff$ $A$ is bijective $\iff$ $\ker{A}={0}$ $\iff$ rank $A$ is full $\iff$ $\det{A}\ne 0$.
Like Thomas says, for any field $F$, a matrix $A\in{\sf M}_{n\times n}(F)$ is invertible if and only if $\det(A)\neq 0$.
If $A$ is invertible, by using the property that $\det(AB)=\det(A)\cdot\det(B)$, $$1=\det(I_n)=\det(AA^{-1})=\det(A)\cdot\det(A^{-1}).$$ This gives $\det(A)\neq 0$. Conversely, if $A$ is not invertible, then the rows of $A$ is linearly dependent. Let the $i$th row of $A$ be obtained by a finite sequence of adding a multiple of row of $A$, then we may obtain $B\in{\sf M}_{n\times n}(F)$ such that the $i$th row of $B$ is a zero vector and other rows are those in $A$. Thus $$\det(A)=\det(B)=0. $$