Let $$A$$ be a square matrix all of whose entries are integers. Then which one of the following is true?

For any square matrix $$A$$ with integer entries, write its inverse by the classical adjugate formula: $$A^{-1}= \dfrac{\operatorname{adj}(A)}{\det A}\,.$$

Step 1. Show that $$\operatorname{adj}(A)$$ has only integer entries.
Each entry of $$\operatorname{adj}(A)$$ is a cofactor, i.e. the determinant of an $$(n-1)\times(n-1)$$ sub-matrix of $$A$$, possibly multiplied by $$\pm1$$. Because every entry of $$A$$ is an integer, every such determinant is an integer. Hence every entry of $$\operatorname{adj}(A)$$ is an integer.

Step 2. Investigate the role of $$\det A$$.
The inverse exists iff $$\det A \neq 0$$. When $$\det A = \pm1$$, we are dividing an integer matrix $$\operatorname{adj}(A)$$ by $$\pm1$$, which does not change the integrality of the entries. Therefore every entry of $$A^{-1}$$ is also an integer.

Step 3. Check the options.
• Option A is wrong because the entries of $$A^{-1}$$ are in fact guaranteed to be integers when $$\det A=\pm1$$.
• Option B is wrong because (i) $$\det A$$ might be $$0$$ so the inverse may fail to exist, and (ii) even when the inverse exists, some entries of $$A^{-1}$$ can still be integers (e.g. $$\begin{pmatrix}2&0\\0&2\end{pmatrix}^{-1}=\begin{pmatrix}\tfrac12&0\\0&\tfrac12\end{pmatrix}$$ has zeros, which are integers).
• Option C correctly states that the inverse exists (since $$\det A\ne0$$) and contains only integer entries.
• Option D is wrong because a non-zero determinant always guarantees the existence of an inverse.

Hence the correct statement is:

Option C which is: If $$\det A = \pm 1$$, then $$A^{-1}$$ exists and all its entries are integers.