# $A \in \mathbb{Z} ^{n \times n}$. Show that $\det(A) = \pm 1$. [duplicate]

Im stuck with this question because I'm probably misunderstanding it.

There is given that $$\mathbb{Z}^{n \times n} = \left\{A \in \mathbb{R}^{n \times n} \mid (A)_{ij} \in \mathbb{Z} \text{ for }i,j \in \left\{1,\cdots,n\right\}\right\}$$ I have to prove that for a matrice $A \in \mathbb{Z}^{n \times n}$ always: $$\det(A) = \pm1 \Longleftrightarrow A^{-1} \in \mathbb{Z}^{n \times n}$$ $A$ is invertible here.

I were thinking that if you take $A \in \mathbb{Z}^{n \times n}$ then $$A^{-1} = \frac{\operatorname{adj}(A)}{\det(A)}$$ Now if $\det(A)$ is different from $\{-1,1\}$ then you would get elements in $A^{-1}$ that are not in $\mathbb{Z}$. Or not because elements in $\operatorname{adj}(A)$ could be multiples of $\det(A)$. The question is: how do I know that $$\operatorname{adj}(A)_{ij} \neq c \cdot \det(A)$$ for all elements in $\operatorname{adj}(A)$ where $c$ is an integer?

If an integer-valued $A$ is invertible, and its inverse is also integer-valued, then $\det(A)$ and $\det(A^{-1})$ must be divisors of 1 (i.e, 1 or -1), as $\det(A) \det(A^{-1}) = \det(I) = 1.$
For a matrix $A\in GL_n(R)$ over a commutative ring $R$ the determinant $\det(A)$ is a unit in $R$. This has been proved at MSE. The units in $R=\mathbb{Z}$ are $\{\pm 1\}$.