Prove that the determinant of an invertible matrix $A$ is equal to $±1$ when all of the entries of $A$ and $A^{−1}$ are integers.

Question

Prove that the determinant of an invertible matrix $A$ is equal to $±1$ when all of the entries of $A$ and $A^{−1}$ are integers.

I can explain the answer but would like help translating it into a concise proof with equations.

My explanation:

The fact that $\det(A) = ±1$ implies that when we perform Gaussian elimination on $A$, we never have to multiply rows by scalars. This means that for each column, the pivot entry is created by the previous column’s row operations and can be brought into place by swapping rows. (And the first column must already contain a $1$). Therefore, we never need to multiply by a non-integral value to perform Gaussian elimination.

Answer 1

Let $A\in\mathbb Z^{n\times n}$ such that $A^{-1}\in\mathbb Z^{n\times n}$. Note that the determinant of an integer matrix is an integer, so $\det\colon\mathbb Z^{n\times n}\to \mathbb Z$. Now, $1=\det(\mathbb I)=\det(A\cdot A^{-1})=\det(A)\cdot\det(A^{-1})$. Since both $\det(A)$ and $\det(A^{-1})$ are integers, they can only be $1$ or $-1$.