# Explain why the determinant of $A$ must be $1$ or $-1$

Assume $$A$$ is an $$n\times n$$ matrix where every entry is an integer.

Suppose $$A$$ is invertible, and that every entry in $$A^{-1}$$ is also an integer. Why must $$\det(A)$$ be only $$1$$ or $$-1$$?

It's clear that any matrix $$A$$ consisting only of integers will produce a determinant that is an integer, but I am unsure of how to show why that integer can only be $$1$$ or $$-1$$ in this case. Is there a specific equation I should consider? Any push in the right direction is appreciated.

• Hint:$\;\det(A)\det(A^{-1})=\;??$. – quasi Oct 27 '18 at 22:37
• If $ab = 1$ and $a, b$ are both integers, then both $a, b$ are $\pm 1$. – Joppy Oct 27 '18 at 22:37

$$1=\det I = \det (A\cdot A^{-1}) = \det A \cdot \det A^{-1}.$$
And by the assumption, both $$\det A$$ and $$\det A^{-1}$$ are integers.
Hence, $$\det A$$ should be $$1$$ or $$-1$$.