# Invertible matrices over a commutative ring and their determinants

Why is it true that a matrix $A \in \operatorname{Mat}_n(R)$, where $R$ is a commutative ring, is invertible iff its determinant is invertible?

Since $\det(A)I_n = A\operatorname{adj}(A) = \operatorname{adj}(A)A$ then I can see why the determinant being invertible implies the inverse exists, since the adjoint always exists, but I can't see why it's true the other way around.

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In essence $$AB = I \implies \det(A)\det(B) = 1$$ so that $\det(A)$ and $\det(B)$ are units.