# Invertible Matrices over a commutative ring and their determinants.

Why is it true that a matrix $A \in Mat_n(R)$ where R is a commutative ring is invertible iff it's determinant is invertible? Since $det(A)A^{-1} = adj(A)$ 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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@ navigetor23 I meant it in the sense that $adj(A)A = Aadj(A) = det(A)I$. If you consider the set of matrices over the field of fractions of $R$ then $A^{-1}$ then does exist and is in the set of matrices over this field. (I think) –  Tom Oldfield Oct 27 '12 at 0:05
Not necessarily, so I admit the field of fractions thing won't always be true, but the first thing about it being a kind of "scaled" inverse will still hold. –  Tom Oldfield Oct 27 '12 at 17:40
In essence $$AB = I \implies \det(A)\det(B) = 1$$ so that $\det(A)$ and $\det(B)$ are units.