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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

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The determinant of the identity is always the multiplicative identity of the underlying ring and consequently the multiplicative property of the determinant implies that the determinant of an invertible matrix is itself invertible.

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

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