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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 $\det(A)I_n = A\operatorname{adj}(A)$ 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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1 Answer 1

up vote 9 down vote accepted

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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