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Suppose A is a matrix over some ring R (might be non-commutative). How to work out the inverse matrix $A^{-1}$?

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Consider $A$ as a homomorphism of a free $R$-module. If this homomorphism is invertible, then you can build $A^{-1}$ which is a matrix of the inverse homomorphism.

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How is this different from just restating the question? – Marc van Leeuwen Aug 12 '13 at 4:52
@Marc van Leeuwen: Of course, it is simply translation into the language of homomorphisms. I think OP ask just this. – Boris Novikov Aug 12 '13 at 6:29

Should you be so lucky that $R$ is commutative, then iff the determinant of $A$ is invertible in $R$, you can take the adjugate matrix and multiply it with the inverse of the determinant squared, see the section named "inverse". This might be possible to apply to some matrices over some non-commutative rings too, I have little experience with those.

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