TITLE: Which is easier to work out: determinant or inverse?
QUESTION [6 upvotes]: Suppose $A\in M_n(R)$ be a $n\times n$ matrix over some ring $R$. Which of the following two tasks is easier?

to work out $\det(A)$;
to work out $A^{-1}$. 

More specifically, I want to know the answers according to the following different settings of $R$: 

$R$ is commutative;
$R$ is non-commutative.
$R$ is ring group $\mathbb{Z}_n[\mathbb{G}]$ for (1) commutative group $\mathbb{G}$, (2) non-commutative group $\mathbb{G}$.

REPLY [2 votes]: For the case of commutative rings, from a complexity theoretic viewpoint, computing the inverse of a matrix, computing the determinant of a matrix, and computing a matrix multiplication all have the same complexity. This was shown in the 70s by Strassen, of the well-known Strassen multiplication. That gives a complexity of $O(n^{2.81})$ for the above operations; the current record is, I believe, $O(n^{2.38})$. This is, by the way, complexity counted in operations on $R$ and (maybe) some logarithmic factors are being ignored.
As mentioned in the comments, for the other 2 cases the notion of determinant doesn't directly make sense, so that needs more clarification.