Which is easier to work out: determinant or inverse? 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 $\mathbb{G}$.   
 A: 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.
A: Determinate comes first, and generally easier to figure out by definition not matter which R we are considering. (special situation exists when R is a field other than an principal ideal domain thus every nonsingular matrix has inverse)
A matrix has inverse only if the determine of the matrix is a unit x (have inverse) in the ring R, and its inverse has determine $x^{-1}$ in R.
When R is not p.i.d.(principal ideal domain), like $Z_n[G]$, then the concept of determinate and inverse is not well-defined, as the normal form of a group does not exist.
A: Recently, I realize that the computing determinant might be harder than computing the inverse. Since if the matrices are defined over a non-commutative ring $R$, the computing $\det(A)$ for $A\in M_n(R)$ seems very hard. However, let $d=|M_n(R)|=|R|^{n^2}$. That is, the semigroup $M_n(R)$ with respect to matrix multiplication has order $d$. Thus, $A^{d-1}=A^{-1}$, which can be efficiently worked out by employing the so-called ``square-and-multiply" method. 
