The determinant and the permanent of a matrix, though deceivingly similar in their definitions, behave very differently with respect to how efficiently one can compute these quantities. The determinant of a matrix over a field can be easily computed via Gaussian elimination while computing the permanent, as shown by Valiant, is at least as hard as counting the number of satisfiable assignments to a Boolean formula. Given this, it is natural to ask ``over which algebras, is the determinant easier to compute than the permament?" Furthermore, since all algorithms for determinant crucially use commutivity of the underlying algebra, we could ask ``is commutativity essential for efficient determinant computation?"
Extending the recent result of Arvind and Srinivasan [STOC 2010], we show that computing the determinant of an $n\times n$ matrix whose entries are themselves $2\times 2$ matrices over a field is as hard as computing the permanent over the field. On the other hand, surprisingly if one restricts the elements to be $d\times d$ upper triangular matrices, then determinant can be computed in $poly(n^d)$ time. Combining this with the decomposition theorem of finite algebras, we get the following dichotomy result: if $A$ is a constant dimensional algebra over a finite field of odd characteristic, then the commutativity of the quotient algebra $A/R(A)$ determines efficient determinant computation (where $R(A)$ is the radical of $A$).