Recursive Determinant
Just a few thoughts:


Your method (IIUC) is conceptually correct, but has computation complexity O(n!) where n is the dimension of a given matrix. If you need better complexity (polynomial in n), you have to use another solution, such as using PLU decomposition described here.
Be aware that since Haskell lists are essentially linked lists, getting i-th element using (!!) takes O(i). So your code

[r !! j | j &lt;- [0..n-1], j /= i] 


has O(n2) complexity. You could express it in O(n) for example using splitAs as

let (left, right) = splitAt i r
 in left ++ (tail right)


The same applies for i &lt;- [0..n-1], let x = (head mat) !! i. You could do something like

(i, x) &lt;- zip [0..] (head mat)


instead.