Let be R a commutative ring with 1 and M a n x n matrix over R.
There exists an inverse M-1 of M with M-1M=M M-1 = I, iff the determinant of M is an unit in R.

Proof: We have the law det(M) I = M adj(M) = adj(M) M, where adj(M) is the adjoint matrix of M.
If det(M) is an unit, then (det(M))-1adj(M) is obviously the inverse of M.
On the other hand if M has an inverse M-1, then 1 = det(I) = det(M M-1) = det(M) det(M-1) = det(M-1) det(M). Therefore det(M) has the inverse det(M-1). Example:
| 1 2 |
| 1 3 |
has in inverse 2 x 2 matrix over Z (the ring of integers).