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

^{-1}M=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))^{-1}adj(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).