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).