Theorem. The following properties hold:

- If
*B*and*C*are inverses of*A*then*B=C*. Thus we can speak about**the**inverse of a matrix*A*,*A*.^{-1} - If
*A*is invertible and*k*is a non-zero scalar then*kA*is invertible and*(kA)*.^{-1}=1/k A^{-1} - If
*A*and*B*are invertible then*AB*is invertible and*(AB)*^{-1}=B^{-1}A^{-1}that is the inverse of the product is the product of inverses in the opposite order. In particular

*(A*.^{n})^{-1}=(A^{-1})^{n} -
*(A*, the inverse of the transpose is the transpose of the inverse.^{T})^{-1}=(A^{-1})^{T} - If
*A*is invertible then*(A*.^{-1})^{-1}=A

Proof.
1. Indeed if *AB*=**I**, *CA*=**I** then

3. We need to prove that if *A* and *B* are invertible square matrices then
*B ^{-1}A^{-1}* is the inverse of

We used the
associativity of the product of matrices, the definition of
an inverse
and the fact that **I***A*=*A***I**=*A* for every matrix *A*.

Other properties were left as exercises.