Here is the theorem that we are proving. Theorem. The following properties hold:

1. If B and C are inverses of A then B=C. Thus we can speak about the inverse of a matrix A, A-1.
2. If A is invertible and k is a non-zero scalar then kA is invertible and (kA)-1=1/k A-1.
3. 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

(An)-1=(A-1)n.

4. (AT)-1=(A-1)T, the inverse of the transpose is the transpose of the inverse.
5. If A is invertible then (A-1)-1=A. Proof. 1. Indeed if AB=I, CA=I then

B=I*B=(CA)B=C(AB)=C*I=C.

3. We need to prove that if A and B are invertible square matrices then B-1A-1 is the inverse of AB. Let us denote B-1A-1 by C (we always have to denote the things we are working with). Then by definition of the inverse we need to show that (AB)C=C(AB)=I. Substituting B-1A-1 for C we get:

(AB)(B-1A-1)=ABB-1A-1=A(BB-1)A-1=AIA-1=AA-1=I.

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

Other properties were left as exercises. 