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


  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


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:


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.