A is invertible iff there exists a matrix B so that AB = BA = I. It is simple to show that: (i) If A is invertible and AB = I, then BA = I as well and B is the unique such matrix. (ii) If A is invertible and BA = I, then AB = I as well and B is the unique such matrix. This shows that if A is invertible, then there is a unique matrix B such that AB = I or BA = I. Call this unique matrix A−1 . The goal here is to show that the assumption “A is invertible” is not needed in (i) or (ii). Prove: Let A and B be square matrices with AB = I. Show that A is invertible and hence B = A−1 . You may refer to Theorem 1.5.2 or Theorem 2.2.2, but be clear and complete in your argument.

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