If A and B are square matrices, and AB=I, then I think it is also true that BA=I. In fact, this Wikipedia page says that this “follows from the associativity of matrix multiplication”. I assume there’s a nice simple one-line proof, but can’t seem to find it.

Nothing exotic, here — assume that the matrices have finite size and their elements are real numbers.

This isn’t homework (if that matters to you). My last homework assignment was about 50 years ago.

**Answer**

Since AB=I then B=B(AB)=(BA)B. Note from AB=I that 1=det so \det(B)\neq0.

So by (BA)B=B we have:

(BA-I)B=0. Since \det(B)\neq0 then B is not a 0 divisor. So BA=I

**Attribution***Source : Link , Question Author : bubba , Answer Author : user71352*