Question

If there are two square matrices A & B of order nxn, and it is given that AB=I(n) where I is the identity matrix of nth order (n rows and n columns), does BA=I hold as well?

Options are:
A) Always holds
B) Never holds
C) Holds sometimes
D) Insufficient information

I am confused since we know that matrix multiplication is not always communicable. And we also know that in general, if AB=I then B is the inverse of A and so BA=I should hold. But this is again limited to cases only when A is a non singular matrix (since the inverse exists only then). Should the answer be holds sometimes?

Answer 1

Always holds

Explanation:

You are right: since `AB = I`, then we know that `B = A^{-1}`, but also `A = B^{-1}`.

So, we can rewrite

`AB = A A^{-1} = I`
`BA = BB^{-1} = I`

You are right when you say that a generic matrix `A` is not always invertible, but this doesn't change the answer: the statement is that if `AB= I`, then also `BA = I`.

What we are claiming is that the implication always holds, not that the hypotesis is always true.

A silly but hopefully useful example: if I say that the following implication always holds:

"If it rains, then I'll take my umbrella with me"

I'm not claiming that it always rains! But simply that everytime it rains, then you can bet that I'll have my umbrella.

Here is the same: we're not claiming that every matrix is invertible, but that everytime a matrix is invertible, then it commutes with its inverse.