If $A$ and $B$ commute so do $A$ and $B^{-1}$

Theorem 1.

Let $A$ and $B$ be commuting matrices.If $B$ is invertible,then $A$ and $B^{\mathrm{-}\mathrm{1}}$ commute,and if $A$ and $B$ are invertible, then $A^{\mathrm{-}\mathrm{1}}$ and $B^{\mathrm{-}\mathrm{1}}$ commute.

Proof.

By assumption

$$AB=BA,$$

multiplying from the left and from the right by $B^{-1}$ yields

$$B^{-1}A=A{B}^{-1}.$$

The second claim follows similarly.∎

The statement and proof of this result can obviously be extended to elements of any monoid. In particular, in the case of a group, we see that two elements commute if and only if their inverses do.