Prove that if two matrices commute iff their inverse also commute Let A, B be two invertible matrices
es; prove that A and B 
commute if and only if $A^{−1}$ and $B^{−1}$ Commute.
Any suggestions? 
 A: Hint: Note that $(AB)(B^{-1}A^{-1}) = I$, so that $(AB)^{-1} = B^{-1}A^{-1}$.  Note that proving the implication in one direction is sufficient, since $(A^{-1})^{-1} = A$.
A: Suppose that $A$ and $B$ commute, where $A$ and $B$ are both invertible matrices. The $AB=BA$. Taking inverses, we have $(AB)^{-1} = (BA)^{-1}$, so $B^{-1}A^{-1} = A^{-1}B^{-1}$, by property of inverses. This tells you that $A^{-1}$ and $B^{-1}$ commute. The exact same argument works going the other way (you can actually do it one go with equivalences). 
A: Let $a,b$ be elements of an arbitrary group $G$. Then $ab=ba$ is equivalent to $a^{-1}b^{-1}=b^{-1}a^{-1}$, because it is equivalent to $(ab)^{-1}=(ba)^{-1}$, and
$(ab)^{-1}=b^{-1}a^{-1}$, $(ba)^{-1}=a^{-1}b^{-1}$.
Here we have the special case $G=GL_n(K)$, where $K$ probably is meant to be a field. It could be more general of course, e.g., $G=GL_n(\mathbb{Z})$.
A: Assuming you know that $$(AB)^{-1}=B^{-1}A^{-1},$$ then you get by interchanging the roles of $A,B$ that also $$(BA)^{-1}=A^{-1}B^{-1}.$$
Now the left hand sides of these equations are equal if and only if their right hand sides are equal. And inverses of two matrices are equal if and only if the matrices themselves are equal.
