Prove or Disprove. If A, B and C are n x n matrices such that A is invertible and AB = CA, then B = C. I know this is false because matrix multiplication is not commutative, but I can't think of any counterexamples.
 A: Basically, you want to compare $B$ and $ A^{-1}CA$; these two matrices are different in general case. As an example, take $C=\begin{pmatrix}1&0\\0&2\end{pmatrix}$, take a nontrivial invertible $A$ and find $  A^{-1}CA$.
A: You are correct, and it's not so hard to find an example - say
$$\left[\begin{matrix}1&0\\1&1\end{matrix}\right]\left[\begin{matrix}1&1\\1&0\end{matrix}\right]=\left[\begin{matrix}0&1\\1&1\end{matrix}\right]\left[\begin{matrix}1&0\\1&1\end{matrix}\right]$$
In general playing around with $2\times2$ matrices should net you a solution pretty quickly - lots of matrices don't commute.
A: You can say $$AB=CA \implies A^{-1}AB=A^{-1}CA \implies A^{-1}ABA^{-1}=A^{-1}CAA^{-1}$$
and so $BA^{-1}=A^{-1}C$. Using the original statement we have:
$$B^2=A^{-1}C^2A$$
and
$$AB^2A^{-1}=C^2$$
but this is about as far as it goes.
A: You know there exist $A,B$ with $BA\neq AB$, and it certainly is possible to have $A$ invertible in such a pair. Then for such a pair multiply the inequation on the right by $A^{-1}$ to obtain $B\neq ABA^{-1}$, so taking $C=ABA^{-1}$ gives you a counterexample.
