Prove the following: If $A,B,C$ are all $n \times n$ matrices where $AB=CA$ and $A$ is invertible, then $B=C$

Here is my attempt at the solution, but I'm stuck

Let $D$ be the $n \times n$ inverse matrix of $A$, then $AD = I = DA$. Then,

$B = (I)B = (DA)B= D(AB) = A(CA)$

After that, I'm stuck. What should I do next?

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    $\begingroup$ The claim is not true in general. $\endgroup$ – Peter Mar 14 '18 at 18:02
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    $\begingroup$ Your last step should be $A^{-1}CA$ $\endgroup$ – Nick Mar 14 '18 at 18:04
  • $\begingroup$ @Peter The questions is a prove or disprove, is there a counter example to disprove this? $\endgroup$ – Mark Dodds Mar 14 '18 at 18:05
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    $\begingroup$ A hint at coming up with a counterexample. Pick $A$ and $C$ to be invertible matrices. Then let $B=A^{-1}CA$. $\endgroup$ – JMoravitz Mar 14 '18 at 18:05
  • $\begingroup$ @Nobody it is true in more cases than that. So long as $A$ commutes with both $B$ and $C$. $\endgroup$ – JMoravitz Mar 14 '18 at 18:10

This is wrong:


where $\left(\matrix{0&1\\1&0}\right)$ is invertible, is valid for any $a$, $b$, $c$, $d$.

What is true is that if $AB=CA=\operatorname{Id}_n$, then $B=C$ (the left inverse is the right inverse).


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