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Let $A$ and $B$ be invertible $n \times n$- matrices and $C$ be an $n \times n$- matrix with $\det C =1$. Prove that $\det A = \det B$ if and only if $A=CB$.

I've got the proof backward but I got stuck on the forward. I got some hints like expressing those matrices with elementary matrices, but I failed to find anything useful.

I just don't know how to prove "if $\det A = \det B$ then $A=CB$".

ps.thanks for all clarifications and that was just my understanding. This is the original question and I'm also confused about it.

Let $A$ and $B$ be invertible $n \times n$- matrices. Prove that $\det A = \det B$ if and only if $A=CB$, where $C$ is an $n \times n$- matrix such that $\det C =1$.

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    $\begingroup$ It is not true when $det A = det B = 0$. But for invertible matrices, clearly $C = AB^{-1}$ works. $\endgroup$ – hardmath Jul 14 '15 at 21:55
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    $\begingroup$ @OP do you have to prove this assertion $\mathrm{det}(AB) = \mathrm{det}(A)\mathrm{det}(B)$ or can you assume that? $\endgroup$ – Chinny84 Jul 14 '15 at 21:57
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    $\begingroup$ I disapprove of this way of using the word "where". That word should explain notation or conventions, thus $abc = def$ where $a$ is the temperature and $b$ is the price of eggs, etc. In this case it appears that "where $C$ is a $n\times n$ matrix" means "FOR SOME $n\times n$ matrix $C$. Whether it means "for some" or "for every" is something the reader may fail to know. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jul 14 '15 at 22:15
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    $\begingroup$ The Statement is obviously false, it implies that there is only one Matrix C with determinant 1 $\endgroup$ – john Jul 14 '15 at 22:47
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    $\begingroup$ @MaggieMak: The correct statement, which was more closely approximated by the original Question, is that if $\det A = \det B \neq 0$ (for matrices $A,B$ of equal dimension), then there exists a matrix $C$ with $\det C = 1$ such that $A = CB$. In the usual understanding, you have specified matrices $A,B,C$ and then predicate a relation $A=CB$ only on the equality of determinants $\det A = \det B$, which is clearly unsupportable. $\endgroup$ – hardmath Jul 14 '15 at 23:54
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Define $C:=AB^{-1}$, which is possible since $B$ is invertible. Then

$$CB=(AB^{-1})B=A(B^{-1}B)=AI=A$$ and

$$\det(C)=\det(A)\det(B^{-1})=\frac{\det(A)}{\det(B)}=1$$

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  • $\begingroup$ Er..what are you proving? I don't think we can assume A=CB $\endgroup$ – Maggie Mak Jul 14 '15 at 22:01
  • $\begingroup$ @Mathgemini: Use $DetADetA^{-1}=DetId=1$ to show $DetC=1$. $\endgroup$ – Gary. Jul 14 '15 at 22:05
  • $\begingroup$ OK, if $C=AB^{-1}$, what can we say about $CB$ ?? $\endgroup$ – GEdgar Jul 14 '15 at 22:16
  • $\begingroup$ @GEdgar He literally answers that question in his answer. $\endgroup$ – user223391 Jul 14 '15 at 22:25
  • $\begingroup$ @avid: thanks. that was in an edit after Maggie doubted it, and after I asked her what can she say about $CB$. $\endgroup$ – GEdgar Jul 14 '15 at 22:28

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