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If $A$ is a given matrix and $B$ is a non-singular matrix(both of order $n$),then show that there exists one and only one matrix $X$ such that $A=BX$ and one and only one matrix $Y$ such that $A=YB$.

I don't know how to prove the above. I know that rank if a matrix remains same after pre-multiplication or post multiplication by a non-singular matrix. Does this fact help at all in proving the above? Or is there some other way?

Please anyone help me solve it. Thanks in advance.

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If $B$ is non-singular, it means that $B$ is invertible: there exists $B^{-1}$ with $B^{-1}B=BB^{-1}=I$. If you multiply $A=BX$ by $B^{-1}$ on the left, you get $B^{-1}A=X$. That's the only possible value of $X$. Similar with the other equation.

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  • $\begingroup$ Ohh...got it. Thank you. $\endgroup$ – user587389 Dec 11 '19 at 5:37

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