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$\begin{pmatrix}1&2&3&4&5&6\\ 4&3&1&5&2&6\end{pmatrix} \circ X = \begin{pmatrix}1&2&3&4&5&6\\ 3&2&4&5&1&6\end{pmatrix}$

I've tried this solution and checked this website, but none of them were helpful or my problem is little bit different and may require more steps to get the solution. Could anybody help me sort this out?

Millaray
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1 Answers1

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Hint: $$X=\begin{pmatrix}1&2&3&4&5&6\\ 4&3&1&5&2&6\end{pmatrix}^{-1}\circ \begin{pmatrix}1&2&3&4&5&6\\ 3&2&4&5&1&6\end{pmatrix}.$$

janmarqz
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    Thank you so much, that worked out! – Millaray Dec 13 '22 at 21:45
  • I'm glad to be inspiring, He-he : ) – janmarqz Dec 13 '22 at 21:58
  • How would you calculate following equation?:

    $\begin{pmatrix}1&2&3&4&5&6\ 3&1&5&2&4&6\end{pmatrix} \circ X \circ \begin{pmatrix}1&2&3&4&5&6\ 2&5&3&6&1&4\end{pmatrix} = \begin{pmatrix}1&2&3&4&5&6\ 1&2&4&3&6&5\end{pmatrix}$

    I tried to find inverse of the first two permutations, then followed hint shown above, but I didn't get expected output. What am I doing wrong?

     – Millaray Dec 14 '22 at 22:25
  • Being of the form $A\circ X\circ B=C$ then $X=A^{-1}\circ C\circ B^{-1}$. – janmarqz Dec 15 '22 at 00:56