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I'm having trouble figuring out how to find matrices such that when I multiply on either side of a given matrix, I get some desired matrix out of it. For example using matrix $A$ find matrix $X$ and $Y$ such that $XAY=B$.

$$ A= \begin{bmatrix} a&b&0&c\\ d&e&f&0\\ 0&r&s&t\\ u&0&v&w\\ \end{bmatrix} \qquad\qquad B=\begin{bmatrix} v&w&u&0\\ s&t&0&r\\ f&0&d&e\\ 0&c&a&b\\ \end{bmatrix} $$

I'm running in circles trying to find the correct matrix. Is there an easier way to do this other than just brute forcing it?

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  • $\begingroup$ Do mean mean "find matrices $X$ and $Y$ such that $XAY$ results in the second matrix"? It's much more common to take a matrix $A$ and transform in on the left by one matrix and on the right by another. $\endgroup$ – Mike Pierce Sep 28 '16 at 3:04
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Notice that

$$ \begin{bmatrix} 0&0&0&1\\ 0&0&1&0\\ 0&1&0&0\\ 1&0&0&0\\ \end{bmatrix} \begin{bmatrix} a&b&0&c\\ d&e&f&0\\ 0&r&s&t\\ u&0&v&w\\ \end{bmatrix} \begin{bmatrix} 0&0&1&0\\ 0&0&0&1\\ 1&0&0&0\\ 0&1&0&0\\ \end{bmatrix} = \begin{bmatrix} v&w&u&0\\ s&t&0&r\\ f&0&d&e\\ 0&c&a&b\\ \end{bmatrix}\;. $$

Does this look familiar? Can you tell me why this is true?

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  • $\begingroup$ I've forgotten all my matrix identities and operations so it really doesn't. Is this somehow related to column and row selection? The first matrix's first row selects the fourth row of A. Second row selects the third and so on... $\endgroup$ – Sammy Sep 28 '16 at 3:56
  • $\begingroup$ Well then this is a good learning opportunity. :) Just verify the equality above. First multiply by the matrix on the left, then by the one on right, and be sure to pay attention to what each matrix is doing. $\endgroup$ – Mike Pierce Sep 28 '16 at 4:21

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