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I have a given matrix A and it is multiplied with a matrix S. The result is also given. How do I get 3x3 matrix S?

$$ A= \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 6 & 7 & 8 & 9 & 10\\ 0 & 0 & 1 & 0 & 0 \\ \end{pmatrix} $$ $$ SA= \begin{pmatrix} 6 & 7 & 0 & 9 & 10 \\ 1 & 2 & 0 & 4 & 5\\ 0 & 0 & 1 & 0 & 0 \\ \end{pmatrix} $$

As from the look of the matrix it has to be something obvious, but I just don't know what I should do.

What are the steps to solve this kind of problems?

Maximilian Kindshofer
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    Well you know that $S\pmatrix{1 \ 6 \ 0} = \pmatrix{6 \ 1 \ 0}$ and that $S\pmatrix{2 \ 7 \ 0} = \pmatrix{7 \ 2 \ 0}$ and that $\dots$ –  Oct 08 '15 at 13:04
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    Take only three linearly independent columns of $A$ (you get an invertible matrix) and the corresponding columns of $SA$. You have then to solve $SX=Y$, with square $S,X,Y$ and invertible $X$: multiply on the right by $X^{-1}$. Of course, it's annoying if you can't find three independent columns in $A$, because then the solution is not unique. – Jean-Claude Arbaut Oct 08 '15 at 13:10
  • Ok I have tried something like this using the falksches schema and using the S as variable but I made a mistake which lead me of track – Maximilian Kindshofer Oct 08 '15 at 13:19

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The first two rows are $(0,1,-8) $ and $(1,0,-3) $. The third one you can find out.

tattwamasi amrutam
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