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The concrete problem is this: Find triplets of distinct matrices $(A,B,C)$ of dimension $6\times 6$ over the field $\mathbb{F}_{2^2}$ such that:

  1. $A^2B=AB^2$
  2. $C^2A=CA^2$
  3. $B^3C=BC^3$

However, I'm also interested in how this can be done in a more general setting at least with other sets of equations; for now I can assume I always need triplets of square matrices of small dimension over a small finite field.

If it can be done with a well-known computer algebra system (esp. SAGE) it will be great.

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Do you have any objection against taking $A=B=C$? –  Marc van Leeuwen Apr 5 '12 at 13:08
    
Yes, I should have said so explicitly. –  Gadi A Apr 5 '12 at 13:22
    
    
You do realize that if you want the matrices to be invertible, then the first equation implies $A=B$, the second implies $C=A$, and the third implies that $B^2=C^2$, which is actually a consequence of the two earlier ones. If you don't insist that the matrices should be invertible, then the easiest way is to follow azimut. –  Jyrki Lahtonen Aug 21 '13 at 9:59

1 Answer 1

The following three matrices give a solution to your initial problem: $$ A = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix},\\ B = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix},\\ C = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix}. $$

Many more solutions can be built in a similar way.

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+1 and a huuuuge grin :-D –  Jyrki Lahtonen Aug 21 '13 at 9:55

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