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**Please, help, nasty algebra got me :((**

I've been going nowhere with induction and possible favorable arrangements for A (see below) and i'm desperate for help...

Let

[tex] M_{3}=\{ A\in \mathcal{M}_{3}\left(\mathbb{Z}\right)| \forall i,j=\overline{1,3}, \ a_{ij}\in \{\pm 1 \} \} [/tex]

Prove that

[tex] A^{2006} \neq 0_{3} , \forall A\in M_{3} [/tex].

I tried using complete induction, i.e. assuming A^k \neq 0_{3} and then trying to see whether A^(k+1) is 0_{3} or not. Of course, for this i have to assume that det A =0 (which fortunately is the case for a part of the 512 elements of M, for others, det A = \pm 4 and the proof is trivial). But this doesn't work, simply because A mixes plus & minus one and A^(k+1) could still be 0.

Then i tried using particular values for A. This lead me nowhere...The only relevant thing i found and hopefully is to be used is that if A\in M, then (-A) \in M, too.

This brought me to the following

[tex] -A^{2006}=(-A) A^{2005} [/tex]

[tex] A^{2006}=(-A) (-A)^{2005} [/tex]

But

[tex] A^{2005} \neq (-A)^{2005} \Leftrightarrow A^{2005} \neq 0_{3} [/tex]

So i can assume that the latter is satisfied. But in proving that [itex] A^{2006} \neq 0_{3} [/itex] from the arguments above leads me to this issue

For matrices with det equal to 0

[tex] BA=C [/tex]

[tex] EA=D [/tex]

I know that [itex] B\neq E [/itex]. However, det of A =0. Can i still prove that [itex] C\neq D[/itex] ??

Daniel.

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