Hello and thanks in advance for taking the time to read this. I have following problem (not homework or anything, personal interest - I will be organzing a course on the Banach-Tarski paradox next week and need this as part of the proof):
We have the two matrices $$ r_x^{+\alpha} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \frac{3}{5} & -\frac{4}{5} \\ 0 & \frac{4}{5} & \frac{3}{5} \end{pmatrix}$$ and
$$ r_z^{+\alpha} = \begin{pmatrix} \frac{3}{5} & -\frac{4}{5} & 0 \\ \frac{4}{5} & \frac{3}{5} & 0 \\ 0 & 0 & 1 \end{pmatrix}$$
and their inverses, $$ r_x^{-\alpha} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \frac{3}{5} & \frac{4}{5} \\ 0 & -\frac{4}{5} & \frac{3}{5} \end{pmatrix} $$ and $$r_z^{-\alpha} = \begin{pmatrix} \frac{3}{5} & \frac{4}{5} & 0 \\ -\frac{4}{5} & \frac{3}{5} & 0 \\ 0 & 0 & 1 \end{pmatrix} $$
(These correspond to rotations around coordinate axes by a set irrational angle).
I need to show that it is impossible to reach the identity matrix by repeatedly multiplying any of these four matrices with each other in any order, but not allowing for two matrices that are inverse to each other directly be next to each other (obviously - or else you'd get it immediately by just by multiplying rx(a) and rx(-a)). This isn't a yes/no question, I know it's true, I need to prove it. I've tried induction and some other stuff, and it simply isn't working out. I hope someone can at least point me in the right direction, I need to have this until next week.
Some pointers I've found out: Supposedly, there exists a proof somewhere that applies the resulting matrix to the vector $ \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$, shows by induction that rotation will always result in a vector in the form $ \begin{pmatrix} \frac{a}{5^k} \\ \frac{b}{5^k} \\ \frac{c}{5^k} \end{pmatrix}$ (which is pretty much trivial), but then somehow shows that b is always incongruent to 0 mod 5, which would then complete the proof - this last bit is what I've failed to show so far.
I really appreciate the help.