Expression from generators of Special Linear Groups II

Question

I wonder whether one can generate this $t$ matrix form the $A_1$ and $A_2$ matrix below. Here $$ t=\begin{pmatrix} 1& 1& 0\\ 0& 1& 0\\ 0& 0& 1 \end{pmatrix} $$ from: $$ A_1=\begin{pmatrix} 0& 0& 1\\ 1& 0& 0\\ 0& 1& 0 \end{pmatrix}, \text{ and }\;\; A_2=\begin{pmatrix} 0& 1& 0\\ -1& 0& 0\\ 0& 0& 1 \end{pmatrix}. $$

Question: How to generate $t$ from $A_1$ and $A_2$? i.e. So what is the exact expression to make $t=\dots A_1 \dots A_2$ as a product of $A_1$ and $A_2$ matrices?

Notice that $(A_1)^3=1$ and $(A_2)^4=1$.

I learned the basics about those generators and SL(N,Z) from the book Coxeter and Moser on "Generators and relations for discrete groups" (published by Springer, 1957).

Linear algebra and special-linear group experts please help. Thank you! :o)

Answer 1

Let me try again, hopefully not telling any needless lies.

The two matrices $A_1$ and $A_2$ are signed permutation matrices, and the group of all signed permutation matrices is the hyperoctohedral group. So the group generated by $A_1$ and $A_2$ is a subgroup of this hyperoctohedral group, which has order $2^3\cdot 3!$. Now, the matrix $t$ has infinite order, and so it is not contained in any finite group of matrices.

Answer 2

From a simple linear algebra perspective, if you multiply two generalized permutation matrices (such as $A_1, A_2$ - assuming $\mathbb{Z}$ is the field of interest) the result is always again a generalized permutation matrix, and since $t$ is not a generalized permutation matrix, it cannot therefore be generated by $A_1$ and $A_2$.

In group theory terms, the generalized permutation matrices form a subgroup of GL$(n,F)$ ($n$ being 3 in this case and $F$ being $\mathbb{Z}$), and $t$ does not belong to this subgroup.