For $k > 1$, is it possible that $\begin{pmatrix} a_1 & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} a_2 & 1 \\ -1 & 0 \end{pmatrix}\ldots \begin{pmatrix} a_k & 1 \\ -1 & 0 \end{pmatrix} = \pm \begin{pmatrix} b & 1 \\ -1 & 0 \end{pmatrix}$ if $a_1,a_2,\ldots a_k,b$ are Eisenstein integers and $|a_i| > 1$ for $i=1,2,\ldots k$?

If the Eisenstein integers are replaced with the Gaussian integers, this is possible. $\begin{pmatrix} 3 & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 1 - i & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 1 + i & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 1 - i & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} -2 & 1 \\ -1 & 0 \end{pmatrix} = -\begin{pmatrix} i & 1 \\ -1 & 0 \end{pmatrix}$

This problem came up in my research; I am primarily interested in the case where $b = (\pm 1 \pm \sqrt{-3})/2$, but I suspect that there might not be a solution for any choice of $b$.

I originally asked this question on math.stackexchange (https://math.stackexchange.com/q/3903567/202799), but it seems to be more difficult than I first thought.