**Lemma 1.** Let $\eta$ denote primitive $m$-th root of 1. Consider all
$m$-tuples $0\leq a_1<a_2<\dots< a_m\leq 2m$ such that
either $a_1>0$ or $a_{m}<2m$. Take the sum of $\eta^{a_1+\dots+a_m}$
over all such $m$-tuples. It equals $3(-1)^{m+1}$.

**Proof.** Consider the polynomial
$$
f(x)=(1-x\eta)(1-x\eta^2)\dots (1-x\eta^{2m-1})\cdot
\left((1-x)(1-x\eta^{2m})-x^2\eta^{2m}\right).
$$
Our sums equals $(-1)^m$ times its coefficient of $x^m$.
That is, we have to prove that this coefficient equals $-3$.
It is straigthforward: since $(1-x)(1-x\eta)\dots (1-x\eta^{m-1})=1-x^{m}$ and
$(1-x\eta)\dots (1-x\eta^{m-1})=1+x+\dots+x^{m-1}$,
$$
f(x)=(1-x)(1-x^m)^2-x^2(1+x+\dots+x^{m-1}),
$$
coefficient of $x^m$ really equals $-3$.

**Lemma 2.** Let $\eta$ denote primitive $m$-th root of 1.
Consider all $m$-tuples $0\leq b_1<b_2<\dots< b_m\leq 3m-1$
such that $b_{i+1}\geq b_i+2$ for $i=1,\dots,m-1$; and either $b_1>0$
or $b_{m}<3m-1$. In other words, pairs $(b_i,b_i+1)$ must be disjont modulo
$3m$. Take the sum of $\eta^{b_1+\dots+b_m}$
over all such $m$-tuples. It equals $3$.

**Proof.** Denote $a_i=b_i-i+1$. It reduces to the previous lemma
(and obvious equality $(-1)^{m+1}\eta^{m(m-1)/2}=1$.)

I claim that Peter Mueller's case $n=3m$, $P(z)=a_n+b_nz$ reduces
to Lemma 2.
Indeed, consider how diagonal element of the product of
our matrices may appear and contain $z$ in exponent $m=n/3$.
For $m$ matrices we should choose elements

$\zeta^j z$, let them be matrices indexed by $b_1<\dots<b_m$. Then for
matrices indexed by $b_1+1,\dots,b_m+1$ we must choose 1,
for other matrices $\zeta^{-j}$. This is so if $b_m<n-1$,
if $b_m=n-1$, then our diagonal element is in second row and column, hence
$b_1\ne 0$. Totally $\zeta$ is taken in a power $$b_1+\dots+b_m-(0+\dots+n-1-
(b_1+\dots+b_m+(b_1+1)+\dots+(b_m+1))=m+3(b_1+\dots+b_m)-n(n-1)/2.$$
It remains to denote $\eta=\zeta^3$ and apply Lemma 2.