Let $A=\begin{pmatrix} 1 & 2 \\ 0 & 1\end{pmatrix}$ and $B=\begin{pmatrix} 1 & 0 \\ -2 & 1\end{pmatrix}$.

Can a product $X_1 X_2...X_n$ be equal to the identity matrix if every factor $X_i$ equals either $A$ or $B$?

I believe that the answer is negative, but I don't really know how to prove it. I thought of two approaches:

  1. We could try doing this by induction. The base case is trivial since neither of $A$ nor $B$ are the identity matrix. However, I don't know how to go from $n$ to $n+1$.

  2. Maybe we should use the fact that if $U, V$ are two square matrices such that $UV$ is the identity, then $VU$ is also the identity. I guess that we should somehow shuffle the order of the factors using this observation, but then again, I don't know how to use this.

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    $\begingroup$ Where did you come across this question? You've tagged your question as constest-math; is this question from a competition? $\endgroup$ Commented Dec 10, 2021 at 17:20
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    $\begingroup$ @BenGrossmann I suspect it is, but I for one got it from a selection test for a competition (to make things clear, that selection test was 6 months ago, so this is not some attempt to cheat). But as I said, I suspect it was given somewhere. $\endgroup$
    – MathIsCool
    Commented Dec 10, 2021 at 18:32

1 Answer 1


Let $s=\begin{bmatrix}0 & -1 \\1&0\end{bmatrix}$ and $t=\begin{bmatrix}1 & -1 \\ 1&0\end{bmatrix}$. Then, in $PSL_2(\mathbb{Z})$, $s^2=t^3=1$, $s,t$ generate this group and have no other nontrivial relation (see https://chiasme.wordpress.com/2015/03/08/an-elementary-application-of-ping-pong-lemma/).

Now, note that in $PSL_2(\mathbb{Z})$, $(ts)^{2}=A$ while $B=(st)^2$.

So we want to see that no word in $stst$ and $tsts$ simplifies to the trivial word when $s^2=t^3=1$. That looks obvious (we can only reduce when $stst$ comes after $tsts$ and then the result is $tst^{-1}st$ which cannot be reduced further).


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