Multiplication in $S_4$ with cycle notation and permutation matrices

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Solve the equation $(23)x^{-1}(124)=(34)^4$ in $S_4$.

I'm a bit confused about which way I should go when multiplying. Using cycle notation I've been told to go from right to left, but I find it much easier to do these using matrix notation. So I write the equation as $$\begin{pmatrix}1&2&3&4\\ 1&3&2&4\end{pmatrix}\begin{pmatrix}1&2&3&4\\ &&&\end{pmatrix}\begin{pmatrix}1&2&3&4\\ 2&4&3&1\end{pmatrix}=\begin{pmatrix}1&2&3&4\\ 1&2&3&4\end{pmatrix}$$ since I figured out that $(34)^4=(1)$.

Now I need to figure out what to put in the blank spots. Going from left to right I figured that I need to have $$\begin{pmatrix}1&2&3&4\\ 1&3&2&4\end{pmatrix}\begin{pmatrix}1&2&3&4\\ 4&3&1&2\end{pmatrix}\begin{pmatrix}1&2&3&4\\ 2&4&3&1\end{pmatrix}=\begin{pmatrix}1&2&3&4\\ 1&2&3&4\end{pmatrix}$$

but when going from right to left I ended up with

$$\begin{pmatrix}1&2&3&4\\ 1&3&2&4\end{pmatrix}\begin{pmatrix}1&2&3&4\\ 4&1&2&3\end{pmatrix}\begin{pmatrix}1&2&3&4\\ 2&4&3&1\end{pmatrix}=\begin{pmatrix}1&2&3&4\\ 1&2&3&4\end{pmatrix}$$ which is wrong. Why is the multiplication reversed in this case?

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You are right that $(34)^4=e.$

What I would do next is take the inverse of the whole of the LHS, since $(ab^{-1}c)^{-1}=c^{-1}ba^{-1}$; thus

$$(124)^{-1}x(23)^{-1}=e^{-1}=e,$$

which gives

$$\begin{align} x&=(124)e(23)\\ &=(1234)\\ &=\begin{pmatrix} 1 & 2 & 3 & 4\\ 2 & 3 & 4 & 1 \end{pmatrix}, \end{align}$$

whose inverse is

$$x^{-1}=\begin{pmatrix} 1 & 2 & 3 & 4\\ 4 & 1 & 2 & 3 \end{pmatrix},$$

so you were right; you just forgot to take inverses.