Multiplication in $S_4$ with cycle notation and permutation matrices 
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?

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.
