Why the multiplication between two Markov Chain transition matrices is commutaive?

Question

Can anyone give some intuitive proof the following proposition?

For any two Markov Chain transition matrices $A$, $B$ ($n$ by $n$ matrix that has every row sum to $1$), we have $A * B = B * A$.

Answer 1

It's false. For example, $$\begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}\begin{bmatrix}1 & 0 \\ 1 & 0\end{bmatrix} = \begin{bmatrix}1 & 0 \\ 1 & 0\end{bmatrix}$$ but $$\begin{bmatrix}1 & 0 \\ 1 & 0\end{bmatrix}\begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix} = \begin{bmatrix}0 & 1 \\ 0 & 1\end{bmatrix}.$$ The zero-one example is just particularly easy to compute with, but $AB$ won't be equal to $BA$ for almost any pair of transition matrices $A,B$.