Is there any way to make matrix multiplication 'act' commutative? I've come across a certain matrix equation that I'm trying to manipulate and solve; made more difficult because the equation is arbitrarily long and interspersed with matrix powers. 
The matrix in question is $Q = B^{c_1}A \ B^{c_2}A \cdots B^{c_n}A$ where 
$A = \left[ \begin{matrix} 3 & 1 \\ 0 & 1 \end{matrix} \right] \\ B = \left[ \begin{matrix} 1/2 & 0 \\ 0 & 1 \end{matrix} \right]$ 
And where $c_i \in \Bbb{Z}^+$ 
Note that $Q$ is invertible, which is nice. Noticing this, I let $Q^{-1} = M$ and I then tried to manipulate the equation $Q = M^{-1}$, hoping to end with $I_2 = B^m M^{-1} A^n$ or something similar. After computing the first few steps of this I realized it would not be this easy, and in general $Q \ne B^m A^n$ because matrix multiplication is not commutative. But is there any way to make it "act" like it is? Is there some way to simplify $Q$ into something nicer?
 A: Note that
$$
B^{c}A=\left[\begin{array}{cc}
1/2^{c}\\
 & 1
\end{array}\right]\left[\begin{array}{cc}
3 & 1\\
 & 1
\end{array}\right]=\left[\begin{array}{cc}
3/2^{c} & 1/2^{c}\\
 & 1
\end{array}\right].
$$
As such, $Q$ can be written as a product of these matrices: $Q=\prod_{i=1}^{n}C_{i}$
where $C_{i}\equiv B^{c_{i}}A$. Use induction (on $n$) to show that
$$
Q=\left[\begin{array}{cc}
3^{n}\cdot2^{-\sum_{i=1}^{n}c_{i}} & \sum_{m=1}^{n}3^{m-1}2^{-\sum_{i=1}^{m}c_{i}}\\
0 & 1
\end{array}\right].
$$

 Proof: the claim is trivially true for $n=1$. Suppose it holds for a fixed $n$. Then, \begin{align*}\prod_{i=1}^{n+1}C_i = \left(\prod_{i=1}^n C_i\right) C_{n+1} & =\left[\begin{array}{cc}3^{n}\cdot2^{-\sum_{i=1}^{n}c_{i}} & \sum_{m=1}^{n}3^{m-1}2^{-\sum_{i=1}^{m}c_{i}}\\0 & 1\end{array}\right]\left[\begin{array}{cc}3/2^{c_{n+1}} & 1/2^{c_{n+1}}\\ & 1\end{array}\right].\\ & =\left[\begin{array}{cc}3/2^{c_{n+1}}\cdot3^{n}\cdot2^{-\sum_{i=1}^{n}c_{i}} & 3^{n}\cdot2^{-\sum_{i=1}^{n+1}c_{i}}+\sum_{m=1}^{n}3^{m-1}2^{-\sum_{i=1}^{m}c_{i}}\\0 & 1\end{array}\right]\\ & =\left[\begin{array}{cc}3^{n+1}\cdot2^{-\sum_{i=1}^{n+1}c_{i}} & \sum_{m=1}^{n+1}3^{m-1}2^{-\sum_{i=1}^{m}c_{i}}\\0 & 1\end{array}\right]\end{align*}

