Since we only have a finite number of \( \alpha \) and \( \beta \) values which are non-zero, we can rewrite the above convolution expressions as a matrix-vector multiplication $$ \boldsymbol{\delta}=\begin{bmatrix}\alpha_0 & 0 & 0 & 0 \\ \alpha_1 & \alpha_0 & 0 & 0 \\ \alpha_2 & \alpha_1 & \alpha_0 & 0 \\ 0 & \alpha_2 & \alpha_1 & \alpha_0 \\ 0 & 0 & \alpha_2 & \alpha_1 \\ 0 & 0 & 0 & \alpha_2 \end{bmatrix}\begin{bmatrix} \beta_0 \\ \beta_1 \\ \beta_2 \\ \beta_3\end{bmatrix}. $$
The process is commutative and we can easily see that we can rewrite the multiplication in terms of a matrix holding \( \beta \) and a vector holding \( \alpha \). In this case we have $$ \boldsymbol{\delta}=\begin{bmatrix}\beta_0 & 0 & 0 \\ \beta_1 & \beta_0 & 0 \\ \beta_2 & \beta_1 & \beta_0 \\ \beta_3 & \beta_2 & \beta_1 \\ 0 & \beta_3 & \beta_2 \\ 0 & 0 & \beta_3 \end{bmatrix}\begin{bmatrix} \alpha_0 \\ \alpha_1 \\ \alpha_2\end{bmatrix}. $$
Note that the use of these matrices is for mathematical purposes only and not implementation purposes. When implementing the above equation we do not encode (and allocate memory) the matrices explicitely. We rather code the convolutions in the minimal memory footprint that they require.
Does the number of floating point operations change here when we use the commutative property?