\begin{align}
AB
&=
\overbrace{\left(a + \vec{b}\cdot\vec{\sigma}\right)}^{A}\,
\overbrace{\left(a' + \vec{b}'\cdot\vec{\sigma}\right)}^{B}
=
aa' + \vec{b}\cdot\vec{b}'
+
\left(a\vec{b}' + a'\vec{b}
+
{\rm i}\,\vec{b}\times\vec{b}'\right)\cdot\vec{\sigma}
\\[3mm]
BA
&=
\left(a' + \vec{b}'\cdot\vec{\sigma}\right)
\left(a + \vec{b}\cdot\vec{\sigma}\right)
=
a'a + \vec{b}'\cdot\vec{b}
+
\left(a'\vec{b} + a\vec{b}'
+
{\rm i}\,\vec{b}'\times\vec{b}\right)\cdot\vec{\sigma}
\end{align}
$$
AB - BA
=
2{\rm i}\,\vec{b}\times\vec{b}'
=
0
\quad\Longrightarrow\quad
\vec{b}' = \mu\vec{b}
$$
Given a $2\times 2$ matrix $A \equiv a + \vec{b}\cdot\vec{\sigma}$, all the matrix
of the form $a' + \mu\vec{b}\cdot\vec{\sigma}$ commutes with $A$.
For example:
\begin{align}
A
&=
\left(%
\begin{array}{cc}
1 & 2
\\
0 & 0
\end{array}\right)
=
\left(%
\begin{array}{cc}
{1 \over 2} & 0
\\
0 & {1 \over 2}
\end{array}\right)
+
\left(%
\begin{array}{cc}
{1 \over 2} & 2
\\
0 & -\,{1 \over 2}
\end{array}\right)
\\[3mm]&=
{1 \over 2}
+
\left(%
\begin{array}{cc}
0 & 1
\\
1 & 0
\end{array}\right)
+
{\rm i}\,\left(%
\begin{array}{cc}
0 & -{\rm i}
\\
{\rm i} & 0
\end{array}\right)
+
{1 \over 2}
\left(%
\begin{array}{cc}
1 & 0
\\
0 & -1
\end{array}\right)
\\[3mm]&=
{1 \over 2}
+
\sigma_{x} + {\rm i}\sigma_{y} + {1 \over 2}\sigma_{z}
=
{1 \over 2} + \left(1, {\rm i}, {1 \over 2}\right)\cdot\vec{\sigma}
\end{align}
Then,
$$
\color{#ff0000}{\large B}
=
\nu + \mu\left(1, {\rm i}, {1 \over 2}\right)\cdot\vec{\sigma}
=
\nu
+
\mu\,\left(%
\begin{array}{cc}
{1 \over 2} & 2
\\
0 & -\,{1 \over 2}
\end{array}\right)
=
\color{#ff0000}{\large%
\left(%
\begin{array}{cc}
\nu + {1 \over 2}\,\mu & 2\mu
\\
0 & \nu -\,{1 \over 2}\,\mu
\end{array}\right)}
$$
$\displaystyle{\vec{\sigma}_{i}}$ is a $\tt\mbox{Pauli matrix}$.
$\displaystyle{i \equiv x, y, z.\quad}$
$\displaystyle{\vec{\sigma} \equiv \sum_{i = x, y, z}\sigma_{i}\,e_{i}\quad}$
is the $\tt\mbox{Pauli matrix vector}.\quad$
$\displaystyle{e_{x} \equiv \hat{x},\quad e_{y} \equiv \hat{y},\quad e_{z} \equiv \hat{z}}$.