Answer
$AB=BA$.
Work Step by Step
We have
\begin{align*}
AB=\left[\begin{array}{cc}{\cos \alpha} &{-\sin \alpha} \\{ \sin \alpha}& {\cos \alpha} \end{array}\right]\left[\begin{array}{cc}{\cos \beta} &{-\sin \beta} \\{ \sin \beta}& {\cos \beta} \end{array}\right]\\
= \left[\begin{array}{cc}{\cos \alpha\cos \beta-\sin \alpha\sin \beta} &{-\cos \alpha\sin\beta-\sin \alpha\cos \beta} \\{ \sin\alpha\cos \beta+\cos \alpha\sin \beta}& {-\sin \alpha\sin\beta+\cos\alpha\cos\beta} \end{array}\right]
\end{align*}
and
\begin{align*} BA=
\left[\begin{array}{cc}{\cos \alpha} &{-\sin \alpha} \\{ \sin \alpha}& {\cos \alpha} \end{array}\right]\left[\begin{array}{cc}{\cos \beta} &{-\sin \beta} \\{ \sin \beta}& {\cos \beta} \end{array}\right]\\
= \left[\begin{array}{cc}{\cos \alpha\cos \beta-\sin \alpha\sin \beta} &{-\cos \alpha\sin\beta-\sin \alpha\cos \beta} \\{ \sin\alpha\cos \beta+\cos \alpha\sin \beta}& {-\sin \alpha\sin\beta+\cos\alpha\cos\beta} \end{array}\right]
\end{align*}.
Which shows that $AB=BA$.
