Answer
$(AB)^T=B^TA^T$.
Work Step by Step
Since we have
$$AB=\left[\begin{array}{ccc}{1} &{2} \\ {0}& {-2} \end{array}\right]\left[\begin{array}{cc}{-3} &{-1} \\{2}& {1} \end{array}\right]=\left[\begin{array}{cc}{1} &{1} \\{-4}& {-2} \end{array}\right]$$
then $$(AB)^T=\left[\begin{array}{cc}{1} &{-4} \\{1}& {-2} \end{array}\right].$$
Now, $$B^TA^T= \left[\begin{array}{ccc}{-3} &{2} \\ {-1} &{1} \end{array}\right]\left[\begin{array}{ccc}{1} &{0} \\{2} & {-2} \end{array}\right]=\left[\begin{array}{cc}{1} &{-4} \\{1}& {-2} \end{array}\right].$$
It is easy to see that $(AB)^T=B^TA^T$.
