Answer
$(AB)^T=B^TA^T$.
Work Step by Step
Since we have
$$AB=\left[\begin{array}{cc}{2} &{1} \\{0}& {1}\\{-2}&{1} \end{array}\right] \left[\begin{array}{ccc}{2} &{3} &{1} \\ {0}& {4} &{-1} \end{array}\right]=\left[\begin{array}{cc}{1} &{1} \\{-4}& {-2} \end{array}\right]
=\left[\begin{array}{ccc}{4} &{10}&{1} \\{0}& {4}&{-1} \\{-4}&{-2}&{-3} \end{array}\right]$$
then $$(AB)^T=\left[\begin{array}{ccc}{4} &{0}&{-4} \\{10}& {4}&{-2}\\{1}&{-1}&{-3} \end{array}\right].$$
Now, $$B^TA^T=\left[\begin{array}{ccc}{2} &{0} \\{3} & {4} \\ {1}&{-1} \end{array}\right] \left[\begin{array}{ccc}{2} &{0}&{-2} \\{1}& {1} &{1} \end{array}\right]=\left[\begin{array}{ccc}{4} &{0}&{-4} \\{10}& {4}&{-2}\\{1}&{-1}&{-3} \end{array}\right].$$
It is easy to see that $(AB)^T=B^TA^T$.
