Answer
$(AB)^T=B^TA^T$.
Work Step by Step
Since we have
$$AB=\left[\begin{array}{ccc}{-1} &{1} & {-2} \\{2} & {0}& {1} \end{array}\right]\left[\begin{array}{cc}{-3} &{0} \\{1}& {2} \\{1} &{-1} \end{array}\right]=\left[\begin{array}{cc}{2} &{4} \\{-5}& {-1} \end{array}\right]$$
then $$(AB)^T=\left[\begin{array}{cc}{2} &{-5} \\{4}& {-1} \end{array}\right].$$
Now, $$B^TA^T=\left[\begin{array}{ccc}{-3} &{1}&{1} \\{0}& {2} &{-1} \end{array}\right]\left[\begin{array}{ccc}{-1} &{2} \\{1} & {0} \\ {-2}&{1} \end{array}\right] =\left[\begin{array}{cc}{2} &{-5} \\{4}& {-1} \end{array}\right].$$
It is easy to see that $(AB)^T=B^TA^T$.
