## Elementary Linear Algebra 7th Edition

$(AB)^T=B^TA^T$.
Since we have $$AB=\left[\begin{array}{ccc}{2} &{1} &{-1}\\{0}& {1}&{3}\\{4}&{0}&{2} \end{array}\right] \left[\begin{array}{ccc}{1} &{0} &{-1} \\ {2}& {1} &{-2}\\{0}&{1}&{3} \end{array}\right]= \left[\begin{array}{ccc}{4} &{0}&{-7} \\{2}& {4}&{7} \\{4}&{2}&{2} \end{array}\right]$$ then $$(AB)^T=\left[\begin{array}{ccc}{4} &{2}&{4} \\{0}& {4}&{2}\\{-7}&{7}&{2} \end{array}\right].$$ Now, $$B^TA^T=\left[\begin{array}{ccc}{1} &{2} &{0} \\{0} & {1}&{1} \\ {-1}&{-2}&{3} \end{array}\right] \left[\begin{array}{ccc}{2} &{0}&{4} \\{1}& {1} &{0}\\{-1}&{3}&{2} \end{array}\right]=\left[\begin{array}{ccc}{4} &{2}&{4} \\{0}& {4}&{2}\\{-7}&{7}&{2} \end{array}\right].$$ It is easy to see that $(AB)^T=B^TA^T$.