Question

If $A=\left[\begin{array}{c}-2\\ 4\\ 5\end{array}\right]$, B = [1 3 −6], verify that (AB)^T = B^T A^T

Answer 1

$$\begin{gathered} \mathrm{Given}:A=\left[\begin{array}{c}-2\\ 4\\ 5\end{array}\right] \\ {A}^T=\left[\begin{array}{ccc}-2& 4& 5\end{array}\right] \\ B=\left[\begin{array}{ccc}1& 3& -6\end{array}\right] \\ {B}^T=\left[\begin{array}{c}1\\ 3\\ -6\end{array}\right] \\ \mathrm{Now}, \\ AB=\left[\begin{array}{c}-2\\ 4\\ 5\end{array}\right]\left[\begin{array}{ccc}1& 3& -6\end{array}\right] \\ \Rightarrow AB=\left[\begin{array}{ccc}-2& -6& 12\\ 4& 12& -24\\ 5& 15& -30\end{array}\right] \\ \Rightarrow {\left(AB\right)}^T=\left[\begin{array}{ccc}-2& 4& 5\\ -6& 12& 15\\ 12& -24& -30\end{array}\right]...\left(1\right) \\ {B}^T{A}^T=\left[\begin{array}{c}1\\ 3\\ -6\end{array}\right]\left[\begin{array}{ccc}-2& 4& 5\end{array}\right] \\ \Rightarrow B^T{A}^T=\left[\begin{array}{ccc}-2& 4& 5\\ -6& 12& 15\\ 12& -24& -30\end{array}\right]...\left(2\right) \\ \therefore {\left(AB\right)}^T=B^T{A}^T\left[\mathrm{From}\mathrm{eqs}.\left(1\right)\mathrm{and}\left(2\right)\right] \end{gathered}$$