Question

(i) For two matrices A and B, $A=\left[\begin{array}{ccc}2& 1& 3\\ 4& 1& 0\end{array}\right],B=\left[\begin{array}{cc}1& -1\\ 0& 2\\ 5& 0\end{array}\right]$ verify that
(AB)^T = B^T A^T.

(ii) For the matrices A and B, verify that (AB)^T = B^T A^T, where
$A=\left[\begin{array}{cc}1& 3\\ 2& 4\end{array}\right],B=\left[\begin{array}{cc}1& 4\\ 2& 5\end{array}\right]$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right) \\ \mathrm{Given}:A=\left[\begin{array}{ccc}2& 1& 3\\ 4& 1& 0\end{array}\right] \\ {A}^T=\left[\begin{array}{cc}2& 4\\ 1& 1\\ 3& 0\end{array}\right] \\ B=\left[\begin{array}{cc}1& -1\\ 0& 2\\ 5& 0\end{array}\right] \\ {B}^T=\left[\begin{array}{ccc}1& 0& 5\\ -1& 2& 0\end{array}\right] \\ \mathrm{Now}, \\ AB=\left[\begin{array}{ccc}2& 1& 3\\ 4& 1& 0\end{array}\right]\left[\begin{array}{cc}1& -1\\ 0& 2\\ 5& 0\end{array}\right] \\ \Rightarrow AB=\left[\begin{array}{cc}2+0+15& -2+2+0\\ 4+0+0& -4+2+0\end{array}\right] \\ \Rightarrow AB=\left[\begin{array}{cc}17& 0\\ 4& -2\end{array}\right] \\ \Rightarrow {\left(AB\right)}^T=\left[\begin{array}{cc}17& 4\\ 0& -2\end{array}\right]...\left(1\right) \\ {B}^T{A}^T=\left[\begin{array}{ccc}1& 0& 5\\ -1& 2& 0\end{array}\right]\left[\begin{array}{cc}2& 4\\ 1& 1\\ 3& 0\end{array}\right] \\ \Rightarrow B^T{A}^T=\left[\begin{array}{cc}2+0+15& 4+0+0\\ -2+2+0& -4+2+0\end{array}\right] \\ \Rightarrow B^T{A}^T=\left[\begin{array}{cc}17& 4\\ 0& -2\end{array}\right]...\left(2\right) \\ \therefore {\left(AB\right)}^T=B^T{A}^T\left[\mathrm{From}\mathrm{eqs}.\left(1\right)\mathrm{and}\left(2\right)\right] \end{gathered}$$

$$\begin{gathered} \left(ii\right) \\ \mathrm{Given}:\hspace{0.17em}A=\left[\begin{array}{cc}1& 3\\ 2& 4\end{array}\right] \\ {A}^T=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right] \\ B=\left[\begin{array}{cc}1& 4\\ 2& 5\end{array}\right] \\ {B}^T=\left[\begin{array}{cc}1& 2\\ 4& 5\end{array}\right] \\ \mathrm{Now}, \\ AB=\left[\begin{array}{cc}1& 3\\ 2& 4\end{array}\right]\left[\begin{array}{cc}1& 4\\ 2& 5\end{array}\right] \\ \Rightarrow AB=\left[\begin{array}{cc}1+6& 4+15\\ 2+8& 8+20\end{array}\right] \\ \Rightarrow AB=\left[\begin{array}{cc}7& 19\\ 10& 28\end{array}\right] \\ \Rightarrow {\left(AB\right)}^T=\left[\begin{array}{cc}7& 10\\ 19& 28\end{array}\right]...\left(1\right) \\ Also, \\ {B}^T{A}^T=\left[\begin{array}{cc}1& 2\\ 4& 5\end{array}\right]\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right] \\ \Rightarrow B^T{A}^T=\left[\begin{array}{cc}1+6& 2+8\\ 4+15& 8+20\end{array}\right] \\ \Rightarrow B^T{A}^T=\left[\begin{array}{cc}7& 10\\ 19& 28\end{array}\right]...\left(2\right) \\ \therefore {\left(AB\right)}^T=B^T{A}^T\left[\mathrm{From}\mathrm{eqs}.\left(1\right)\mathrm{and}\left(2\right)\right] \end{gathered}$$