Question

For the matrices A and B, verify that (AB)^T = B^T A^T, where

\[A = \begin{bmatrix}1 & 3 \\ 2 & 4\end{bmatrix}, B = \begin{bmatrix}1 & 4 \\ 2 & 5\end{bmatrix}\]

Answer 1

\[Given: \hspace{0.167em} A = \begin{bmatrix}1 & 3 \\ 2 & 4\end{bmatrix}\] 
\[ A^T = \begin{bmatrix}1 & 2 \\ 3 & 4\end{bmatrix}\] 
\[B = \begin{bmatrix}1 & 4 \\ 2 & 5\end{bmatrix}\] 
\[ B^T = \begin{bmatrix}1 & 2 \\ 4 & 5\end{bmatrix}\] 

\[Now, \] 
\[AB = \begin{bmatrix}1 & 3 \\ 2 & 4\end{bmatrix} \begin{bmatrix}1 & 4 \\ 2 & 5\end{bmatrix}\] 
\[ \Rightarrow AB = \begin{bmatrix}1 + 6 & 4 + 15 \\ 2 + 8 & 8 + 20\end{bmatrix}\] 
\[ \Rightarrow AB = \begin{bmatrix}7 & 19 \\ 10 & 28\end{bmatrix}\] 
\[ \Rightarrow \left( AB \right)^T = \begin{bmatrix}7 & 10 \\ 19 & 28\end{bmatrix} . . . \left( 1 \right)\] 

\[Also, \] 
\[ B^T A^T = \begin{bmatrix}1 & 2 \\ 4 & 5\end{bmatrix}\begin{bmatrix}1 & 2 \\ 3 & 4\end{bmatrix}\] 
\[ \Rightarrow B^T A^T = \begin{bmatrix}1 + 6 & 2 + 8 \\ 4 + 15 & 8 + 20\end{bmatrix}\] 
\[ \Rightarrow B^T A^T = \begin{bmatrix}7 & 10 \\ 19 & 28\end{bmatrix} . . . \left( 2 \right)\] 
\[ \therefore \left( AB \right)^T = B^T A^T \left[ \text{From eqs} . (1) and (2) \right]\]