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Question
Solution
\[Given: \hspace{0.167em} A = \begin{bmatrix}1 & 3 \\ 2 & 4\end{bmatrix}\]
\[ A^T = \begin{bmatrix}1 & 2 \\ 3 & 4\end{bmatrix}\]
\[B = \begin{bmatrix}1 & 4 \\ 2 & 5\end{bmatrix}\]
\[ B^T = \begin{bmatrix}1 & 2 \\ 4 & 5\end{bmatrix}\]
\[Now, \]
\[AB = \begin{bmatrix}1 & 3 \\ 2 & 4\end{bmatrix} \begin{bmatrix}1 & 4 \\ 2 & 5\end{bmatrix}\]
\[ \Rightarrow AB = \begin{bmatrix}1 + 6 & 4 + 15 \\ 2 + 8 & 8 + 20\end{bmatrix}\]
\[ \Rightarrow AB = \begin{bmatrix}7 & 19 \\ 10 & 28\end{bmatrix}\]
\[ \Rightarrow \left( AB \right)^T = \begin{bmatrix}7 & 10 \\ 19 & 28\end{bmatrix} . . . \left( 1 \right)\]
\[Also, \]
\[ B^T A^T = \begin{bmatrix}1 & 2 \\ 4 & 5\end{bmatrix}\begin{bmatrix}1 & 2 \\ 3 & 4\end{bmatrix}\]
\[ \Rightarrow B^T A^T = \begin{bmatrix}1 + 6 & 2 + 8 \\ 4 + 15 & 8 + 20\end{bmatrix}\]
\[ \Rightarrow B^T A^T = \begin{bmatrix}7 & 10 \\ 19 & 28\end{bmatrix} . . . \left( 2 \right)\]
\[ \therefore \left( AB \right)^T = B^T A^T \left[ \text{From eqs} . (1) and (2) \right]\]
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