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Question
Solution
\[\left( i \right) \]
\[ Given: A = \begin{bmatrix}2 & 1 & 3 \\ 4 & 1 & 0\end{bmatrix} \]
\[ A^T = \begin{bmatrix}2 & 4 \\ 1 & 1 \\ 3 & 0\end{bmatrix}\]
\[\]
\[B = \begin{bmatrix}1 & - 1 \\ 0 & 2 \\ 5 & 0\end{bmatrix}\]
\[ B^T = \begin{bmatrix}1 & 0 & 5 \\ - 1 & 2 & 0\end{bmatrix}\] \[Now, \]
\[AB = \begin{bmatrix}2 & 1 & 3 \\ 4 & 1 & 0\end{bmatrix}\begin{bmatrix}1 & - 1 \\ 0 & 2 \\ 5 & 0\end{bmatrix} \]
\[ \Rightarrow AB = \begin{bmatrix}2 + 0 + 15 & - 2 + 2 + 0 \\ 4 + 0 + 0 & - 4 + 2 + 0\end{bmatrix}\]
\[ \Rightarrow AB = \begin{bmatrix}17 & 0 \\ 4 & - 2\end{bmatrix}\]
\[ \Rightarrow \left( AB \right)^T = \begin{bmatrix}17 & 4 \\ 0 & - 2\end{bmatrix} . . . \left( 1 \right)\] \[ B^T A^T = \begin{bmatrix}1 & 0 & 5 \\ - 1 & 2 & 0\end{bmatrix}\begin{bmatrix}2 & 4 \\ 1 & 1 \\ 3 & 0\end{bmatrix}\]
\[ \Rightarrow B^T A^T = \begin{bmatrix}2 + 0 + 15 & 4 + 0 + 0 \\ - 2 + 2 + 0 & - 4 + 2 + 0\end{bmatrix}\]
\[ \Rightarrow B^T A^T = \begin{bmatrix}17 & 4 \\ 0 & - 2\end{bmatrix} . . . \left( 2 \right)\]
\[\]
\[ \therefore \left( AB \right)^T = B^T A^T \left[ \text{From eqs} . (1) and (2) \right]\]
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