Question

A = \begin{bmatrix}1 & - 1 & 0 \\ 2 & 1 & 3 \\ 1 & 2 & 1\end{bmatrix} and B = \begin{bmatrix}1 & 2 & 3 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{bmatrix} . Find A^T, B^T and verify that , 

(A B)^T = B^T + A^T

Answer 1

 

\[AB = \begin{bmatrix}1 & - 1 & 0 \\ 2 & 1 & 3 \\ 1 & 2 & 1\end{bmatrix} \begin{bmatrix}1 & 2 & 3 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{bmatrix} \]
\[ \Rightarrow AB = \begin{bmatrix}1 - 2 + 0 & 2 - 1 + 0 & 3 - 3 + 0 \\ 2 + 2 + 0 & 4 + 1 + 3 & 6 + 3 + 3 \\ 1 + 4 + 0 & 2 + 2 + 1 & 3 + 6 + 1\end{bmatrix}\]
\[ \Rightarrow AB = \begin{bmatrix}- 1 & 1 & 0 \\ 4 & 8 & 12 \\ 5 & 5 & 10\end{bmatrix}\]
\[ \Rightarrow \left( AB \right)^T = \begin{bmatrix}- 1 & 4 & 5 \\ 1 & 8 & 5 \\ 0 & 12 & 10\end{bmatrix} . . . \left( 1 \right)\]
\[Now, \]
\[ B^T A^T = \begin{bmatrix}1 & 2 & 0 \\ 2 & 1 & 1 \\ 3 & 3 & 1\end{bmatrix}\begin{bmatrix}1 & 2 & 1 \\ - 1 & 1 & 2 \\ 0 & 3 & 1\end{bmatrix}\]
\[ \Rightarrow B^T A^T = \begin{bmatrix}1 - 2 + 0 & 2 + 2 + 0 & 1 + 4 + 0 \\ 2 - 1 + 0 & 4 + 1 + 3 & 2 + 2 + 1 \\ 3 - 3 + 0 & 6 + 3 + 3 & 3 + 6 + 1\end{bmatrix}\]
\[ \Rightarrow B^T A^T = \begin{bmatrix}- 1 & 4 & 5 \\ 1 & 8 & 5 \\ 0 & 12 & 10\end{bmatrix} . . . \left( 2 \right)\]
\[ \Rightarrow \left( AB \right)^T = B^T A^T \left[ {\text{From eqs}} . (1) \hspace{0.167em} and (2) \right]\]