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Question
Let `A= [[1,-1,0],[2,1,3],[1,2,1]]` And `B=[[1,2,3],[2,1,3],[0,1,1]]` Find `A^T,B^T` and verify that (A + B)T = AT + BT
Solution
\[Given: A = \begin{bmatrix}1 & - 1 & 0 \\ 2 & 1 & 3 \\ 1 & 2 & 1\end{bmatrix} \text{and B }= \begin{bmatrix}1 & 2 & 3 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{bmatrix}\]
`A^T = [[1 2 1],[-1 1 2 ],[0 3 1]]` and `B^T = [[1 2 0],[2 1 1 ],[3 3 1]]`
\[\left( i \right) \]
\[A + B = \begin{bmatrix}1 & - 1 & 0 \\ 2 & 1 & 3 \\ 1 & 2 & 1\end{bmatrix} + \begin{bmatrix}1 & 2 & 3 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{bmatrix} \]
\[ \Rightarrow A + B = \begin{bmatrix}1 + 1 & - 1 + 2 & 0 + 3 \\ 2 + 2 & 1 + 1 & 3 + 3 \\ 1 + 0 & 2 + 1 & 1 + 1\end{bmatrix}\]
\[ \Rightarrow A + B = \begin{bmatrix}2 & 1 & 3 \\ 4 & 2 & 6 \\ 1 & 3 & 2\end{bmatrix}\]
\[ \Rightarrow \left( A + B \right)^T = \begin{bmatrix}2 & 4 & 1 \\ 1 & 2 & 3 \\ 3 & 6 & 2\end{bmatrix} . . . \left( 1 \right)\]
\[Now, \]
\[ A^T + B^T = \begin{bmatrix}1 & 2 & 1 \\ - 1 & 1 & 2 \\ 0 & 3 & 1\end{bmatrix} + \begin{bmatrix}1 & 2 & 0 \\ 2 & 1 & 1 \\ 3 & 3 & 1\end{bmatrix}\] \[ \Rightarrow A^T + B^T = \begin{bmatrix}1 + 1 & 2 + 2 & 1 + 0 \\ - 1 + 2 & 1 + 1 & 2 + 1 \\ 0 + 3 & 3 + 3 & 1 + 1\end{bmatrix}\]
\[ \Rightarrow A^T + B^T = \begin{bmatrix}2 & 4 & 1 \\ 1 & 2 & 3 \\ 3 & 6 & 2\end{bmatrix} . . . \left( 2 \right)\]
\[ \Rightarrow \left( A + B \right)^T = A^T + B^T \left[ \text{From eqs} . \left( 1 \right) and \left( 2 \right) \right]\]
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