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Answer: $\left ( A+B\right )^{T}=A^{T}+B^{T}$

Given:$A=\begin{bmatrix} 1 & -1 &0 \\ 2&1 & 3\\ 1 &2 & 1 \end{bmatrix}, B=\begin{bmatrix} 1 & 2 &3 \\ 2 & 1&3 \\ 0 & 1 & 1 \end{bmatrix}$

To prove: $\left ( A+B\right )^{T}=A^{T}+B^{T}$

Hint: The $A^{T}$ of matrix$A$ can be obtained by reflecting the elements along it’s main diagonal.

Solution:

$\left ( A+B\right )^{T}=A^{T}+B^{T}$

$\left ( \begin{bmatrix} 1 &-1 &0 \\ 2 & 1 &3 \\ 1 & 2 & 1 \end{bmatrix}+\begin{bmatrix} 1 & 2 &3 \\ 2 &1 &3 \\ 0 & 1 & 1 \end{bmatrix} \right )^{T}= \begin{bmatrix} 1 &-1 &0 \\ 2 &1 &3 \\ 1 &2 &1 \end{bmatrix}^{T}+\begin{bmatrix} 1 & 2 & 3\\ 2 & 1 &3 \\ 0 & 1 & 1 \end{bmatrix}^{T}$

$\begin{bmatrix} 1+1 &-1+2 &0+3 \\ 2+2 & 1+1 &3+3 \\ 1+0 & 2+1 & 1+1 \end{bmatrix} ^{T}= \begin{bmatrix} 1 &2 &1 \\ -1 &1 &2 \\ 0 &3 &1 \end{bmatrix}+\begin{bmatrix} 1 & 2 & 0\\ 2 & 1 &1 \\ 3 & 3 & 1 \end{bmatrix}$

$\begin{bmatrix} 2 &1 &3 \\ 4 & 2 &6 \\ 1 & 3 & 2 \end{bmatrix} ^{T}= \begin{bmatrix} 2 &4 &1 \\ 1 &2 &3 \\ 3&6 &2 \end{bmatrix}$

$\begin{bmatrix} 2 &4 &1 \\ 1 &2 &3 \\ 3&6 &2 \end{bmatrix}= \begin{bmatrix} 2 &4 &1 \\ 1 &2 &3 \\ 3&6 &2 \end{bmatrix}$

LHS=RHS

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