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=A^T+B^T

Given: A= 1&-1&0 2&1&3 1&2&1 and B= 1&2&3 2&1&3 0&1&1 A^T= 1&2&1 -1&1&2 0&3&1 and B^T= 1&2&0 2&1&1 3&3&1 A+B= 1&-1&0 2&1&3 1&2&1 + 1&2&3 2&1&3 0&1&1 +B= 1+1&-1+2&0+3 2+2&1+1&3+3 1+0&2+1&1+1 +B= 2&1&3 4&2&6 1&3&2 (A+B)^T= 2&4&1 1&2&3 3&6&2 (1) Now, A^T+B^T= 1&2&1 -1&1&2 0&3&1 + 1&2&0 2&1&1 3&3&1 ^T+B^T= 1+1&2+2&1+0 -1+2&1+1&2+1 0+3&3+3&1+1 ^T+B^T= 2&4&1 1&2&3 3&6&2 (2) (A+B)^T=A^T+B^T [From eqs. (1) and (2)]

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^…

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Question

Let $\text{A}=\begin{bmatrix}1&-1&0\\2&1&3\\1&2&1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&2&3\\2&1&3\\0&1&1\end{bmatrix},$ Find $A^T, B^T$ and verify that.$(\text{A}+\text{B})^\text{T}=\text{A}^\text{T}+\text{B}^\text{T}$

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Answer

Given: $\text{A}=\begin{bmatrix}1&-1&0\\2&1&3\\1&2&1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&2&3\\2&1&3\\0&1&1\end{bmatrix}$
$\text{A}^\text{T}=\begin{bmatrix}1&2&1\\-1&1&2\\0&3&1\end{bmatrix}$ and $\text{B}^\text{T}=\begin{bmatrix}1&2&0\\2&1&1\\3&3&1\end{bmatrix}$
$\text{A}+\text{B}=\begin{bmatrix}1&-1&0\\2&1&3\\1&2&1\end{bmatrix}+\begin{bmatrix}1&2&3\\2&1&3\\0&1&1\end{bmatrix}$
$\Rightarrow\text{A}+\text{B}=\begin{bmatrix}1+1&-1+2&0+3\\2+2&1+1&3+3\\1+0&2+1&1+1\end{bmatrix}$
$\Rightarrow\text{A}+\text{B}=\begin{bmatrix}2&1&3\\4&2&6\\1&3&2\end{bmatrix}$
$\Rightarrow(\text{A}+\text{B})^\text{T}=\begin{bmatrix}2&4&1\\1&2&3\\3&6&2\end{bmatrix}\ \dots(1)$
Now,
$\text{A}^\text{T}+\text{B}^\text{T}=\begin{bmatrix}1&2&1\\-1&1&2\\0&3&1\end{bmatrix}+\begin{bmatrix}1&2&0\\2&1&1\\3&3&1\end{bmatrix}$
$\Rightarrow\text{A}^\text{T}+\text{B}^\text{T}=\begin{bmatrix}1+1&2+2&1+0\\-1+2&1+1&2+1\\0+3&3+3&1+1\end{bmatrix}$
$\Rightarrow\text{A}^\text{T}+\text{B}^\text{T}=\begin{bmatrix}2&4&1\\1&2&3\\3&6&2\end{bmatrix}\ \dots(2)$
$\Rightarrow(\text{A}+\text{B})^\text{T}=\text{A}^\text{T}+\text{B}^\text{T}$ [From eqs. (1) and (2)]

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