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.$(2\text{A})^\text{T}=2\text{A}^\text{T}$

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 $\text{B}^\text{T}=\begin{bmatrix}1&2&0\\2&1&1\\3&3&1\end{bmatrix}$
$(2\text{A})^\text{T}=2\text{A}^\text{T}$
$\Rightarrow\begin{pmatrix}2\begin{bmatrix}1&-1&0\\2&1&3\\1&2&1\end{bmatrix}\end{pmatrix}^\text{T}=2\begin{bmatrix}1&-1&0\\2&1&3\\1&2&1\end{bmatrix}^\text{T}$
$\Rightarrow\begin{bmatrix}2&-2&0\\4&2&6\\2&4&2\end{bmatrix}^\text{T}=2\begin{bmatrix}1&2&1\\-1&1&2\\0&3&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}2&4&2\\-2&2&4\\0&6&2\end{bmatrix}=\begin{bmatrix}2&4&2\\-2&2&4\\0&6&2\end{bmatrix}$
$\Rightarrow\text{LHS}=\text{RHS}$
So,
$(2\text{A})^\text{T}=2\times\text{A}^\text{T}$

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