Question
Let $\text{A}=\begin{bmatrix}2&-3\\-7&5\end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&0\\2&-4\end{bmatrix},$ verify that
$(\text{A}\text{B})^\text{T}=\text{B}^\text{T}\text{A}^\text{T}$

Answer

Given: $\text{A}=\begin{bmatrix}2&-3\\-7&5\end{bmatrix}$
$\text{A}^\text{T}=\begin{bmatrix}2&-7\\-3&5\end{bmatrix}$
$\text{B}=\begin{bmatrix}1&0\\2&-4\end{bmatrix}$
$\text{B}^\text{T}=\begin{bmatrix}1&2\\0&-4\end{bmatrix}$
Given,
$(\text{A}\text{B})^\text{T}=\text{B}^\text{T}\text{A}^\text{T}$
$\Rightarrow\begin{pmatrix} \begin{bmatrix} 2&-3\\-7&5\end{bmatrix}-\begin{bmatrix} 1&0\\2&-4\end{bmatrix}\end{pmatrix}^\text{T}$ $=\begin{bmatrix}1&2\\0&-4\end{bmatrix}\begin{bmatrix}2&-7\\-3&5\end{bmatrix}$
$\Rightarrow\begin{pmatrix}\begin{bmatrix}2-6&0+12\\-7+10&0-20\end{bmatrix}\end{pmatrix}^\text{T}$ $=\begin{bmatrix}2-6&-7+10\\0+12&0-20\end{bmatrix}$
$\Rightarrow\begin{pmatrix}\begin{bmatrix}-4&12\\3&-20\end{bmatrix}\end{pmatrix}^\text{T}=\begin{bmatrix}-4&3\\12&-20\end{bmatrix}$
$\Rightarrow\begin{bmatrix}-4&3\\12&-20\end{bmatrix}=\begin{bmatrix}-4&3\\12&-20\end{bmatrix}$
$\therefore\ \text{LHS}=\text{RHS}$

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