Question
If $\text{A}=\begin{bmatrix} 3\\5\\2\end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&0&4\end{bmatrix},$ verify that $(AB)^T = B^TA^T.$

Answer

Given,
$\text{A}=\begin{bmatrix}3\\5\\2 \end{bmatrix},\text{B}=\begin{bmatrix}1&0&4\end{bmatrix}$
$\text{AB}^\text{T}=\text{B}^\text{T}\text{A}^\text{T }$
$\Rightarrow\begin{pmatrix}\begin{bmatrix}3\\5\\2\end{bmatrix}\begin{bmatrix}1&0&4 \end{bmatrix}\end{pmatrix}^\text{T}=\begin{bmatrix}1&0&4\end{bmatrix}^\text{T}\begin{bmatrix}3\\5\\2\end{bmatrix}^\text{T}$
$\Rightarrow\begin{bmatrix}3&0&12\\5&0&20\\2&0&8\end{bmatrix}^\text{T}=\begin{bmatrix}1\\0\\4\end{bmatrix}\begin{bmatrix}3&5&2 \end{bmatrix}$
$\Rightarrow\begin{bmatrix} 3&5&2\\0&0&0\\12&20&8\end{bmatrix}=\begin{bmatrix} 3&5&2\\0&0&0\\12&20&8\end{bmatrix}$
$\Rightarrow\text{LHS}=\text{RHS}$
So,
$(\text{AB})^\text{T}=\text{B}^\text{T}\text{A}^\text{T}$

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