Question
If $\text{A}=\begin{bmatrix}-2\\4\\5\end{bmatrix},\text{B}=\begin{bmatrix}1&3&-6\end{bmatrix},$ verify that $(AB)^T = B^TA^T$

Answer

Given: $\text{A}=\begin{bmatrix}-2\\4\\5\end{bmatrix}$
$\text{A}^\text{T}=\begin{bmatrix}-2&4&5\end{bmatrix}$
$\text{B}=\begin{bmatrix}1&3&-6\end{bmatrix}$
$\text{B}^\text{T}=\begin{bmatrix}1\\3\\-6\end{bmatrix}$
Now,
$\text{AB}=\begin{bmatrix}-2\\4\\5\end{bmatrix}\begin{bmatrix}1&3&-6\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}-2&-6&12\\4&12&-24\\5&15&-30\end{bmatrix}$
$\Rightarrow(\text{AB})^\text{T}=\begin{bmatrix}-2&4&5\\-6&12&15\\12&-24&-30\end{bmatrix}\ \dots(1)$
$\text{B}^\text{T}\text{A}^\text{A}=\begin{bmatrix}1\\3\\-6\end{bmatrix}\begin{bmatrix}-2&4&5\end{bmatrix}$
$\Rightarrow\text{B}^\text{T}\text{A}^\text{T}=\begin{bmatrix}-2&4&5\\-6&12&15\\12&-24&-30\end{bmatrix}\ \dots(2)$
$\therefore\ (\text{AB})^\text{T}=\text{B}^\text{T}\text{A}^\text{T}$ [From eqs. (1) and (2)]

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