If $A = \begin{bmatrix} -2 \\ 4 \\ 5 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 3 & -6 \end{bmatrix}$,verify that $(AB)^{\prime} = B^{\prime} A^{\prime}$.

(A) Given $A = \begin{bmatrix} -2 \\ 4 \\ 5 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 3 & -6 \end{bmatrix}$.
First,calculate the product $AB$:
$AB = \begin{bmatrix} -2 \\ 4 \\ 5 \end{bmatrix} \begin{bmatrix} 1 & 3 & -6 \end{bmatrix} = \begin{bmatrix} -2 & -6 & 12 \\ 4 & 12 & -24 \\ 5 & 15 & -30 \end{bmatrix}$.
Now,find the transpose of the product $(AB)^{\prime}$:
$(AB)^{\prime} = \begin{bmatrix} -2 & 4 & 5 \\ -6 & 12 & 15 \\ 12 & -24 & -30 \end{bmatrix}$.
Next,find the transposes of $A$ and $B$:
$A^{\prime} = \begin{bmatrix} -2 & 4 & 5 \end{bmatrix}$ and $B^{\prime} = \begin{bmatrix} 1 \\ 3 \\ -6 \end{bmatrix}$.
Now,calculate the product $B^{\prime} A^{\prime}$:
$B^{\prime} A^{\prime} = \begin{bmatrix} 1 \\ 3 \\ -6 \end{bmatrix} \begin{bmatrix} -2 & 4 & 5 \end{bmatrix} = \begin{bmatrix} -2 & 4 & 5 \\ -6 & 12 & 15 \\ 12 & -24 & -30 \end{bmatrix}$.
Comparing the results,we see that $(AB)^{\prime} = B^{\prime} A^{\prime}$. Hence,the property is verified.