For the matrices $A$ and $B$,verify that $(AB)^{\prime} = B^{\prime}A^{\prime}$ where $A = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix}$ and $B = \begin{bmatrix} -1 & 2 & 1 \end{bmatrix}$.
(A) First,calculate the product $AB$:
$AB = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix} \begin{bmatrix} -1 & 2 & 1 \end{bmatrix} = \begin{bmatrix} -1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3 \end{bmatrix}$
Now,find the transpose $(AB)^{\prime}$:
$(AB)^{\prime} = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}$
Next,find $A^{\prime}$ and $B^{\prime}$:
$A^{\prime} = \begin{bmatrix} 1 & -4 & 3 \end{bmatrix}$
$B^{\prime} = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix}$
Now,calculate the product $B^{\prime}A^{\prime}$:
$B^{\prime}A^{\prime} = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix} \begin{bmatrix} 1 & -4 & 3 \end{bmatrix} = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}$
Since $(AB)^{\prime} = B^{\prime}A^{\prime}$,the property is verified.
$AB = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix} \begin{bmatrix} -1 & 2 & 1 \end{bmatrix} = \begin{bmatrix} -1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3 \end{bmatrix}$
Now,find the transpose $(AB)^{\prime}$:
$(AB)^{\prime} = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}$
Next,find $A^{\prime}$ and $B^{\prime}$:
$A^{\prime} = \begin{bmatrix} 1 & -4 & 3 \end{bmatrix}$
$B^{\prime} = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix}$
Now,calculate the product $B^{\prime}A^{\prime}$:
$B^{\prime}A^{\prime} = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix} \begin{bmatrix} 1 & -4 & 3 \end{bmatrix} = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}$
Since $(AB)^{\prime} = B^{\prime}A^{\prime}$,the property is verified.
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