If $A = \begin{bmatrix} 3 & \sqrt{3} & 2 \\ 4 & 2 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & -1 & 2 \\ 1 & 2 & 4 \end{bmatrix}$,verify that $(A+B)^{\prime} = A^{\prime} + B^{\prime}$.

(N/A) Given $A = \begin{bmatrix} 3 & \sqrt{3} & 2 \\ 4 & 2 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & -1 & 2 \\ 1 & 2 & 4 \end{bmatrix}$.
First,calculate $A+B$:
$A+B = \begin{bmatrix} 3+2 & \sqrt{3}-1 & 2+2 \\ 4+1 & 2+2 & 0+4 \end{bmatrix} = \begin{bmatrix} 5 & \sqrt{3}-1 & 4 \\ 5 & 4 & 4 \end{bmatrix}$.
Now,find the transpose $(A+B)^{\prime}$:
$(A+B)^{\prime} = \begin{bmatrix} 5 & 5 \\ \sqrt{3}-1 & 4 \\ 4 & 4 \end{bmatrix}$.
Next,find $A^{\prime}$ and $B^{\prime}$:
$A^{\prime} = \begin{bmatrix} 3 & 4 \\ \sqrt{3} & 2 \\ 2 & 0 \end{bmatrix}$,$B^{\prime} = \begin{bmatrix} 2 & 1 \\ -1 & 2 \\ 2 & 4 \end{bmatrix}$.
Finally,calculate $A^{\prime} + B^{\prime}$:
$A^{\prime} + B^{\prime} = \begin{bmatrix} 3+2 & 4+1 \\ \sqrt{3}-1 & 2+2 \\ 2+2 & 0+4 \end{bmatrix} = \begin{bmatrix} 5 & 5 \\ \sqrt{3}-1 & 4 \\ 4 & 4 \end{bmatrix}$.
Since $(A+B)^{\prime} = A^{\prime} + B^{\prime}$,the property is verified.