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 $(kB)^{\prime} = kB^{\prime}$,where $k$ is any constant.
(N/A) We have $B = \begin{bmatrix} 2 & -1 & 2 \\ 1 & 2 & 4 \end{bmatrix}$.
Multiplying by a constant $k$,we get $kB = k \begin{bmatrix} 2 & -1 & 2 \\ 1 & 2 & 4 \end{bmatrix} = \begin{bmatrix} 2k & -k & 2k \\ k & 2k & 4k \end{bmatrix}$.
Now,taking the transpose of $kB$,we get $(kB)^{\prime} = \begin{bmatrix} 2k & k \\ -k & 2k \\ 2k & 4k \end{bmatrix}$.
We can factor out $k$ from the matrix: $(kB)^{\prime} = k \begin{bmatrix} 2 & 1 \\ -1 & 2 \\ 2 & 4 \end{bmatrix}$.
Since $B^{\prime} = \begin{bmatrix} 2 & 1 \\ -1 & 2 \\ 2 & 4 \end{bmatrix}$,it follows that $(kB)^{\prime} = kB^{\prime}$.
Thus,the property is verified.
Multiplying by a constant $k$,we get $kB = k \begin{bmatrix} 2 & -1 & 2 \\ 1 & 2 & 4 \end{bmatrix} = \begin{bmatrix} 2k & -k & 2k \\ k & 2k & 4k \end{bmatrix}$.
Now,taking the transpose of $kB$,we get $(kB)^{\prime} = \begin{bmatrix} 2k & k \\ -k & 2k \\ 2k & 4k \end{bmatrix}$.
We can factor out $k$ from the matrix: $(kB)^{\prime} = k \begin{bmatrix} 2 & 1 \\ -1 & 2 \\ 2 & 4 \end{bmatrix}$.
Since $B^{\prime} = \begin{bmatrix} 2 & 1 \\ -1 & 2 \\ 2 & 4 \end{bmatrix}$,it follows that $(kB)^{\prime} = kB^{\prime}$.
Thus,the property is verified.
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