If then verify that:

Given, $A=\left[\begin{array}{ll}1 & 2 \\ 4 & 1 \\ 5 & 6\end{array}\right]$ and $B=\left[\begin{array}{ll}1 & 2 \\ 6 & 4 \\ 7 & 3\end{array}\right]$

(i) $(2 A+B)=\left[\begin{array}{cc}2 & 4 \\ 8 & 2 \\ 10 & 12\end{array}\right]+\left[\begin{array}{cc}1 & 2 \\ 6 & 4 \\ 7 & 3\end{array}\right]=\left[\begin{array}{cc}3 & 6 \\ 14 & 6 \\ 17 & 15\end{array}\right]$

And, $(2 A+B)^{\prime}=\left[\begin{array}{ccc}3 & 14 & 17 \\ 6 & 6 & 15\end{array}\right]$

Also, $2 A^{\prime}+B^{\prime}$

$=\left[\begin{array}{lll}2 & 8 & 10 \\ 4 & 2 & 12\end{array}\right]+\left[\begin{array}{lll}1 & 6 & 7 \\ 2 & 4 & 3\end{array}\right]=\left[\begin{array}{ccc}3 & 14 & 17 \\ 6 & 6 & 15\end{array}\right]=(2 A+B)^{\prime}$

Hence, $2 A^{\prime}+B^{\prime}=(2 A+B)^{\prime}$

(ii) $A-B=\left[\begin{array}{ll}1 & 2 \\ 4 & 1 \\ 5 & 6\end{array}\right]-\left[\begin{array}{ll}1 & 2 \\ 6 & 4 \\ 7 & 3\end{array}\right]=\left[\begin{array}{cc}0 & 0 \\ -2 & -3 \\ -2 & 3\end{array}\right]$

And, $\quad(A-B)^{\prime}=\left[\begin{array}{ccc}0 & -2 & -2 \\ 0 & -3 & 3\end{array}\right]$

Also, $\quad A^{\prime}-B^{\prime}=\left[\begin{array}{lll}1 & 4 & 5 \\ 2 & 1 & 6\end{array}\right]-\left[\begin{array}{lll}1 & 6 & 7 \\ 2 & 4 & 3\end{array}\right]=\left[\begin{array}{ccc}0 & -2 & -2 \\ 0 & -3 & 3\end{array}\right]=(A-B)^{\prime}$

Thus, $\quad A^{\prime}-B^{\prime}=(A-B)^{\prime}$

- Hence proved