If then verify that:

Solution:

Given, $A=\left[\begin{array}{ccc}0 & -1 & 2 \\ 4 & 3 & -4\end{array}\right]$ and $B=\left[\begin{array}{ll}4 & 0 \\ 1 & 3 \\ 2 & 6\end{array}\right]$

(i) We have to verify that, $\left(A^{\prime}\right)^{\prime}=A$

So, $A^{\prime}=\left[\begin{array}{cc}0 & 4 \\ -1 & 3 \\ 2 & -4\end{array}\right]$

And, $\left(A^{\prime}\right)^{\prime}=\left[\begin{array}{ccc}0 & -1 & 2 \\ 4 & 3 & -4\end{array}\right]=A$

(ii) We have to verify that, $(A B)^{\prime}=B^{\prime} A^{\prime}$

So, $\quad A B=\left[\begin{array}{ccc}0^{\circ} & -1 & 2 \\ 4 & 3 & -4\end{array}\right]\left[\begin{array}{ll}4 & 0 \\ 1 & 3 \\ 2 & 6\end{array}\right]=\left[\begin{array}{cc}3 & 9 \\ 11 & -15\end{array}\right]$

$(A B)^{\prime}=\left[\begin{array}{cc}3 & 11 \\ 9 & -15\end{array}\right]$

and, $B^{\prime} A^{\prime}=\left[\begin{array}{ccc}4 & 1 & 2 \\ 0 & 3 & 6\end{array}\right]\left[\begin{array}{cc}0 & 4 \\ -1 & 3 \\ 2 & -4\end{array}\right]=\left[\begin{array}{cc}3 & 11 \\ 9 & -15\end{array}\right]=(A B)^{\prime}$

Hence proved.

(iii) We have to verify that, $(k A)^{\prime}=\left(k A^{\prime}\right)$

Now, $(k A)=\left[\begin{array}{ccc}0 & -k & 2 k \\ 4 k & 3 k & -4 k\end{array}\right]$

And,  $(k A)^{\prime}=\left[\begin{array}{cc}0 & 4 k \\ -k & 3 k \\ 2 k & -4 k\end{array}\right]$

Also, $\quad k A^{\prime}=\left[\begin{array}{cc}0 & 4 k \\ -k & 3 k \\ 2 k & -4 k\end{array}\right]=(k A)^{\prime}$

Hence proved.