If $A = \left[ {\begin{array}{cccccccccccccccccccc}0&{ - 1}&2\\4&3&{ - 4}\end{array}} \right]$ and $B = \left[ {\begin{array}{llllllllllllllllllll}4&0\\1&3\\2&6\end{array}} \right]$, then verify that
(i) ${\left( {{A^\prime }} \right)^\prime } = A$
(ii) ${(AB)^\prime } = {B^\prime }{A^\prime }$
(iii) ${(kA)^\prime } = \left( {k{A^\prime }} \right)$.

We have, $A = \left[ {\begin{array}{cccccccccccccccccccc}0&{ - 1}&2\\4&3&{ - 4}\end{array}} \right]$ and $B = \left[ {\begin{array}{llllllllllllllllllll}4&0\\1&3\\2&6\end{array}} \right]$

(i) We have to verify that,

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

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

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

$\therefore$ $AB = \left[ {\begin{array}{cccccccccccccccccccc}3&9\\{11}&{ - 15}\end{array}} \right]$
$\Rightarrow$ ${(AB)^\prime } = \left[ {\begin{array}{cccccccccccccccccccc}3&{11}\\9&{ - 15}\end{array}} \right]$

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

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

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

and ${(kA)^\prime } = \left[ {\begin{array}{cccccccccccccccccccc}0&{4k}\\{ - k}&{3k}\\{2k}&{ - 4k}\end{array}} \right]$

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