For the matrices A and B, verify that (AB)’ =B’A’, where

(i) $A = \left[ {\begin{array}{cccccccccccccccccccc}1\\{ - 4}\\3\end{array}} \right],B = [\begin{array}{cccccccccccccccccccc}{ - 1}&2&{1]}\end{array}$

(ii) $A = \left[ {\begin{array}{cccccccccccccccccccc}0\\1\\2\end{array}} \right],B = [\begin{array}{cccccccccccccccccccc}1&5&{7]}\end{array}$

.:

(i) $A = \left[ {\begin{array}{cccccccccccccccccccc}1\\{ - 4}\\3\end{array}} \right] \Rightarrow A' = [\begin{array}{cccccccccccccccccccc}1&{ - 4}&{3]}\end{array}$

and $B = [\begin{array}{cccccccccccccccccccc}{ - 1}&2&{1]}\end{array} \Rightarrow B' = \left[ {\begin{array}{cccccccccccccccccccc}{ - 1}\\2\\1\end{array}} \right]$

$AB = \left[ {\begin{array}{cccccccccccccccccccc}1\\{ - 4}\\3\end{array}} \right][\begin{array}{cccccccccccccccccccc}{ - 1}&2&{1]}\end{array}$

$= \left[ {\begin{array}{cccccccccccccccccccc}{1 \times ( - 1)}&{1 \times 2}&{1 \times 1}\\{ - 4 \times ( - 1)}&{( - 4) \times 2}&{( - 4) \times 1}\\{3 \times ( - 1)}&{3 \times 2}&{3 \times 1}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{ - 1}&2&1\\4&{ - 8}&{ - 4}\\{ - 3}&6&3\end{array}} \right]$

L.H.S. = $(AB)’ =\left[ {\begin{array}{cccccccccccccccccccc}{ - 1}&4&{ - 3}\\2&{ - 8}&6\\1&{ - 4}&3\end{array}} \right]$

R.H.S. = $B’A’ =\left[ {\begin{array}{cccccccccccccccccccc}{ - 1}\\2\\1\end{array}} \right]\left[ {\begin{array}{cccccccccccccccccccc}1&{ - 4}&3\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{ - 1 \times 1}&{ - 1 \times ( - 4)}&{ - 1 \times 3}\\{2 \times 1}&{2 \times ( - 4)}&{2 \times 3}\\{1 \times 1}&{1 \times ( - 4)}&{1 \times 3}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{ - 1}&4&{ - 3}\\2&{ - 8}&6\\1&{ - 4}&3\end{array}} \right]$

Hence, $(AB)’ = B’A’.$

(ii) $A = \left[ {\begin{array}{cccccccccccccccccccc}0\\1\\2\end{array}} \right],B = [\begin{array}{cccccccccccccccccccc}1&5&7\end{array}] \Rightarrow A' = [\begin{array}{cccccccccccccccccccc}0&1&2\end{array}]$

and $B' = \left[ {\begin{array}{cccccccccccccccccccc}1\\5\\7\end{array}} \right]$

Now, $AB = \left[ {\begin{array}{cccccccccccccccccccc}0\\1\\2\end{array}} \right][\begin{array}{cccccccccccccccccccc}1&5&7\end{array}]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{0 \times 1}&{0 \times 5}&{0 \times 7}\\{1 \times 1}&{1 \times 5}&{1 \times 7}\\{2 \times 1}&{2 \times 5}&{2 \times 7}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}0&0&0\\1&5&7\\2&{10}&{14}\end{array}} \right]$

$\Rightarrow$ $(AB)'$

% $= {\left[ {\begin{array}{cccccccccccccccccccc}0&0&0\\1&5&7\\2&{10}&{14}\end{array}} \right]^{\prime}}$

$= \left[ {\begin{array}{cccccccccccccccccccc}0&1&2\\0&5&{10}\\0&7&{14}\end{array}} \right]$

Now, $B'A' = \left[ {\begin{array}{cccccccccccccccccccc}1\\5\\7\end{array}} \right][\begin{array}{cccccccccccccccccccc}0&1&2\end{array}]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{1 \times 0}&{1 \times 1}&{1 \times 2}\\{5 \times 0}&{5 \times 1}&{5 \times 2}\\{7 \times 0}&{7 \times 1}&{7 \times 2}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}0&1&2\\0&5&{10}\\0&7&{14}\end{array}} \right]$

$\therefore$ $(AB)' = B'A'$.