Show that

(i)$\left[ {\begin{array}{cccccccccccccccccccc}5&{ - 1}\\6&7\end{array}} \right]\left[ {\begin{array}{cccccccccccccccccccc}2&1\\3&4\end{array}} \right] \ne \left[ {\begin{array}{cccccccccccccccccccc}2&1\\3&4\end{array}} \right]\left[ {\begin{array}{cccccccccccccccccccc}5&{ - 1}\\6&7\end{array}} \right]$

(ii) $\left[ {\begin{array}{cccccccccccccccccccc}1&2&3\\0&1&0\\1&1&0\end{array}} \right]\left[ {\begin{array}{cccccccccccccccccccc}{ - 1}&1&0\\0&{ - 1}&1\\2&3&4\end{array}} \right] \ne \left[ {\begin{array}{cccccccccccccccccccc}{ - 1}&1&0\\0&{ - 1}&1\\2&3&4\end{array}} \right]\left[ {\begin{array}{cccccccccccccccccccc}1&2&3\\0&1&0\\1&1&0\end{array}} \right]$

.:

(i)

Let A$= \left[ {\begin{array}{cccccccccccccccccccc}5&{ - 1}\\6&7\end{array}} \right]$and $B = \left[ {\begin{array}{cccccccccccccccccccc}2&1\\3&4\end{array}} \right]$
$\Rightarrow$ $AB = \left[ {\begin{array}{cccccccccccccccccccc}5&{ - 1}\\6&7\end{array}} \right]\left[ {\begin{array}{cccccccccccccccccccc}2&1\\3&4\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{10 - 3}&{5 - 4}\\{12 + 21}&{6 + 28}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}7&1\\{33}&{34}\end{array}} \right]$

and

$BA = \left[ {\begin{array}{cccccccccccccccccccc}2&1\\3&4\end{array}} \right]\left[ {\begin{array}{cccccccccccccccccccc}5&{ - 1}\\6&7\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{10 + 6}&{ - 2 + 7}\\{15 + 24}&{ - 3 + 28}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{16}&5\\{39}&{25}\end{array}} \right]$

Hence, AB $\ne$ BA.

(ii)

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

$\Rightarrow$ $AB = \left[ {\begin{array}{cccccccccccccccccccc}1&2&3\\0&1&0\\1&1&0\end{array}} \right]\left[ {\begin{array}{cccccccccccccccccccc}{ - 1}&1&0\\0&{ - 1}&1\\2&3&4\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{1 \times ( - 1) + 2 \times 0 + 3 \times 2}&{1 \times 1 + 2 \times ( - 1) + 3 \times 3}&{1 \times 0 + 2 \times 1 + 3 \times 4}\\{0 \times ( - 1) + 1 \times 0 + 0 \times 2}&{0 \times 1 + 1 \times ( - 1) + 0 \times 3}&{0 \times 0 + 1 \times 1 + 0 \times 4}\\{1 \times ( - 1) + 1 \times 0 + 0 \times 2}&{1 \times 1 + 1 \times ( - 1) + 0 \times 3}&{1 \times 0 + 1 \times 1 + 0 \times 4}\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{ - 1 + 0 + 6}&{1 - 2 + 9}&{0 + 2 + 12}\\{0 + 0 + 0}&{0 - 1 + 0}&{0 + 1 + 0}\\{ - 1 + 0 + 0}&{1 - 1 + 0}&{0 + 1 + 0}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}5&8&{14}\\0&{ - 1}&1\\{ - 1}&0&1\end{array}} \right]$

and $BA = \left[ {\begin{array}{cccccccccccccccccccc}{ - 1}&1&0\\0&{ - 1}&1\\2&3&4\end{array}} \right]\left[ {\begin{array}{cccccccccccccccccccc}1&2&3\\0&1&0\\1&1&0\end{array}} \right]$
$= \left[ {\begin{array}{cccccccccccccccccccc}{ - 1 \times 1 + 1 \times 0 + 0 \times 1}&{ - 1 \times 2 + 1 \times 1 + 0 \times 1}&{ - 1 \times 3 + 1 \times 0 + 0 \times 0}\\{0 \times 1 - 1 \times 0 + 1 \times 1}&{0 \times 2 - 1 \times 1 + 1 \times 1}&{0 \times 3 - 1 \times 0 + 1 \times 0}\\{2 \times 1 + 3 \times 0 + 4 \times 1}&{2 \times 2 + 3 \times 1 + 4 \times 1}&{2 \times 3 + 3 \times 0 + 4 \times 0}\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{ - 1 + 0 + 0}&{ - 2 + 1 + 0}&{ - 3 + 0 + 0}\\{0 - 0 + 1}&{0 - 1 + 1}&{0 - 0 + 0}\\{2 + 0 + 4}&{4 + 3 + 4}&{6 + 0 + 0}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{ - 1}&{ - 1}&{ - 3}\\1&0&0\\6&{11}&6\end{array}} \right]$

Hence, L.H.S.$\ne$ R.H.S.