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

(i) $(AB)C = A(BC)$.

(ii) $A(B + C) = AB + AC$.

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

and $C = \left[ {\begin{array}{cccccccccccccccccccc}1&0\\{ - 1}&0\end{array}} \right]$

(i) $(AB) = \left[ {\begin{array}{cccccccccccccccccccc}1&2\\{ - 2}&1\end{array}} \right]\left[ {\begin{array}{cccccccccccccccccccc}2&3\\3&{ - 4}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{2 + 6}&{3 - 8}\\{ - 4 + 3}&{ - 6 - 4}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}8&{ - 5}\\{ - 1}&{ - 10}\end{array}} \right]$

and $(AB)C = \left[ {\begin{array}{cccccccccccccccccccc}8&{ - 5}\\{ - 1}&{ - 10}\end{array}} \right]\left[ {\begin{array}{cccccccccccccccccccc}1&0\\{ - 1}&0\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{8 + 5}&0\\{ - 1 + 10}&0\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{13}&0\\9&0\end{array}} \right]$

……(i)
Again, $(BC) = \left[ {\begin{array}{cccccccccccccccccccc}2&3\\3&{ - 4}\end{array}} \right]\left[ {\begin{array}{cccccccccccccccccccc}1&0\\{ - 1}&0\end{array}} \right]$

$= \left[ {\begin{array}{llllllllllllllllllll}{2 - 3}&0\\{3 + 4}&0\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{ - 1}&0\\7&0\end{array}} \right]$

and $A(BC) = \left[ {\begin{array}{cccccccccccccccccccc}1&2\\{ - 2}&1\end{array}} \right]\left[ {\begin{array}{cccccccccccccccccccc}{ - 1}&0\\7&0\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{ - 1 + 14}&0\\{ + 2 + 7}&0\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{13}&0\\9&0\end{array}} \right]$

…….(ii)
$\therefore$ $(AB)C = A(BC)$

[using Eqs. (i) and (ii)]
(ii)$(B + C) = \left[ {\begin{array}{cccccccccccccccccccc}2&3\\3&{ - 4}\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}1&0\\{ - 1}&0\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}3&3\\2&{ - 4}\end{array}} \right]$

and $A \cdot (B + C) = \left[ {\begin{array}{cccccccccccccccccccc}1&2\\{ - 2}&1\end{array}} \right]\left[ {\begin{array}{cccccccccccccccccccc}3&3\\2&{ - 4}\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{3 + 4}&{3 - 8}\\{ - 6 + 2}&{ - 6 - 4}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}7&{ - 5}\\{ - 4}&{ - 10}\end{array}} \right]$

….(iii)
Also, $AB = \left[ {\begin{array}{cccccccccccccccccccc}1&2\\{ - 2}&1\end{array}} \right] \cdot \left[ {\begin{array}{cccccccccccccccccccc}2&3\\3&{ - 4}\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{2 + 6}&{3 - 8}\\{ - 4 + 3}&{ - 6 - 4}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}8&{ - 5}\\{ - 1}&{ - 10}\end{array}} \right]$

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

$= \left[ {\begin{array}{cccccccccccccccccccc}{2 + 6}&{3 - 8}\\{ - 4 + 3}&{ - 6 - 4}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}8&{ - 5}\\{ - 1}&{ - 10}\end{array}} \right]$

and $AC = \left[ {\begin{array}{cccccccccccccccccccc}1&2\\{ - 2}&1\end{array}} \right]\left[ {\begin{array}{cccccccccccccccccccc}1&0\\{ - 1}&0\end{array}} \right]$

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

$\therefore$ $AB + AC = \left[ {\begin{array}{cccccccccccccccccccc}8&{ - 5}\\{ - 1}&{ - 10}\end{array}} \right] + \left[ {\begin{array}{llllllllllllllllllll}{ - 1}&0\\{ - 3}&0\end{array}} \right]$

$\Rightarrow$ $AB + AC = \left[ {\begin{array}{cccccccccccccccccccc}7&{ - 5}\\{ - 4}&{ - 10}\end{array}} \right]$

……..(iv)
From Eqs. (iii) and (iv),
$A(B + C) = AB + AC$