If $A = \left[ {\begin{array}{cccccccccccccccccccc}1&2\\{ - 1}&3\end{array}} \right],B = \left[ {\begin{array}{llllllllllllllllllll}4&0\\1&5\end{array}} \right],C = \left[ {\begin{array}{cccccccccccccccccccc}2&0\\1&{ - 2}\end{array}} \right],a = 4$, and $b = - 2$, then show that
(i) $A + (B + C) = (A + B) + C$

(ii) $A(BC) = (AB)C$

(iii) $(a + b)B = aB + bB$

(iv) $a(C - A) = aC - aA$

(v) ${\left( {{A^T}} \right)^T} = A$

(vi) ${(bA)^T} = b{A^T}$

(vii) ${(AB)^T} = {B^T}{A^T}$

(viii) $(A - B)C = AC - BC$

(ix) ${(A - B)^T} = {A^T} - {B^T}$

We have, $A = \left[ {\begin{array}{cccccccccccccccccccc}1&2\\{ - 1}&3\end{array}} \right],B = \left[ {\begin{array}{llllllllllllllllllll}4&0\\1&5\end{array}} \right]$

$C = \left[ {\begin{array}{cccccccccccccccccccc}2&0\\1&{ - 2}\end{array}} \right]$ and $a = 4,b = - 2$

(i) $A + (B + C) = \left[ {\begin{array}{cccccccccccccccccccc}1&2\\{ - 1}&3\end{array}} \right] + \left[ {\begin{array}{llllllllllllllllllll}6&0\\2&3\end{array}} \right] = \left[ {\begin{array}{llllllllllllllllllll}7&2\\1&6\end{array}} \right]$

and $(A + B) + C = \left[ {\begin{array}{llllllllllllllllllll}5&2\\0&8\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}2&0\\1&{ - 2}\end{array}} \right]$

$= \left[ {\begin{array}{llllllllllllllllllll}7&2\\1&6\end{array}} \right] = A + (B + C)$

(ii) $(BC) = \left[ {\begin{array}{llllllllllllllllllll}4&0\\1&5\end{array}} \right]\left[ {\begin{array}{cccccccccccccccccccc}2&0\\1&{ - 2}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}8&0\\7&{ - 10}\end{array}} \right]$

and $A(BC) = \left[ {\begin{array}{cccccccccccccccccccc}1&2\\{ - 1}&3\end{array}} \right]\left[ {\begin{array}{cccccccccccccccccccc}8&0\\7&{ - 10}\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{8 + 14}&{0 - 20}\\{ - 8 + 21}&{0 - 30}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{22}&{ - 20}\\{13}&{ - 30}\end{array}} \right]$

$(AB) = \left[ {\begin{array}{cccccccccccccccccccc}1&2\\{ - 1}&3\end{array}} \right] \cdot \left[ {\begin{array}{llllllllllllllllllll}4&0\\1&5\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}6&{10}\\{ - 1}&{15}\end{array}} \right]$

$(AB)C = \left[ {\begin{array}{cccccccccccccccccccc}6&{10}\\{ - 1}&{15}\end{array}} \right]\left[ {\begin{array}{cccccccccccccccccccc}2&0\\1&{ - 2}\end{array}} \right]$

$= \left[ {\begin{array}{llllllllllllllllllll}{22}&{ - 20}\\{13}&{ - 30}\end{array}} \right] = A(BC)$

(iii) $(a + b)B = (4 - 2)\left[ {\begin{array}{llllllllllllllllllll}4&0\\1&5\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}8&0\\2&{10}\end{array}} \right]$
and $aB + bB = 4B - 2B$

$= \left[ {\begin{array}{cccccccccccccccccccc}{16}&0\\4&{20}\end{array}} \right] - \left[ {\begin{array}{cccccccccccccccccccc}8&0\\2&{10}\end{array}} \right]$
$= \left[ {\begin{array}{cccccccccccccccccccc}8&0\\2&{10}\end{array}} \right]$

$= (a + b)B$

(iv) $(C - A) = \left[ {\begin{array}{cccccccccccccccccccc}{2 - 1}&{0 - 2}\\{1 + 1}&{ - 2 - 3}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}1&{ - 2}\\2&{ - 5}\end{array}} \right]$

and $a(C - A) = \left[ {\begin{array}{cccccccccccccccccccc}4&{ - 8}\\8&{ - 20}\end{array}} \right]$

$aC - aA = \left[ {\begin{array}{cccccccccccccccccccc}8&0\\4&{ - 8}\end{array}} \right] - \left[ {\begin{array}{cccccccccccccccccccc}4&8\\{ - 4}&{12}\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}4&{ - 8}\\8&{ - 20}\end{array}} \right]$

$= a(C - A)$

(v) ${A^T} = {\left[ {\begin{array}{cccccccccccccccccccc}1&2\\{ - 1}&3\end{array}} \right]^T}$

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

Now, ${\left( {{A^T}} \right)^T} = {\left[ {\begin{array}{cccccccccccccccccccc}1&2\\{ - 1}&3\end{array}} \right]^ \top }$

$= A$

(v) ${A^T} = {\left[ {\begin{array}{cccccccccccccccccccc}1&2\\{ - 1}&3\end{array}} \right]^T} = \left[ {\begin{array}{cccccccccccccccccccc}1&{ - 1}\\2&3\end{array}} \right]$

Now, ${\left( {{A^T}} \right)^T} = {\left[ {\begin{array}{cccccccccccccccccccc}1&2\\{ - 1}&3\end{array}} \right]^T}$
$= A$

(vi) ${(bA)^T} = {\left[ {\begin{array}{cccccccccccccccccccc}{ - 2}&{ - 4}\\2&{ - 6}\end{array}} \right]^T}$

$= \left[ {\begin{array}{llllllllllllllllllll}{ - 2}&2\\{ - 4}&{ - 6}\end{array}} \right]$

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

$\therefore$ $b{A^T} = \left[ {\begin{array}{llllllllllllllllllll}{ - 2}&2\\{ - 4}&{ - 6}\end{array}} \right] = {(bA)^T}$

(vii) $AB = \left[ {\begin{array}{cccccccccccccccccccc}1&2\\{ - 1}&3\end{array}} \right]\left[ {\begin{array}{llllllllllllllllllll}4&0\\1&5\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{4 + 2}&{0 + 10}\\{ - 4 + 3}&{0 + 15}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}6&{10}\\{ - 1}&{15}\end{array}} \right]$

$\therefore$ ${(AB)^T} = \left[ {\begin{array}{cccccccccccccccccccc}6&{ - 1}\\{10}&{15}\end{array}} \right]$

Now, ${B^T}{A^T} = \left[ {\begin{array}{llllllllllllllllllll}4&1\\0&5\end{array}} \right]\left[ {\begin{array}{cccccccccccccccccccc}1&{ - 1}\\2&3\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}6&{ - 1}\\{10}&{15}\end{array}} \right]$

$= {(AB)^T}$

(viii) $(A - B) = \left[ {\begin{array}{cccccccccccccccccccc}{1 - 4}&{2 - 0}\\{ - 1 - 1}&{3 - 5}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{ - 3}&2\\{ - 2}&{ - 2}\end{array}} \right]$

$(A - B)C = \left[ {\begin{array}{cccccccccccccccccccc}{ - 3}&2\\{ - 2}&{ - 2}\end{array}} \right]\left[ {\begin{array}{cccccccccccccccccccc}2&0\\1&{ - 2}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{ - 4}&{ - 4}\\{ - 6}&4\end{array}} \right]$

……(i)
Now, $AC = \left[ {\begin{array}{cccccccccccccccccccc}1&2\\{ - 1}&3\end{array}} \right]\left[ {\begin{array}{cccccccccccccccccccc}2&0\\1&{ - 2}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}4&{ - 4}\\1&{ - 6}\end{array}} \right]$

……..(ii)
and $BC = \left[ {\begin{array}{llllllllllllllllllll}4&0\\1&5\end{array}} \right]\left[ {\begin{array}{llllllllllllllllllll}2&0\\1&{ - 2}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}8&0\\7&{ - 10}\end{array}} \right]$

…..(iii)
$\therefore$ $AC - BC = \left[ {\begin{array}{cccccccccccccccccccc}{4 - 8}&{ - 4 - 0}\\{1 - 7}&{ - 6 + 10}\end{array}} \right]$

[using Eqs. (ii) and (iii)]
$= \left[ {\begin{array}{cccccccccccccccccccc}{ - 4}&{ - 4}\\{ - 6}&4\end{array}} \right]$

$= (A - B)C$

[using Eq. (i)]

(ix) ${(A - B)^T} = {\left[ {\begin{array}{cccccccccccccccccccc}{1 - 4}&{2 - 0}\\{ - 1 - 1}&{3 - 5}\end{array}} \right]^T}$

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

${A^T} - {B^T} = \left[ {\begin{array}{cccccccccccccccccccc}1&{ - 1}\\2&3\end{array}} \right] - \left[ {\begin{array}{cccccccccccccccccccc}4&1\\0&5\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{ - 3}&{ - 2}\\2&{ - 2}\end{array}} \right] = {(A - B)^T}$