Find X and Y, if

(i) $X + Y = \left[ {\begin{array}{cccccccccccccccccccc}7&0\\2&5\end{array}} \right]$and $X - Y = \left[ {\begin{array}{cccccccccccccccccccc}3&0\\0&3\end{array}} \right]$

(ii) $2X + 3Y = \left[ {\begin{array}{cccccccccccccccccccc}2&3\\4&0\end{array}} \right]$ and $3X + 2Y = \left[ {\begin{array}{cccccccccccccccccccc}2&{ - 2}\\{ - 1}&5\end{array}} \right]$

.:

(i)

We are given that $X + Y = \left[ {\begin{array}{cccccccccccccccccccc}7&0\\2&5\end{array}} \right]$ …..(i)

and X $-$ Y $= \left[ {\begin{array}{cccccccccccccccccccc}3&0\\0&3\end{array}} \right]$ …..(ii)

Adding (i) and (ii),

we get
$(X + Y) + (X - Y) = \left[ {\begin{array}{cccccccccccccccccccc}7&0\\2&5\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}3&0\\0&3\end{array}} \right]$
$\Rightarrow$ $2X = \left[ {\begin{array}{cccccccccccccccccccc}{10}&0\\2&8\end{array}} \right] \Rightarrow X = \left[ {\begin{array}{cccccccccccccccccccc}5&0\\1&4\end{array}} \right]$

Subtracting (ii) from (i),

we get
$\Rightarrow$ $2Y = \left[ {\begin{array}{cccccccccccccccccccc}7&0\\2&5\end{array}} \right] - \left[ {\begin{array}{cccccccccccccccccccc}3&0\\0&3\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}4&0\\2&2\end{array}} \right] \Rightarrow Y = \left[ {\begin{array}{cccccccccccccccccccc}2&0\\1&1\end{array}} \right]$

Hence, $X = \left[ {\begin{array}{cccccccccccccccccccc}5&0\\1&4\end{array}} \right]$ and $Y = \left[ {\begin{array}{cccccccccccccccccccc}2&0\\1&1\end{array}} \right]$

(ii) We have, $2X + 3Y = \left[ {\begin{array}{cccccccccccccccccccc}2&3\\4&0\end{array}} \right]$ ….(i)

and $3X + 2Y = \left[ {\begin{array}{cccccccccccccccccccc}2&{ - 2}\\{ - 1}&5\end{array}} \right]$

Adding (i) \& (ii),

we get
$5X + 5Y = \left[ {\begin{array}{cccccccccccccccccccc}2&3\\4&0\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}2&{ - 2}\\{ - 1}&5\end{array}} \right]$

$\Rightarrow$ $5X + 5 = \left[ {\begin{array}{cccccccccccccccccccc}4&1\\3&5\end{array}} \right] \Rightarrow X + Y = \cfrac{1}{5}\left[ {\begin{array}{cccccccccccccccccccc}4&1\\3&5\end{array}} \right]$ ….(iii)

Subtracting (i) from (ii),

we get
(3X + 2y)$-$(2X + 3Y)$= \left[ {\begin{array}{cccccccccccccccccccc}2&{ - 2}\\{ - 1}&5\end{array}} \right] - \left[ {\begin{array}{cccccccccccccccccccc}2&3\\4&0\end{array}} \right]$

$\Rightarrow$ $X - Y = \left[ {\begin{array}{cccccccccccccccccccc}0&{ - 5}\\{ - 5}&5\end{array}} \right]$ ….(iv)

Finally, adding (iii) and (iv), we get

$2X = \cfrac{1}{5}\left[ {\begin{array}{cccccccccccccccccccc}4&1\\3&5\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}0&{ - 5}\\{ - 5}&5\end{array}} \right]$

$\Rightarrow$ $2X = \left[ {\begin{array}{cccccccccccccccccccc}{\cfrac{4}{5}}&{\cfrac{1}{5}}\\{\cfrac{3}{5}}&1\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}0&{ - 5}\\{ - 5}&5\end{array}} \right]$

$\Rightarrow$ $2X = \left[ {\begin{array}{cccccccccccccccccccc}{\cfrac{4}{5}}&{\cfrac{1}{5} - 5}\\{\cfrac{3}{5} - 5}&{1 + 5}\end{array}} \right]$ $= \left[ {\begin{array}{cccccccccccccccccccc}{\cfrac{4}{5}}&{\cfrac{{ - 24}}{5}}\\{\cfrac{{ - 22}}{5}}&6\end{array}} \right]$

$\Rightarrow$ $X = \cfrac{1}{2}\left[ {\begin{array}{cccccccccccccccccccc}{\cfrac{4}{5}}&{\cfrac{{ - 24}}{5}}\\{\cfrac{{ - 22}}{5}}&6\end{array}} \right] \Rightarrow X = \left[ {\begin{array}{cccccccccccccccccccc}{\cfrac{2}{5}}&{\cfrac{{ - 12}}{5}}\\{\cfrac{{ - 11}}{5}}&3\end{array}} \right]$

Subtracting (iii) and (iv),

we get
$2Y = \cfrac{1}{5}\left[ {\begin{array}{cccccccccccccccccccc}4&1\\3&5\end{array}} \right] - \left[ {\begin{array}{cccccccccccccccccccc}0&{ - 5}\\{ - 5}&5\end{array}} \right]$

$\Rightarrow$ $2Y = \left[ {\begin{array}{cccccccccccccccccccc}{\cfrac{4}{5}}&{\cfrac{1}{5}}\\{\cfrac{3}{5}}&1\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}0&5\\5&{ - 5}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{\cfrac{4}{5}}&{\cfrac{{26}}{5}}\\{\cfrac{{28}}{5}}&{ - 4}\end{array}} \right]$

$\Rightarrow$ $Y = \cfrac{1}{2}\left[ {\begin{array}{cccccccccccccccccccc}{\cfrac{4}{5}}&{\cfrac{{26}}{5}}\\{\cfrac{{28}}{5}}&{ - 4}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{\cfrac{2}{5}}&{\cfrac{{13}}{5}}\\{\cfrac{{14}}{5}}&{ - 2}\end{array}} \right]$

Hence, $X - \left[ {\begin{array}{cccccccccccccccccccc}{\cfrac{2}{5}}&{\cfrac{{ - 12}}{5}}\\{\cfrac{{ - 11}}{5}}&3\end{array}} \right],Y = \left[ {\begin{array}{cccccccccccccccccccc}{\cfrac{2}{5}}&{\cfrac{{13}}{5}}\\{\cfrac{{14}}{5}}&{ - 2}\end{array}} \right]$