If $X$ and $Y$ are $2 \times 2$ matrices, then solve the following matrix equations for $X$ and $Y$

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

We have,
$2X + 3Y = \left[ {\begin{array}{llllllllllllllllllll}2&3\\4&0\end{array}} \right]$

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

……(ii)
On subtracting Eq. (i) from Eq. (ii),

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

$(X - Y) = \left[ {\begin{array}{llllllllllllllllllll}{ - 4}&{ - 1}\\{ - 3}&{ - 5}\end{array}} \right]$

….(iii)
On adding Eqs. (i) and (ii),

we get
$(5X + 5Y) = \left[ {\begin{array}{llllllllllllllllllll}0&5\\5&{ - 5}\end{array}} \right]$

$\Rightarrow$ $(X + Y) = \frac{1}{5}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&5\\5&{ - 5}\end{array}} \right] = \left[ {\begin{array}{llllllllllllllllllll}0&1\\1&{ - 1}\end{array}} \right]$

……(iv)
On adding Eqs. (iii) and (iv),

we get
$(X - Y) + (X + Y) = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 4}&0\\{ - 2}&{ - 6}\end{array}} \right]$

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

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

From Eq. (iv),
$\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 2}&0\\{ - 1}&{ - 3}\end{array}} \right] + Y = \left[ {\begin{array}{llllllllllllllllllll}0&1\\1&{ - 1}\end{array}} \right]$

$\therefore$ $Y = \left[ {\begin{array}{llllllllllllllllllll}2&1\\2&2\end{array}} \right]$ and $X = \left[ {\begin{array}{llllllllllllllllllll}{ - 2}&0\\{ - 1}&{ - 3}\end{array}} \right]$