If $X = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&1&{ - 1}\\5&{ - 2}&{ - 3}\end{array}} \right]$ and $Y = \left[ {\begin{array}{cccccccccccccccccccc}2&1&{ - 1}\\7&2&4\end{array}} \right]$, then find
(i) $X + Y$.
(ii) $2X - 3Y$.
(iii) a matrix $Z$ such that $X + Y + Z$ is a zero matrix.

We have, $X = {\left[ {\begin{array}{cccccccccccccccccccc}3&1&{ - 1}\\5&{ - 2}&{ - 3}\end{array}} \right]_{2 \times 3}}$ and $Y = {\left[ {\begin{array}{cccccccccccccccccccc}2&1&{ - 1}\\7&2&4\end{array}} \right]_{2 \times 3}}$

(i) $X + Y = \left[ {\begin{array}{cccccccccccccccccccc}{3 + 2}&{1 + 1}&{ - 1 - 1}\\{5 + 7}&{ - 2 + 2}&{ - 3 + 4}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}5&2&{ - 2}\\{12}&0&1\end{array}} \right]$

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

and $3Y = 3\left[ {\begin{array}{cccccccccccccccccccc}2&1&{ - 1}\\7&2&4\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}6&3&{ - 3}\\{21}&6&{12}\end{array}} \right]$

$\therefore$ $2X - 3Y = \left[ {\begin{array}{cccccccccccccccccccc}{6 - 6}&{2 - 3}&{ - 2 + 3}\\{10 - 21}&{ - 4 - 6}&{ - 6 - 12}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}0&{ - 1}&1\\{ - 11}&{ - 10}&{ - 18}\end{array}} \right]$

(iii) $X + Y = \left[ {\begin{array}{cccccccccccccccccccc}{3 + 2}&{1 + 1}&{ - 1 - 1}\\{5 + 7}&{ - 2 + 2}&{ - 3 + 4}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}5&2&{ - 2}\\{12}&0&{ + 1}\end{array}} \right]$

Hence, $X + Y + Z = \left[ {\begin{array}{llllllllllllllllllll}0&0&0\\0&0&0\end{array}} \right]$

We see that $Z$ is the additive inverse of $(X + Y)$ or negative of $(X + Y)$.

$\therefore$ $Z = \left[ {\begin{array}{cccccccccccccccccccc}{ - 5}&{ - 2}&2\\{ - 12}&0&{ - 1}\end{array}} \right]$