If (i) X + Y

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

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

(ii) $2 X-3 Y=2\left[\begin{array}{ccc}3 & 1 & -1 \\ 5 & -2 & -3\end{array}\right]-3\left[\begin{array}{ccc}2 & 1 & -1 \\ 7 & 2 & 4\end{array}\right]$

$=\left[\begin{array}{ccc}6 & 2 & -2 \\ 10 & -4 & -6\end{array}\right]-\left[\begin{array}{ccc}6 & 3 & -3 \\ 21 & 6 & 12\end{array}\right]$

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

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

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

So, $Z$ is the additive inverse of $(X+Y)$ or negative of $(X+Y)$.

Therefore, $Z=-(X+Y)=\left[\begin{array}{ccc}-5 & -2 & 2 \\ -12 & 0 & -1\end{array}\right]$