Answer
$$0=
\left[\begin{array}{rrr}{0} & {0} & {0} \\ {0} & {0} & {0} \end{array}\right].
$$
$$-u=\left[\begin{array}{rrr}{-a_{11}} & {-a_{12}} & {-a_{13}} \\ {-a_{21}} & {-a_{22}} & {-a_{23}} \end{array}\right].$$
Work Step by Step
The zero vector of the vector space $M_{2,2}$ is given by
$$0=
\left[\begin{array}{rrr}{0} & {0} & {0} \\ {0} & {0} & {0} \end{array}\right].
$$
Let $u$ be an vector in $M_{2,3}$ such that
$$u=\left[\begin{array}{rrr}{a_{11}} & {a_{12}} & {a_{13}} \\ {a_{21}} & {a_{22}} & {a_{23}} \end{array}\right].$$
Now, the additive inverse of $u$ is given by
$$-u=\left[\begin{array}{rrr}{-a_{11}} & {-a_{12}} & {-a_{13}} \\ {-a_{21}} & {-a_{22}} & {-a_{23}} \end{array}\right].$$