Suppose $UT$ is invertible, then $UTZ = I$, where $I$ is the identity on $\mathbb R^3$. However, this implies that $U(TZ) = I$ , so that $U$ is invertible. But $U$ is not invertible, since by the rank-nullity theorem, its rank must be atmost two, hence it is not surjective.
You can see how to generalize this : see that $3 \geq 2$ played a role here. Now, how can you modify $3$ and $2$ so that the argument above still works?