Answer
See answer below
Work Step by Step
The vector $S$ has 5 independent elements
and we also have $dim (M_{m n}(R))=m\times n=3 \times 2 =6$
For $S$ to be a basis of $M_{3 \times 2}(R)$, $S$ should have 6 linearly independent elements.
Hence, $S$ is not a basis of $M_{3 \times 2}(R)$
