Answer
$S$ is a basis for $R^4$.
Work Step by Step
The set $S=\{(-1,2,0,0),(2,0,-1,0),(3,0,0,4),(0,0,5,0)\}$
is not a basis for $R^4$ because $S$ is linearly independent set of vectors. Indeed, assume that
$$a(-1,2,0,0)+b(2,0,-1,0)+c(3,0,0,4)+d (0,0,5,0)=(0,0,0,0), \quad a,b,c,d\in R.$$
Which yields the following system of equations
\begin{align*}
-a+2b+3c&=0\\
2a&=0\\
-b+5d&=0\\
4c&=0.
\end{align*}
Solving the above system of equations, we find that $a=0,b=0,c=0,d=0$ and hence $S$ is linearly independent set of vectors. Now, since $R^4$ is vector space of dimension $4$, then by Theorem 4.12, $S$ is a basis for $R^4$.
