## Elementary Linear Algebra 7th Edition

$S$ is a basis for $R^4$.
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 $R^4$.