## Elementary Linear Algebra 7th Edition

$S$ does not span $R^3$.
The set $S=\{(7,0,3),(8,-4,1)\}$ is not a basis for $R^3$ because $S$ does not span $R^3$. For example, suppose the vector $(1,0,0)\in R^2$ and writing it as a linear combination of the elements of $S$, we get $$(1,0,0)=c_1(7,0,3)+c_2(8,-4,1), \quad c_1,c_2\in R.$$ Which yields the following system of equations \begin{align*} 7c_1+8c_2&=1\\ -4c_2&=0\\ 3c_1+c_2&=0. \end{align*} It is easy to note that the above system is in-consistent, since the second and third equations lead to $c_1=0,c_2=0$ and these values do not satisfy the first equation.