## Elementary Linear Algebra 7th Edition

$S$ is a basis for $R^3$.
The set $S=\{(1,5,3),(0,1,2),(0,0,6)\}$ is a linearly independent set of vectors. Indeed, assume that $$a(1,5,3)+b(0,1,2)+c(0,0,6)=(0,0,0), \quad a,b,c\in R.$$ Then, we have the following system of equations \begin{align*} a&=0\\ 5a+b&=0\\ 3a+2b+6c&=0. \end{align*} Solving the above equations, we find that $a=0,b=0,c=0$ and hence $S$ is linearly independent. Now, since $R^3$ is vector space of dimension $2$, then by Theorem 4.12, $S$ is a basis for $R^3$.