## Elementary Linear Algebra 7th Edition

The set $S=\{(1,3,0),(4,1,2),(-2,5,-2)\}$ is not a basis for $R^3$.
The set $S=\{(1,3,0),(4,1,2),(-2,5,-2)\}$ is not a basis for $R^3$ because the vectors in $S$ are not linearly independent. For example, one can see that $$(-2,5,-2)=2(1,3,0)-(4,1,2).$$ So we have the non trivial combination $$(-2,5,-2)-2(1,3,0)+(4,1,2)=(0,0,0).$$ Which means that the vectors $\{(1,3,0),(4,1,2),(-2,5,-2)\}$ are not Linearly independent. Hence, the set $S=\{(1,3,0),(4,1,2),(-2,5,-2)\}$ is not a basis for $R^3$.