## Elementary Linear Algebra 7th Edition

The set $S=\{(-1,2),(1,-2),(2,4)\}$ is not a basis for $R^2$.
The set $S=\{(-1,2),(1,-2),(2,4)\}$ is not a basis for $R^2$ because the vectors in $S$ are not linearly independent. Indeed, considering the linear combination $$a (-1,2)+b(1,-2)+c(2,4)=(0,0),$$ then we have the system \begin{align*} -a+b+2c&=0\\ 2a-2b+4c&=0 \end{align*} It is easy to see that the above system has the solution $a=t, b=t , c=0$ . Hence, the vectors of $S$ are not linearly independent. Hence, the set $S=\{(-1,2),(1,-2),(2,4)\}$ is not a basis for $R^2$