# Chapter 4 - Vector Spaces - 4.5 Basis and Dimension - 4.5 Exercises - Page 187: 16

The set $S=\{(2,1,-2),(-2,-1,2),(4,2,-4)\}$ is not a basis for $R^3$ .

#### Work Step by Step

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

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