Elementary Linear Algebra 7th Edition

Published by Cengage Learning
ISBN 10: 1-13311-087-8
ISBN 13: 978-1-13311-087-3

Chapter 4 - Vector Spaces - 4.5 Basis and Dimension - 4.5 Exercises - Page 187: 11

Answer

the set $S=\{(6,-5),(12,-10)\}$ is not a basis for $R^2$.

Work Step by Step

The set $S=\{(6,-5),(12,-10)\}$ is not a basis for $R^2$ because the vectors in $S$ are not linearly independent. Because one can see that $(12,-10)=2(6,-5)$, and hence we have the non trivial combination $$2(6,-5)-(12,-10)=(0,0)$$ Which means that the vectors $(6,-5),(12,-10)$ are not Linearly independent. Hence, the set $S=\{(6,-5),(12,-10)\}$ is not a basis for $R^2$
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