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 - 4.4 Spanning Sets and Linear Independence - 4.4 Exercises - Page 179: 49

Answer

$$(0,0)=-\frac{2}{7}t(3,4)+\frac{8}{7}t(-1,1)+t(2,0) .$$ $$(2,0)=\frac{2}{7}(3,4)-\frac{8}{7}(-1,1) .$$

Work Step by Step

Consider the combination $$a(3,4)+b(-1,1)+c(2,0) =0, \quad a,b,c \in R.$$ Which yields the following system of equations \begin{align*} 3a-b+2c&=0\\ 4a+b &=0. \end{align*} The augmented matrix matrix is given by $$\left[\begin {array}{cccc} 3&-1&2\\ 4&1&0\end {array} \right]$$ The reduced row echelon form can be calculated as follows $$\left[\begin {array}{cccc}1&0&\frac{2}{7}\\ 0&1&-{\frac {8}{7}} \end {array} \right],$$ then we have the solution $$a=-\frac{2}{7}t,\quad b=\frac{8}{7}t, \quad c=t.$$ Hence, we have the combinations $$(0,0)=-\frac{2}{7}t(3,4)+\frac{8}{7}t(-1,1)+t(2,0) .$$ $$(2,0)=\frac{2}{7}(3,4)-\frac{8}{7}(-1,1) .$$
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