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: 51


$$(0,0,0)=-t(1,1,1)+t(1,1,0)+0(0,1,1) +t(0,0,1) .$$ $$(1,1,1)=(1,1,0)+0(0,1,1) + (0,0,1) .$$

Work Step by Step

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