Differential Equations and Linear Algebra (4th Edition)

Published by Pearson
ISBN 10: 0-32196-467-5
ISBN 13: 978-0-32196-467-0

Chapter 4 - Vector Spaces - 4.4 Spanning Sets - Problems - Page 282: 9

Answer

See below

Work Step by Step

Given the set of vectors $\{(-4,1,3),(5,1,6),(6,0,2)\}$ Obtain: $\begin{vmatrix} -4 & 5 & 6\\ 5 & 1 & 6 \\ 6 & 0 &2 \end{vmatrix}=-4.\begin{vmatrix} 1 & 0 \\ 6 &2 \end{vmatrix}-5.\begin{vmatrix} 1 & 0 \\ 3 &2 \end{vmatrix}+6.\begin{vmatrix} 1 & 1\\3 & 6 \end{vmatrix}=-4.2-5.2+6.3=0$ Thus, the set of vectors are coplanar and therefore set $\{(-4,1,3),(5,1,6),(6,0,2)\}$ does not span $R^3$ We can write set $S$ as $S=\{(x,y,z)\in M_R^3: x+13y-3z=0\}$ We have $x+13y-3z=0 \rightarrow x=3z-13y \forall (x,y,z) \in S$ The elements now can be written as: $(x,y,z)=(3z-13y,y,z)=y(-13,1,0)+z(3,0,1)$ Notice that: $(-13,1,0)=2(-4,1,3)-(5,1,6)\\ (3,0,1)=\frac{1}{2}(6,0,2)$ Substitute: $y(-13,1,0)+z(3,0,1)\\ =y[2(-4,1,3)-(5,1,6)]+z.\frac{1}{2}(6,0,2)\\ =2y(-4,1,3)-y(5,1,6)+\frac{1}{2}(6,0,2)$ Every vector $(x,y,z) \in S$ can be written as a linear combination of vectors $\{(-4,1,3),(5,1,6),(6,0,2)\}$ Hence, the set of vectors $\{(-4,1,3),(5,1,6),(6,0,2)\}$ span subspace $S$
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