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 - 4.7 Change of Basis - Problems - Page 318: 9

Answer

$(6,0,-15)$

Work Step by Step

We know that $a(5-3x)+b(1)+c(1+2x^2)=15-18x-30x^2$. Thus $5a-3ax+b+c+2cx^2\\ =(5a+b+c)-3ax+2cx^2\\ =15-18x-30x^2$ Thus $5a+b+c=15\\ -3a=-18 \\ 2c=-30$ Thus $5.6+b-15=15 \\ a=6 \\ c=-15$ Thus $a=6 \\ b= 0 \\ c=-15$ Thus the vector is $(6,0,-15)$.
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