Answer
$y=(\frac{1}{3},\frac{4}{3},\frac{2}{3},3,4)$
Work Step by Step
$2x+(-3)y=-z\\2(1, 2, 3, 4, 5)+(-3)y=-(−1, 0, −4, 1, 2)\\(2,4,6,8,10)+(-3)y=(1,0,4,-1,-2)\\(-3)y=(-1,-4,-2,-9,-12)\\y=(\frac{1}{3},\frac{4}{3},\frac{2}{3},3,4)$
Differential Equations and Linear Algebra (4th Edition)
by Goode, Stephen W.; Annin, Scott A.
Published by
Pearson
ISBN 10:
0-32196-467-5
ISBN 13:
978-0-32196-467-0
Chapter 4 - Vector Spaces - 4.1 Vectors in Rn - Problems - Page 252: 5
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