24 14Let v = | -9 | v2 =V3 = -8and H = Span (v,,V2,V3}It can be verified that 4v, +2v, – 3v, = 0. Use this information to find a basis for H.%3D4-52.4)

Name: 24 14Let v = | -9 | v2 =V3 = -8and H = Span (v,,V2,V3}It can be verifi
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Description: 24 14Let v = | -9 | v2 =V3 = -8and H = Span (v,,V2,V3}It can be verified that 4v, +2v, – 3v, = 0. Use this information to find a basis for H.%3D4-52.4)

Answered: 1 4 Let v = | -9 | v2 = V3 = -8 and H =…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Let v = | -9 | v2 =
V3 = -8
and H = Span (v,,V2,V3}It can be verified that 4v, +2v, – 3v, = 0. Use this information to find a basis for H.
%3D
4
-5
2.
4)

Expert Solution

Step 1

As it is given that $4 v_{1} + 2 v_{2} - 3 v_{3} = 0 . . . (i)$

implies $v_{1}, v_{2}, v_{3}$ are linearly dependent

from (i), we get

$v_{3} = \frac{1}{3} (4 v_{1} + 2 v_{2})$

that is $v_{3}$ can be written as linear combination of $v_{1} a n d v_{2}$ $v_{1} and v_{2}$

$\Rightarrowv_{1} a n d v_{2}$ are linearly indepependent

as given

$H = s p a n {v_{1}, v_{2}, v_{3}}$

thus basis for $H = {v_{1}, v_{2}}$

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