Express one vector as a linear combination of others

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SUMMARY

The discussion focuses on demonstrating that the set of vectors {(1,2,3), (3,2,1), (5,5,5)} is linearly dependent by expressing one vector as a linear combination of the others. The user systematically set up the equations a(1,2,3) + b(3,2,1) + c(5,5,5) = 0 and derived relationships between the coefficients a, b, and c. Ultimately, they confirmed linear dependence by substituting values and verifying the identity holds true, concluding that the vectors are indeed multiples of each other.

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Homework Statement



Show that the set of vectors is linearly dependent (LD) by expressing one vector as a linear combination (LC) of the others:

{(1,2,3), (3,2,1), (5,5,5)}


The Attempt at a Solution



I would like to do this systematically (without guess and check). So I assumed that if the set is LD, then there exists some values of a,b,c not all zero such that:

a(1,2,3) + b(3,2,1) + c(5,5,5) = 0

hence,

a + 3b +5c = 0
2a + 2b +5c = 0
3a + 1b +5c = 0

Now I should be able to solve for 2 of the parameters (assuming only one of the equations is not independent, else I an solve for 1 ...)) a,b,c in terms of the third which would remain arbitrary.

Eliminating c from the first and second equations, we find that a = b .

Eliminating b, from the second and third equations we find that c = -4a/5 .

Now, if I plug these values back into anyone of the equations, I simply get the identity. What am I missing here? Am I messing something up? Or are they ALL multiples of each other?

Thanks!
 
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I'm confused. Haven't you done what you wanted to? You seem to have found everything in terms of a.
 
hgfalling said:
I'm confused. Haven't you done what you wanted to? You seem to have found everything in terms of a.

Oh. Maybe I have :redface: So here, a is my arbitrary parameter. Say, I let a = 1, then b = 1 and c = -4/5. Plugging these into the original equation, yields:

(1,2,3) + (3,2,1) - (4,4,4) = 0 which is indeed true.

I'm a little slow tonight. :smile:
 

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