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Expressing a vector as a linear combination 
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#1
Feb1313, 10:23 PM

P: 13

1. The problem statement, all variables and given/known data
express the vector (6,4,7) as a linear combination of (4,2,1) and (3,1,2) 2. Relevant equations 3. The attempt at a solution I am not quite sure how to go at it but this is what I am assuming to do: (C1*4,C2*2,C3*1)=(6,4,7) C1=3/2 C2=2 C3=7 and the same method for the 2nd one? I know how to show linear dependence and dependence as a vector is independent if c1=c2=c3 and not if constants aren't equal. Any help in how to solve this is appreciated. 


#2
Feb1313, 10:26 PM

P: 132

All you did was express ##(6,4,7)## as a scalar multiple of the ELEMENTS of the vector ##(4,2,1)##. You cannot do that.



#3
Feb1313, 10:27 PM

P: 13




#4
Feb1313, 10:30 PM

P: 132

Expressing a vector as a linear combination
Basically, what you need to do is find scalars ##r, s \in \mathbb{R}## such that
##(6,4,7) = r(4,2,1) + s(3,1,2)## What your teacher gave you was the general form of a linear combination for any ##n## vectors. In this case, you only have 2 to work with. 


#5
Feb1313, 10:30 PM

HW Helper
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PF Gold
P: 7,562

C_1(4,2,1) + C_2(3,1,2) = (6,4,7)$$That will give you 3 equations in 2 unknowns. 


#6
Feb1313, 10:42 PM

P: 13

(1)4c13c2=6 (2)2c1+c2=4 (3)c1+2c2=7 >(3) becomes c1=72C2 > into (2) 2(72C2)+c2=4 C2=2 C1=3 3(4,2,1)+2(3,1,2)=(6,4,7) (12,6,3)+(6,2,4)=(6,4,7)? And these are linearly dependent since c1 doesn't equal c2 correct? For part B. I have to determine the linear dependence of vectors (6,4,7),(4,2,1), an (3,1,2) but not quite sure if I set it up the same way since I dont know which of those vectors I set equal to. 


#7
Feb1313, 10:52 PM

HW Helper
Thanks
PF Gold
P: 7,562




#8
Feb1313, 11:00 PM

P: 13

I assume I'd do something like this (let vectors be v1,v2,v3 to save time) c1v1+c2v2+c3v3=0 and if it does =0, then that implied c1=c2=c3=0 and therefore they are linearly independent? Im just curious if it its just all the constants have to be zero or if they just all have to be the same? The definition I have written down is basically what I just wrote with c1=c2=c3=0 and then it states if this is the case then the vectors are linearly independent. 


#9
Feb1413, 10:57 AM

HW Helper
Thanks
PF Gold
P: 7,562




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