# Expressing a vector as a linear combination

by zabumafu
Tags: combination, expressing, linear, vector
 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.
 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.
P: 13
 Quote by Karnage1993 All you did was express ##(6,-4,7)## as a scalar multiple of only the vector ##(4,-2,1)##. You need to use ## (-3,1,2)## as well.
So would I need to find the same c1,c2,c3 that satisfy it for both? The only equation my teacher gave us was V=c1v1+c2v2+CnVn

 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.
HW Helper
Thanks
PF Gold
P: 7,719
 Quote by zabumafu 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.
No. That isn't how you do it. You have to find ##C_1,\, C_2## such that$$C_1(4,-2,1) + C_2(-3,1,2) = (6,-4,7)$$That will give you 3 equations in 2 unknowns.
P: 13
 Quote by LCKurtz No. That isn't how you do it. You have to find ##C_1,\, C_2## such that$$C_1(4,-2,1) + C_2(-3,1,2) = (6,-4,7)$$That will give you 3 equations in 2 unknowns.
Okay that makes much more sense I think I have it:

(1)4c1-3c2=6
(2)-2c1+c2=-4
(3)c1+2c2=7

->(3) becomes c1=7-2C2 -> into (2) -2(7-2C2)+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.
HW Helper
Thanks
PF Gold
P: 7,719
 Quote by zabumafu Okay that makes much more sense I think I have it: (1)4c1-3c2=6 (2)-2c1+c2=-4 (3)c1+2c2=7 ->(3) becomes c1=7-2C2 -> into (2) -2(7-2C2)+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?
What do you mean by these? You aren't testing anything for linear independence. You are expressing (6,-4,7) as a linear combination of the other two, which you have done correctly at the step I highlighted in blue.

 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.
Do you mean (4,-2,1) as the second vector so they are the same three as in part A? If so, look at the definition of linear dependence or independence and think about what you have above in blue.
P: 13
 Quote by LCKurtz What do you mean by these? You aren't testing anything for linear independence. You are expressing (6,-4,7) as a linear combination of the other two, which you have done correctly at the step I highlighted in blue. Do you mean (4,-2,1) as the second vector so they are the same three as in part A? If so, look at the definition of linear dependence or independence and think about what you have above in blue.
I was just asking trying to think ahead for part b but I think I might get it now. And no they switched it up for part be its not -2 its (4,2,1).

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.
HW Helper
Thanks
PF Gold
P: 7,719
 Quote by zabumafu I was just asking trying to think ahead for part b but I think I might get it now. And no they switched it up for part be its not -2 its (4,2,1). 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.
I think you have the idea. The equation ##c_1\vec v_1 + c_2\vec v_2 + c_3\vec v_3=\vec 0## gives you 3 equations in 3 unknowns. It always works if all the ##c##'s are zero. If those are the only values for which it works, the three vectors are linearly independent. But if there are any non-zero values that work, they are linearly dependent. You have to solve the system and see.

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