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Determine if vector b is a linear combination of vectors a1, a2, a3.by roids
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#1
Feb1813, 05:05 PM

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Hi guys. I've solved an exercise but the solution sheet says what doesn't make sense to me. Could you please help with this problem?
Determine if vector b is a linear combination of vectors a1, a2, a3. a1=[1, 2, 0], a2=[0, 1, 2], a3=[5, 6, 8], b=[2, 1, 6]. b is a linear combination when there exist scalars x1, x2, x3 such that x1*a1 + x2*a2 + x3*a3 = b. right? I put a's in a coefficient matrix and b in the augmented column. [a1 a2 a3  b]. Rowreduced it produces a consistent system (although I get x3 a free variable  third row all zeroes). But the solution sheet says b is not a linear combination of the a vectors. Where is the catch? Should the RREF have a unique solution? Thank you. 


#2
Feb1813, 05:14 PM

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hi roids! welcome to pf!



#3
Feb1813, 05:24 PM

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Your solution sheet is wrong: ##2 a_1 + 3 a_2 + 0 a_3 = b##.



#4
Feb1813, 05:36 PM

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Determine if vector b is a linear combination of vectors a1, a2, a3.
Argh: jbunnii beat me to it.



#5
Feb1813, 06:23 PM

P: 3

Thanks for the responses fine gentlemen. The book with the solutions is David Lay  Linear Algebra, fourth edition.
Can you please take a look at the same problem someone asked here, where the answerer said that no, b is not a linear combination? http://www.physicsforums.com/showthread.php?t=531233 While you and this someone's from Berkeley document say that b is indeed is a linear combination: http://math.berkeley.edu/~honigska/M54HW01Sols.pdf (page 6, exercise 14) I attached the original problem and solution from the book. Can it be that the book is asking to not combine the a vectors and just test them one by one with b to see if they separately are linear combinations of b? 


#6
Feb1813, 06:54 PM

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#7
Feb1813, 06:58 PM

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To elaborate on what Alchemista said in this thread:
http://www.physicsforums.com/showthread.php?t=531233 He is saying that for any ##t \in \mathbb{R}##, if we set ##x_1 = 2  5t##, ##x_2 = 3  4t##, and ##x_3 = t##, then we will have ##x_1 a_1 + x_2 a_2 + x_3 a_3 = b##. Thus, not only is there a solution, there are infinitely many solutions. The solution I mentioned above in post #3 is a special case of this, with ##t = 0##. The reason for this is that, for any ##t##, $$5t a_1  4t a_2 + t a_3 = 0$$ 


#8
Feb1813, 07:21 PM

P: 3

Thanks so much jbunnii, you made my day. Love the notation by alchemista.
Have a good day! 


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