# Determine if vector b is a linear combination of vectors a1, a2, a3.

by roids
Tags: linear combination
 P: 3 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]. Row-reduced 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.
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PF Gold
P: 26,127
hi roids! welcome to pf!
 Quote by roids I put a's in a coefficient matrix and b in the augmented column. [a1 a2 a3 | b]. Row-reduced it produces a consistent system (although I get x3 a free variable - third row all zeroes).
show us what you got
 Sci Advisor HW Helper PF Gold P: 2,655 Your solution sheet is wrong: ##2 a_1 + 3 a_2 + 0 a_3 = b##.
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PF Gold
P: 10,269

## Determine if vector b is a linear combination of vectors a1, a2, a3.

Argh: jbunnii beat me to it.
 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? Attached Thumbnails
 Sci Advisor HW Helper PF Gold P: 2,655 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##, $$-5t a_1 - 4t a_2 + t a_3 = 0$$