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

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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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

 Blog Entries: 1 Recognitions: Gold Member Homework Help Science Advisor Your solution sheet is wrong: ##2 a_1 + 3 a_2 + 0 a_3 = b##.

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

Argh: jbunnii beat me to it.

 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

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 Blog Entries: 1 Recognitions: Gold Member Homework Help Science Advisor 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##, $$-5t a_1 - 4t a_2 + t a_3 = 0$$