# 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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tiny-tim
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welcome to pf!

hi roids! welcome to pf! 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 jbunniii
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Your solution sheet is wrong: ##2 a_1 + 3 a_2 + 0 a_3 = b##.

Simon Bridge
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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? https://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 [Broken] (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?

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jbunniii
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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? https://www.physicsforums.com/showthread.php?t=531233
 OK, I had to re-read the thread, and actually it is agreeing with us. I'll elaborate below.
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?
No, the question is clearly worded. I'm sure your interpretation was correct. The solution is simply wrong.

jbunniii
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To elaborate on what Alchemista said in this thread:

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##,
$$-5t a_1 - 4t a_2 + t a_3 = 0$$

Thanks so much jbunnii, you made my day. Love the notation by alchemista.

Have a good day!