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

roids

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.

tiny-tim

hi roids! welcome to pf!
show us what you got

jbunniii

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

Simon Bridge

Argh: jbunnii beat me to it.

roids

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

   

jbunniii

[edit] OK, I had to re-read the thread, and actually it is agreeing with us. I'll elaborate below.
No, the question is clearly worded. I'm sure your interpretation was correct. The solution is simply wrong.

jbunniii

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

roids

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

Have a good day!