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]

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]

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? 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 (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?

 

[jbunniii]

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

[edit] 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]

To elaborate on what Alchemista said in this thread:

https://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!