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Homework Help: Linear algebra problem

  1. Jan 26, 2005 #1
    I am a bit confused with this problem:

    Given AX = B, B != 0; X and Y satisfy the system. Find constants a and b such that aX + bY also satisfy the system.

    The hint was: does (1/3)X+(2/3)Y work? Which would mean (1/3 + 2/3)(X + Y), which means X + Y. So, then I have X+Y as a potential solution, which is a solution? So, then does it mean that if a system has solutions, sum of those solutions is still a valid solution, but is any linear combination a solution? Then it would make a,b either such that a+b=1 or a,b any number.
    Could someone explain, please?

    Thank you very much.
     
    Last edited: Jan 26, 2005
  2. jcsd
  3. Jan 26, 2005 #2
    No, you mean (1/3)(X + 2Y), (1/3 + 2/3)(X + Y) isn't the same as X/3 + 2Y/3.
     
  4. Jan 26, 2005 #3
    Is there some extradata in the problem, like a relation between X and Y. Because if you don't have one you can prove that these constants exist
     
  5. Jan 26, 2005 #4
    Ooops, sorry you are right...basic algebra...
    I am still not clear if a system has several solutions, when combination of those solutions is a valid solution...if ever.

    No, there is no extra info.
    But maybe if one of (a,b) is zero then it's possible?
     
    Last edited: Jan 26, 2005
  6. Jan 26, 2005 #5
    If X is solution and Y is solution, then X=cY where c is a constant. then

    c*AY=B
    AY=B

    or: ((c+1)/2)*AY=B

    By linearity:

    A(X/2)+A(Y/2)=B

    then a=1/2 b=1/2
     
  7. Jan 26, 2005 #6
    That's always true? Is solution set a vector space?

    Where does that follow from?

    Thanks.
     
    Last edited: Jan 26, 2005
  8. Jan 26, 2005 #7
    this condition is true if the X and the Y are lineary dependent, then there is unique solution, the X or the Y. and that's the problem. for that system, exist unique solution or infinite.

    The expression comes from summing the eqs:

    cAY+AY=2B
    ((c+1)/2)*AY=B
     
  9. Jan 26, 2005 #8
    note the expression comes from summing the eqs i initially post. The eqs. i post latter is the reductions
     
  10. Jan 26, 2005 #9
    Z = aX + bY

    AZ = B
    A(aX + bY) = B
    AaX + AbY = B
    a(Ax) + b(AY) = B
    aB + bB = B
    a + b = 1

    --J
     
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