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Linear algebra proof problem

  1. Aug 30, 2011 #1
    Hi, first post here. I need help with a proof from linear algebra.
    It states:
    suppose that (x,y)=(r,s) is a solution of:

    system of equations #1
    ax+by=p
    cx+dy=q

    and that (x,y)=(u,v) is a solution of:

    system of equations #2
    ax+by=0
    cx+dy=0

    prove that (x,y)=(r+u , s+v) is a solution for system of equations #1

    . The attempt at a solution

    ar+bs=p
    cr+dy=q

    au+bv=0
    cu+dv=0

    au+bv=cu+dv

    i then tried solving for a,b,c and d and plugging them back into the first system of equations, however after doing so my equations become very long and confusing. I tried working it backwards but i still seem to get stuck. There has to be a simpler way of solving it but i cant seem to figure it out.
     
  2. jcsd
  3. Aug 30, 2011 #2

    micromass

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    Just plug in (r+u,s+v) in your equations and see if it works out.
     
  4. Aug 30, 2011 #3
    I attempted to do that and working backwards from there but i get stuck here:

    a(r+u)+b(s+v)=p
    c(r+u)+d(s+v)=q

    ar+au+bs+bv=p
    cr+cu+ds+sv=q
     
  5. Aug 30, 2011 #4

    micromass

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    Yes, and now use that au+bv=0 and cu+dv=0
     
  6. Aug 30, 2011 #5

    Mark44

    Staff: Mentor

    You should not be trying to solve for these numbers. The variables in your problem are x and y. Everything else is a constant.
     
  7. Aug 30, 2011 #6
    Oh i see now, thank you!
     
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