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Linear algebra 3 lines in r2 Unique solution

  1. Aug 16, 2011 #1
    1. The problem statement, all variables and given/known data
    ax+by=k
    cx+dy=l
    ex+fy=m

    If in Exercise 12 k=l=m=0, explain why the system must be consistent. What can be said about
    the point of intersection of the three lines if the system has exactly one solution?
    2. Relevant equations



    3. The attempt at a solution
    ofcourse the system is consisten because x,y=0 is always a solution
    but for the second part
    all i did was try to get it to reduced row echelon form and what i got is that in the last colum and last row
    of augmented matrix m-(f(l-kc/a)/(d-bc/a))
    so i said f=0 , l- kc/a = 0 , m-(f(l-kc/a)/(d-bc/a)) = 0 for there to be unique solution i dont know what that would mean about the point but most likely i did mistake in reduced row echelon because too many terms is there any other way to do this?

    i know we must have a 0 because if not then 0x +0y = number which cant be
     
    Last edited: Aug 16, 2011
  2. jcsd
  3. Aug 16, 2011 #2
    oh and i got x= k- b(l-kc/a)/d-bc/a
    y= (l-kc/a )/ (d - bc/a)
     
  4. Aug 16, 2011 #3
    oh one thing i see is that l-kc/a = 0 , l/k = c/a , which means there must be proportionality between the x coefficient and the y intercept
     
  5. Aug 16, 2011 #4

    HallsofIvy

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    The line ax+ by+ cz= 0 passes through the point (0, 0, 0) for any a, b, c. So if the system is solvable for one point of intersection, that point must be ?
     
  6. Aug 16, 2011 #5
    The origin? but ax+ by+ cz= 0 is a plane i thought? and we are in r2 not r3 which makes me more confused
    but we have this
    ax+by=k
    cx+dy=l
    ex+fy=m
    which is different?
     
  7. Aug 16, 2011 #6
    wait wait so are we still considering the first part where k=l=m=0 in the second part? cause i mean ofcourse we will get x=y=0 if thats the case
     
  8. Aug 16, 2011 #7

    SammyS

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    As you said, x = y = 0 is a solution to the system. If the system has exactly one solution, then x = y = 0 must be the only solution.
     
  9. Aug 16, 2011 #8
    yea sorry i didnt know that we were still considering k=l=m=0 thats why i got confused
     
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