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Matrix Row Reduction Question

  1. Jan 28, 2014 #1
    1. The problem statement, all variables and given/known data

    Write and row reduce the augmented matrix to find out whether the given set of equations has exactly one solution, no solutions, or infinitely many solutions.

    -x+y-z=4
    x-y+2z=3
    2x-2y+4z=6

    2. Relevant equations



    3. The attempt at a solution

    I saw right away that Row 3 and Row 2 are the same equation, off by a factor of 2. Because of this, I was able to make the matrix with a zero row for Row 2, which shows that row 2 and row 3 are linearly dependent. However, my question arises here. I know that a zero row and linear dependence of these two equations means that there is either 0 solutions or infinitely many solutions. Since they are linearly dependent, there is not one unique solution. However, How can I tell whether there is 0 solutions or infinitely many?
     
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  3. Jan 28, 2014 #2

    Ray Vickson

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    Have you carried through the row-reduction procedure, as the question asked you to do? Doing that, or some equivalent thing, is necessary if you want a final answer.
     
  4. Jan 28, 2014 #3
    Yes, I continued a bit. Here is what I have done and what I think so far: I have simplified the matrix down to:

    Row 1: -1 1 -1 4
    Row 2: 0 0 0 0
    Row 3: 0 0 2 14

    Once I do this, it tells me in the bottom equation that z=7. However, I'm not too sure how to tell anything about how many solutions there are, besides the fact that there is NOT just one.
     
  5. Jan 28, 2014 #4

    Ray Vickson

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    OK, so now what is y? What is x?
     
  6. Jan 28, 2014 #5
    So for row 3, x = 0, y = 0, z = 7, but for row 1 you have -x+y-z=4, so i'm a little stuck as what to do next.
     
  7. Jan 28, 2014 #6
    No. For R3
    0x+0y+2z=14
     
  8. Jan 28, 2014 #7

    HallsofIvy

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    Row 3 reduced to 2z= 14 which tells you that z= 7 but says nothing about what x and y are! Knowing that z= 7, -x+ y- z= 4 becomes -x+ y= 7+4= 11 or y= x+ 11.
     
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