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Homework Help: Solving linear systems + gaussian elimination

  1. Jan 21, 2010 #1
    1. The problem statement, all variables and given/known data
    http://img.rflz.net/problems.jpg [Broken]
    2. Relevant equations
    3. The attempt at a solution
    Question 1
    x4 = 1 + 4x5
    but from there I am not sure where to go or if there is even a solution?

    Question 2
    I eventually got the matrix down to this:

    Code (Text):
    1  2 -3  1   4     1
    0  1  1  0   1     1
    0  0  1  0  -1/3  -1/3
    0  0  0  1  -1/3   4/3
    but again I am not sure if that is enough? I don't understand how back substitution is possible.

    Question 3
    Similar to 3. I sort of got it to row-echelon, but am having trouble. Any help on the previous question should help here.

    Thanks
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jan 21, 2010 #2

    Mark44

    Staff: Mentor

    For questions 1 and 2 you have more variables than equations, so you're going to have an infinite number of solutions.

    For 1, you can get the system to reduced row-echelon form, which is a step beyond Gaussian elimination. In RREF, each row that has a nonzero first entry has 1 in that position, and all rows above and below have 0 in the same position. Doing this, I was able to get the system in this form:
    Code (Text):

    [1 0 -5 0 36 | 3]
    [0 1 1 0 -15 |-3]
    [0 0 0 1 -4 | 1]
     
    From this you can write x1 in terms of x3 and x5, x2 in terms of x3 and x5, and x4 in terms of x5.

    If you do this by back substitution, you'll start with the 3rd equation, and write x4 in terms of x5. Substitute for x4 in the 2nd equation, so you'll eventually get to x2 in terms of x3 and x5. Substitute what you have here for x2 in the 1st equation to get x1 in terms of x3 and x5. IOW, the same as what I did before using RREF, which is essentially Gauss elimination with back substitution.

    For question 2, do the same thing.
     
  4. Jan 21, 2010 #3
    Thanks for the reply. I managed to get 2 and 3 on my own, but your advice on question 1 should be very useful. :)
     
  5. Jan 21, 2010 #4
    I tried putting the system in the form you have, and I think you have made a mistake. The -3 in the last column should be a -1, no?
     
  6. Jan 21, 2010 #5

    Mark44

    Staff: Mentor

    Yes, you're right. The last entry in the 2nd row should be -1.
    Code (Text):

    [1 0 -5 0 36 | 3]
    [0 1 1 0 -15 |-1]
    [0 0 0 1 -4 | 1]
     
     
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