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Homework Help: Finding values to make a linear system consistent

  1. Nov 10, 2017 #1
    1. The problem statement, all variables and given/known data
    Given the following matrix:
    I need to determine the conditions for b1, b2, and b3 to make the system consistent. In addition, I need to check if the system is consistent when:
    a) b1 = 1, b2 = 1, b3 = 3
    b) b1 = 1, b2 = 0., b3 = -1
    c) b1 = 1, b2 = 2, b3 = 3

    2. Relevant equations

    Gaussian elimination method I used here:

    3. The attempt at a solution
    For the matrix to be consistent, I knew that the number of non-zero rows had to be less than the number of columns. Hence I tried to get the last row to be 0 0 0 | *; however while I managed to get the last row to become 0 0 1, I don't know how to make it the zero row I want.

    I'm wondering if my thought process to make this consistent is correct, or if there is another way I can make this system consistent.
  2. jcsd
  3. Nov 10, 2017 #2


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    Deduct multiples of the first row from the second and third rows, where the multipliers used are chosen to make the first element of the modified second and third rows be zero.

    From there it is easy to make the last row zero (in the first three columns). If it doesn't look easy, post what you get on here.
  4. Nov 10, 2017 #3

    This is what I got from simplifying my matrix. But since the rows all have zeroes at different columns, how can I reduce one of the rows to 0 0 0?
  5. Nov 10, 2017 #4


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    There's an arithmetic error. You have got a sign wrong in the operation you perform on row 3. Fix that and it should all fall into place.
  6. Nov 10, 2017 #5
    This is the method I used to find out how the system could be consistent. Is my process for determining the condition and testing it for the given values the correct method?
  7. Nov 11, 2017 #6


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    The condition and the way you are applying it is correct. It looks like you used the wrong values of b in (iii).
  8. Nov 11, 2017 #7
    No that was my bad, I copied it down wrong, the actual values were 0 1 and 2.
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