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Matrix operation

  1. Mar 24, 2007 #1
    In my note, it said that

    Counting multiplication and division only, in solving linear equations (matrix operation),

    Elimination of first row: total n^2 operations

    So, forward elimination operations for the matrix is Σ(2 to n) k^2 = n*(n+1)*(2n+1)/6

    I have tried to solve the equations but it seem do not need n^2 steps.

    Can anyone tell me conceptually why it needs n^2 operations to eliminate the first row?

  2. jcsd
  3. Mar 24, 2007 #2

    matt grime

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    You have to eliminate the firsty entry, so you add a multiple of another row - that is n multiplications. Then you need to do the second entry in the row. That is another n multiplications in another row. You do this n times, so that is n*n operations.
  4. Mar 24, 2007 #3

    But i have the following interpretation

    Eliminate the first entry and this is n multiplication
    Then, I do it n-1, including the first time.

    So, I think it is n*(n-1).

    I am quite not sure about this. :confused:
  5. Mar 24, 2007 #4

    matt grime

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    To be honest, I'd like you to say what it is that you're doing precisely. I'm not aware of anytime I'd actually want to eliminate the entire first row (of what, by the way? nxn matrix? Why?)
  6. Mar 24, 2007 #5
    The following link is a picture which shows what my note says.

    Last edited by a moderator: Apr 22, 2017
  7. Mar 24, 2007 #6

    matt grime

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    Doesn't really answer the questions I asked.

    1) you're trying to solve simultaneous equations
    2) in how many unknowns and how many equations? I presume n of each.

    at least it corrects your first sentence - elimination *for* first row.

    Strictly speaking you can do it n*(n-1) operations, I agree. Though you could be supposed to multiply every row by somethings so that they all have the same first entry (eg, 1), and that would be n^2 operations, generically. Unless you describe the algorithm you're attempting to cost, there's no way for anyone else to say what is really going on.
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