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Matrix operation |
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| Mar24-07, 04:53 AM | #1 |
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Matrix operation
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? Thanks. |
| Mar24-07, 06:51 AM | #2 |
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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.
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| Mar24-07, 07:32 AM | #3 |
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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.
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| Mar24-07, 08:58 AM | #4 |
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Matrix operation
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?)
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| Mar24-07, 09:33 AM | #5 |
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| Mar24-07, 09:38 AM | #6 |
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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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