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Operation Count w/ LU Decomposition
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[QUOTE="brmath, post: 4517489, member: 486151"] I guess it comes down to what you call a "multiplication". Do you consider multiplying a row through by a value to be 1 multiplication ? (which is what I was assuming when I said "4"). Or since it is 5 across do you consider it to be 5 multiplications? Or do you consider it to be as many multiplications per row as you have non-zero elements in the row? If it is the first, then zeroing out the 4 elements in the lower triangle would take 4 multiplications. If it is the second, then you are multiplying 4 rows of 5 each and the answer is 20. It does seem as if you are using the third definition, in which case you have the right answer for creating an lower triangular matrix by getting all zeros into the upper triangle. Of course, you are performing the same operation no matter how you choose to count the steps. Now about the upper matrix. Yes you use the same steps with the identity. So you can consider that you used the same number of multiplications. Another way of looking at it, however, is simply to substitute into the identity the factors you worked out when producing the lower matrix, in which case you have no further multiplication. (This is written up nicely here: [PLAIN]http://www.math.ust.hk/~macheng/math111/LU_Decomposition.pdf[/PLAIN] ) Which way you count the multiplications at this time comes down to what is required for your class -- i.e. what is in your book or what your teacher said. And if nothing was said clearly, then you should state your assumptions when you write up the problem. It seems to me that you have gotten the concept of LU down pretty well, and that's probably what this problem was trying to get at. [/QUOTE]
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