anonymous
  • anonymous
Does anyone understand the explination of the costs of elimination. I have both of his books, and different versions, and it is really not clear to me. He is very ambigiuos ...
MIT 18.06 Linear Algebra, Spring 2010
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schrodinger
  • schrodinger
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anonymous
  • anonymous
In lecture 4 he sad that a n x n matrix has a 1/3*n^3 cost, because the matrix has n*n=n^2 elements. The sum of all the elements from 1^2 =1 to n^2 is (1/3)*n^3 for any n. In calculus you could solve this with integration where you take a function f is n^2, than you integrate function f and an undifinite integral of n^2 is 1/3*n^3.\[\sum_{n=1}^{n}\ n^{2}=\frac{ 1 }{ 3 }*n^{3}\rightarrow \int\limits_{}^{}n^{2} dn= \frac{ n^{3} }{ 3 }\]
anonymous
  • anonymous
On top of what M.J. said, the meaning of that is that the number of operations that takes to do elimination can be bounded by n^3/3, this means that no matter how much n grows, it will never require more than n^3/3 computations
banda_mohammod_al_helal
  • banda_mohammod_al_helal
I don't understand your question clearly if you mean target of elimination then i say you we do elimination to find the solution of a system.

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