anonymous
  • anonymous
I wonder if the Prof is correct in Lecture 2 at minute 8:19 where he said that the 3rd pivot is 5. I think that it is -3
MIT 18.06 Linear Algebra, Spring 2010
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anonymous
  • anonymous
I wonder if the Prof is correct in Lecture 2 at minute 8:19 where he said that the 3rd pivot is 5. I think that it is -3
MIT 18.06 Linear Algebra, Spring 2010
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
The matrix after the first step of elimination looks like: \[\left[\begin{matrix}1 & 2 & 1\\ 0 & 2 & -2\\ 0 & 4 & 1\end{matrix}\right]\] We want to eliminate element 32 (that is 4). To do that, we multiply second row by 2 (to get 4 in 22 position): \[\left[\begin{matrix}1 & 2 & 1\\ 0 & 4 & -4\\ 0 & 4 & 1\end{matrix}\right]\] Finally, we subtract second line from the third: \[\left[\begin{matrix}1 & 2 & 1\\ 0 & 2 & -2\\ 0-0 & 4-4 & 1-(-4)\end{matrix}\right] =\left[\begin{matrix}1 & 2 & 1\\ 0 & 2 & -2\\ 0& 0& 5\end{matrix}\right]. \]

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