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I have a question with something written by David C Lay in his textbook "Linear Algebra and its Applications". Here's a picture of the excerpt in the textbook:

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My question has to do with when he labels $x_3 = 3$ as "new equation 3". Why is it the new equation for equation 3? What explains why it can't be the new equation for equation 2? Because that means he replaced the bottom matrix row with the new one. Why couldn't it replace the second row, for example?

sangstar
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1 Answers1

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The goal appears to be getting to the triangular form (lower left triangle of zeroes), and $1$'s on the adjacent diagonal.

Presumably the manipulations prior gave a $1$ in the upper left, and $0$ everywhere else in that column.

This step gives you the $1$ in the center row, second from left, and the $0$ in the bottom row, second from the left.

You can add to one row some linear combination of any of the other rows at will, but doing what you suggest doesn't help you to develop the triangular form.

John
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  • So I could've replaced any of the previous row with the $z = 3$ row? Not necessarily the third equation? It just is more convenient to replace the third? And that makes sense because all of those rows are independent of eachother? – sangstar Aug 29 '17 at 22:53
  • I was a bit breezy (I'm thinking incorrect) in my last sentence, which I've edited. There are lots of ways to arrive at the solution for the system, but this is an organized way to do it, and one where you can see your progress immediately as you go along. – John Aug 30 '17 at 15:55
  • ah, so the rows aren't dependent upon each other explicitly, and are true on their own, but doing this way is just a helpful way to solve it? – sangstar Aug 30 '17 at 22:16
  • That's a good way of putting it! – John Aug 30 '17 at 23:21