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anonymous

  • one year ago

Thee augmented matrix of a linear system has been reduced by row operations to the form shown. in each case, continue the appropriate row operations and describe the solution set of the original system The matrix is 1 3 0 -2 -7 0 1 0 3 6 0 0 1 0 2 0 0 0 1 -2

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  1. anonymous
    • one year ago
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    That is reduce echelon form already. To get the row reduce echelon form from here, you need: -3R2 + R1 to get rid off 3 on entry \(A_{1,2}\) . Show me what you get ?

  2. anonymous
    • one year ago
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    Got it?

  3. anonymous
    • one year ago
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    next is -3 R3 + R2 to get a new row 2

  4. anonymous
    • one year ago
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    oops, I meant 1 0 0 -11 -25 0 1 0 3 6 0 0 1 0 2 0 0 0 1 -2

  5. anonymous
    • one year ago
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    oh, gosh, my bad,

  6. anonymous
    • one year ago
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    ok, I redo it. let clear them all.

  7. anonymous
    • one year ago
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    I cleared mine!

  8. anonymous
    • one year ago
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    \(\left[\begin{matrix}1&3&0&-2&7\\0&1&0&3&6\\0&0&1&0&2\\0&0&0&1&-2\end{matrix}\right]\)

  9. anonymous
    • one year ago
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    1 0 0 -11 -25 (after -3R2 + R1) 0 1 -3 3 0 (after -3R3 + R2) 0 0 1 0 2 0 0 0 1 -2 Is that correct?

  10. anonymous
    • one year ago
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    ok?

  11. anonymous
    • one year ago
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    No... Oh my... how did you end up with -11 as your row one column five answer? It used to be -25...

  12. anonymous
    • one year ago
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    step1 it is -3R2 + R1 right? -3* (0 1 0 3 6) = (0 -3 0 -9 -18) + Row1 (1 3 0 -2 7) that is -------------------------------- new row 1: (1 0 0 -11 -11)

  13. anonymous
    • one year ago
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    ok??

  14. anonymous
    • one year ago
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    Oh my ^_^; I see what the problem is. The seven at the end of row one is a -7 in the problem (a simple typo. -7 + - 18 = -25.

  15. anonymous
    • one year ago
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    aaaaaaaaaaaaah, your last one is -7, not 7, ok hence it is -25. I am so so sorry

  16. anonymous
    • one year ago
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    \[\left[\begin{matrix}1&0&0&-11&-25\\0&1&0&3&6\\0&0&1&0&2\\0&0&0&1&-2\end{matrix}\right]\]

  17. anonymous
    • one year ago
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    Step2: 11R4 + R1 to get new row 1

  18. anonymous
    • one year ago
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    So, multiply row 4 by 11 and add to row one to get the new row one? Why row 4?

  19. anonymous
    • one year ago
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    \[\left[\begin{matrix}1&0&0&0&-45\\0&1&0&3&6\\0&0&1&0&2\\0&0&0&1&-2\end{matrix}\right]\]

  20. anonymous
    • one year ago
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    Good question: Why R4, because R4 has all entry before 1 is 0's, hence when you multiple to any number, it doesn't change other entry.

  21. anonymous
    • one year ago
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    Ah! I see. Good point. So when I go through this process I should choose the row that is the closest and/or most convenient for the row I am trying to change?

  22. anonymous
    • one year ago
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    So, what is the next step?

  23. anonymous
    • one year ago
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    \[\left[\begin{matrix}1&0&0&\color{red}{-11}&-25\\0&1&0&3&6\\0&0&1&0&2\\0&0&0&1&-2\end{matrix}\right]\] I choose the row 4, because I want to get rid of the number right in above its 1.

  24. anonymous
    • one year ago
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    Ah, I see.

  25. anonymous
    • one year ago
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    To the very brand new matrix, \[\left[\begin{matrix}1&0&0&0&-45\\0&1&0&\color{red}{3}&6\\0&0&1&0&2\\0&0&0&1&-2\end{matrix}\right]\] This guy is the number I want to get rid off, and it stay right above 1 of R4, right? Hence again, -3 R4+R2 to get new R2

  26. anonymous
    • one year ago
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    So the new row2 would be [0 1 0 0 0]?

  27. anonymous
    • one year ago
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    yes, and you are done.

  28. anonymous
    • one year ago
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    \[\left[\begin{matrix}1&0&0&0&-45\\0&1&0&0&0\\0&0&1&0&2\\0&0&0&1&-2\end{matrix}\right]\]

  29. anonymous
    • one year ago
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    So wait... The entire goal of this problem is basically to get all of the columns before the last one to be filled with ones and zeros.

  30. anonymous
    • one year ago
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    yes, it is.

  31. anonymous
    • one year ago
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    unless the last column, all of the entries are either 1 or 0.

  32. anonymous
    • one year ago
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    with 1's are in diagonal line.

  33. anonymous
    • one year ago
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    Thank you so much! I really appreciate your help! I "fanned" you or whatever the follow thing on here is.

  34. anonymous
    • one year ago
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    ok, don't forget: I am crazy one.

  35. anonymous
    • one year ago
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    Crazy One?

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