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anonymous

  • one year ago

"A system in echelon form can be inconsistent." Can someone give me an explanation as to why this is false?

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  1. dan815
    • one year ago
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    any system in echelon form can be rewritten in row reduced echelon form

  2. dan815
    • one year ago
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    |dw:1439931493101:dw|

  3. dan815
    • one year ago
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    so u cannot have inconsistency

  4. anonymous
    • one year ago
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    Isn't a matrix in echelon form if it has a row of zeroes on the bottom? Like, is the following not in echelon form? \[\begin{bmatrix}1&&1&&1\\0&&0&&1\end{bmatrix}\](The system in this case would be \(x + y = 1\) and \(0x + 0y = 1\))

  5. anonymous
    • one year ago
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    I just didn't see anything in my book's definition of echelon form that prevented this, but I'm probably wrong on that.

  6. dan815
    • one year ago
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    that has 3 variables x ,y ,z we dont know the constants

  7. dan815
    • one year ago
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    |dw:1439931839159:dw|

  8. anonymous
    • one year ago
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    I meant for there to to be the line for an augmented matrix between columns 2 and 3, but I don't know how to do that in \(\LaTeX\) just yet.

  9. dan815
    • one year ago
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    |dw:1439931879525:dw|

  10. dan815
    • one year ago
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    this is not echelon form there is simply no echelon form for this

  11. dan815
    • one year ago
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    http://prntscr.com/866zua

  12. dan815
    • one year ago
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    actually wait i dont know, what does inconsistent mean again?

  13. dan815
    • one year ago
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    there is actually an example there where RREF has all 0s in the bottom soo

  14. anonymous
    • one year ago
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    Yeah, that's what's confusing me.

  15. dan815
    • one year ago
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    i was thining inconsistent is same as dependant but thats not it

  16. anonymous
    • one year ago
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    They said \[\begin{bmatrix}1~~~1~~~3\\0~~~0~~~2\end{bmatrix}\]isn't in ehcelon form but \[\begin{bmatrix}1~~~1~~~3\\0~~~0~~~1\end{bmatrix}\]is?\[\\\]And my book cites inconsistent as having no solution.

  17. dan815
    • one year ago
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    gotta have it as 1 i guess

  18. anonymous
    • one year ago
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    Hmm...but if the second one is allowed, then is it assumed that the matrix has 3 unknowns, but we don't know the constant matrix since there is no divider?

  19. dan815
    • one year ago
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    what is the definition of inconsistent, that should clear this up

  20. anonymous
    • one year ago
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    My book's definition of inconsistent is that there is no solution that satisfies all parts of the system simultaneously. "Inconsistent: No pair of numbers (s1,s2) satisfies all three equations simultaneously."

  21. anonymous
    • one year ago
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    And to clarify, the question on my assignment was asking if the above statement was true or false. The answer ended up being false, and I just can't seem to figure out why.

  22. dan815
    • one year ago
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    Ohh okay

  23. anonymous
    • one year ago
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    Is it just something weird based on definition or am I way off? Also, here's my book's definition of echelon form if it helps. "In each row of a system, the first variable with a nonzero coefficient is the row’s leading variable. A system is in echelon form if each leading variable is to the right of the leading variable in the row above it, except for the leading variable in the first row, and any all-zero rows are at the bottom."

  24. anonymous
    • one year ago
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    They use those examples to show you how a matrix is called echelon. \(\left[\begin{matrix}1&1&3\\0&0&2\end{matrix}\right]\) is not an echelon matrix because the leading 1 of the second row is 2, not 1. But you can get the echelon form by divided row2 by 2 to get \(\left[\begin{matrix}1&1&3\\0&0&1\end{matrix}\right]\)

  25. anonymous
    • one year ago
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    That is just NOTATION.

  26. anonymous
    • one year ago
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    That shows you how "an echelon form can be inconsistent"

  27. anonymous
    • one year ago
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    Ok, I'm pretty sure the leading one's thing is just differences in definition between texts (I cross checked multiple books, and some said they HAD to have leading one's and others didn't care.)

  28. anonymous
    • one year ago
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    But the question answer was that the statement was false, indicating that it cannot be inconsistent...That's what I was confused on.

  29. anonymous
    • one year ago
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    It's above my head!! :)

  30. anonymous
    • one year ago
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    lol ok. I'm just as confused as you are, and this is the assignment from my first Linear Algebra class XD

  31. anonymous
    • one year ago
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    I have to go, but @ganeshie8 , if you could take a look at this whenever you get online, that'd be great!

  32. ganeshie8
    • one year ago
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    The question is about "system of equations", not about a "matrix" being inconsistent. We say a "system of equations" is consistent or inconsistent. But never say a matrix is inconsistent or inconsistent. If the system of equations is inconsistent, then after eliminations, the augmented matrix in echelon form will have at least one row in the bottom with all 0's except for the last entry. For example : \(\begin{bmatrix}1&2&5&|&5\\0&0&0&|&\color{red}{3}\end{bmatrix}\).

  33. ganeshie8
    • one year ago
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    The given statement, "A system in echelon form can be inconsistent." means that the "starting" system of equations itself are in echelon form, for example : \[ \begin{array}{} x&+&3y&+&z&=&5\\ &&y&+&5z&=&3\\ &&&&z&=&8 \end{array}\] In other words, the given system of equations itself is in some kind of triangular/staircase form. Our task is to prove that any system like this can never be inconsistent..

  34. anonymous
    • one year ago
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    So, when it said "in echelon form", it meant that the system was given in echelon form, not manipulated into it?

  35. ganeshie8
    • one year ago
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    Exactly! the "given" set of equations are in "staircase" looking form

  36. ganeshie8
    • one year ago
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    The given statement is false because the system of equations in that staircase form "always has a solution" by back substitution.

  37. ganeshie8
    • one year ago
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    heard of "back substitution" before ?

  38. anonymous
    • one year ago
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    Oh! That makes a lot more sense. I guess it was just the wording that I misinterpreted. And yes, I know what back substitution is. I've self-studied Linear Algebra before, but I never formally studied it in a classroom setting (which I am now doing).

  39. ganeshie8
    • one year ago
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    Fact : you can never learn linear algebra fully, it is a vast subject and there is always something we never thought of..

  40. anonymous
    • one year ago
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    haha seems like it. My teacher started off lecture stressing a similar point, bringing in lots of different fields where it is applied. Anyway, thanks so much for the help. :)

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