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Echelon Matrices

  1. Feb 22, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the general solution
    69ca6177fa.png

    2. Relevant equations
    Row operations
    Gaussian elimination

    3. The attempt at a solution
    5d39a77963.jpg

    This is has happened twice now and I'm not too sure how to deal with it. The last row ends up being all zeros except in that spot. I need to make this into an echelon matrix and for that I need a 1 in the 3rd entry of row 3. Not sure how to get it like that.
     
  2. jcsd
  3. Feb 22, 2016 #2

    Samy_A

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    What is the definition of an echelon matrix or row echelon form?
     
  4. Feb 22, 2016 #3
    We say that an m × n matrix is a row echelon matrix if it has the following three properties.
    1 All zero rows, if there are any, are at the bottom of the matrix.
    2 The leading entry of each non-zero row equals 1.
    3 If rows numbered i and i + 1 are two successive non-zero rows, the leading entry of row i + 1 is in a column strictly to the right of the column containing the leading entry of row i.
     
  5. Feb 22, 2016 #4

    Samy_A

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    Correct, although property 2 is not always included, and easily fixed anyway.
    Does your last matrix satisfy properties 1 and 3?

    Do you notice that these properties do not imply what you wrote in the first post (bolding mine)?
     
  6. Feb 22, 2016 #5
    For the first condition, there are no zero rows so that satisfied.
    It doesn't satisfy property 3, assuming row 2 to be row i and row 3 to be i+1, then I would still need a 1 in the third entry of the row 3 as that is the column to the right of the leading entry of row 2.
     
  7. Feb 22, 2016 #6

    Samy_A

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    No, this is not a correct interpretation of property 3: "If rows numbered i and i + 1 are two successive non-zero rows, the leading entry of row i + 1 is in a column strictly to the right of the column containing the leading entry of row i."
    "column strictly to the right of ..." doesn't necessarily mean "first column to the right of ...".
     
  8. Feb 22, 2016 #7
    Hm. I understand what youre saying. In other parts of the notes he says:
    Echelon Matrices:
    All zero rows are at the bottom
    Leading entry in each row is 1
    Leading entries are moving to the right as we move down the rows.

    So what youre saying is basically at this point in time I can solve that matrix by just make that -3 in row 2 to a 1(not even necessary)?
     
  9. Feb 22, 2016 #8

    Samy_A

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    Yes. Your last matrix is in row echelon form (except for property 2, which you can safely ignore here).
     
  10. Feb 22, 2016 #9
    Are there like certain weird exceptions to this ? Also when can you not ignore the fact that the leading entry isnt 1?
     
  11. Feb 22, 2016 #10

    Samy_A

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    I don't know. As property 2 is often included in the definition, I guess there must be a reason for that.
    For solving the system of equations from the bottom up using the row echelon form I don't see what benefit we get from having specifically 1 as leading row entry.

    Another take on this:
     
  12. Feb 22, 2016 #11
    I mean through the entire class, he has always told us to pivot at a 1 in order to make the values below the 1 in that column a 0, and then we move to the second column and pivot at a 1.
     
  13. Feb 22, 2016 #12

    Samy_A

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    Maybe you can ask him for the reason.
     
  14. Feb 22, 2016 #13

    Ray Vickson

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    Many discussions of echelon matrices do not require that the leading coefficients in rows are = 1; they are satisfied with having them ≠ 0 (so you can divide by them).
     
  15. Feb 23, 2016 #14

    HallsofIvy

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    Are you required to use matrices here? It looks to me like simple "elimination" would work better. For example, adding twice the first equation to the second equation will eliminate both z and w and give 3x+ y= 5 so y= 5- 3x. Putting that into the last equation, -x+ 10- 6x+ 3z= 10- 7x+ 3z= 4 so 3z= -6+ 7x so z= -2+ (7/3)x. Finally, putting both y= 5- 3x and z= -2+ (7/3)x into the first equation gives x+ 5- 3x- 2+ (7/3)x- w= (1/3)x+ 3- w= 4 so w= 1- (1/3)x. Since you have only three equations in four unknowns, the "general solution" will be three of the unknowns in terms of the fourth.
     
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