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Math Help - Row echelon form or Reduced row echelon form?

  1. #1
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    Row echelon form or Reduced row echelon form?



    I am very new to linear algebra and am having a bit of trouble understanding exactly what reduced row echelon form and row echelon form mean. From what I know, if a matrix is reduced row echelon form, it is always also row echelon form?

    a) It follows the rule that "any row that consists of only zeroes is at the bottom of the matix", so so far it follows the definition for reduced row echelon form.

    The first non-zero entry in each other row is 1: It follows this too, so so far so good.

    The leading 1 of each row, after the first row, lies to the right of the leading 1 of the previous row: This is where I get confused. The whole second column is filled with zeroes, but I thought it should always be completely diagonal along the matrix (the leading 1s that is)?

    What really confuses me is I thought there is supposed to be a leading 1 on every row? If the bottom row is all zeroes that is not the case. Is the bottom an exception or something, it can be all zeroes and not have a leading 1?


    For a matix to be reduced row echelon form, is it correct to say that all numbers above and below the leading 1 in a column have to be 0?

    What also really confuses me is the last column, isn't this the constant column? Do the same rules apply to this column?

    For d), it looks like the leading 1 doesn't start until row 2. Or it is in the constant column, which makes it out of the diagonal line.
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  2. #2
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    Re: Row echelon form or Reduced row echelon form?

    it is possible for a matrix to have fewer leading 1's than rows in reduced row echelon form. in this case, some number of rows at the bottom will be all 0's. the leading 1's may not always be on the diagonal.

    for your matrix (a) it is indeed in reduced row echelon form. the 2nd leading 1 in row 2 is indeed rightward of the leading 1 in row 1 (which is always on the diagonal, unless the matrix is the 0-matrix). more importantly, the entire 3rd column is 0's except for the 1 in the 2nd row. the only relevant entry in the last column of matrix (a) is the one in the 3rd row. since the leading 1 of row 2 is in column 3, the first 3 entries of row 3 MUST be 0 for (a) to be in either rref form or echelon form, and the 4th entry must either be 0, or 1 if non-zero. since the 4th entry in row 3 is 0, the entries above it might be any number. if the 4th entry had been 1, in order for the matrix (a) to be in rref form, the entries above it would then have to be 0.

    think of it like this: no matter what multiple of row 3 we add/subtract from rows 1 or 2, nothing will change. this matrix is equivalent to the following set of equations:

    x = -5w
    z = -4w

    if you were solving for (x,y,z,w), you could pick any values you like for y and w, then x and z are determined. the matrix has 2 non-zero rows in rref form, and it is no coincidence that you must make 2 choices to solve the system of equations. this number, 2, is the rank of the system (or of the associated matrix).

    the main difference between reduced row echelon form and row ecehelon form, is that the numbers above the leading 1 may be non-zero in row echelon form, but must be 0 in reduced row echelon form. in both cases, the entries in a column containing a leading 1 must be 0 below the leading 1 (or else the leading 1 in a subsequent row wouldn't be to the right of some previous leading 1).

    in matrix (d) the leading 1 in row 2 is NOT to the right of the leading 1 in row 1. it is 2 places to the LEFT. thus this matrix is neither in reduced row echelon form nor row ecehlon form.
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  3. #3
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    Re: Row echelon form or Reduced row echelon form?

    Thanks! So so far I have

    a) both reduced row echelon and row echelon form

    b) row echelon form (not reduced because there is a non zero above leading 1 in column 2)

    c) (not quite sure)

    d) neither because the leading 1 in columns 1 and 2 are to the left of the leading 1 in column 3

    e) row echelon form (not reduced because of non zeroes above leading 1 in column 3)


    I'm not quite sure of c) though. Would that be reduced row echelon form.. even though there is only one leading 1 in the entire matrix :S
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    Re: Row echelon form or Reduced row echelon form?

    C wouldn't be in reduced row echelon form because of that 4 in the 4th column, 2nd row. Every leading coefficient must be 1.
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  5. #5
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    Re: Row echelon form or Reduced row echelon form?

    The last column is what always confuses me. There always seems to be random numbers in there. So c) has a 4 in the last column which makes it not reduced, however in a) there are also random numbers (5 and 4) yet that is reduced?
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  6. #6
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    Re: Row echelon form or Reduced row echelon form?

    C is not in RREF only because that 4 is not a 1. If you divide that row by 4 (or multiply by 1/4), you will end up with a RREF matrix.

    As for random numbers in the last column, that's okay. Not every column needs to have a pivot.
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