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A matrix question(simple one i suppose)

  1. Feb 23, 2006 #1
    I have th following matrix and I am suppose to determine whether the matrix is in row-echelon form(R-E), reduced row-echelon(R-RE) form, both, or neither..

    I know the following about R-R-E
    * If there is a row of all 0's there at the bottom of the matrix
    * The upper left most number has to be a 1
    * the following number below it has to be to the right of the leading 1 has to be a 1..
    * in R-RE any number above the 1 and below have to be 0..
    - For R-E they have to be only below the 1 not above it..
    all right,
    So here is the matrix:
    1 2 0 3 0
    0 0 1 1 0
    0 0 0 0 1
    0 0 0 0 0

    The book shows the answer as being both a R-R-E and R-E.
    I disagree because the 1 below the 3, if it were in R-R-E, that 3 would have to be a 0. Therefore, I believe it is in R-E form. Am I correct in my reasoning?
    thanks in advance..
     
  2. jcsd
  3. Feb 23, 2006 #2
    This is what PlanetMath has to say about R-E:

    A matrix is said to be in row echelon form if each non-zero row has more leading zeros than the previous row

    So, your argument that the 3 has to be a 0 would even kill that and you would have neither.

    Further more, PlanetMath says this about R-R-E:

    For a matrix to be in reduced row echelon form it has to first satisfy the requirements to be in row echelon form and additionally satisfy the following requirements:

    The first non-zero element in any row must be 1.
    The first element of value 1 in any row must the only non-zero value in its column.


    Now, look at these arguments. Can you tell why your matrix is indeed R-R-E?
     
  4. Feb 23, 2006 #3

    HallsofIvy

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    Notice the word the in "the 1". It does not say "each 1 in the row". That clearly refers to the leading 1 in the row, not the second 1.
     
  5. Feb 23, 2006 #4
    doh! I see said the blind man to his deaf wife.
     
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