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HOw many solutions does this echelon matrix have? Mine isn't right! :\

  1. Sep 21, 2005 #1
    Hello everyone
    The reduced row-echelon forms of the augmented matrices of four systems are given below. How many solutions does each system have?
    Here is the matrices:
    1 0 -12 0
    0 1 0 0
    0 0 0 1
    0 0 0 0

    A. Infinitely many solutions
    B. No solutions
    C. Unique solution
    D. None of the above
    I said No solutions because 0 does not equal 1

    0 1 0 -15
    0 0 1 7

    A. No solutions
    B. Unique solution
    C. Infinitely many solutions
    D. None of the above

    I said Unqiue solution because y = 1, z = 7.

    1 0 0 8
    0 0 1 0

    A. Unique solution
    B. Infinitely many solutions
    C. No solutions
    D. None of the above

    I said unique solution because, y = 8, and z = 0;

    1 0 11
    0 1 9
    0 0 0
    A. Unique solution
    B. No solutions
    C. Infinitely many solutions
    D. None of the above

    I said Infinitely many solutions because you have a line of 0 0 0.
    NOw i submitted the answer but it said at least 1 is wrong, so i don't know iif they are all wrong or just 1 of them, any help would be great.
     
  2. jcsd
  3. Sep 22, 2005 #2

    TD

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    Homework Helper

    Since these are the reduced row-echelon forms of the augmented matrices, remember that the last column represents the constant and that every row is still an equation.

    After having reduced them, you can leave out all the 0 row's, that means row that have all 0's in every column. These are superfluous.

    When you end up with exactly the same number of equations (eq) as variables (var), then there is a unique solution.
    If you end up with less eq than var, there will be infinitely many solutions.

    But! You have to be careful if you get rows which have 0's for all the coëfficiënts but not 0 in the last column, of the constant. Back translated into an equation, this means something like [tex]0x+0y+0z=c[/tex], with c a constant different from 0. That is of course, not possible. In this case, your system has no solutions.
     
  4. Sep 22, 2005 #3
    thanks!! I think i got this right...
    So for
    1 0 -12 0
    0 1 0 0
    0 0 0 1

    no solutions because 0 != 1

    0 1 0 -15
    0 0 1 7

    unqiue solution

    1 0 0 8
    0 0 1 0
    unqiue solution


    1 0 11
    0 1 9
    0 0 0
    Infin. many solutions becuase we got a 0 0 0
     
  5. Sep 23, 2005 #4

    TD

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    Although you have a 0-row in the last one, you still end up with an equal ammount of unknowns and equations, so that yields a unique solution. You only have infinite solutions if your system is underdeterminate, that means that you end up with more variables than equations so you get to "choose" one or more variables (let x = s etc...)
     
  6. Sep 23, 2005 #5
    Thanks again TD! it worked fine after a few tries!!
     
  7. Sep 23, 2005 #6

    TD

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    Homework Helper

    Great :smile:
     
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