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How tdo I solve a Linear Equation with an unknown

  1. Sep 3, 2009 #1
    1. I am having a trouble with this math problem and would like help finding the answer. Here goes: Let (the ~~ indicates where a line should straight line should be going downward and /// are like a space between the numbers)
    1 /// (-1) /// 1 ~~ 2
    a /// 0 /// a+1 ~~ 2a+3
    2a /// (-2a) /// (a^2)-3 ~~ 5a-3

    be the augmented matrix of a linear system. For what values of a does the system have no solution? One solution? Infinitely many solutions?





    So, if I correctly reduced it, that would mean that
    1 /// 0 /// 1+1/a ~~ 2+3/a
    0 /// 1 /// 1/a ~~ 3/a
    0 /// 0 ///(a-3)(a+1) ~~ a-3


    0 solutions when a=-1
    infinite solutions when a=3
    one solution when a/=/(doest not equal) 1, 3
    And also a/=/ 0 because that would be dividing by 0.

    Is this the way you are supposed to do this type of problem?
     
  2. jcsd
  3. Sep 3, 2009 #2
    I don't really understand the matrix. Can you not write it a little clearer? then maybe I can help.
     
  4. Sep 3, 2009 #3
    Hmm... I don't really know how to write it clearer does anyone know how to write a matrix on a forum?
     
  5. Sep 4, 2009 #4
    [tex ] \left(

    \begin{array}{ccc|c}

    a & b & c & u\\

    d & e & f & v\\

    g & h & j & w\\

    \end{array}

    \right)

    [/ tex]

    (except leave out the spaces in the "tex" tags. Substitute the values you need.

    It looks like this:

    [tex] \left( \begin{array}{ccc|c}
    a & b & c & u\\
    d & e & f & v\\
    g & h & j & w\\
    \end{array} \right)
    [/tex]

    --Elucidus
     
  6. Sep 4, 2009 #5
    The OP meant:
    [tex]\begin{vmatrix}
    1 & -1 & 1 & 2\\
    a & 0 & a+1 & 2a+3 \\
    2a & -2a & a^2-3 & 5a-3
    \end{vmatrix}[/tex]

    Anyway you did it right, except you should mention that for a=0 there are also 0 solutions.

    Also, there is one solution when a/=/(doest not equal) -1, 3

    I guess it was typo.
     
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