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Finding all solutions of Ax = y

  1. Oct 20, 2004 #1
    Hi everyone...First post here so be kind plz

    In one of my assignments, I need to find all the solutions of the following system:

    2 3 -5 1 x1 0
    1 3 -2 5 x2 = 1
    1 4 1 0 x3 0

    A X = Y

    I can easilly find the range space R{A}(Ind. Columns) and Null space N{A} (Ax =A~x= 0) using the echelon form of A. Since we have 3 leading entrys in A~ (Echelon), R{A} = R3, Dim(R{A}) = 3 thus Dim(N{a}) = 1 (single vector) which tells us that at least one solution is present. My problem is that I have no idea how to find that solution...I was thinking of taking the inverse of A and solving for A but since A is non-square...A is uninvertible.

    I know that all the solution will have the following structure:

    x = (the solution I cant find) + alpha(variable in R) * vector spanning N{A}

    Please help me, Im really desperate!! I have an exam tomorow morning on this and I want to have a good grade ;)

    thanks

    Mike
     
  2. jcsd
  3. Oct 21, 2004 #2

    Galileo

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    Science Advisor
    Homework Helper

    Row-reduce the equation untill A is in row-echelon form.
    You can basically read the answer from that form.
     
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