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Solutions to linear equations

  1. Jun 17, 2007 #1
    For which values (a,b,c) does the above equation have a solution where a,b,c,x,y,z belong to R?
    Initially I found the determinant of the above matrix and it was O. From this I know that there will be nontrivial solution solutions for (a,b,c) = (0,0,0).
    But I am confused as to the possibility of having any other solutions for nonhomogenous system. I know that it will not have a uniques solution. Can any body tell me if there is going to be any other solutions for values of (a,b,c) other than the homogenous one.
  2. jcsd
  3. Jun 24, 2007 #2
    Any multiple of (1,-2,1) is sent to (0,0,0) by your matrix.
    In other words, <{(1,-2,1)}> represents its one-dimensional nullspace.

    The rhs has to lie in the two-dimensional columnspace for there to be any solutions.
    The constraint on a,b,c for existence of solutions is a-2b+c=0.

    For any rhs that satisfies the constraint, there exits a one-dimensional linear manifold that
    constitutes the solution set.

    Simple example:

    Let (a,b,c) = (-1,-1,-1). Then <{(1,-2,1)}> + (1,-1,0) is the solution set.
    Last edited: Jun 24, 2007
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