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How to get the 4th equation using this data(lenear algebra)

  1. Feb 13, 2009 #1
    this is the data that i was given about the question:
    http://img23.imageshack.us/img23/6041/49022610ps1.th.gif [Broken]

    i need to find the matrix of T
    i know that T is of 2X4 size
    so i built T as
    a1 a2 a3 a4
    b1 b2 b3 b4

    and multiplied by the given vector
    i built a matrix for "a" parameters and "b" parameters
    but i got only 3 equation instead of the needed 4
    in order to get the values of "a"'s and "b"'s

    how to use the independent vectors data to construct the 4th equation??
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 13, 2009 #2


    Staff: Mentor

    Let's call your vectors u1 = [1 1 0 1], u2 = [0 1 0 1], and u3 = [1 1 0 0].
    Since Tu1 = Tu2 = Tu3 = [1 1], then
    Tu1 - Tu2 = [0 0] ==> T(u1 - u2) = [0 0]
    Tu1 - Tu3 = [0 0] ==> T(u1 - u3) = [0 0]
    Tu2 - Tu3 = [0 0] ==> T(u2 - u3) = [0 0]

    These equations tell us that u1 - u2, u1 - u3, and u2 - u3 are in the nullspace of T.
    Find these vectors, and find a subset of them that is a linearly independent set. From those vectors you can get values for the entries of the matrix corresponding to the transformation T.
  4. Feb 13, 2009 #3
    i solved it differently
    i said
    T=(a1 a2 a3 a4)
    (b1 b2 b3 b4)

    and built an equations T*U1=(1,1) T*U2=(1,1) T*U3=(1,1)
    so i get for "a" parameters only 3 equations
    which is not enough
    how to use the data that those vectors are independant
    in order to build the 4th "a" parameter equation

    ( the same thing goes for "b" parameters)

    Last edited: Feb 13, 2009
  5. Feb 13, 2009 #4


    Staff: Mentor

    You say you solved it differently, but you're still asking questions, so it doesn't sound like the problem is actually solved. On the other hand, I actually found a matrix A for which A*u1 = [1 1], A*u2 = [1 1] and A*u3 = [1 1], so which is the better solution?
  6. Feb 13, 2009 #5
    you are not using he fact that two vectors are independent
  7. Feb 13, 2009 #6


    Staff: Mentor

    Yes, I am. Look at what I said in post #2.
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