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Homework Help: Why this equation can't solve

  1. Aug 19, 2010 #1
    I would like to solve equation with x=a-1*b but when I inverse matrix A . It is bad solution to solve (I use mathematica to solve it)(Attach matrix A in excel file).
    I want to know why matrix A is bad solution when I inverse???????
    If I want to solve this equation. How i can do?

    Attached Files:

    Last edited: Aug 19, 2010
  2. jcsd
  3. Aug 19, 2010 #2


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    Is your matrix a square? If not - no inverse. (Although being square is not itself sufficient for an inverse to exist.)
  4. Aug 19, 2010 #3
    -I use Pseudo method but answer is bad solution
  5. Aug 19, 2010 #4


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    Okay, for which item are you trying to solve.

    your equation was
    x = a - 1b

    That has nothing to do at all with your "solution" - your result doesn't follow from the equation.
  6. Aug 19, 2010 #5


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    Roger Penrose worked on the inverses on non-square matries before relativity.
  7. Aug 19, 2010 #6


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    I believe the OP meant x = A-1b.

    Then he/she revised the above to this: x=(aT*a)-1*B,
    which I believe means this: x = (ATA)-1b.
  8. Aug 20, 2010 #7


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    Aha - good catch. Of course, this means that the original equation is nowhere to be seen, so it remains impossible to see why this approach fails.
  9. Aug 20, 2010 #8


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    Need to pay attention to dimension. The matrix A has 35 rows and 20 columns (i.e. 35 x 20) and B has 35 rows (i.e. 35 x 1). ATA is a 20 x 20 matrix and so is (ATA)-1. Multiplying a 35 x 1 matrix by a 20 x 20 matrix isn't valid.

    This smells like a least squares problem, which means the solution would be x = (ATA)-1 ATb.
  10. Aug 21, 2010 #9


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    What in the world do you mean by this? Roger Penrose wasn't born until 26 years after relativity was developed! Do you mean he worked on them before they were used in relativity?
  11. Aug 21, 2010 #10
    Penrose worked those inverses before he worked on relativity
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