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Existance of a matrix that satisfies a condition

  1. Jan 20, 2009 #1

    fluidistic

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    Gold Member

    1. The problem statement, all variables and given/known data
    A=(2 3, 1 2, 2 5) where the coma separates the rows of the matrix.
    Does there exist a matric C such that AC=I? Where I is the 3x3 identity matrix.
    2. The attempt at a solution
    No. First I note that if it exists then C is a 2x3 matrix. I also note that if AC=I, then C is the inverse of A. But as A is not a square matrix, it doesn't have an inverse, so C cannot exist.
    Am I right?
     
  2. jcsd
  3. Jan 20, 2009 #2
    Not exactly.
    2 3
    1 2
    2 5
    C should be 2x3, i.e of the form:
    a b c
    d e f

    so you need to solve the system of equations:
    2a+3d=1
    2b+3e=0
    2c+3f=0
    2a+3d=0
    b+2c=1
    c+2f=0
    2a+5d=0
    2b+5e=0
    2c+5f=1

    I don't know if the answer is affirmative or not, you need to solve these equations, in general it's possible that even a non square matrix will have a right reciprocal or left one, and both of them need not be the same necessarily.
     
  4. Jan 20, 2009 #3

    Dick

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    Homework Helper

    You could also note that your product looks like column(v1,v2,v3)*row(w1,w2,w3) where the v's (rows) and w's (columns) are 2-vectors. You know one of the v's can be written as a linear combination of the other two, say v3=a*v1+b*v2. You want v3.w3=1. What do other matrix entries in the product AC tell you about v1.w3 and v2.w3?
     
    Last edited: Jan 20, 2009
  5. Jan 20, 2009 #4

    fluidistic

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    Gold Member

    Ok I think I know where I made an error, it was there :" I also note that if AC=I, then C is the inverse of A."
    I'm not really used to linear combinations so I don't get all what you said Dick, but very soon I'll be in it.
    Doing the method of loop quantum gravity, I realized from the beggining that the system has no solution.
    Thanks both for your help.
     
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