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Homework Help: Concept of Invertible Matrices

  1. Oct 30, 2008 #1
    1. The problem statement, all variables and given/known data

    If A is an invertible square matrix, then Ax = b is consistent for each b in R^n

    2. The attempt at a solution

    If A multiplied by A inverse is identity, then it would always be consistent. So I thought , if A is just randomly multiplied by some x, then it will still be consistent right? I can't seem to find anything wrong with the statement above.
     
  2. jcsd
  3. Oct 30, 2008 #2
    I think Ax = b holds true regardless of A being invertible or not
     
  4. Oct 30, 2008 #3

    Dick

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    Ax=b is not exactly 'true' if there no solutions x to the equation given a b. If A is invertible then A^(-1)Ax=Ix=x=A^(-1)b. So given any b, x=A^(-1)b. Yes, it's consistent.
     
  5. Oct 31, 2008 #4

    HallsofIvy

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    I have no idea what you mean by this. What is "Ax= b" that it could "hold true"? Certainly if A is a matrix and x a vector with as many components as A has columns, then there exist a vector b such that Ax= b. Is that what you meant?

    If A is invertible, then given any such vector b, there exist a vector x such that Ax= b.

    If A is not invertible then, given b, there may not exist such an x or there may exist an infinite number of the them.
     
  6. Oct 31, 2008 #5
    Thanks for the help guys. That little algebra helped me see it properly Dick.
     
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