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Homework Help: Invertible Matrix Theorem

  1. May 7, 2008 #1
    General question regarding the Inv. Matrix Thm:

    One part of the theorem states that for an nxn invertible matrix, then there exists at least one solution for each b in Ax=b. Why wouldn't it be "there exists at MOST one solution for each b" since every column/row has a pivot. How would there exist more than one solution for each b if the columns span R_n?
     
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  3. May 7, 2008 #2

    Defennder

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    Are you sure that's what the theorem says? My book doesn't say "at least one". It says "exactly one". Here's what Wikipedia says:

     
    Last edited by a moderator: May 3, 2017
  4. May 7, 2008 #3
    Oh, weird. Yeah my book does say "at least one solution". Thanks for showing me the wiki entry though.
     
  5. May 8, 2008 #4

    HallsofIvy

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    Well, the statement is still true: if a matrix is invertible, then the equation Ax= b has exactly one solution so it is certainly true that there is at least one solution.

    If A is not invertible then Ax= b may have no solutions or an infinite number of solutions.

    You book may have some reason for emphasizing "the solution exists" right now rather than "the solution is unique"- both of which are true for A invertible.
     
  6. May 17, 2010 #5

    HallsofIvy

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    First, don't "hijack" someone else's thread for your own question- that's rude. Use the "new topic" button to start your own thread.

    Second, go back and reread the question. You can't prove any of those, they are all false. For example, if A is the 0 matrix, "Ax= b" has NO soution for b non-zero and has an infinite number of solutions if b is 0.
     
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