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Invertible matrix

  1. Mar 17, 2007 #1

    Ami

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    1. The problem statement, all variables and given/known data
    Let A be a square matrix.
    If B is a square matrix satisfying AB=I


    2. Relevant equations
    Proof that B=A^-1


    3. The attempt at a solution
     
  2. jcsd
  3. Mar 17, 2007 #2

    radou

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    Well, you can first use the relation det(AB) = det(A)det(B). What does it tell you?
     
  4. Mar 18, 2007 #3

    mjsd

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    you can show this by showing that the inverse is unique.. ie. if AB=AC=I then B=C
     
  5. Mar 18, 2007 #4

    radou

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    Or, you have to show that, for a regular matrix A, and matrices B, C (where C is the inverse of A, the existence of which you have the right to assume after working out the hint in post #2), the cancellation law holds.
     
  6. Mar 18, 2007 #5

    AlephZero

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    You seem to be making this more complicated than it is. You don't know anything about A and B except that A has an inverse. So the only thing that you can use is the definition of what the inverse is:

    A^-1.A = A.A^-1 = I

    So just use that definition, to get from the equation AB = I to the equation B = A^-1.

    Hint: the associative law for multiplication says (XY)Z = X(YZ) for any matrices X Y and Z.
     
  7. Mar 18, 2007 #6

    HallsofIvy

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    I imagine that the point of this exercise is to show that the inverse is unique. You know that AB= I. Can you use that to prove that BA= I?
    Suppose AC= CA= I. Can you prove that B= C (hint, if AB= I = AC, multiply on both sides, on the left, by C.)
     
  8. Mar 21, 2007 #7

    Ami

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    Thanks so much to all of for replying

    The main problem is to show that A is invertible

    than I can show That [B=A^-1] easily
     
    Last edited: Mar 21, 2007
  9. Mar 21, 2007 #8

    Ami

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    Can you please show me how I can prove this???
     
  10. Mar 21, 2007 #9

    radou

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    Actually, the problem is already solved for you. Just re-read the replies.
     
  11. Mar 21, 2007 #10

    Ami

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    Thanks.
    But can you show me the solution more clearly ,please???
    I still confused about it.
     
  12. Mar 21, 2007 #11

    radou

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    No. You show us exactly which part you're confused about. :smile:
     
  13. Mar 21, 2007 #12

    AlephZero

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    It A is not invertible, then det(A) = 0

    That leads to a contradiction - see Radou's first hint.
     
  14. Mar 22, 2007 #13

    Ami

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    I'm sorry.I need to slove this problem without using the determinants.

    My attempt at the solution is:_
    First: If A is invertible:-
    By multiplying by A^-1 on both sides on the left:_
    (A^-1)(AB)=(A^-1)I
    IB=A^-1
    B=A^-1

    Second: I need to show now, that A is invertible, to complete the solution.
    This is the part which I'm confused about.

    please,I need the solution very quickly.And sorry for the disturbance.
     
  15. Mar 22, 2007 #14

    radou

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    If you need to show that A is invertible, the upper attempt is invalid, since you can't multiply an equation with something you don't even know exists (A^-1).
     
  16. Mar 22, 2007 #15

    Ami

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    Yes.but it can be valid if we show that A is invertible.

    There was another part of this problem,but it was about BA=I,The reference book solved it like the way I've written above.
    But It was easier:_

    To show that A is invertible: We can show that AX=0, have only the trivial solution:_
    So by multypling AX=0 by B on the left:_
    B(AX)=B0
    (BA)X=0 And, since BA=I
    IX=0
    X=0 And this is the trivial solution
    So A is invertible.

    But in our case,The problem is that we have [AB=I],not [BA=I],so we can't multiply by B, and derive that easily.
     
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