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Homework Help: Prove that A is reversible matrix if B is reversible

  1. May 6, 2013 #1
    Using [itex]A\widetilde{A}=(detA)I[/itex] prove that [itex]\widetilde{A}[/itex] is reversible matrix if and only if [itex]A[/itex] is reversible. Also prove [itex]det(\widetilde{A})=(detA)^{n-1}[/itex] for any square matrix [itex]A[/itex].

    First part:
    1. direction:
    Lets say [itex]\widetilde{A}[/itex] is reversible, this means that [itex]\widetilde{A}\widetilde{A}^{-1}=I[/itex] and [itex]det\widetilde{A}\neq 0[/itex].

    [itex]A\widetilde{A}=(detA)I[/itex] can than be written as:
    [itex]A=(detA)\widetilde{A}^{-1}[/itex] but now I don't know what else I can do here to prove that A is reversible?
     
  2. jcsd
  3. May 6, 2013 #2

    mfb

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    2017 Award

    Staff: Mentor

    Try to find a matrix, which, if multiplied by [itex](detA)\widetilde{A}^{-1}[/itex], gives I.
     
  4. May 6, 2013 #3
    Something like [itex]((detA)\widetilde{A}^{-1})^{-1}[/itex]? But I have absoulutely no reason to believe that [itex](detA)\neq 0[/itex] (that's what I am trying to prove).

    [itex]((detA)\widetilde{A}^{-1})^{-1}[/itex]
    [itex](detA)^{-1}\widetilde{A}[/itex]
     
  5. May 6, 2013 #4

    HallsofIvy

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    Science Advisor

    Then do it as two separate cases:
    Suppose det(A)= 0. Then what?

    Suppose [itex]det(A)\ne 0[/itex]. Then what?
     
  6. May 6, 2013 #5
    Hmm, why don't I have brains like you? That would really help me! :D

    Case No.1: suppose [itex]det(A)\ne 0[/itex]
    [itex]A=(detA)\widetilde{A}^{-1}[/itex] I can than multiply both sides with [itex](detA)^{-1}\widetilde{A}[/itex] which gives me
    [itex](detA)^{-1}\widetilde{A}A=((detA)\widetilde{A}^{-1})((detA)^{-1}\widetilde{A})[/itex]
    [itex](detA)^{-1}\widetilde{A}A=I[/itex]
    So [itex]A[/itex] is reversible and [itex]A^{-1}=(detA)^{-1}\widetilde{A}[/itex]

    Case No.2: suppose [itex]detA=0[/itex]
    [itex]A=(detA)\widetilde{A}^{-1}[/itex]
    [itex]A=0[/itex]

    So if [itex]detA=0[/itex] than A is not reversible by definiton?

    Does that sound right?
     
  7. May 6, 2013 #6
    Right

    It is correct that ##A## will not be reversible. But it's not good enough. You are proving that if ##\tilde{A}## is invertible, then ##A## is invertible.
    You have now found that if ##det(A)=0## and ##\tilde{A}## is invertible, then ##A## is not invertible. So you have found a situation where ##\tilde{A}## is invertible and ##A## is not invertible, but you had to prove that ##A## is always inverible whenever ##\tilde{A}## was. This is an apparent contradiction!! To solve it, you must prove that the situation ##det(A)=0## and ##\tilde{A}## invertible never occurs! So you need to prove that if ##det(A)=0## and ##\tilde{A}## is invetible, then some contradiction occurs!
     
  8. May 6, 2013 #7
    Hmmm, I can see where this is going but up to this moment I haven't got a good idea yet. We often proved this kind of statement using paradox:
    Let's say that ##A## is invertible and show that this statement will lead us to paradox and prove that ##A## is NOT invertible.

    Hmm, if ##A## is invertible, than ##detA\neq 0## but one of the previous condition says that ##detA=0## and this is paradox, so ##A## is NOT invertible.

    I tried using the same idea in equation: (so ##detA=0##, ##AA^{-1}=I## and ##\tilde{A}\tilde{A}^{-1}=I##)

    ##A\tilde{A}=(detA)I## but I didn't get really far this way... all I found out is that ##I=(detA)\tilde{A}^{-1}A^{-1}## where I can see that ##A=(detA)\tilde{A}^{-1}## which was obvious before too...
     
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