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A matrix arithmetic T or F

  1. Feb 18, 2013 #1
    I suppose the title should be "Matrix polynomial T or F" but whatever.

    1. The problem statement, all variables and given/known data
    True or false: if A2-2A+I=0, then A-1=2I2-I

    3. The attempt at a solution
    My thought:

    A2-2A+I=0 becomes (A-I)(A-I)=0
    A = I

    The inverse of I is I.

    So the second equation:



    I = 2I2-I


    2I-I = I

    So the answer is True.

    Where did I go wrong?
  2. jcsd
  3. Feb 18, 2013 #2


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    XX=0 doesn't mean X=0 for matrices. Try to find a counter example for your claim. Try A=[[1,1],[0,1]].
  4. Feb 18, 2013 #3
    Gaaah, thanks. Although the TorF question specified, now that I think of it, a 4x4 matrix, making figuring it out for a fraction of a point even harder. Jesus, the entire lack of class or book examples is really screwing everyone in this course over.

    One other, if you have a moment.

    A = PBP-1

    and thus A = B and detA=detB

    Most people put "false" thinking that because B was between P and P-1, they did not always constitute I and thus both A=B and detA=detB would be wrong. But the answer was true.

    Sorry if I'm breaking any rules, I just felt silly make an entirely new thread for this.
  5. Feb 18, 2013 #4


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    Not breaking any rules. A doesn't have to be equal to B. At all. But det(PBP^(-1))=det(P)*det(B)*det(P^(-1)). Now you can rearrange that to det(P)*det(P^(-1))*det(B). So det(A) does equal det(B). If you assume det(P) is nonzero. Otherwise even that's not even true.
    Last edited: Feb 18, 2013
  6. Feb 18, 2013 #5
    Thank you. So basically, we have half of a test question that is wrong and another half that was not covered in class notes, nor in the textbook we're using - we were never told and there were never any questions suggesting we could rearrange determinants like that.

    Any recommendations on textbooks that DO cover these totally important topics?
  7. Feb 18, 2013 #6
    I mean, they're just numbers, but showing that we can DO it would have been nice.

    Again, it's late for me.
  8. Feb 18, 2013 #7


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    I'm not really in teaching these days. So I can't really recommend one. But I can't imagine any of them don't mention det(AB)=det(A)det(B). Determinants are just numbers. You can rearrange them any way you anyway you can rearrange numbers.
  9. Feb 18, 2013 #8


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    As noted, this is false. However, what can be said is that
    $$0 = A^2 - 2A + I = A(A - 2I) + I$$
    $$A(2I-A) = I$$
    and therefore ##A^{-1} = 2I - A##.
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