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Math Help - Show A and B are invertible and find their determinants

  1. #1
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    Algebra

    Let
    A and B be square matrices of the same dimensions. Suppose that

    det(
    AB2) = 1
    det(
    AB) = 3

    Show that A and B are invertible and find their determinants.


    I'm not even sure where to start with this!
    Last edited by Arita; January 14th 2011 at 08:16 PM.
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  2. #2
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    What is B2 here?
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  3. #3
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    The 2 is suppose to represent squared. So it would be det(AB(squared))=1
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  4. #4
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    Would it be helpful to say \det(AB)= \det(A)\det(B) ?
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  5. #5
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    Start with these facts:

    1) det(AB)= det(A)det(B)

    2) A is invertible if and only det(A) is not 0.

    3) xy= 0 if and only if either x= 0 or y= 0. So that if xy is not 0 then neither x nor y is 0.

    det(AB)= 3 which is not 0 so neither det(A) nor det(B)= 0 and both A and B are invertible.

    det(AB^2)= det(AB)det(B) and you are told both det(AB) and det(AB^2).
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  6. #6
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    So far what I have is det(AB)=det(A)det(B)=3 and det(AB(squared))= det(A)det(B(squared))
    (square root of 1) = det(A)(square root (det(B)))
    1 = det(A)(square root (det(B)))
    1 = 1/3(square root of 9)
    1 = 1/3(3)
    1 = 1
    det(AB)= det(A)det(B)
    3 = 1/3(9)
    3 = 3

    But I'm not sure if this is right?
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  7. #7
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    Now you're on the right track! Hall's hint was the killer here!

    \det(AB^2)= \det(AB)\det(B)

    1= 3\det(B)

    \det(B)= \frac{1}{3}

    now find \det(A) given \det(A)\det(B)=3
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  8. #8
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    So det(AB^2) = det(AB)det(B)?

    I thought it would be det(AB^2) = det(A)det(B)^2
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