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Math Help - Find det(X) given BXA^2 + 2I = 0.

  1. #1
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    Find det(X) given BXA^2 + 2I = 0.

    hello, this look rather simple, but I struggle, I need some help, or at least a final answer so I can check myself....thank you

    A and B are 5X5 matrices. it is known that:
    det(A)=2
    det(B)=-3

    X is a square matrix so BXA^2 + 2I = 0
    (multiplication of B, X and the square of A, plus twice the identity matrix)

    what is det(X) ?
    Last edited by mr fantastic; December 9th 2011 at 02:40 PM. Reason: Re-titled.
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  2. #2
    MHF Contributor Amer's Avatar
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    Re: simple question...

    the determinant function is multiplicative which means det(AB) =det(A)*det(B)

    so first i suggest to add -2I to both sides then take the det

    BXA^2 = -2I

    can you continue ?
    Last edited by Amer; December 9th 2011 at 09:07 AM.
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    Re: Find det(X) given BXA^2 + 2I = 0.

    thank you, I think I got it, I got det(X)=2.666, hope this is correct....

    thanks again !
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  4. #4
    MHF Contributor Amer's Avatar
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    Re: Find det(X) given BXA^2 + 2I = 0.

    det(BXA^2) = det(-2I)

    det(B) \; det(X) \; det(A)\; det(A) = -2

    -3 det(X) (4) = -2

    det(X) = \frac{-2}{-12}
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  5. #5
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    Re: Find det(X) given BXA^2 + 2I = 0.

    Quote Originally Posted by Amer View Post
    det(BXA^2) = det(-2I)

    det(B) \; det(X) \; det(A)\; det(A) = -2

    -3 det(X) (4) = -2

    det(X) = \frac{-2}{-12}
    No! These are 5x5 matrices, so \det(-2I) = (-2)^5 = -32.
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  6. #6
    MHF Contributor Amer's Avatar
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    Re: Find det(X) given BXA^2 + 2I = 0.

    Quote Originally Posted by Opalg View Post
    No! These are 5x5 matrices, so \det(-2I) = (-2)^5 = -32.
    thanks I did not noticed that
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