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Math Help - Determinant proof

  1. #1
    Senior Member Danneedshelp's Avatar
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    Determinant proof

    Q: Prove that if |A|=|B|\neq{0} and A and B are of the same size, then there exists a matrix C such that |C|=1 and A=CB.

    A: Since |A|=|B|\neq{0} we know A^{-1} and B^{-1} exists. So,

    |A^{-1}||A|=|A^{-1}||B|

    1=|A^{-1}||B|

    Now, let |C|=1

    |C|=|A^{-1}||B|

    Since |A|=|B| and |C|=|A^{-1}||A|

    |A^{-1}||A|=|A^{-1}||B|

    (|A||A^{-1}|)|A|=(|A||A^{-1}|)|B|

    1|A|=(|A||A^{-1}|)|B|

    |A|=|C||B|

    \therefore A=CB

    Does that work? I guess A=I, B=I, and C=I. Do I need to show more?
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  2. #2
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    I presume from the thread title that |A| is the determinant of the matrix A. I'm afraid your arguments only allow you to say that if  A,B,C are matrices with |A|=|B|\ne 0 and |C|=1, then |A|=|C||B|=|B| which is vacuously true. That does not imply A=CB. Example: A=B=I, C=\begin{pmatrix}1 &1 \\ 0 & 1\end{pmatrix}.

    You started at the right place: Since |A|,|B| are nonzero, they are invertible. Then A=AI=A(B^{-1}B)=(AB^{-1})B, so if we can show |AB^{-1}|=1, then we have found our  C!

    Now we need some properties of the determinant, namely that |A^{-1}|=|A|^{-1} and |AB|=|A||B|. Then it is easy to see that |AB^{-1}|=|A||B^{-1}|=|A||B|^{-1}=1 as required.
    Last edited by siclar; July 14th 2009 at 05:03 PM. Reason: typo
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  3. #3
    Senior Member Danneedshelp's Avatar
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    Then it is easy to see that |A^{-1}B|=|A^{-1}||B|=|A|^{-1}|A|=1 as required.
    I'm assuming you meant to show |AB^{-1}|=|A||B^{-1}|=|A||A|^{-1}=1 which gives us the |C| we were searching for. Correct? I guess it doesn't really matter what order since A=B and these are square matrices, but it looks cleaner.
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  4. #4
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    Quote Originally Posted by Danneedshelp View Post
    I'm assuming you meant to show |AB^{-1}|=|A||B^{-1}|=|A||A|^{-1}=1 which gives us the |C| we were searching for. Correct? I guess it doesn't really matter what order since A=B and these are square matrices, but it looks cleaner.
    Precisely
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