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Thread: Determinant proof

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

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

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

    $\displaystyle |A^{-1}||A|=|A^{-1}||B|$

    $\displaystyle 1=|A^{-1}||B|$

    Now, let $\displaystyle |C|=1$

    $\displaystyle |C|=|A^{-1}||B|$

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

    $\displaystyle |A^{-1}||A|=|A^{-1}||B|$

    $\displaystyle (|A||A^{-1}|)|A|=(|A||A^{-1}|)|B|$

    $\displaystyle 1|A|=(|A||A^{-1}|)|B|$

    $\displaystyle |A|=|C||B|$

    $\displaystyle \therefore$ $\displaystyle A=CB$

    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 $\displaystyle |A|$ is the determinant of the matrix $\displaystyle A$. I'm afraid your arguments only allow you to say that if $\displaystyle A,B,C$ are matrices with $\displaystyle |A|=|B|\ne 0$ and $\displaystyle |C|=1$, then $\displaystyle |A|=|C||B|=|B|$ which is vacuously true. That does not imply $\displaystyle A=CB$. Example: $\displaystyle A=B=I, C=\begin{pmatrix}1 &1 \\ 0 & 1\end{pmatrix}$.

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

    Now we need some properties of the determinant, namely that $\displaystyle |A^{-1}|=|A|^{-1}$ and $\displaystyle |AB|=|A||B|$. Then it is easy to see that $\displaystyle |AB^{-1}|=|A||B^{-1}|=|A||B|^{-1}=1$ as required.
    Last edited by siclar; Jul 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 $\displaystyle |A^{-1}B|=|A^{-1}||B|=|A|^{-1}|A|=1$ as required.
    I'm assuming you meant to show $\displaystyle |AB^{-1}|=|A||B^{-1}|=|A||A|^{-1}=1$ which gives us the $\displaystyle |C|$ 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 $\displaystyle |AB^{-1}|=|A||B^{-1}|=|A||A|^{-1}=1$ which gives us the $\displaystyle |C|$ 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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