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Thread: Matrix Identity

  1. #1
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    Matrix Identity

    Hey all,

    Given $\displaystyle \[
    B =
    \left( {\begin{array}{ccc}
    1 & 0 & 2 \\
    3 & 1 & 0 \\
    1 & 0 & 1 \\
    \end{array} } \right)
    \]$ and $\displaystyle \[
    B^{-1} =
    \left( {\begin{array}{ccc}
    -1 & 0 & 2 \\
    3 & 1 & -6 \\
    1 & 0 & -1 \\
    \end{array} } \right)
    \]$
    find C, where $\displaystyle BCB^{T} = I$
    Thanks.
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  2. #2
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    What ideas have you had so far?
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  3. #3
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    $\displaystyle BCB^{T}=I \Rightarrow BB^{T}C = I \\ $
    so $\displaystyle C = I(BB^{T})^{-1}\\$
    which is $\displaystyle
    \[
    M =
    \left( {\begin{array}{ccc}
    11 & 3 & -2 \\
    3 & 1 & -6 \\
    -21 & -6 & 41 \\
    \end{array} } \right)
    \]$
    That seems to work.. not sure if its right though...
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  4. #4
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    You might just get the right answer for the wrong reasons. *slaps your wrist*

    You don't know that $\displaystyle B$ and $\displaystyle C$ commute! You can't just slip $\displaystyle B^{T}$ past the $\displaystyle C$ like that. However, you can left- and right-multiply by inverses, if you know they exist. In this case, you've been given $\displaystyle B^{-1}$, so you know that exists. How about this first step:

    $\displaystyle BCB^{T}=I$

    $\displaystyle B^{-1}BCB^{T}=B^{-1}I=B^{-1}$

    $\displaystyle CB^{T}=B^{-1}.$

    What else could you do?
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  5. #5
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    (In regards to my second post - Not ackbeets)
    Sorry thats wrong... I think it should be that
    The product of $\displaystyle BC$ should give the inverse of $\displaystyle B^{T}$
    so $\displaystyle B \cdot C = B^{-1} \Rightarrow C = B^{T-1} \cdot B^{-1}$
    Which is

    $\displaystyle \[
    M =
    \left[ {\begin{array}{ccc}
    5 & -15 & -3 \\
    -15 & 46 & 9 \\
    -3 & 9 & 2 \\
    \end{array} } \right]
    \]$
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  6. #6
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    A few comments:

    1. What is $\displaystyle M?$ No need to introduce another variable if you don't need it.

    2. The notation $\displaystyle B^{T-1}$ should really be $\displaystyle (B^{T})^{-1}.$ Otherwise, someone might think you're raising $\displaystyle B$ to the power $\displaystyle T-1.$

    3. You've got the order wrong. You need to right-multiply by $\displaystyle (B^{T})^{-1}.$ That is, you've got to do this (picking up from where I left off in my last post):

    $\displaystyle CB^{T}=B^{-1}$

    $\displaystyle CB^{T}(B^{T})^{-1}=B^{-1}(B^{T})^{-1}$

    $\displaystyle C=B^{-1}(B^{T})^{-1}.$

    How can you compute the RHS here?
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  7. #7
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    Ah, thanks for the pointers. RHS would just be the product of the inverses of B and its Transpose..
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  8. #8
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    Ah, but you're not directly given $\displaystyle (B^{T})^{-1}.$ How can you get it?
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  9. #9
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    Wouldn't that just be:

    $\displaystyle \[
    B =
    \left( {\begin{array}{ccc}
    1 & 0 & 2 \\
    3 & 1 & 0 \\
    1 & 0 & 1 \\
    \end{array} } \right)
    \]$ Then $\displaystyle \[
    (B^{T})^{-1} =
    \left( {\begin{array}{ccc}
    1 & 3 & 1 \\
    0 & 1 & 0 \\
    2 & 0 & 1 \\
    \end{array} } \right)^{-1}
    \]$
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  10. #10
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    True, but I think you can get what you need faster than that, if you use an identity.
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