Results 1 to 10 of 10

Math Help - Matrix Identity

  1. #1
    Member
    Joined
    Jan 2011
    Posts
    87

    Matrix Identity

    Hey all,

    Given \[<br />
B =<br />
\left( {\begin{array}{ccc}<br />
 1 & 0 & 2  \\<br />
 3 & 1 & 0 \\<br />
 1 & 0 & 1 \\<br />
 \end{array} } \right)<br />
\] and \[<br />
B^{-1} =<br />
\left( {\begin{array}{ccc}<br />
 -1 & 0 & 2  \\<br />
 3 & 1 & -6 \\<br />
 1 & 0 & -1 \\<br />
 \end{array} } \right)<br />
\]
    find C, where BCB^{T} = I
    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    What ideas have you had so far?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jan 2011
    Posts
    87
    BCB^{T}=I \Rightarrow BB^{T}C = I \\
    so C = I(BB^{T})^{-1}\\
    which is <br />
\[<br />
M =<br />
\left( {\begin{array}{ccc}<br />
 11 & 3 & -2  \\<br />
 3 & 1 & -6  \\<br />
 -21 & -6 & 41 \\<br />
 \end{array} } \right)<br />
\]
    That seems to work.. not sure if its right though...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    You might just get the right answer for the wrong reasons. *slaps your wrist*

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

    BCB^{T}=I

    B^{-1}BCB^{T}=B^{-1}I=B^{-1}

    CB^{T}=B^{-1}.

    What else could you do?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jan 2011
    Posts
    87
    (In regards to my second post - Not ackbeets)
    Sorry thats wrong... I think it should be that
    The product of BC should give the inverse of B^{T}
    so B \cdot C = B^{-1} \Rightarrow C = B^{T-1} \cdot B^{-1}
    Which is

    \[<br />
M =<br />
\left[ {\begin{array}{ccc}<br />
 5 & -15 & -3 \\<br />
 -15 & 46 & 9  \\<br />
 -3 & 9 & 2 \\<br />
 \end{array} } \right]<br />
\]
    Follow Math Help Forum on Facebook and Google+

  6. #6
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    A few comments:

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

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

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

    CB^{T}=B^{-1}

    CB^{T}(B^{T})^{-1}=B^{-1}(B^{T})^{-1}

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

    How can you compute the RHS here?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Jan 2011
    Posts
    87
    Ah, thanks for the pointers. RHS would just be the product of the inverses of B and its Transpose..
    Follow Math Help Forum on Facebook and Google+

  8. #8
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    Ah, but you're not directly given (B^{T})^{-1}. How can you get it?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Jan 2011
    Posts
    87
    Wouldn't that just be:

    \[<br />
B =<br />
\left( {\begin{array}{ccc}<br />
 1 & 0 & 2  \\<br />
 3 & 1 & 0  \\<br />
 1 & 0 & 1 \\<br />
 \end{array} } \right)<br />
\] Then \[<br />
(B^{T})^{-1} =<br />
\left( {\begin{array}{ccc}<br />
 1 & 3 & 1 \\<br />
 0 & 1 & 0  \\<br />
 2 & 0 & 1 \\<br />
 \end{array} } \right)^{-1}<br />
\]
    Follow Math Help Forum on Facebook and Google+

  10. #10
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    True, but I think you can get what you need faster than that, if you use an identity.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Identity Matrix
    Posted in the Algebra Forum
    Replies: 3
    Last Post: June 3rd 2010, 10:39 PM
  2. [SOLVED] Elementary matrix, restore to identity matrix
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: February 13th 2010, 10:04 AM
  3. Inverse of an Identity Matrix
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: July 3rd 2009, 10:23 AM
  4. Identity Matrix
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: October 5th 2008, 04:00 PM
  5. Prove identity matrix
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: May 5th 2008, 06:30 AM

Search Tags


/mathhelpforum @mathhelpforum