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Math Help - formula for matrix determinant

  1. #1
    MHF Contributor kalagota's Avatar
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    formula for matrix determinant

    please compute the determinant of this matrix..

    <br />
\left[ { \begin{array}{ccccc} k & 1 & 0 & \ldots & 0 \\ 1 & k & 1 & \ldots & 0 \\ 0 & 1 & k & \ldots & 0 \\ \vdots & \, & \ddots & \ddots & \vdots \\ 0 & 0 & \ldots & 1 & k \\ \end{array} } \right]<br /> <br />

    can someone give the general formula.. thanks!!
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  2. #2
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    Quote Originally Posted by kalagota View Post
    please compute the determinant of this matrix..

    <br />
\left[ { \begin{array}{ccccc} k & 1 & 0 & \ldots & 0 \\ 1 & k & 1 & \ldots & 0 \\ 0 & 1 & k & \ldots & 0 \\ \vdots & \, & \ddots & \ddots & \vdots \\ 0 & 0 & \ldots & 1 & k \\ \end{array} } \right]<br /> <br />

    can someone give the general formula.. thanks!!
    Develop a recursive form for this determinant, for the n \times n case put::

    <br />
K_n=\left[ { \begin{array}{ccccc} k & 1 & 0 & \ldots & 0 \\ 1 & k & 1 & \ldots & 0 \\ 0 & 1 & k & \ldots & 0 \\ \vdots & \, & \ddots & \ddots & \vdots \\ 0 & 0 & \ldots & 1 & k \\ \end{array} } \right]<br /> <br />

    The expand the determinant about the 1 st row:

    \det(K_n) = k \det(K_{n-1}) - 1 \det (L_{n-1})

    where:

    <br />
L_n=\left[ { \begin{array}{ccccc} 1 & 1 & 0 & \ldots & 0 \\ 0 & k & 1 & \ldots & 0 \\ 0 & 1 & k & \ldots & 0 \\ \vdots & \, & \ddots & \ddots & \vdots \\ 0 & 0 & \ldots & 1 & k \\ \end{array} } \right]<br /> <br />

    Now expand the determinant of L_n about the first column:

    \det(L_n) = \det(K_{n-1})

    so we have:

    \det(K_n) = k \det(K_{n-1}) - 1 \det (K_{n-2})

    and hopefully you can take it from there.

    ZB
    Last edited by Constatine11; December 6th 2007 at 06:46 AM. Reason: error in subscripting
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  3. #3
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Constatine11 View Post
    Develop a recursive form for this determinant, for the n \times n case put::

    <br />
K_n=\left[ { \begin{array}{ccccc} k & 1 & 0 & \ldots & 0 \\ 1 & k & 1 & \ldots & 0 \\ 0 & 1 & k & \ldots & 0 \\ \vdots & \, & \ddots & \ddots & \vdots \\ 0 & 0 & \ldots & 1 & k \\ \end{array} } \right]<br /> <br />

    The expand the determinant about the 1 st row:

    \det(K_n) = k \det(K_{n-1}) - 1 \det (L_{n-1})

    where:

    <br />
L_n=\left[ { \begin{array}{ccccc} 1 & 1 & 0 & \ldots & 0 \\ 0 & k & 1 & \ldots & 0 \\ 0 & 1 & k & \ldots & 0 \\ \vdots & \, & \ddots & \ddots & \vdots \\ 0 & 0 & \ldots & 1 & k \\ \end{array} } \right]<br /> <br />

    Now expand the determinant of L_n about the first column:

    \det(L_n) = \det(K_n-1)

    so we have:

    \det(K_n) = k \det(K_{n-1}) - 1 \det (K_{n-2})

    and hopefully you can take it from there.

    ZB

    ? i can't see how you got L_n.. can you show it? thanks
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  4. #4
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    Quote Originally Posted by kalagota View Post
    ? i can't see how you got L_n.. can you show it? thanks
    Delete the second column and first row from K_{n+1}.

    ( -\det(L_n) is the cofactor of the 1 in the first row of K_{n+1} )

    ZB
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  5. #5
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Constatine11 View Post
    Delete the second column and first row from K_{n+1}.

    ( -\det(L_n) is the cofactor of the 1 in the first row of K_{n+1} )

    ZB
    ok.. i'm just used to do the expansion about rows.. thanks anyway.. (although its too late.. )
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  6. #6
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    Quote Originally Posted by kalagota View Post
    ok.. i'm just used to do the expansion about rows.. thanks anyway.. (although its too late.. )
    By the way I have just coded up the solution to the recurrence and compared it to the determinant of some low dimensional examples and they agree.

    ZB
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