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Math Help - Determinant calculation

  1. #1
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    Determinant calculation

    Haw to calculate this determinant with 2012 columns and 2012 rows?
    Determinant calculation-codecogseqn.gif
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  2. #2
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    Lightbulb Re: Determinant calculation

    Quote Originally Posted by mathmagix View Post
    Haw to calculate this determinant with 2012 columns and 2012 rows?
    Click image for larger version. 

Name:	CodeCogsEqn.gif 
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    Det =
    35529225841856483043651808603739698746526147540091 25781343678823103964817334008922598220766872045514 43818875406235659520138976614372162888864682450773 56809484825684704986491920597144636004496957589321 45807071799620915629374867308911091630619589522654 20876697189137171661445806764230295352215847197485 40965173460832558185841042472018236366467686259189 20543195609189247214180736087814328750732314127804 25828702376026086993776454381667343776846151461024 32549811417270673109240013618063599797009890166540 39928418776207564993809101133409522367753645705409 46732318890156068952709993893776999699786363056503 576854528
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  3. #3
    Senior Member DeMath's Avatar
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    Re: Determinant calculation

    Quote Originally Posted by mathmagix View Post
    Haw to calculate this determinant with 2012 columns and 2012 rows?
    Click image for larger version. 

Name:	CodeCogsEqn.gif 
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ID:	24684
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    Use this

    \det(A+x) = \begin{vmatrix}a_{1,1} + x&a_{1,2} + x& \ldots &a_{1,n} + x \\ a_{2,1} + x&a_{2,2} + x& \ldots &a_{2,n} + x \\ \vdots & \vdots & \ddots & \vdots  \\ a_{n,1}+ x&a_{n,2} + x& \ldots &a_{n,n}+ x \end{vmatrix}= \det A + x \cdot \sum_{i = 1}^n \sum_{j = 1}^n A_{i,j}, where A_{i,j} - cofactors of the matrix A.

    Let A+x= \begin{bmatrix} -3&1&1& \cdots &1 \\ 1&2&1& \cdots &1 \\ 1&1& -3& \cdots &1 \\ \vdots & \vdots & \vdots & \ddots & \vdots  \\ 1&1&1& \cdots &2 \end{bmatrix}, where x=1 and A=\begin{bmatrix} -4&0&0& \cdots &0 \\ 0&1&0& \cdots &0 \\ 0&0& -4& \cdots &0 \\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 0&0&0& \cdots &1 \end{bmatrix}.

    Since \det A =(-4)^{\frac{2012}{2}}= 2^{2012} and A_{i,j}= \begin{cases} 0,&\text{if}\quad i \ne j,\\ (-4)^{\frac{2012}{2}}} = 2^{2012},&\text{if}\quad i=j=2k,\\ (-4)^{\frac{2012}{2}-1}= -2^{2010},&\text{if}\quad i=j=2k-1,\end{cases}\!\!k=1,2,\ldots,1006, it follows that

    {\det(A+x)= \det A + x \cdot \sum\limits_{i=1}^{2012} \sum\limits_{j = 1}^{2012} A_{i,j}= 2^{2012} + 1 \cdot \frac{2012}{2}\bigl(2^{2012}-2^{2010}\bigr) = 1511 \cdot 2^{2011}}
    Last edited by DeMath; September 5th 2012 at 08:30 AM.
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