1. ## Determinant calculation

Haw to calculate this determinant with 2012 columns and 2012 rows?

Thanks

2. ## Re: Determinant calculation

Originally Posted by mathmagix
Haw to calculate this determinant with 2012 columns and 2012 rows?

Thanks
Det =
35529225841856483043651808603739698746526147540091 25781343678823103964817334008922598220766872045514 43818875406235659520138976614372162888864682450773 56809484825684704986491920597144636004496957589321 45807071799620915629374867308911091630619589522654 20876697189137171661445806764230295352215847197485 40965173460832558185841042472018236366467686259189 20543195609189247214180736087814328750732314127804 25828702376026086993776454381667343776846151461024 32549811417270673109240013618063599797009890166540 39928418776207564993809101133409522367753645705409 46732318890156068952709993893776999699786363056503 576854528

3. ## Re: Determinant calculation

Originally Posted by mathmagix
Haw to calculate this determinant with 2012 columns and 2012 rows?

Thanks
Use this

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

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

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

$\displaystyle {\det(A+x)= \det A + x \cdot \sum\limits_{i=1}^{2012} \sum\limits_{j = 1}^{2012} A_{i,j}= 2^{2012} + 1 \cdot \frac{2012}{2}\bigl(2^{2012}-2^{2010}\bigr) = 1511 \cdot 2^{2011}}$