1. Apr 12, 2009 #1

   Derill03

   

   This matrix is said to be quickly solvable but im not sure of the trick:
   
   -1 1 1 1 1
   1 -1 1 1 1
   1 1 -1 1 1
   1 1 1 -1 1
   1 1 1 1 -1
   
   the matrix is all 1's with -1's on the main diagonal, im not sure if i should row reduce until i have an upper triangular matrix and use product of diagonal entries or if there is a simple trick im missing

    

   Derill03, Apr 12, 2009
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3. Apr 12, 2009 #2

   maze

   

   One way is to realize that it is a rank-1 update of a matrix you know how to invert
   [tex]\left(\begin{matrix}-1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1\end{matrix}\right) = \left(\begin{matrix}1 \\ 1 \\ 1\end{matrix}\right)\left(\begin{matrix}1 & 1 & 1\end{matrix}\right) - 2 \left(\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}\right)[/tex]
   
   Apply the Sherman-Morrison formula:
   http://en.wikipedia.org/wiki/Sherman–Morrison_formula
   
   There may be other ways.

    

   maze, Apr 12, 2009

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