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 November 22nd, 2013, 01:26 PM #1 Newbie   Joined: Nov 2013 Posts: 26 Thanks: 0 determinant by Laplace Hi, I want to show : $\sum\limits_{j=1}^{n}a_{i,j}c_{k,j}=\delta_{i,k}de t(A)$,... the case i=k was not the problem, but how i can show that if i $\ne$k, the result of the sum is 0 ? hope you can help me
 November 22nd, 2013, 01:30 PM #2 Newbie   Joined: Nov 2013 Posts: 26 Thanks: 0 Re: determinant by Laplace maybe i have to add some information: c are the cofactors delta is the Kronecker symbol
 November 23rd, 2013, 06:49 AM #3 Newbie   Joined: Nov 2013 Posts: 26 Thanks: 0 Re: determinant by Laplace no ideas???
 November 23rd, 2013, 09:36 AM #4 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: determinant by Laplace Just an observation. The result is true for any invertible matrix A by the normal inverse process. What it says, therefore, is that when det(A) = 0, the normal process for finding an inverse results in the zero matrix when you mutiply it by A. If you've shown this for j = k, then perhaps you can use the normal inverse calculations to show it for i not = j. In other words the inverse process shows that for any matrix: $AC^T= det(A)I$ Where C is the cofactor matrix.

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