Linear Simultaneous Eqns - Gauss Elim Problem

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Hello, I'm stuck on a simultaneous eqns problem. From what I can see it seems the easiest way would be to get the matrix into row echelon form, but I'm not sure if another way would be better. I can see a pattern here but not sure what it means. I attached the problem to the page. Any help would be greatly appreciated.
 

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Fjolvar said:
Hello, I'm stuck on a simultaneous eqns problem. From what I can see it seems the easiest way would be to get the matrix into row echelon form, but I'm not sure if another way would be better. I can see a pattern here but not sure what it means. I attached the problem to the page. Any help would be greatly appreciated.

It is a symmetric square matrix, so if it is invertible it satisfies A-1AT = I.

Take a look here, http://en.wikipedia.org/wiki/Symmetric_matrix.
 
Interesting, how would I use this to solve for the variables X1, X2, X3.. etc?
 
Fjolvar said:
Interesting, how would I use this to solve for the variables X1, X2, X3.. etc?

So, what you really have is [A|B], where B is the column vector of solutions, right? So when you multiply by the matrix you have by its inverse what you really get is [I|A-1B]. Remember when you right it in the form [A|B], it is really just shorthand saying Ax = B. When A is invertible, you have x = A-1B. Does that make sense?
 
I do understand how to find X by taking the inverse of A and multiplying by B, however I'm still not quite sure of the significance of this matrix being symmetric. So A = A transpose, does that help us in finding the inverse of A?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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