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?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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