Recent content by Alex1812

Thanks for your help. I got it now.

That's true. Always good to keep in mind the impact of nul denominators. Anyway, I think we've disproven this more ways than Femme_Physics has time to read about lol

You're right Ouabache, I didn't bother equating the pairs of equations to reduce it to two cases. As written, if (xy^2)=(ab^2) then (x/a) = (b^2/y^2) if xy=ab then (x/a)=(b/y) Then the original statement is true only iff (b/y)=(b^2/y^2)=((b/y)^2) which, as previously pointed out, is...

I tried writing out the solution in terms of the six free variables, but my solution set is different thant the one given (sorry for poor pensmanship). I don't understand how the given solution set uses no more than two free varaibles for each of (x1, ..., x8) when x4 depends on 4 other variables.

Assuming you mean (xy)^2 = (ab)^2, xy does not neccesarilly equal ab: sqrt[(xy)^2] = sqrt[(ab)^2] then (+/-)xy = (+/-)ab which is true for only two out of four cases: +xy=+ab -xy=-ab But, you wind up with problems when -xy = +ab and +xy=-ab. Hope that helps a little.

Thanks Karen. I verified that the provided solution set is valid in that it satisfies the given linear equations. But I have no idea how to obtain this solution set if it were not already given. Is there something I'm missing?

Homework Statement Show how the following solution set: [x1,x2,x3,x4,x5,x6,x7,x8]^T =[1,-1,0,0,0,0,0,0]^T, [0,-1,1,0,0,0,0,0]^T,[0,0,0,1,-1,0,0,0]^T,[0,0,0,-1,0,1,0,0]^T,[0,0,0,0,-1,0,1,0]^T,[0,0,0,0,0,-1,0,1] is obtained from the two linear equations x1+x2+x3-x4-x5-x6-x7-x8=0 and...