October 16th, 2007, 11:52 PM  #1 
Member Joined: Dec 2006 Posts: 65 Thanks: 0  Linear equations
Consider n linear equations a11* x1+ a12 * x2 +…+ a1n * xn = b1 a21* x1+ a22 * x2 +…+ a2n * xn = b2 . . . an1* x1+ an2 * x2 +…+ ann * xn = bn There are several methods to asses and solve these equations . Finally by any method the solutions x1, x2 ,…, xn are linear combinations of b1, b2 ,…, bn . But is it possible without any algebraic (or simpler) calculations to show that x1, x2 ,…, xn are such linear combinations? I am thinking about concepts which predict solutions without a direct effort to solve. That may be useful in some difficult situations ,of course no this simple example. 
October 19th, 2007, 09:13 AM  #2 
Senior Member Joined: Oct 2007 From: France Posts: 121 Thanks: 1 
You can try by induction on n.

January 7th, 2008, 12:44 AM  #3 
Member Joined: Dec 2006 Posts: 65 Thanks: 0  Linear equations
At first I apologize all forgetting to notify that existence and uniqueness of answers has been supposed .Consider other equations with answers x'1, x'2 ,…, x'n as following : a11* x'1+ a12 * x'2 +…+ a1n * x'n = b'1 a21* x'1+ a22 * x'2 +…+ a2n * x'n = b'2 . . . an1* x'1+ an2 * x'2 +…+ ann * x'n = b'n Then regard the following equations : a11* x"1+ a12 * x"2 +…+ a1n * x"n = C* b1 + D * b'1 a21* x"1+ a22 * x"'2 +…+ a2n * x"n = C* b2 + D * b'2 . . . an1* x"1+ an2 * x"2 +…+ ann * x"n = C* bn + D * b'n Uniqueness of answers shows that : x"i = C* xi + D * x'i , i =1 ,2,…,n We can regard b1,b2,…, bn as a function on X ={1,2,…,n} such that : bi = f(i) , i=1,2,…,n f: X  > C (complex values) X along its all subsets make a locally compact Hausdorff topological space. All functions X  > C are continuous by this topology . For each function such as f,there are x1,x2,….,xn that satisfy the linear equations . If we regard one of them ,e.g. xm ,xm is a linear functional of f. This functional is bounded and is defined on Cc(X).So according to Friesz theorem (real and complex analysis Walter Rudin , third edition ,19.6) ,there is a measure such as M (along some properties)that xm= int (fdM)on X . Clearly this integral can regard as a linear combination of b1,b2,…, bn . 
January 7th, 2008, 08:24 AM  #4 
Senior Member Joined: Oct 2007 From: France Posts: 121 Thanks: 1 
If there is existence and unicity of solutions, it suffices to use Cramer's rule to obtain your result.

January 11th, 2008, 08:07 PM  #5 
Member Joined: Dec 2006 Posts: 65 Thanks: 0  Linear equations
Yes,you are right,but as mentioned that is just an idea.For example,some primary problems are considered by algebra to represent its abilities of theorems and axioms although they can be easily solved by elementary methods.That also represents abilities of the theorem and encourages us to use it for furter applications.On the other hand to solve some equations,e.g. differential equations,some properties of answer are assumed to be true.Having that the equation is solved then the properties will be checked.


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