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# Solution of a general linear system of equations: 4-term n-equations

Posted 10 years ago
 I have the following system of equations.... y1 = c11 * x11 + c12 * x12 + c13 * x13 + c14 * x14 y2 = c21 * x21 + c22 * x22 + c23 * x23 + c24 * x24 y3 = c31 * x31 + c32 * x32 + c33 * x33 + c34 * x34 ................. ................ yn = cn1 * xn1 + cn2 * xn2 + cn3 * xn3 + cn4 * xn4 How to solve for x?
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Posted 10 years ago
 Dear Srinivasan Anand,is this a general math question or you looking for an implemention with Wolfram Technologies. This forum is ONLY about topics related to Wolfram technologies and is NOT a general science forum. For Wolfram Language see: For a general math discussion please find an appropriate forum, for example: Mathematics Stack Exchange
Posted 10 years ago
 Dear Moderator, I would like to know the implementation of the solution of linear systems in Mathematica.....Thanks and Regards, Anand
Posted 10 years ago
 Thank you for the clarification. Did you follow the links we posted above? - they have many examples on solving such systems.
Posted 10 years ago
 Hi Srinivasan Anand,are you sure that the system is posted correctly. You seem to suggest that you are interested in a linear system, which yours is. But usually I would think of it as "vector = matrix times vector". Now the y is a vector, the c seems to be a matrix of coefficients and the x in your case is also a matrix, is it not? It that case the linear system would be: Solve[{ y1 == c11*x1 + c12*x2 + c13*x3 + c14*x4, y2 == c21*x1 + c22*x2 + c23*x3 + c24*x4, y3 == c31*x1 + c32*x2 + c33*x3 + c34*x4, y4 == c41*x1 + c42*x2 + c3*x3 + c44*x4}, {x1, x2, x3, x4}] which gives a solution. In the case that you have posted things look differently. If I read your system correctly, all equations are independent. None of the variables occurs in more than one equation. Also, the $x_{nj}$ will be underdetermined: one equation four variables. That means that you can get this relation: Solve[yn == cn1*xn1 + cn2*xn2 + cn3*xn3 + cn4*xn4, {xn1, xn2, xn3, xn4}] (*{{xn4 -> -((cn1 xn1)/cn4) - (cn2 xn2)/cn4 - (cn3 xn3)/cn4 + yn/cn4}}*) The result is a relationship which allows you to determine one of the components of $x$ based on three others and the coefficients of $c$.Perhaps I am missing something here...Cheers,Marco