Mathematical Methods for Physics and Engineering - Matematica.NET

27.3 SIMULTANEOUS LINEAR EQUATIONSvariables (unknowns), x i , i =1, 2,...,N. The equations take the general formA 11 x 1 + A 12 x 2 + ···+ A 1N x N = b 1 ,A 21 x 1 + A 22 x 2 + ···+ A 2N x N = b 2 , (27.21)..A N1 x 1 + A N2 x 2 + ···+ A NN x N = b N ,where the A ij are constants and form the elements of a square matrix A. Theb iare given and form a column matrix b. IfA is non-singular then (27.21) can besolved for the x i using the inverse of A, according to the formulax = A −1 b.This approach was discussed at length in chapter 8 and will not be consideredfurther here.27.3.1 Gaussian eliminationWe follow instead a continuation of one of the earliest techniques acquired by astudent of algebra, namely the solving of simultaneous equations (initially onlytwo in number) by the successive elimination of all the variables but one. This(known as Gaussian elimination) is achieved by using, at each stage, one of theequations to obtain an explicit expression for one of the remaining x i in termsof the others and then substituting for that x i in all other remaining equations.Eventually a single linear equation in just one of the unknowns is obtained. Thisis then solved and the result is resubstituted in previously derived equations (inreverse order) to establish values for all the x i .This method is probably very familiar to the reader, and so a specific exampleto illustrate this alone seems unnecessary. Instead, we will show how a calculationalong such lines might be arranged so that the errors due to the inherent lack ofprecision in any calculating equipment do not become excessive. This can happenif the value of N is large and particularly (and we will merely state this) if theelements A 11 ,A 22 ,...,A NN on the leading diagonal of the matrix in (27.21) aresmall compared with the off-diagonal elements.The process to be described is known as Gaussian elimination with interchange.The only, but essential, difference from straightforward elimination is that beforeeach variable x i is eliminated, the equations are reordered to put the largest (inmodulus) remaining coefficient of x i on the leading diagonal.We will take as an illustration a straightforward three-variable example, whichcan in fact be solved perfectly well without any interchange since, with simplenumbers and only two eliminations to perform, rounding errors do not havea chance to build up. However, the important thing is that the reader should995