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Method to Solve Simultaneous Linear Equations

IP.com Disclosure Number: IPCOM000073938D
Original Publication Date: 1971-Feb-01
Included in the Prior Art Database: 2005-Feb-23
Document File: 3 page(s) / 43K

Publishing Venue

IBM

Related People

Chignoli, C: AUTHOR

Abstract

This is a method to solve simultaneous linear equations in which the number of equations is not necessarily equal to the number of unknowns. If the system has more than one solution, they are all computed and given in parametric form. If the system has no solution the method recognizes which are the equations which contradict the other ones.

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Method to Solve Simultaneous Linear Equations

This is a method to solve simultaneous linear equations in which the number of equations is not necessarily equal to the number of unknowns. If the system has more than one solution, they are all computed and given in parametric form. If the system has no solution the method recognizes which are the equations which contradict the other ones.

Given the system of m linear equations in n unknowns

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In fact all the solutions of the original system (1) also clearly verify the transformated system (that with the h-th equation replaced by (2) ). Besides the transformated system cannot get any extraneous solution, since we could apply the inverse transformation to it: subtract from its h-th equation (which is the (2) ) the k-th (left unalterated) multiplied by -c; and we should again get the original system (1) with just its solutions.

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In the following speech we shall meet fractions with denominator a((k))/kk/. If a((k))/kk/ = 0 we would overcome this, if possible, by changing the k-th equation with one of the subsequent ones, and if that would not be sufficient we would also change the position of the unknown x(k) with one of the subsequent ones; in other words we would permute two rows and two columns of the system so that, after the exchange of subscripts, we would have a((k))/kk/ not= 0. In the practice of numerical calculus it is not felt that a((k))/kk/ not= 0 is sufficient; we prefer that the maximum, in absolute value, among a((k))/ij/ (i = k, ....., m; j = k, ...., n) will be
transferred, with exchanges of rows and columns, to the place of a((k))/kk/.

Let us start by transforming the equations of (1), from the second to the m-th, subtracting from Each of these the first one multiplied by an appropriate constant; the constant for the i-th equation will be:

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These systems, one for each value of k(k= 2,3, .....) are equivalent to system
(1) because they are obtained with transformations of type (2) which leave the solutions unaltered. If, with the advance of iterations, we get a system (...