3 Equations which have an infinite number of solutions

Consider the following system of equations

x 1 + x 2 3 x 3 = 3 2 x 1 3 x 2 + 4 x 3 = 4 x 1 x 2 + x 3 = 1

In augmented form we have:

1 1 3 3 2 3 4 4 1 1 1 1

Now performing the usual Gauss elimination operations we have

1 1 3 3 2 3 4 4 1 1 1 1 R 2 2 × R 1 R 3 R 1 1 1 3 3 0 5 10 10 0 2 4 4

Now applying R 2 ÷ 5 and R 3 ÷ 2 gives

1 1 3 3 0 1 2 2 0 1 2 2

Then R 2 R 3 gives

1 1 3 3 0 1 2 2 0 0 0 0

We see that all the elements in the last row are zero. This means that the variable x 3 can take any value whatsoever, so let x 3 = t then using back substitution the second row now implies

x 2 = 2 + 2 x 3 = 2 + 2 t

and then the first row implies

x 1 = 3 x 2 + 3 x 3 = 3 ( 2 + 2 t ) + 3 ( t ) = 1 + t

In this example the system of equations has an infinite number of solutions:

x 1 = 1 + t , x 2 = 2 + 2 t , x 3 = t or [ x 1 , x 2 , x 3 ] T = [ 1 + t , 2 + 2 t , t ] T

where t can be assigned any value. For every value of t these expressions for x 1 , x 2 and x 3 will simultaneously satisfy each of the three given equations.

Systems of linear equations arise in the modelling of electrical circuits or networks. By breaking down a complicated system into simple loops, Kirchhoff’s law can be applied. This leads to a set of linear equations in the unknown quantities (usually currents) which can easily be solved by one of the methods described in this Workbook.