Gauss-Jordan method for solving systems of linear equations

in english •  4 months ago

In a system of linear equations, addition operations between equations can be performed, which will lead us to a valid equation. Let's see through an example the procedure.

Let's suppose that we want to solve the following system of linear equations:

gaussjordan1.jpg

The solution of the system is a triad of real numbers (x, y, z) whose coordinates x, y, z satisfy the three equations of the system simultaneously.

What we want is to simplify the system so that in equation 1 we only have the variable x in the corresponding position; in the second equation and, in the position that corresponds to it; and in the third the z in the corresponding position.
As in equation 1 of our system the variable x has coefficient 1, then there is nothing to do, but in the second equation, yes; we see that the coefficient of x in this equation is 2, to eliminate it we multiply this equation by ½.

gaussjordan2.jpg



The third equation is multiplied by -1, to eliminate the negative sign of the x.

gaussjordan3.jpg



Let's use our first equation to eliminate the x from the second and third equations: multiplying equation 1 by (-1) and the result we add to equation 2.

gaussjordan4.jpg.png



Similarly, we will use equation 1 to eliminate x from the third equation.

gaussjordan5.jpg.jpg



Let the coefficient of y in the second equation be equal to 1, multiplying equation 2 by (-2/3).

gaussjordan6.jpg.jpg



Let's eliminate y from the third equation and use the resulting equation to eliminate y from the first equation.

gaussjordan7.jpg.jpg



Let the coefficient of z in the third equation be equal to 1, multiplying the third equation by (-1/3):

gaussjordan8.jpg.jpg



Let's eliminate z from the second equation, multiplying equation 3 by (1/3), and we add the result to equation 2.

gaussjordan9.jpg.jpg



We eliminate y from the first equation, multiplying by (-1) to equation 2, and we add the result to equation 1.

gaussjordan10.jpg.jpg


Let's eliminate z from the first equation. Multiplying by (-1) to equation 3 and then adding to equation 1:

gaussjordan11.jpg.jpg



The obtained values: x = 1, y = 2 and, z = 3 correspond to the unique solution of the system.

This method is called the Gauss-Jordan Removal Method.

In summary, these are the steps to follow:

1.- We divide to make the coefficient of x1 in the first equation equal to 1.

2.- We eliminate the terms x1 in the second and third equation. That is, make them equal to 0. To do this, we use the first equation, multiplying it by the appropriate number and then adding it to the corresponding equation.

3.- Divide to make the coefficient of x2 in the second equation equal to 1.

4.- We use this second equation to eliminate x2 from the second and third equation.

5.- Divide to make the term in x3 equal to 1 in the third equation.

6.- Then we use this third equation to eliminate x3 from the first and third equation.


Use of matrices to describe the Gauss-Jordan method.

An easier way to describe the steps of this method to solve systems of linear equations is through the use of matrices. The first thing that is identified is the matrix determined by the coefficients of the variables of the equations.
In this case it is a 3x3 square matrix, since there are three variables and three equations.

gaussjordan12.jpg.jpg



We transform it into an augmented matrix by introducing the constants to which each equation is equated, in an additional column.

gaussjordan13.jpg.jpg
gaussjordan14.jpg.jpg

Reference:
Stanley I. Grossman. Linear Algebra.1983. Iberoamericana Editorial Group.

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