The following example will demonstrate the Gaussian method by example.

# Example: The Gaussian Method in Practice

### Example

We want to solve the system of linear equations with three unknowns

$$\begin{array}{rcr} x_1 -3x_2 +2x_3&=&1\\ 5x_1 + 4x_2 -3x_3&=&4\\ 2x_1 -8x_2 +4x_3&=& -2\\ \end{array}\quad\quad( * )$$

This system has the following extended coefficient matrix:

$$A|\beta:= \left(\begin{array}{rrr|r} 1&-3&2&1\\ 5&4&-3&4\\ 2&-8&4&-2\\ \end{array}\right)$$

In the following, we use SageMath. You will have to click the evaluate buttons to see the results.

print("Original extended matrix:\n") A=matrix(QQ,[[1,-3,2,1],[5,4,-3,4],[2,-8,4,-2]]) print(A) print("") print("STEP 1: Bring the matrix to the upper-triangular form") print("\nAdding the -5-fold multiple of the first row to the second:\n") A1=A A1=A1-5*A1 print(A1) print("\nAdding the -2-fold multiple of the first row to the third:\n") A1=A1-2*A1 print(A1) print("\nAdding the 2/19-fold multiple of the second row to the third:\n") A1=A1+2/19*A1 print(A1)

The resulting upper-triangular matrix is

$$A|\beta:= \left(\begin{array}{rrr|r} 1&-3&2&1\\ 0&19&-13&-1\\ 0&0&-\frac{26}{19}&-\frac{78}{19}\\ \end{array}\right)$$

Now we can use the backward substitution to solve the system

print("STEP 2: Backward substitution:\n") A=matrix(QQ,[[1,-3,2,1],[0,19,-13,-1],[0,0,-26/19,-78/19]]) print(A) print("\nSetting x3=-78/19*(-19/26), substituting x3 in second row, setting x2=...etc....\n") x3=A/A x2=(A-A*x3)/A x1=(A-A*x3-A*x2)/A print("x3=", x3) print("x2=", x2) print("x1=", x1)

Therefore, $x_1=1, x_2=2, x_3=3$ is the solution of the system $( * ).$

Github: ### References

#### Bibliography

1. Knabner, P; Barth, W.: "Lineare Algebra - Grundlagen und Anwendungen", Springer Spektrum, 2013