The following example will demonstrate the Gaussian method by 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:"
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 "Adding the -5-fold multiple of the first row to the second:"
A1=A
A1[1]=A1[1]-5*A1[0]
print(A1);print "Adding the -2-fold multiple of the first row to the third:"
A1[2]=A1[2]-2*A1[0]
print(A1);print "Adding the 2/19-fold multiple of the second row to the third:"
A1[2]=A1[2]+2/19*A1[1]
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:"
A=matrix(QQ,[[1,-3,2,1],[0,19,-13,-1],[0,0,-26/19,-78/19]])
print(A); print; print "Setting x3=-78/19*(-19/26), substituting x3 in second row, setting x2=...etc...."
x3=A[2][3]/A[2][2]
x2=(A[1][3]-A[1][2]*x3)/A[1][1]
x1=(A[0][3]-A[0][2]*x3-A[0][1]*x2)/A[0][0]
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 $( * ).$

| | | | created: 2018-04-15 18:13:46 | modified: 2018-04-15 20:37:16 | by: *bookofproofs* | references: [7937]

[7937] **Knabner, P; Barth, W.**: “Lineare Algebra – Grundlagen und Anwendungen”, Springer Spektrum, 2013