Question - Solving a System of Linear Equations Using the Gauss-Jordan Method

Given the system of equations:

\[\begin{cases} 3x_1 - 0.1x_2 - 0.2x_3 = 7.85 \\ 0.1x_1 + 7x_2 - 0.3x_3 = -19.3 \\ 0.3x_1 - 0.2x_2 + 10x_3 = 71.4 \end{cases}\]

First, we form the augmented matrix:

\[\begin{bmatrix} 3 & -0.1 & -0.2 & | & 7.85 \\ 0.1 & 7 & -0.3 & | & -19.3 \\ 0.3 & -0.2 & 10 & | & 71.4 \end{bmatrix}\]

Next, apply Gauss-Jordan elimination steps to reduce the matrix to reduced row echelon form.

1. Make the first element of the first row a 1 by dividing the first row by 3:

\[\begin{bmatrix} 1 & -0.0333 & -0.0667 & | & 2.6167 \\ 0.1 & 7 & -0.3 & | & -19.3 \\ 0.3 & -0.2 & 10 & | & 71.4 \end{bmatrix}\]

2. Make the first element of the second and third rows a 0:

\[\begin{bmatrix} 1 & -0.0333 & -0.0667 & | & 2.6167 \\ 0 & 7.0033 & -0.2933 & | & -19.5617 \\ 0 & -0.1 & 9.98 & | & 70.61 \end{bmatrix}\]

3. Make the second element of the second row a 1 by dividing the second row by 7.0033:

\[\begin{bmatrix} 1 & -0.0333 & -0.0667 & | & 2.6167 \\ 0 & 1 & -0.0418 & | & -2.7936 \\ 0 & -0.1 & 9.98 & | & 70.61 \end{bmatrix}\]

4. Make the second element of the first and third rows a 0:

\[\begin{bmatrix} 1 & 0 & -0.0677 & | & 2.5234 \\ 0 & 1 & -0.0418 & | & -2.7936 \\ 0 & 0 & 9.982 & | & 70.8886 \end{bmatrix}\]

5. Make the third element of the third row a 1 by dividing the third row by 9.982:

\[\begin{bmatrix} 1 & 0 & -0.0677 & | & 2.5234 \\ 0 & 1 & -0.0418 & | & -2.7936 \\ 0 & 0 & 1 & | & 7.1047 \end{bmatrix}\]

6. Make the third element of the first and second rows a 0:

\[\begin{bmatrix} 1 & 0 & 0 & | & 2.9969 \\ 0 & 1 & 0 & | & -2.5035 \\ 0 & 0 & 1 & | & 7.1047 \end{bmatrix}\]

We now have the reduced row echelon form, which corresponds to the solutions for \(x_1\), \(x_2\), and \(x_3\):

\[x_1 \approx 2.9969\]

\[x_2 \approx -2.5035\]

\[x_3 \approx 7.1047\]

So, the solution to the system of equations is approximately \(x_1 \approx 2.9969\), \(x_2 \approx -2.5035\), \(x_3 \approx 7.1047\).