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

To solve the system of equations using the Gauss-Jordan method, we need to represent the system as an augmented matrix and then use row operations to get the matrix in reduced row echelon form.

The augmented matrix for the system is:

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

Step 1: Use the first row to eliminate the entries in the first column of the second and third rows.

\[\begin{bmatrix} 3 & -0.1 & -0.2 & | & 7.85 \\ 0 & 7.0033 & -0.2933 & | & -19.475 \\ 0 & -0.13 & 9.94 & | & 70.345 \end{bmatrix}\]

Step 2: Make the leading coefficient of the second row as 1.

\[\begin{bmatrix} 3 & -0.1 & -0.2 & | & 7.85 \\ 0 & 1 & -0.0419 & | & -2.7817 \\ 0 & -0.13 & 9.94 & | & 70.345 \end{bmatrix}\]

Step 3: Use the second row to eliminate the second column entries in the first and third rows.

\[\begin{bmatrix} 3 & 0 & -0.1983 & | & 7.7239 \\ 0 & 1 & -0.0419 & | & -2.7817 \\ 0 & 0 & 9.9355 & | & 70.0891 \end{bmatrix}\]

Step 4: Make the leading coefficient of the third row as 1.

\[\begin{bmatrix} 3 & 0 & -0.1983 & | & 7.7239 \\ 0 & 1 & -0.0419 & | & -2.7817 \\ 0 & 0 & 1 & | & 7.0522 \end{bmatrix}\]

Step 5: Use the third row to eliminate the third column entries in the first and second rows.

\[\begin{bmatrix} 3 & 0 & 0 & | & 8.4617 \\ 0 & 1 & 0 & | & -2.5108 \\ 0 & 0 & 1 & | & 7.0522 \end{bmatrix}\]

Step 6: Make the leading coefficient of the first row as 1.

\[\begin{bmatrix} 1 & 0 & 0 & | & 2.8206 \\ 0 & 1 & 0 & | & -2.5108 \\ 0 & 0 & 1 & | & 7.0522 \end{bmatrix}\]

Now, the matrix is in reduced row echelon form, and we can read the solutions for \(x_1\), \(x_2\), and \(x_3\) directly:

\[x_1 = 2.8206\]

\[x_2 = -2.5108\]

\[x_3 = 7.0522\]

The solution to the system of linear equations using the Gauss-Jordan method is \(x_1 = 2.8206\), \(x_2 = -2.5108\), and \(x_3 = 7.0522\).