To solve the system of equations using the Gauss-Jordan elimination method, we can use the following steps:
Step 1: Write the augmented matrix for the system of equations.
The augmented matrix for the given system of equations is:
```
[ 0.99 -0.04 -0.2 | 4 ]
[-0.1 0.97 0 | 6 ]
[-0.2 -0.01 0.96| 3 ]
```
Step 2: Use row operations to transform the matrix into row echelon form.
We can start by using row operations to create zeros below the first entry in the first column:
R2 = R2 + 0.1010R1
R3 = R3 + 0.2020R1
The new matrix is:
```
[ 0.99 -0.04 -0.20 | 4 ]
[ 0 0.9696 -0.0196| 6.04 ]
[ 0 -0.0602 0.9192| 3.808 ]
```
We can now use row operations to create a zero in the second row and second column:
R3 = R3 + 0.0620R2
The new matrix is:
```
[ 0.99 -0.04 -0.20 | 4 ]
[ 0 0.9696 -0.0196| 6.04 ]
[ 0 0 0.9142| 4.0088 ]
```
Step 3: Use row operations to transform the matrix into reduced row echelon form.
We can start by using row operations to create a one in the third row and third column:
R3 = (1/0.9142)R3
The new matrix is:
```
[ 0.99 -0.04 -0.20 | 4 ]
[ 0 0.9696 -0.0196| 6.04 ]
[ 0 0 1 | 4.3852 ]
```
We can now use row operations to create zeros above and below the pivot in the third column:
R1 = R1 + 0.20R3
R2 = R2 + 0.0196R3
The new matrix is:
```
[ 0.99 -0.04 0 | 4.87704 ]
[ 0 0.9696 0 | 6.1304 ]
[ 0 0 1 | 4.3852 ]
```
We can now use row operations to create zeros above and below the pivot in the second column:
R1 = R1 + 0.0412R2
The new matrix is:
```
[ 0.99 0 0 | 5.0038 ]
[ 0 0.9696 0 | 6.1304 ]
[ 0 0 1 | 4.3852 ]
```
Step 4: Write the solution in terms of the variables.
The reduced row echelon form of the augmented matrix corresponds to the system of equations:
```
0.99x = 5.0038
0.9696y = 6.1304
z = 4.3852
```
Solving for the variables, we get:
x = 5.0565
y = 6.3215
z = 4.3852
Therefore, the solution to the system of equations is (5.0565, 6.3215, 4.3852).