solve by the gauss-jordan method
x+y+z=6
3x+2y-z=4
3x+2y+2z=11
y-3=11
To solve the system of equations using the Gauss-Jordan method, we will use an augmented matrix.
Step 1: Set up the augmented matrix
The augmented matrix for the system of equations is:
```
[ 1 1 1 | 6 ]
[ 3 2 -1 | 4 ]
[ 3 2 2 | 11 ]
```
Step 2: Perform row operations to reduce the matrix
Start by performing row operations to create zeros below the first element of the first column.
Multiply the second row by -3 and add it to the first row to eliminate the x coefficient.
```
[ 1 1 1 | 6 ]
[ 0 -1 4 | -14 ]
[ 3 2 2 | 11 ]
```
Multiply the third row by -3 and add it to the first row to eliminate the x coefficient.
```
[ 1 1 1 | 6 ]
[ 0 -1 4 | -14 ]
[ 0 -1 -1 | -23 ]
```
Step 3: Perform additional row operations to further reduce the matrix
Multiply the second row by -1 and add it to the third row to eliminate the y coefficient.
```
[ 1 1 1 | 6 ]
[ 0 -1 4 | -14 ]
[ 0 0 -5 | -9 ]
```
Step 4: Perform row operations to obtain a diagonal matrix
Multiply the third row by -1/5 to make the coefficient of z equal to 1.
```
[ 1 1 1 | 6 ]
[ 0 -1 4 | -14 ]
[ 0 0 1 | 9/5 ]
```
Multiply the second row by -1/5 and add it to the first row to eliminate the z coefficient in the first row.
```
[ 1 1 0 | 6 - (4/5)(-14) ]
[ 0 -1 4 | -14 ]
[ 0 0 1 | 9/5 ]
```
Multiply the second row by 1 and add it to the third row to eliminate the z coefficient in the third row.
```
[ 1 1 0 | 6 - (4/5)(-14) ]
[ 0 -1 0 | -14 - (4/5)(-14) ]
[ 0 0 1 | 9/5 ]
```
Step 5: Solve for the variables
From the row-reduced form of the augmented matrix, we have the following equations:
```
x + y = 6 - (4/5)(-14)
-y = -14 - (4/5)(-14)
z = 9/5
```
Simplifying the equations, we get:
```
x + y = 13.2
y = -11.2
z = 9/5
```
Therefore, the solution to the system of equations is:
x = 13.2 - y
y = -11.2
z = 9/5
To solve the given system of equations using the Gauss-Jordan method, we need to perform a series of row operations on the augmented matrix.
Step 1: Write the augmented matrix for the system of equations:
[1 1 1 | 6]
[3 2 -1 | 4]
[3 2 2 | 11]
Step 2: Apply row operations to transform the augmented matrix into row-echelon form.
First, let's use row 1 as a pivot row and try to eliminate the coefficients below the pivot position in column 1.
[R2 = R2 - 3R1]
[R3 = R3 - 3R1]
The new matrix becomes:
[1 1 1 | 6]
[0 -1 -4 | -14]
[0 -1 -1 | -13]
Next, we can use row 2 as a pivot row and eliminate the coefficient below the pivot position in column 2.
[R3 = R3 - R2]
The new matrix becomes:
[1 1 1 | 6]
[0 -1 -4 | -14]
[0 0 3 | 1]
Now, we have obtained the row-echelon form of the augmented matrix.
Step 3: Apply backward substitution to find the solution.
From the last row of the matrix, we can see that 3z = 1, so z = 1/3.
Substituting the value of z back into the second row, we have -y - 4z = -14. Since z = 1/3, we can solve for y:
-y - 4(1/3) = -14
-y - 4/3 = -14
-y = -14 + 4/3
-y = -42/3 + 4/3
-y = -38/3
y = 38/3
Finally, substituting the values of y and z into the first row, we can solve for x:
x + y + z = 6
x + 38/3 + 1/3 = 6
x + 39/3 = 6
x + 13 = 6
x = 6 - 13
x = -7
Therefore, the solution to the system of equations is x = -7, y = 38/3, and z = 1/3.