solve gauss jordan elimination
6w-2x-4y+2z=2
3w-3x-6y+z=-4
-12w+8x+21y-8z=8
-6w-10y+7z=43
hard to type up on here.
Try this nice applet: http://www.gregthatcher.com/Mathematics/GaussJordan.aspx
It shows the operations taken each time
Careful when you enter the data, the last equation has no x term, so enter 0
To solve the system of equations using the Gauss-Jordan elimination method, follow these steps:
Step 1: Write the augmented matrix for the system of equations. The augmented matrix is a matrix that includes both the coefficients of the variables and the constants on the right-hand side of the equations.
The augmented matrix for your system of equations is:
| 6 -2 -4 2 | 2 |
| 3 -3 -6 1 | -4 |
| -12 8 21 -8 | 8 |
| -6 0 -10 7 | 43 |
Step 2: Perform row operations to simplify the matrix. You need to use the following row operations:
- Multiply a row by a nonzero scalar.
- Add or subtract a multiple of one row to/from another row.
- Swap two rows.
The goal is to transform the matrix into the reduced row-echelon form, where each leading coefficient (the first nonzero entry in each row) is equal to 1, and all other entries in the column with a leading coefficient are zero.
Step 3: Perform the row operations to achieve the reduced row-echelon form.
Start with the first column:
1. Divide the first row (R1) by 6:
|R1| -> | 1 -1/3 -2/3 1/3 | 1/3 |
2. Perform a row operation to get a zero in the (1,2) position:
|R2| -> | 0 -1/3 -2/3 1/3 | -11/3 |
3. Perform a row operation to get a zero in the (3,1) position:
|R3| -> | 0 10/3 19/3 -10/3 | 62/3 |
4. Perform a row operation to get a zero in the (4,1) position:
|R4| -> | 0 1/2 5/6 -7/6 | 43/2 |
Next, focus on the second column:
5. Perform a row operation to get a zero in the (2,3) position:
|R2| -> | 0 1/3 2/3 -1/3 | 11/3 |
6. Perform a row operation to get a zero in the (3,2) position:
|R3| -> | 0 0 1 -1 | 10 |
7. Perform a row operation to get a zero in the (4,2) position:
|R4| -> | 0 0 -2/3 0 | 1/2 |
Now, focus on the third column:
8. Perform a row operation to get a zero in the (3,3) position:
|R3| -> | 0 0 1 -1 | 10 |
9. Perform a row operation to get a zero in the (4,3) position:
|R4| -> | 0 0 0 1 | -7/9 |
The matrix is now in reduced row-echelon form.
Step 4: Read the solutions from the reduced row-echelon form.
From the row-echelon form, we can see that the variables correspond to the columns. The values in the rightmost column represent the solutions.
The solutions for the variables are:
w = -7/9
x = 0
y = 10
z = -7/9
Therefore, the solution to the system of equations is:
w = -7/9
x = 0
y = 10
z = -7/9