Write the augmented matrix, and then solve the system, using Gauss Jordan elimination on the
augmented matrix.
x + 2y - z = 4
-2x + y - 4z = -6
4x - 3y + 2z = -10
lots of writing...
Here is a neat page
http://www.gregthatcher.com/Mathematics/GaussJordan.aspx
tell it you have 3 rows and 4 columns, then enter the values
Notice it takes it all the way to the matrix
1 0 0 ?
0 1 0 ?
0 0 1 ?
you only have to go as far as
1 x x x
0 1 x x
0 0 1 x
get the answer from the last row, sub that back into the second, and finally into the first.
To solve the system using Gauss Jordan elimination, we need to represent the system as an augmented matrix. Each equation will correspond to a row in the matrix, and the coefficients of the variables will be the entries in each row. The right side of each equation will be included in the last column of the augmented matrix.
The augmented matrix for the given system is:
```
1 2 -1 | 4
-2 1 -4 | -6
4 -3 2 | -10
```
Now, we will perform row operations on the augmented matrix to transform it to reduced row-echelon form. The goal is to replace the entries below and above each leading coefficient (the leftmost non-zero entry in a row) with zeroes, and make all the leading coefficients equal to 1.
For convenience, I'll use the notation R1, R2, and R3 to represent the rows of the augmented matrix.
Step 1: Swap R1 and R2
```
-2 1 -4 | -6
1 2 -1 | 4
4 -3 2 | -10
```
Step 2: Use R1 to eliminate the first entry (1) in R2 and R3
Multiply R1 by 1/(-2) and add it to R2. Replace R2 with the result.
Multiply R1 by -4 and add it to R3. Replace R3 with the result.
```
1 2 -1 | 4
0 5 -3 | 8
0 5 -2 | 6
```
Step 3: Use R2 to eliminate the second entry (5) in R3
Multiply R2 by 1/5 and add it to R3. Replace R3 with the result.
```
1 2 -1 | 4
0 5 -3 | 8
0 0 1 | -2
```
Step 4: Use R3 to eliminate the second entry (2) in R1 and the first entry (2) in R2
Multiply R3 by -2 and add it to R1. Replace R1 with the result.
Multiply R3 by 3 and add it to R2. Replace R2 with the result.
```
1 0 -3 | 8
0 5 -3 | 8
0 0 1 | -2
```
Step 5: Use R2 to eliminate the second entry (5) in R1
Multiply R2 by -3 and add it to R1. Replace R1 with the result.
```
1 0 0 | 17
0 5 -3 | 8
0 0 1 | -2
```
Step 6: Use R2 to eliminate the third entry (-3) in R2
Multiply R2 by 1/5 and add it to R2. Replace R2 with the result.
```
1 0 0 | 17
0 1 0 | 3
0 0 1 | -2
```
The augmented matrix is now in reduced row-echelon form. Each column represents the coefficients of the variables in the system.
From the augmented matrix, we can conclude the following solutions:
x = 17
y = 3
z = -2
Therefore, the system is consistent and has a unique solution.