Use the Gauss-Jordan method to find all the solutions of the system
-3x1 +2x2 -x3 +6x4 = -7
7x1 -3x2 +2x3 -11x4 = 14
x1 -x4 = 1
It is a 3x5 matrix
-3 2 -3 6 -7
7 -3 2 -11 14
1 0 0 -1 1
Using the Gauss-Jordan method,
I got
1 0 0 -1 1
0 1 0 2 5/3
0 0 1 1 2
Yet when I verify it using an online calculator it give me this
1 0 0 -1 1
0 1 0 2 -1
0 0 1 1 2
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I retry doing it and I still cant get the -1 on the 2nd row last column
You have 4 variables but only 3 equations, so you can't expect a unique solution.
Since the steps to follow in GJ are not unique, I can't know how you arrived at it.
Here is a neat webpage that let's you see the steps
https://matrix.reshish.com/gauss-jordanElimination.php
notice the make x4 the independant variable, that is, let x4 be anything you want.
Check it out.
I found the error.. I keep doing those calculation error its irritating .__.
Thx Reiny
Solve the following system of linear equation by Gauss Jorden Method
3x1 + x2 + x3 + x4 = 0
5x1 − x2 + x3 − x4 = 0
3x+y=16
2x-27=4
3x+2y=16
2x-27=4
To find all the solutions of the system using the Gauss-Jordan method, we can perform row operations on the augmented matrix until it is in reduced row echelon form (RREF). Here are the steps:
Step 1: Write down the augmented matrix for the given system of equations:
-3 2 -1 6 -7
7 -3 2 -11 14
1 0 0 -1 1
Step 2: Perform row operations to simplify the matrix.
- Divide the first row by -3:
1 -2/3 1/3 -2 7/3
- Multiply the first row by 7 and add it to the second row:
0 -7/3 23/3 -25 14/3
- Multiply the first row by -1 and add it to the third row:
0 2/3 -1/3 1 -4/3
- Divide the second row by -7/3:
0 1 -7/23 25/7 -2/3
- Multiply the second row by 2/3 and add it to the first row:
1 0 -5/23 8/7 -1/3
- Multiply the second row by -2/3 and add it to the third row:
0 0 1/23 0 4/3
Step 3: Continue performing row operations to simplify the matrix further.
- Multiply the third row by 5/23 and add it to the first row:
1 0 0 -1/7 1/3
- Multiply the third row by -23/7 and add it to the second row:
0 1 0 2 -1/3
- Divide the third row by 1/23:
0 0 1 1 2/3
Now, the matrix is in reduced row echelon form (RREF).
The solutions of the system are the values in the last column corresponding to the non-zero rows in the RREF matrix.
Therefore, the solutions are:
x1 = -1/7
x2 = 2
x3 = 1
x4 = 2/3
As for the discrepancy between your hand calculation and the online calculator, it's possible that there was an error in one of the row operations. Double-checking your calculations can help identify the mistake.