solve the system of linear equations using the Gauss- Jordan elimination method

x1- x2+ 4x3 = 15
x1+ x2 = X3 = 4
-3X1-X2 + X3 = -2

There is probably a typo in the second equation. The first equal sign (between x2 and x3) should be a + or a -.

x1+ x2 = X3 = 4

Curiously both corrections using a + or - give integer solutions.

Next, Gauss-Jordan elimination requires the matrix be transformed into a diagonal matrix using elementary row operations. After that, the solution should be evident.

If you are not yet familiar with the method, try reading your textbook, or visit:
http://ceee.rice.edu/Books/CS/chapter2/linear44.html

If you have questions about the method, give it a try and post up to the point where you have a problem or question.

To solve the system of linear equations using the Gauss-Jordan elimination method, follow these steps:

Step 1: Write the augmented matrix
In order to apply the Gauss-Jordan elimination method, we need to write the system of equations in the form of an augmented matrix. The augmented matrix is formed by the coefficients of the variables, with an additional column for the constants.

The augmented matrix for the given system of equations is:
1 -1 4 | 15
1 1 1 | 4
-3 -1 1 | -2

Step 2: Perform row operations
The goal is to transform the augmented matrix into its reduced row-echelon form, where each leading entry (first non-zero entry) in each row is 1, and every column containing a leading entry has zeros elsewhere.

We'll start by making the leading entry in the first row (1,1) equal to 1:
R1 = R1 / 1 (Dividing the first row by 1)

The new augmented matrix after this operation is:
1 -1 4 | 15
1 1 1 | 4
-3 -1 1 | -2

Next, we use the first row to eliminate the first entry (1,2) in the second row:
R2 = R2 - R1 (Subtracting R1 from R2)

The new augmented matrix after this operation is:
1 -1 4 | 15
0 2 -3 | -11
-3 -1 1 | -2

To eliminate the first entry (1,3) in the third row, we use the first row:
R3 = R3 + 3R1 (Adding 3 times R1 to R3)

The new augmented matrix after this operation is:
1 -1 4 | 15
0 2 -3 | -11
0 2 13 | 43

Next, we focus on the second row and make its leading entry (2,2) equal to 1:
R2 = R2 / 2 (Dividing the second row by 2)

The new augmented matrix after this operation is:
1 -1 4 | 15
0 1 -3/2 | -11/2
0 2 13 | 43

To eliminate the second entry (2,3) in the third row, we use the second row:
R3 = R3 - 2R2 (Subtracting 2 times R2 from R3)

The new augmented matrix after this operation is:
1 -1 4 | 15
0 1 -3/2 | -11/2
0 0 19 | 54

Finally, we make the leading entry in the third row (3,3) equal to 1:
R3 = R3 / 19 (Dividing the third row by 19)

The new augmented matrix after this operation is:
1 -1 4 | 15
0 1 -3/2 | -11/2
0 0 1 | 54/19

Step 3: Back-substitution
Starting from the last row and working our way up, we substitute the found values into the equations to solve for the remaining variables.

From the third row, we have: x3 = 54/19

Substituting x3 = 54/19 into the second row, we can solve for x2:
x2 - (3/2)(54/19) = -11/2
x2 - 81/19 = -11/2
x2 = -11/2 + 81/19
x2 = (-(11/2) * (19/19)) + (81/19)
x2 = (-209/38) + (81/19)
x2 = -209/38 + 342/38
x2 = 133/38

Substituting x2 = 133/38 and x3 = 54/19 into the first row, we can solve for x1:
x1 - (133/38) + 4(54/19) = 15
x1 - 133/38 + (216/19) = 15
x1 = 15 + (133/38) - (216/19)
x1 = (570/38) + (133/38) - (216/19)
x1 = (570 + 133 - 228)/38
x1 = 475/38

Therefore, the solution to the system of linear equations is:
x1 = 475/38
x2 = 133/38
x3 = 54/19