Use Gauss Jordan Elimination to write the solution of the system of equations:
x1+4x2+2x3=17
3x1+x2-5x3=7
2x1-3x2-7x3=-10
http://www.gregthatcher.com/Mathematics/GaussJordan.aspx
1 4 2 17
3 1 -5 7
2 -3 -7 -10
1
4
0
To use Gauss-Jordan elimination to solve the system of equations, we can follow these steps:
Step 1: Write the system of equations in augmented matrix form.
The augmented matrix is formed by writing down the coefficients of each variable along with the constants on the right side of the equal sign. The given system of equations can be written as:
| 1 4 2 | 17 |
| 3 1 -5 | 7 |
| 2 -3 -7 | -10 |
Step 2: Perform row operations to transform the matrix into row-echelon form.
The goal is to create a diagonal of 1's in the leftmost column, and zeros below and above them. This is achieved by applying row operations to eliminate the coefficients below and above the diagonal 1's.
First, we want to create a leading 1 in the first row by dividing the entire first row by the first element, which is 1.
| 1 4 2 | 17 |
| 3 1 -5 | 7 |
| 2 -3 -7 | -10 |
After dividing the first row by 1, we obtain:
| 1 4 2 | 17 |
| 3 1 -5 | 7 |
| 2 -3 -7 | -10 |
Next, we want to eliminate the coefficients below the leading 1 in the first column. To do this, we multiply the first row by the appropriate scalar and subtract it from the corresponding row. In this case, we will multiply the first row by -3 and add it to the second row, and multiply the first row by -2 and add it to the third row.
| 1 4 2 | 17 |
| 0 -11 -11 | -44 |
| 0 -11 -11 | -44 |
Step 3: Continue performing row operations to transform the matrix into reduced row-echelon form.
We still have two nonzero rows below the leading 1 in the second column. To eliminate the coefficients below the leading 1 in the second column, we need to multiply the second row by 1 and add it to the third row.
| 1 4 2 | 17 |
| 0 -11 -11 | -44 |
| 0 0 0 | 0 |
Step 4: Convert the reduced row-echelon form back into a system of equations.
The reduced row-echelon form of the augmented matrix is:
| 1 4 2 | 17 |
| 0 -11 -11 | -44 |
| 0 0 0 | 0 |
This corresponds to the system of equations:
x1 + 4x2 + 2x3 = 17
-11x2 - 11x3 = -44
0 = 0
The third equation, 0 = 0, represents an identity and doesn't provide any new information.
Step 5: Solve for the variables.
From the second equation, we can directly solve for x2 in terms of x3:
-11x2 - 11x3 = -44
-11x2 = -44 + 11x3
x2 = (44 - 11x3) / -11
x2 = -4 + x3
We can now substitute this expression for x2 into the first equation:
x1 + 4(-4 + x3) + 2x3 = 17
x1 - 16 + 4x3 + 2x3 = 17
x1 + 6x3 = 33
So, the solution to the system of equations is:
x1 = t (a free parameter)
x2 = -4 + t
x3 = -4