Solve the system of linear equations using the Gauss-Jordan elimination method.
3x−2y+ 4z = 22
2x+y− 2z = 3
x+ 4y − 8z = −16
try this
http://www.gregthatcher.com/Mathematics/GaussJordan.aspx
How do you interpret the result of the last line?
Step 1: Write down the augmented matrix:
[[3, -2, 4, 22],
[2, 1, -2, 3],
[1, 4, -8, -16]]
Step 2: Perform row operations to get a leading 1 in the first column, first row:
R2 = R2 - (2/3) * R1
R3 = R3 - (1/3) * R1
[[3, -2, 4, 22],
[0, 7/3, -10/3, -7/3],
[0, 14/3, -12/3, -38/3]]
Step 3: Perform row operations to get zeros in the first column, second row and third row:
R3 = R3 - (2/7) * R2
[[3, -2, 4, 22],
[0, 7/3, -10/3, -7/3],
[0, 0, -8/3, -32/7]]
Step 4: Perform row operations to get a leading 1 in the second column, second row:
R2 = (3/7) * R2
[[3, -2, 4, 22],
[0, 1, -10/9, -1/3],
[0, 0, -8/3, -32/7]]
Step 5: Perform row operations to get a leading 1 in the third column, third row:
R3 = (-3/8) * R3
[[3, -2, 4, 22],
[0, 1, -10/9, -1/3],
[0, 0, 1, 4/7]]
Step 6: Perform row operations to get zeros in the second column, third row:
R2 = R2 + (10/9) * R3
[[3, -2, 4, 22],
[0, 1, 0, 11/21],
[0, 0, 1, 4/7]]
Step 7: Perform row operations to get zeros in the first column, third row:
R1 = R1 - 4 * R3
[[3, -2, 0, 6],
[0, 1, 0, 11/21],
[0, 0, 1, 4/7]]
Step 8: Perform row operations to get zeros in the second column, first row:
R1 = R1 + 2 * R2
[[3, 0, 0, 26/7],
[0, 1, 0, 11/21],
[0, 0, 1, 4/7]]
Step 9: The final row-echelon form of the augmented matrix is:
[[3, 0, 0, 26/7],
[0, 1, 0, 11/21],
[0, 0, 1, 4/7]]
Step 10: Convert the augmented matrix back into equations:
3x = 26/7
x = 26/21
y = 11/21
z = 4/7
Therefore, the solution to the system of linear equations is x = 26/21, y = 11/21, and z = 4/7.
To solve the system of linear equations using the Gauss-Jordan elimination method, we need to follow these steps:
Step 1: Set up the augmented matrix
- The augmented matrix is formed by writing down the coefficients of the variables and the constant terms in a matrix form.
- For the given system of equations:
3x - 2y + 4z = 22 -> [3 -2 4 | 22]
2x + y - 2z = 3 -> [2 1 -2 | 3]
x + 4y - 8z = -16 -> [1 4 -8 | -16]
Step 2: Perform row operations to modify the matrix into row-echelon form
- Use row operations to transform the matrix into row-echelon form, where the leading coefficient in each row is 1 and the column below a leading coefficient is filled with zeros.
- The goal is to create zeros below each leading coefficient.
- The row operations include:
- Multiply a row by a nonzero constant.
- Add or subtract one row from another.
- Swap two rows.
Starting with the augmented matrix:
[ 3 -2 4 | 22 ]
[ 2 1 -2 | 3 ]
[ 1 4 -8 | -16 ]
Perform row operations:
- Multiply Row 1 by 1/3 to make the leading coefficient 1:
[ 1 -2/3 4/3 | 22/3 ]
[ 2 1 -2 | 3 ]
[ 1 4 -8 | -16 ]
- Multiply Row 1 by -2 and add it to Row 2:
[ 1 -2/3 4/3 | 22/3 ]
[ 0 7/3 -10/3 | -49/3 ]
[ 1 4 -8 | -16 ]
- Multiply Row 1 by -1 and add it to Row 3:
[ 1 -2/3 4/3 | 22/3 ]
[ 0 7/3 -10/3 | -49/3 ]
[ 0 14/3 -20/3 | -58/3 ]
- Multiply Row 2 by 3/7 to make the leading coefficient 1:
[ 1 -2/3 4/3 | 22/3 ]
[ 0 1 -10/7 | -7 ]
[ 0 14/3 -20/3 | -58/3 ]
- Multiply Row 2 by 2/3 and add it to Row 1:
[ 1 0 6/7 | 16/3 ]
[ 0 1 -10/7 | -7 ]
[ 0 14/3 -20/3 | -58/3 ]
- Multiply Row 2 by -14/3 and add it to Row 3:
[ 1 0 6/7 | 16/3 ]
[ 0 1 -10/7 | -7 ]
[ 0 0 0 | 0 ]
Step 3: Convert the matrix to reduced row-echelon form
- Continue performing row operations until the matrix is in reduced row-echelon form, where each leading coefficient is 1 and every other entry in the column is 0.
In the current matrix:
[ 1 0 6/7 | 16/3 ]
[ 0 1 -10/7 | -7 ]
[ 0 0 0 | 0 ]
Since the last row consists of all zeros, we can conclude that the system of equations is dependent, and therefore, there are infinitely many solutions.
Step 4: Express the solution
To express the solution, rewrite the matrix back into equation form:
x + (6/7)z = (16/3)
y - (10/7)z = -7
Here, z is the free variable. Choose a value for z, substitute it back into the equations, and find the corresponding values for x and y.
For example, let's choose z = 7. Substituting this value into the equations:
x + (6/7)(7) = (16/3)
y - (10/7)(7) = -7
Simplifying:
x + 6 = 16/3
y - 10 = -7
x = 4/3
y = 3
Therefore, the solution to the system of equations is:
x = 4/3
y = 3
z can take any value since it's the free variable.
Solve the following system of linear equations using the method of Gauss Jordan Elimination.
𝑥 + 𝑦 + 2𝑧 = 8
−𝑥 − 2𝑦 + 3𝑧 = 1
3𝑥 − 7𝑦 + 4𝑧 = 10