 | Service for Solving Linear Programming Problemsand other interesting typical problems |
Русский |
Example №4. Solving of a System of Linear Equations by the Gauss Jordan Elimination (One Solution)
This solution was made using the calculator presented on the site.
Example №1. Solving of a system of linear equations by the Gauss elimination (one solution)
Example №2. Solving of a system of linear equations by the Gauss elimination (many solutions)
Example №3. Solving of a system of linear equations by the Gauss elimination (no solution)
Example №5. Solving of a system of linear equations by the Gauss Jordan elimination (many solutions)
Example №2. Solving of a system of linear equations by the Gauss elimination (many solutions)
Example №3. Solving of a system of linear equations by the Gauss elimination (no solution)
Example №5. Solving of a system of linear equations by the Gauss Jordan elimination (many solutions)
Please note that the coefficients will disappear which located in the "red" positions.
 | 4 | x1 | + | 2 | x2 | - | 3 | x3 | = | - 3 | |
| 5 | x1 | + | 3 | x2 | - | 5 | x3 | = | - 8 | ||
| 4 | x1 | + | x2 | + | 5 | x3 | = | 22 |
The equation 1 multiplied by -1 is added to the equation 2. more info
( 5 x1 + 4 x1 * ( -1) )
+ ( 3 x2 + 2 x2 * ( -1) )
+ ( -5 x3 + ( -3 x3) * ( -1) )
= -8 + ( -3) * ( -1)
This transformation will allow us to count without fractions for some time.
| | 4 | x1 | + | 2 | x2 | - | 3 | x3 | = | - 3 | |
| x1 | + | x2 | - | 2 | x3 | = | - 5 | ||||
| 4 | x1 | + | x2 | + | 5 | x3 | = | 22 |
The equation 2 and equation 1 are reversed.
| | x1 | + | x2 | - | 2 | x3 | = | - 5 | |||
| 4 | x1 | + | 2 | x2 | - | 3 | x3 | = | - 3 | ||
| 4 | x1 | + | x2 | + | 5 | x3 | = | 22 |
The equation 1 multiplied by -4 is added to the equation 2. more info
( 4 x1 + x1 * ( -4) )
+ ( 2 x2 + x2 * ( -4) )
+ ( -3 x3 + ( -2 x3) * ( -4) )
= -3 + ( -5) * ( -4)
The "red" coefficient is zero.
| | x1 | + | x2 | - | 2 | x3 | = | - 5 | |||
| - | 2 | x2 | + | 5 | x3 | = | 17 | ||||
| 4 | x1 | + | x2 | + | 5 | x3 | = | 22 |
The equation 1 multiplied by -4 is added to the equation 3. more info
( 4 x1 + x1 * ( -4) )
+ ( x2 + x2 * ( -4) )
+ ( 5 x3 + ( -2 x3) * ( -4) )
= 22 + ( -5) * ( -4)
The "red" coefficient is zero.
| | x1 | + | x2 | - | 2 | x3 | = | - 5 | |||
| - | 2 | x2 | + | 5 | x3 | = | 17 | ||||
| - | 3 | x2 | + | 13 | x3 | = | 42 |
The equation 3 multiplied by -1 is added to the equation 2. more info
( -2 x2 + ( -3 x2) * ( -1) )
+ ( 5 x3 + 13 x3 * ( -1) )
= 17 + 42 * ( -1)
This transformation will allow us to count without fractions for some time.
| | x1 | + | x2 | - | 2 | x3 | = | - 5 | |||
| x2 | - | 8 | x3 | = | - 25 | ||||||
| - | 3 | x2 | + | 13 | x3 | = | 42 |
The equation 2 multiplied by 3 is added to the equation 3. more info
( -3 x2 + x2 * 3 )
+ ( 13 x3 + ( -8 x3) * 3 )
= 42 + ( -25) * 3
The "red" coefficient is zero.
| | x1 | + | x2 | - | 2 | x3 | = | - 5 | |||
| x2 | - | 8 | x3 | = | - 25 | ||||||
| - | 11 | x3 | = | - 33 |
The equation 3 is divided by -11.
| | x1 | + | x2 | - | 2 | x3 | = | - 5 | |||
| x2 | - | 8 | x3 | = | - 25 | ||||||
| x3 | = | 3 |
The equation 3 multiplied by 8 is added to the equation 2. more info
x2
+ ( -8 x3 + x3 * 8 )
= -25 + 3 * 8
The "red" coefficient is zero.
| | x1 | + | x2 | - | 2 | x3 | = | - 5 | |||
| x2 | = | - 1 | |||||||||
| x3 | = | 3 |
The equation 3 multiplied by 2 is added to the equation 1. more info
x1
+ x2
+ ( -2 x3 + x3 * 2 )
= -5 + 3 * 2
The "red" coefficient is zero.
| | x1 | + | x2 | = | 1 | ||||||
| x2 | = | - 1 | |||||||||
| x3 | = | 3 |
The equation 2 multiplied by -1 is added to the equation 1. more info
x1
+ ( x2 + x2 * ( -1) )
= 1 + ( -1) * ( -1)
The "red" coefficient is zero.
| | x1 | = | 2 | ||||||||
| x2 | = | - 1 | |||||||||
| x3 | = | 3 |
Result:
x1 = 2
x2 = - 1
x3 = 3
© 2010-2024
If you have any comments, please write to matematika1974@yandex.ru