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Example №2. Solving of a System of Linear Equations by the Gauss Elimination (Many Solutions)
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 №3. Solving of a system of linear equations by the Gauss elimination (no solution)
Example №4. Solving of a system of linear equations by the Gauss Jordan elimination (one solution)
Example №5. Solving of a system of linear equations by the Gauss Jordan elimination (many solutions)
Example №3. Solving of a system of linear equations by the Gauss elimination (no solution)
Example №4. Solving of a system of linear equations by the Gauss Jordan elimination (one 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.
 | 2 | x1 | + | 3 | x2 | - | x3 | + | x4 | = | 1 | |||
| 8 | x1 | + | 12 | x2 | - | 9 | x3 | + | 8 | x4 | = | 3 | ||
| 4 | x1 | + | 6 | x2 | + | 3 | x3 | - | 2 | x4 | = | 3 | ||
| 2 | x1 | + | 3 | x2 | + | 9 | x3 | - | 7 | x4 | = | 3 |
The equation 1 multiplied by -4 is added to the equation 2. more info
( 8 x1 + 2 x1 * ( -4) )
+ ( 12 x2 + 3 x2 * ( -4) )
+ ( -9 x3 + ( - x3) * ( -4) )
+ ( 8 x4 + x4 * ( -4) )
= 3 + 1 * ( -4)
The "red" coefficient is zero.
| | 2 | x1 | + | 3 | x2 | - | x3 | + | x4 | = | 1 | |||
| - | 5 | x3 | + | 4 | x4 | = | - 1 | |||||||
| 4 | x1 | + | 6 | x2 | + | 3 | x3 | - | 2 | x4 | = | 3 | ||
| 2 | x1 | + | 3 | x2 | + | 9 | x3 | - | 7 | x4 | = | 3 |
The equation 1 multiplied by -2 is added to the equation 3. more info
( 4 x1 + 2 x1 * ( -2) )
+ ( 6 x2 + 3 x2 * ( -2) )
+ ( 3 x3 + ( - x3) * ( -2) )
+ ( -2 x4 + x4 * ( -2) )
= 3 + 1 * ( -2)
The "red" coefficient is zero.
| | 2 | x1 | + | 3 | x2 | - | x3 | + | x4 | = | 1 | |||
| - | 5 | x3 | + | 4 | x4 | = | - 1 | |||||||
| 5 | x3 | - | 4 | x4 | = | 1 | ||||||||
| 2 | x1 | + | 3 | x2 | + | 9 | x3 | - | 7 | x4 | = | 3 |
The equation 1 multiplied by -1 is added to the equation 4. more info
( 2 x1 + 2 x1 * ( -1) )
+ ( 3 x2 + 3 x2 * ( -1) )
+ ( 9 x3 + ( - x3) * ( -1) )
+ ( -7 x4 + x4 * ( -1) )
= 3 + 1 * ( -1)
The "red" coefficient is zero.
| | 2 | x1 | + | 3 | x2 | - | x3 | + | x4 | = | 1 | |||
| - | 5 | x3 | + | 4 | x4 | = | - 1 | |||||||
| 5 | x3 | - | 4 | x4 | = | 1 | ||||||||
| 10 | x3 | - | 8 | x4 | = | 2 |
The equation 2 is added to the equation 3. more info
( 5 x3 + ( -5 x3) )
+ ( -4 x4 + 4 x4 )
= 1 + ( -1)
The "red" coefficient is zero.
| | 2 | x1 | + | 3 | x2 | - | x3 | + | x4 | = | 1 | |||
| - | 5 | x3 | + | 4 | x4 | = | - 1 | |||||||
| 0 | = | 0 | ||||||||||||
| 10 | x3 | - | 8 | x4 | = | 2 |
The equation 2 multiplied by 2 is added to the equation 4. more info
( 10 x3 + ( -5 x3) * 2 )
+ ( -8 x4 + 4 x4 * 2 )
= 2 + ( -1) * 2
The "red" coefficient is zero.
| | 2 | x1 | + | 3 | x2 | - | x3 | + | x4 | = | 1 | |||
| - | 5 | x3 | + | 4 | x4 | = | - 1 | |||||||
| 0 | = | 0 | ||||||||||||
| 0 | = | 0 |
| | 2 | x1 | + | 3 | x2 | - | x3 | + | x4 | = | 1 | |||
| - | 5 | x3 | + | 4 | x4 | = | - 1 |
We will find the variable x3 from equation 2 of the system.
- 5 x3 + 4 x4 = - 1
x3 = 1/5 + 4/5 x4
We will find the variable x1 from equation 1 of the system.
2 x1 + 3 x2 - x3 + x4 = 1
2 x1 = 1 - 3 x2 + x3 - x4
2 x1 = 1 - 3 x2 + ( 1/5 + 4/5 x4 ) - x4
x1 = 3/5 - 3/2 x2 - 1/10 x4
Result:
x1 = 3/5 - 3/2 x2 - 1/10 x4
x3 = 1/5 + 4/5 x4
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