The Gaussian algorithm of linear systems of equations explained in a nutshell

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You encounter linear systems of equations for the first time in middle school at high school. From then on you will come across systems of linear equations again and again, provided you decide on a technical profession or are frequently confronted with mathematical problems. The Gauss algorithm is used for the simple and unambiguous solution of systems of equations.

With linear systems of equations you can lose track of things!
With linear systems of equations you can lose track of things!

What you need:

  • Solution scheme
  • basic mathematical knowledge
  • Pen
  • paper

Interesting facts about linear systems of equations

If you tear the term "linear system of equations" apart into the individual word components, you already get a simple idea of ​​what an LGS is.

  • An LGS consists of several linear ones Equations, in which various initially unknown parameters occur. Linear means that the parameters are not in any Potencies respectively root occurrence. For example, the equation x1+ 2x22 = 3 cannot be part of a linear system of equations, since the parameter x2 occurs in the second power.
  • The different equations can be set up by modeling or they are simply given in the task. An example: In a truck delivery, three parts (x1, x2, x3) delivered, which the prices p1 = 1 euro, p2 = 2 euros and p3 = Have 3 euros. The total value of the delivery is 1,000 euros. This information can be summarized in an equation 1x1+ 2x2+ 3x3 = 1,000, where x1, x2 and x3 correspond to the initially unknown quantities of the three parts.
  • In this way further equations can be set up. In this example, the space requirements of the parts and the volume of a truck would be conceivable.
  • After all linear equations have been set up, the LGS can be solved, i.e. the determination of the unknown parameter x1, x2 and x3. This is where the Gaussian algorithm comes into play, with which you can solve the LGS step by step according to a clearly defined scheme.
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  • There are three options for solving a system of linear equations. If you are a little more experienced, you will already see before applying the solution scheme whether an LGS has one, no or an infinite number of solutions.
  • The LGS with the two equations x1+ x2 = 1 and x1+ 2x2 For example, = 1 has no solution because both equations cannot be satisfied at the same time. There is exactly one solution if the number of unknown parameters is equal to the number of equations, there is no contradiction and all equations (each in pairs) are linearly independent. The rank of the matrix belonging to the LGS is then exactly equal to the number of unknowns. If the rank is smaller, there are infinitely many solutions (see example).

Example of the application of the Gaussian algorithm

  1. By modeling a problem, you have the three equations 2x1+ x2-3x3 = 6, x1-2x2-x3 = 2 and -4x1-2x2+ 6x3 = -12 set up.
  2. Now write these three equations one below the other. When applying the Gaussian algorithm, you gradually eliminate the variables. They know that elementary line transformations do not change the solution space.
  3. Now write down the first equation unchanged. Multiply the second and third equations so that when added to the first row, these new equations do not have an x1 contain more. So you multiply the second equation by -2 (because of x1 in the second equation and 2x1 in the first equation) and add them to the first line. Likewise, divide the third equation by two and add it to the first equation.
  4. In the next step you have two equations in which only the parameters x2 and x3 Pop up. Now write down the second equation and multiply the third equation in such a way that when added to the second equation, x2 is eliminated. If you had other equations, proceed in the same way.
  5. In the last equation you then only have the variable x3 that you can now determine. Plugging the result into the other two equations gives you the values ​​for x2 and x1.
  6. In this example, however, there is a special case. In step 3, if you divide the third equation by 2 and add it to the first equation, you get just 0x1+ 0x2+ 0x3 = 0. The reason for this is simple: Equation 1 and Equation 3 are linearly dependent, because the third equation is obtained by multiplying the first equation by -2.
  7. You can cross out the zero line and know that the rank is only 2 and the LGS has an infinite number of solutions, provided that there is no contradiction.
  8. So after steps 3 and 6 you have the two equations 2x1+ x2-3x3 = 6 and 5x2-x3 = 2. You have a degree of freedom. So give x1 and x2 depending on x3 and you are there.
  9. The second equation implies x2 = 2/5 + 1 / 5x3.
  10. If you put x2 into the first equation, we get: 2x1+ 2/5 + 1 / 5x3-3x3 = 6. Resolution to x1 results in: x1 = 14/5 + 7 / 5x3.
  11. The solution space thus lets through L = {(14/5 + 7 / 5x3; 2/5 + 1 / 5x3; x3)} indicate. There are an infinite number of solutions. For x3 = 1, for example, the solution (21/5; 3/5; 1). As a test, you can plug this solution into the original equations and you will find that this solution is actually a solution of the LGS.

Run the Gaussian algorithm in further examples to internalize it. You can specify the numerical values ​​yourself.

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