Linear systems of three equations

Linear systems of three equations . They are systems that can have a single solution, an infinite number of solutions or have no solution.

Summary

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• 1 Story
• 2 Solution of the systems of three equations with three variables
• 3 Elimination method to solve a system of equations
• 4 Matrices method to solve systems of equations of three variables
• 5 Sources

History

The linear equation systems were already solved by the Babylonians, who called the unknowns with words such as length, width, area, or volume, without being related to measurement problems.

An example taken from a Babylonian tablet proposes the solution of a system of equations in the following terms:

1/4 width + length = 7 hands

length + width = 10 hands

To solve it, they begin by assigning the value 5 to a hand and observed that the solution could be: width = 20, length = 30. To verify this, they used a method similar to that of elimination. In notation, it would be:

y + 4x = 28

y + x = 10

Subtracting the second from the first, we obtain 3x = 18, that is: x = 6 and y = 4.

They also solved systems of equations, where some of them were quadratic. The Greeks also solved some systems of equations, but using geometric methods. Thymaridas (400 BC) had found a formula to solve a certain system of n equations with n unknowns.

Diophante also solves problems in which systems of equations appeared, but transforming them into a linear equation. Diophante only accepted positive solutions, because what he was looking for was to solve problems and not equations. You already used syncopated algebra as noted above. However, one of the difficulties we encounter in solving equations by Diophante is that it lacks a general method and uses sometimes excessively ingenious methods in each problem. Equation systems also appear in Indian documents. However, they fail to obtain general solving methods, but solve special types of equations.

The book ” The Mathematical Art “, by an unknown Chinese author (3rd century BC), contains some problems where equations are solved. In them we find an outline of the method of matrices to solve systems of linear equations. One of these problems is equivalent to solving a system of three linear equations by said matrix method.

Solution of the systems of three equations with three variables

In a system of three equations with three unknowns. Each of the equations represents a plane. According to the possible relationships that exist between the three planes, the type of solution that the system has is determined:

Systems of three equations with three unknowns that have no solution:

1. Three parallel planes.
2. Two parallel planes and another cuts them.
3. Plane parallel to the cut line of the other two.
4. Two overlapping planes and the other parallel.

Systems of three equations with three unknowns that have infinite solutions

1. Three overlapping planes.
2. Make plans.
3. Two overlapping planes and another that cuts them.

Systems of three equations with three unknowns that have a unique solution

1. Three planes that meet at one point.

Elimination method to solve a system of equations

Next, through an example it will be shown how to solve a linear system of three equations.

Example:

Solve the system:

x + 2y + 3z = 9 ………………………. (first equation)

4x + 5y + 6z = 24 …………………….. (second equation)

3x + y – 2z = 4 ………………………. (third equation)

Solution:

Add (-4) times the “first equation” to the “second”:

x + 2y + 3z = 9

-3y – 6z = -12

3x + y – 2z = 4

Add (-3) times the “first equation” to the “third”:

x + 2y + 3z = 9

-3y – 6z = -12

-5y – 11z = -23

Multiply by – (1/3) the “second equation”:

x + 2y + 3z = 9

y + 2z = 4

-5y -11z = -23

Multiply by (-1) the “third equation”:

x + 2y + 3z = 9

y + 2z = 4

5y + 11z = 23

Add (-5) to the “second equation” and “third equation”:

x + 2y + 3z = 9

y + 2z = 4

z = 3

The solutions of the last system are easy to find by substitution.

From the “third equation”, we see that z = 3.

Substituting “z” with 3 in the “second equation”, y + 2z = 4 gives y = (- 2)

Finally, we find the value of “x” by substituting y = (- 2) and z = 3, in the “first equation”, x + 2y + 3z = 9 whereby x = 4.

So there is a solution:

x = 4

y = (- 2)

z = 3

Matrices method to solve systems of equations of three variables

Analyzing the method used in the previous example to solve a system of equations of three unknowns, we were able to realize that the symbols used for the variables are unimportant, and only the coefficients of the variables were taken into account. But it is possible to simplify the process if a scheme with the coefficients is introduced, in such a way that there is no need to write the variables.

First, it was verified that the variables appeared in the same order in each equation and that the terms without variables were to the right of the equality signs. To do this, we will use the same example from the previous section, remaining as follows:

An arrangement of numbers of this type is called an array .

The rows (or rows) of the matrix are the numbers that appear one after the other horizontally:

The columns of the matrix are the numbers that appear next to each other vertically.

The matrix obtained from the system of linear equations in the previous way is the matrix of the system. If we delete the last column, the remaining ordering is the coefficient matrix. Since we can have the system matrix from the coefficient matrix by adding a column, we call it an increased coefficient matrix or simply an increased matrix. Then, when matrices are used to find solutions to a system of linear equations, a vertical line segment will be inserted into the augmented matrix to indicate where the equals would appear in the corresponding system of equations.

To solve the system:
x + 2y + 3z = 9
4x + 5y + 6z = 24
3x + y – 2z = 4 We will
start with the system matrix, that is, the augmented matrix

We then apply elementary row transformations to obtain another (simpler) matrix from a system of equivalent equations. We will put suitable symbols between equivalent matrices.

With the final matrix we return to the system of equations:

Which is equivalent to the original system. The solution x = 4, y = -2, z = 3 can now be found by substitution.

The final matrix of the solution is a stepped form