# 3 Systems of Linear Equations and How to Solve Them

The linear equations Are polynomial equations with in one or several unknowns. In this case, the unknowns are not raised to powers, nor are they multiplied with each other (in this case the equation is said to be grade 1 or first degree).

An equation is a mathematical equality where there is one or more of an element that we do not know which we will call incognito or unknown if there is more than one. To solve this equation it is necessary to find out the value of the unknowns.

A linear equation has the following structure:

to · 1 + a 1 · X 1 A 2 · X 2 +... + a N · X N = B

Where to , to 1 , to 2 ,…,to N Are real numbers of which we know their value and are called coefficients, b is also a known real number that is called independent term. And finally there are X 1 , X 2, X N Which are known as unknowns. These are the variables whose value is unknown.

A system of linear equations is a set of linear equations where the value of the unknowns is the mime in each equation.

Logically, the way to solve a system of linear equations is to assign values ​​to the unknowns, so that equality can be verified. That is, you must calculate the unknowns so that all the equations of the system are fulfilled simultaneously. We represent a system of linear equations as follows

to · 1 + a 1 · X 1 A 2 · X 2 +... + a N · X N A N + 1

B · 1 + b 1 · X 1 + B 2 · X 2 +... + b N · X N = B N + 1

C · 1 + c 1 · X 1 + C 2 · X 2 +... + c N · X N = C N + 1

....

D · 1 + d 1 · X 1 + D 2 · X 2 +... + d N · X N = D N + 1

where to 0, to 1 ,…, to N B B 1 B N , C , C 1 A N Etc in real numbers and the unknowns to solve are X ,..., X N , X N + 1 .

Each linear equation represents a straight line and therefore a system of equations of N linear equations represents N straight drawn in space.

Depending on the number of unknowns that each linear equation has, the line representing that equation will be represented in a different dimension, that is, an equation with two unknowns (for example, 2 · X 1 + X 2 = 0) represents a line in a two-dimensional space, an equation with three unknowns (for example 2 · X 1 + X 2 - 5 · X 3 = 10) would be represented in a three-dimensional space and so on.

When solving a system of equations, the values ​​of X ,..., X N , X N + 1 Turn out to be the cut-off points between the lines.

When solving a system of equations we can arrive at different conclusions. Depending on the type of result we can distinguish between 3 types of systems of linear equations:

### 1- Unsupported compatibility

Although it may sound like a joke, it is possible that when we try to solve the system of equations, we arrive at a style 0 = 0 obviousness.

This type of situation happens when there are infinite solutions for the system of equations, and this happens when it turns out that in our system of equations the equations represent the same line. We can see it graphically:

As a system of equations we take: Having 2 equations with 2 unknowns to solve can represent the lines in a two-dimensional plane As we can see the lines with the same therefore all points of the first equation coincide with those of the second equation, therefore it has as many cut points as points have the line, that is, infinites.

### 2- Incompatible

By reading the name we can imagine that our next set of equations will have no solution.

If we try to solve, for example, this system of equations Graphically it would be: If we multiply all the terms of the second equation we obtain that X + Y = 1 equals 2 · X + 2 · Y = 2. And if this last expression we subtract it to the first equation we obtain

2 · X-2 · X + 2 · Y -2 · Y = 3-2

Or what is the same

0 = 1

When we find ourselves in this situation it means that the lines that are represented in the system of equations are parallel, which means that by definition they are never cut and there is no cut-off point. When a system is presented this way it is said to be inconsistent independent.

### 3- Determined Compatible

Finally we come to the case where our system of equations has a single solution, the case where we have lines that intersect and generate a point of intersection. Let's look at an example: To solve it we can sum the two equations so that we obtain

(3 · X-4 · Y) + (2 · X + 4 · Y) = -6 + 16

If we simplify, we have

5 · X + 0 · Y = 5 · X = 10

From where we easily deduce that X = 2 and substituting or X = 2 in any of the original equations we get Y = 3.

Visually it would be: ## Methods of solving systems of linear equations

As we have seen in the previous section, for systems with 2 unknowns and 2 equations, based on simple operations such as addition, subtraction, multiplication, division and substitution we can solve them in a matter of minutes. But if we try to apply this methodology to systems with more equations and more unknown calculations become tedious and we can easily go wrong.

To simplify the calculations there are several methods of resolution, but without doubt the most widespread methods are the Rule of Cramer and the Elimination of Gauss-Jordan.

### Cramer Method

In order to explain how this method is applied it is essential to know what its matrix is ​​and to know its determinant, let's make a parenthesis to be able to define these two concepts.

A matrix Is nothing more than a set of algebraic numbers or symbols placed in horizontal and vertical lines and arranged in the shape of a rectangle. For our theme we will use the matrix as a more simplified way of expressing our system of equations.

Let's look at an example:

It will be the system of linear equations This simple system of equations we can summarize is the operation of two 2 × 2 matrices which results in a 2 × 1 matrix. The first matrix corresponds to all the coefficients, the second matrix is ​​the unknowns to solve and the matrix located after the equality is identified with the independent terms of the equations

He determinant Is an operation that is applied to a matrix whose result is a real number.

In the case of the matrix that we have found in our previous example, its determinant would be: Once the concepts of matrix and determinant are defined we can explain what Cramer's method consists of.

By this method, we can easily solve a system of linear equations as long as the system does not exceed the three equations with three unknowns since the calculation of the determinants of a matrix is ​​very difficult for matrices of 4 × 4 or higher. In the case of having a system with more than three linear equations the Gauss-Jordan elimination method is recommended.

Following the example above, using Cramer we simply have to calculate two determinants and with it we will find the value of our two unknowns.

We have our system: And we have a system represented by matrices: The value of X is: Simply in the calculation of the determinant located in the denominator of the division, we have replaced the first commune by the matrix of independent terms. And in the denominator of division we have the determinant of our original matrix.

Carrying the same calculations to find the Y we obtain: ### Elimination of Gauss-Jordan

We define Extended matrix To the matrix that results from a system of equations where we add at the end of the matrix the independent terms.

If we continue with our example Our expanded matrix would be: The method by eliminating Gauss-Jordan consists of, through operations between rows of the matrix, transform our enlarged matrix into a much simpler matrix where I have zeros in all fields except the diagonal, where I must obtain some. In the following way: Where X and Y would be real numbers that correspond with our unknowns.

Let us solve this system by eliminating Gauss-Jordan: We multiply by 2 the first row and by 3 the second row If we subtract the second row we get the first We have already managed to get a zero in the lower left of our array, the next step is to get a 0 in the upper right side of it. Divide the first row between 2 and the second row between 10 to simplify the numbers Multiplied the second row by 2 I add to the first row the second We have achieved a 0 in the upper left of the matrix, now we just have to convert the diagonal into some and we have solved our system by Gauss-Jordan. I divide the first row by 3 and the second divide by 4 Therefore, we conclude that: #### References

1. Vitutor.com
2. Algebra.us.es
3. Systems of linear equations (undated). Recovered from uco.es.
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5. Linear Algebra and Geometry (2010/2011). Systems of linear equations. Chapter 1. Department of Algebra. Sevilla University. Spain. Retrieved from algebra.us.es.