How to solve a system of equations. Systems of Linear Equations: Basic Concepts

Lesson and presentation on the topic: "Systems of equations. Substitution method, addition method, method of introducing a new variable"

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Methods for solving systems of inequalities

Substitution method

How should you proceed when making a decision?
1. Express one of the variables in terms of another. The variables most often used in equations are x and y. In one of the equations we express one variable in terms of another. Tip: Look at both equations carefully before you start solving, and choose the one where it is easier to express the variable.
2. Substitute the resulting expression into the second equation, instead of the variable that was expressed.
3. Solve the equation that we got.
4. Substitute the resulting solution into the second equation. If there are several solutions, then you need to substitute them sequentially so as not to lose a couple of solutions.
5. As a result, you will receive a pair of numbers $(x;y)$, which must be written down as an answer.

Example.
Solve a system with two variables using the substitution method: $\begin(cases)x+y=5, \\xy=6\end(cases)$.

Solution.
Let's take a close look at our equations. Obviously, expressing y in terms of x in the first equation is much simpler.
$\begin(cases)y=5-x, \\xy=6\end(cases)$.
Let's substitute the first expression into the second equation $\begin(cases)y=5-x, \\x(5-2x)=6\end(cases)$.
Let's solve the second equation separately:
$x(5-x)=6$.
$-x^2+5x-6=0$.
$x^2-5x+6=0$.
$(x-2)(x-3)=0$.
We obtained two solutions to the second equation $x_1=2$ and $x_2=3$.
Substitute sequentially into the second equation.
If $x=2$, then $y=3$. If $x=3$, then $y=2$.
The answer will be two pairs of numbers.
Answer: $(2;3)$ and $(3;2)$.

Algebraic addition method

Example.
Solve the system: $\begin(cases)2x+xy-1=0, \\4y+2xy+6=0\end(cases)$.

Solution.
Let's multiply the first equation by 2.
$\begin(cases)4x+2xy-2=0, \\4y+2xy+6=0\end(cases)$.
Let's subtract the second from the first equation.
$4x+2xy-2-4y-2xy-6=4x-4y-8$.
As you can see, the form of the resulting equation is much simpler than the original one. Now we can use the substitution method.
$\begin(cases)4x-4y-8=0, \\4y+2xy+6=0\end(cases)$.
Let's express x in terms of y in the resulting equation.
$\begin(cases)4x=4y+8, \\4y+2xy+6=0\end(cases)$.
$\begin(cases)x=y+2, \\4y+2(y+2)y+6=0\end(cases)$.
$\begin(cases)x=y+2, \\4y+2y^2+4y+6=0\end(cases)$.
$\begin(cases)x=y+2, \\2y^2+8y+6=0\end(cases)$.
$\begin(cases)x=y+2, \\y^2+4y+3=0\end(cases)$.
$\begin(cases)x=y+2, \\(y+3)(y+1)=0\end(cases)$.
We got $y=-1$ and $y=-3$.
Let's substitute these values ​​sequentially into the first equation. We get two pairs of numbers: $(1;-1)$ and $(-1;-3)$.
Answer: $(1;-1)$ and $(-1;-3)$.

Method for introducing a new variable

Example.
Solve the system: $\begin(cases)\frac(x)(y)+\frac(2y)(x)=3, \\2x^2-y^2=1\end(cases)$.

Solution.
Let us introduce the replacement $t=\frac(x)(y)$.
Let's rewrite the first equation with a new variable: $t+\frac(2)(t)=3$.
Let's solve the resulting equation:
$\frac(t^2-3t+2)(t)=0$.
$\frac((t-2)(t-1))(t)=0$.
We got $t=2$ or $t=1$. Let us introduce the reverse change $t=\frac(x)(y)$.
We got: $x=2y$ and $x=y$.

For each of the expressions, the original system must be solved separately:
$\begin(cases)x=2y, \\2x^2-y^2=1\end(cases)$.   
$\begin(cases)x=2y, \\8y^2-y^2=1\end(cases)$.   
$\begin(cases)x=y, \\2y^2-y^2=1\end(cases)$.
$\begin(cases)x=2y, \\7y^2=1\end(cases)$.      
$\begin(cases)x=2y, \\y^2=1\end(cases)$.
$\begin(cases)x=2y, \\y=±\frac(1)(\sqrt(7))\end(cases)$.     
$\begin(cases)x=y, \\y=±1\end(cases)$.

Example.
$\begin(cases)x=±\frac(2)(\sqrt(7)), \\y=±\frac(1)(\sqrt(7))\end(cases)$.    

Solution.
$\begin(cases)x=±1, \\y=±1\end(cases)$.
We received four pairs of solutions.
Answer: $(\frac(2)(\sqrt(7));\frac(1)(\sqrt(7)))$; $(-\frac(2)(\sqrt(7));-\frac(1)(\sqrt(7)))$; $(1;1)$; $(-1;-1)$.
Solve the system: $\begin(cases)\frac(2)(x-3y)+\frac(3)(2x+y)=2, \\\frac(8)(x-3y)-\frac(9 )(2x+y)=1\end(cases)$.
Let us introduce the replacement: $z=\frac(2)(x-3y)$ and $t=\frac(3)(2x+y)$.
Let's rewrite the original equations with new variables:
$\begin(cases)z+t=2, \\4z-3t=1\end(cases)$.
Let's use the algebraic addition method:
$\begin(cases)3z+3t=6, \\4z-3t=1\end(cases)$.
$\begin(cases)3z+3t+4z-3t=6+1, \\4z-3t=1\end(cases)$.
$\begin(cases)7z=7, \\4z-3t=1\end(cases)$.
$\begin(cases)z=1, \\-3t=1-4\end(cases)$.
$\begin(cases)z=1, \\t=1\end(cases)$.
Let's introduce the reverse substitution:
$\begin(cases)\frac(2)(x-3y)=1, \\\frac(3)(2x+y)=1\end(cases)$.
$\begin(cases)x-3y=2, \\2x+y=3\end(cases)$.
Let's use the substitution method:
$\begin(cases)x=2+3y, \\4+6y+y=3\end(cases)$.

$\begin(cases)x=2+3y, \\7y=-1\end(cases)$.

3. $\begin(cases)xy+y^2=3,\\y^2-xy=5\end(cases)$.
4. $\begin(cases)\frac(2)(x)+\frac(1)(y)=4, \\\frac(1)(x)+\frac(3)(y)=9\ end(cases)$.

5. $\begin(cases)\frac(5)(x^2-xy)+\frac(4)(y^2-xy)=-\frac(1)(6), \\\frac(7 )(x^2-xy)-\frac(3)(y^2-xy)=\frac(6)(5)\end(cases)$. Let us analyze two types of solutions to systems of equations: 1. Solving the system using the substitution method.
2. Solving the system by term-by-term addition (subtraction) of the system equations.
In order to solve the system of equations
by substitution method

you need to follow a simple algorithm: 1. Express. From any equation we express one variable. 2. Substitute. We substitute the resulting value into another equation instead of the expressed variable.
1. Select a variable for which we will make identical coefficients.
2. We add or subtract equations, resulting in an equation with one variable.
3. Solve the resulting linear equation. We find a solution to the system.

The solution to the system is the intersection points of the function graphs.

Let us consider in detail the solution of systems using examples.

Example #1:

Let's solve by substitution method

2x+5y=1 (1 equation)
x-10y=3 (2nd equation)

1. Express
It can be seen that in the second equation there is a variable x with a coefficient of 1, which means that it is easiest to express the variable x from the second equation.
x=3+10y

2.After we have expressed it, we substitute 3+10y into the first equation instead of the variable x.
2(3+10y)+5y=1

3. Solve the resulting equation with one variable.
2(3+10y)+5y=1 (open the brackets)
6+20y+5y=1
25y=1-6
25y=-5 |: (25)
y=-5:25
y=-0.2

The solution to the equation system is the intersection points of the graphs, therefore we need to find x and y, because the intersection point consists of x and y. Let's find x, in the first point where we expressed it we substitute y.
x=3+10y
x=3+10*(-0.2)=1

It is customary to write points in the first place we write the variable x, and in the second place the variable y.
Answer: (1; -0.2)

Example #2:

Let's solve using the term-by-term addition (subtraction) method.

3x-2y=1 (1 equation)
2x-3y=-10 (2nd equation)

1. We choose a variable, let’s say we choose x. In the first equation, the variable x has a coefficient of 3, in the second - 2. We need to make the coefficients the same, for this we have the right to multiply the equations or divide by any number. We multiply the first equation by 2, and the second by 3 and get a total coefficient of 6.

3x-2y=1 |*2
6x-4y=2

2x-3y=-10 |*3
6x-9y=-30

2. Subtract the second from the first equation to get rid of the variable x. Solve the linear equation.
__6x-4y=2

5y=32 | :5
y=6.4

3. Find x. We substitute the found y into any of the equations, let’s say into the first equation.
3x-2y=1
3x-2*6.4=1
3x-12.8=1
3x=1+12.8
3x=13.8 |:3
x=4.6

The intersection point will be x=4.6; y=6.4
Answer: (4.6; 6.4)

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Solving systems of linear algebraic equations (SLAEs) is undoubtedly the most important topic in a linear algebra course. A huge number of problems from all branches of mathematics come down to solving systems of linear equations. These factors explain the reason for this article. The material of the article is selected and structured so that with its help you can

• choose the optimal method for solving your system of linear algebraic equations,
• study the theory of the chosen method,
• solve your system of linear equations by considering detailed solutions to typical examples and problems.

Brief description of the article material.

First, we give all the necessary definitions, concepts and introduce notations.

Next, we will consider methods for solving systems of linear algebraic equations in which the number of equations is equal to the number of unknown variables and which have a unique solution. Firstly, we will focus on Cramer’s method, secondly, we will show the matrix method for solving such systems of equations, and thirdly, we will analyze the Gauss method (the method of sequential elimination of unknown variables). To consolidate the theory, we will definitely solve several SLAEs in different ways.

After this, we will move on to solving systems of linear algebraic equations of general form, in which the number of equations does not coincide with the number of unknown variables or the main matrix of the system is singular. Let us formulate the Kronecker-Capelli theorem, which allows us to establish the compatibility of SLAEs. Let us analyze the solution of systems (if they are compatible) using the concept of a basis minor of a matrix. We will also consider the Gauss method and describe in detail the solutions to the examples.

We will definitely dwell on the structure of the general solution of homogeneous and inhomogeneous systems of linear algebraic equations. Let us give the concept of a fundamental system of solutions and show how the general solution of a SLAE is written using the vectors of the fundamental system of solutions. For a better understanding, let's look at a few examples.

In conclusion, we will consider systems of equations that can be reduced to linear ones, as well as various problems in the solution of which SLAEs arise.

Page navigation.

Definitions, concepts, designations.

We will consider systems of p linear algebraic equations with n unknown variables (p can be equal to n) of the form

Unknown variables, - coefficients (some real or complex numbers), - free terms (also real or complex numbers).

This form of recording SLAE is called coordinate.

IN matrix form writing this system of equations has the form,
Where - the main matrix of the system, - a column matrix of unknown variables, - a column matrix of free terms.

If we add a matrix-column of free terms to matrix A as the (n+1)th column, we get the so-called extended matrix systems of linear equations. Typically, an extended matrix is ​​denoted by the letter T, and the column of free terms is separated by a vertical line from the remaining columns, that is,

Solving a system of linear algebraic equations called a set of values ​​of unknown variables that turns all equations of the system into identities. The matrix equation for given values ​​of the unknown variables also becomes an identity.

If a system of equations has at least one solution, then it is called joint.

If a system of equations has no solutions, then it is called non-joint.

If a SLAE has a unique solution, then it is called certain; if there is more than one solution, then – uncertain.

If the free terms of all equations of the system are equal to zero , then the system is called homogeneous, otherwise - heterogeneous.

Solving elementary systems of linear algebraic equations.

If the number of equations of a system is equal to the number of unknown variables and the determinant of its main matrix is ​​not equal to zero, then such SLAEs will be called elementary. Such systems of equations have a unique solution, and in the case of a homogeneous system, all unknown variables are equal to zero.

We started studying such SLAEs in high school. When solving them, we took one equation, expressed one unknown variable in terms of others and substituted it into the remaining equations, then took the next equation, expressed the next unknown variable and substituted it into other equations, and so on. Or they used the addition method, that is, they added two or more equations to eliminate some unknown variables. We will not dwell on these methods in detail, since they are essentially modifications of the Gauss method.

The main methods for solving elementary systems of linear equations are the Cramer method, the matrix method and the Gauss method. Let's sort them out.

Solving systems of linear equations using Cramer's method.

Suppose we need to solve a system of linear algebraic equations

in which the number of equations is equal to the number of unknown variables and the determinant of the main matrix of the system is different from zero, that is, .

Let be the determinant of the main matrix of the system, and - determinants of matrices that are obtained from A by replacement 1st, 2nd, …, nth column respectively to the column of free members:

With this notation, unknown variables are calculated using the formulas of Cramer’s method as . This is how the solution to a system of linear algebraic equations is found using Cramer's method.

Example.

Cramer's method .

Solution.

The main matrix of the system has the form . Let's calculate its determinant (if necessary, see the article):

Since the determinant of the main matrix of the system is nonzero, the system has a unique solution that can be found by Cramer's method.

Let's compose and calculate the necessary determinants (we obtain the determinant by replacing the first column in matrix A with a column of free terms, the determinant by replacing the second column with a column of free terms, and by replacing the third column of matrix A with a column of free terms):

Finding unknown variables using formulas :

Answer:

The main disadvantage of Cramer's method (if it can be called a disadvantage) is the complexity of calculating determinants when the number of equations in the system is more than three.

Solving systems of linear algebraic equations using the matrix method (using an inverse matrix).

Let a system of linear algebraic equations be given in matrix form, where the matrix A has dimension n by n and its determinant is nonzero.

Since , matrix A is invertible, that is, there is an inverse matrix. If we multiply both sides of the equality by the left, we get a formula for finding a matrix-column of unknown variables. This is how we obtained a solution to a system of linear algebraic equations using the matrix method.

Example.

Solve system of linear equations matrix method.

Solution.

Let's rewrite the system of equations in matrix form:

Because

then the SLAE can be solved using the matrix method. Using the inverse matrix, the solution to this system can be found as .

Let's construct an inverse matrix using a matrix from algebraic additions of elements of matrix A (if necessary, see the article):

It remains to calculate the matrix of unknown variables by multiplying the inverse matrix to a matrix-column of free members (if necessary, see the article):

Answer:

or in another notation x 1 = 4, x 2 = 0, x 3 = -1.

The main problem when finding solutions to systems of linear algebraic equations using the matrix method is the complexity of finding the inverse matrix, especially for square matrices of order higher than third.

Solving systems of linear equations using the Gauss method.

Suppose we need to find a solution to a system of n linear equations with n unknown variables
the determinant of the main matrix of which is different from zero.

The essence of the Gauss method consists of sequentially eliminating unknown variables: first, x 1 is excluded from all equations of the system, starting from the second, then x 2 is excluded from all equations, starting from the third, and so on, until only the unknown variable x n remains in the last equation. This process of transforming system equations to sequentially eliminate unknown variables is called direct Gaussian method. After completing the forward stroke of the Gaussian method, x n is found from the last equation, using this value from the penultimate equation, x n-1 is calculated, and so on, x 1 is found from the first equation. The process of calculating unknown variables when moving from the last equation of the system to the first is called inverse of the Gaussian method.

Let us briefly describe the algorithm for eliminating unknown variables.

We will assume that , since we can always achieve this by rearranging the equations of the system. Let's eliminate the unknown variable x 1 from all equations of the system, starting with the second. To do this, to the second equation of the system we add the first, multiplied by , to the third equation we add the first, multiplied by , and so on, to the nth equation we add the first, multiplied by . The system of equations after such transformations will take the form

where , and .

We would have arrived at the same result if we had expressed x 1 in terms of other unknown variables in the first equation of the system and substituted the resulting expression into all other equations. Thus, the variable x 1 is excluded from all equations, starting from the second.

Next, we proceed in a similar way, but only with part of the resulting system, which is marked in the figure

To do this, to the third equation of the system we add the second, multiplied by , to the fourth equation we add the second, multiplied by , and so on, to the nth equation we add the second, multiplied by . The system of equations after such transformations will take the form

where , and . Thus, the variable x 2 is excluded from all equations, starting from the third.

Next, we proceed to eliminating the unknown x 3, while we act similarly with the part of the system marked in the figure

So we continue the direct progression of the Gaussian method until the system takes the form

From this moment we begin the reverse of the Gaussian method: we calculate x n from the last equation as , using the obtained value of x n we find x n-1 from the penultimate equation, and so on, we find x 1 from the first equation.

Example.

Solve system of linear equations Gauss method.

Solution.

Let us exclude the unknown variable x 1 from the second and third equations of the system. To do this, to both sides of the second and third equations we add the corresponding parts of the first equation, multiplied by and by, respectively:

Now we eliminate x 2 from the third equation by adding to its left and right sides the left and right sides of the second equation, multiplied by:

This completes the forward stroke of the Gauss method; we begin the reverse stroke.

From the last equation of the resulting system of equations we find x 3:

From the second equation we get .

From the first equation we find the remaining unknown variable and thereby complete the reverse of the Gauss method.

Answer:

X 1 = 4, x 2 = 0, x 3 = -1.

Solving systems of linear algebraic equations of general form.

In general, the number of equations of the system p does not coincide with the number of unknown variables n:

Such SLAEs may have no solutions, have a single solution, or have infinitely many solutions. This statement also applies to systems of equations whose main matrix is ​​square and singular.

Kronecker–Capelli theorem.

Before finding a solution to a system of linear equations, it is necessary to establish its compatibility. The answer to the question when SLAE is compatible and when it is inconsistent is given by Kronecker–Capelli theorem:
In order for a system of p equations with n unknowns (p can be equal to n) to be consistent, it is necessary and sufficient that the rank of the main matrix of the system be equal to the rank of the extended matrix, that is, Rank(A)=Rank(T).

Let us consider, as an example, the application of the Kronecker–Capelli theorem to determine the compatibility of a system of linear equations.

Example.

Find out whether the system of linear equations has solutions.

Solution.

. Let's use the method of bordering minors. Minor of the second order different from zero. Let's look at the third-order minors bordering it:

Since all the bordering minors of the third order are equal to zero, the rank of the main matrix is ​​equal to two.

In turn, the rank of the extended matrix is equal to three, since the minor is of third order

different from zero.

Thus, Rang(A), therefore, using the Kronecker–Capelli theorem, we can conclude that the original system of linear equations is inconsistent.

Answer:

The system has no solutions.

So, we have learned to establish the inconsistency of a system using the Kronecker–Capelli theorem.

But how to find a solution to an SLAE if its compatibility is established?

To do this, we need the concept of a basis minor of a matrix and a theorem about the rank of a matrix.

The minor of the highest order of the matrix A, different from zero, is called basic.

From the definition of a basis minor it follows that its order is equal to the rank of the matrix. For a non-zero matrix A there can be several basis minors; there is always one basis minor.

For example, consider the matrix .

All third-order minors of this matrix are equal to zero, since the elements of the third row of this matrix are the sum of the corresponding elements of the first and second rows.

The following second-order minors are basic, since they are non-zero

Minors are not basic, since they are equal to zero.

Matrix rank theorem.

If the rank of a matrix of order p by n is equal to r, then all row (and column) elements of the matrix that do not form the chosen basis minor are linearly expressed in terms of the corresponding row (and column) elements forming the basis minor.

What does the matrix rank theorem tell us?

If, according to the Kronecker–Capelli theorem, we have established the compatibility of the system, then we choose any basis minor of the main matrix of the system (its order is equal to r), and exclude from the system all equations that do not form the selected basis minor. The SLAE obtained in this way will be equivalent to the original one, since the discarded equations are still redundant (according to the matrix rank theorem, they are a linear combination of the remaining equations).

As a result, after discarding unnecessary equations of the system, two cases are possible.

If the number of equations r in the resulting system is equal to the number of unknown variables, then it will be definite and the only solution can be found by the Cramer method, the matrix method or the Gauss method.

Example.

.

Solution.

Rank of the main matrix of the system is equal to two, since the minor is of second order different from zero. Rank of the extended matrix is also equal to two, since the only third order minor is zero

and the second-order minor considered above is different from zero. Based on the Kronecker–Capelli theorem, we can assert the compatibility of the original system of linear equations, since Rank(A)=Rank(T)=2.

As a basis minor we take . It is formed by the coefficients of the first and second equations:


The third equation of the system does not participate in the formation of the basis minor, so we exclude it from the system based on the theorem on the rank of the matrix:


This is how we obtained an elementary system of linear algebraic equations. Let's solve it using Cramer's method:


Answer:

x 1 = 1, x 2 = 2.

If the number of equations r in the resulting SLAE is less than the number of unknown variables n, then on the left sides of the equations we leave the terms that form the basis minor, and we transfer the remaining terms to the right sides of the equations of the system with the opposite sign.

The unknown variables (r of them) remaining on the left sides of the equations are called main.

Unknown variables (there are n - r pieces) that are on the right sides are called free.

Now we believe that free unknown variables can take arbitrary values, while the r main unknown variables will be expressed through free unknown variables in a unique way. Their expression can be found by solving the resulting SLAE using the Cramer method, the matrix method, or the Gauss method.

Let's look at it with an example.

Example.

Solve a system of linear algebraic equations .

Solution.

Let's find the rank of the main matrix of the system by the method of bordering minors. Let's take a 1 1 = 1 as a non-zero minor of the first order. Let's start searching for a non-zero minor of the second order bordering this minor:


This is how we found a non-zero minor of the second order. Let's start searching for a non-zero bordering minor of the third order:


Thus, the rank of the main matrix is ​​three. The rank of the extended matrix is ​​also equal to three, that is, the system is consistent.

We take the found non-zero minor of the third order as the basis one.

For clarity, we show the elements that form the basis minor:


We leave the terms involved in the basis minor on the left side of the system equations, and transfer the rest with opposite signs to the right sides:


Let's give the free unknown variables x 2 and x 5 arbitrary values, that is, we accept , where are arbitrary numbers. In this case, the SLAE will take the form


Let us solve the resulting elementary system of linear algebraic equations using Cramer’s method:


Hence, .

In your answer, do not forget to indicate free unknown variables.

Answer:

Where are arbitrary numbers.

Summarize.

To solve a system of general linear algebraic equations, we first determine its compatibility using the Kronecker–Capelli theorem. If the rank of the main matrix is ​​not equal to the rank of the extended matrix, then we conclude that the system is incompatible.

If the rank of the main matrix is ​​equal to the rank of the extended matrix, then we select a basis minor and discard the equations of the system that do not participate in the formation of the selected basis minor.

If the order of the basis minor is equal to the number of unknown variables, then the SLAE has a unique solution, which can be found by any method known to us.

If the order of the basis minor is less than the number of unknown variables, then on the left side of the system equations we leave the terms with the main unknown variables, transfer the remaining terms to the right sides and give arbitrary values ​​to the free unknown variables. From the resulting system of linear equations we find the main unknown variables using the Cramer method, the matrix method or the Gauss method.

Gauss method for solving systems of linear algebraic equations of general form.

The Gauss method can be used to solve systems of linear algebraic equations of any kind without first testing them for consistency. The process of sequential elimination of unknown variables makes it possible to draw a conclusion about both the compatibility and incompatibility of the SLAE, and if a solution exists, it makes it possible to find it.

From a computational point of view, the Gaussian method is preferable.

See its detailed description and analyzed examples in the article Gauss method for solving systems of general linear algebraic equations.

Writing a general solution to homogeneous and inhomogeneous linear algebraic systems using vectors of the fundamental system of solutions.

In this section we will talk about simultaneous homogeneous and inhomogeneous systems of linear algebraic equations that have an infinite number of solutions.

Let us first deal with homogeneous systems.

Fundamental system of solutions homogeneous system of p linear algebraic equations with n unknown variables is a collection of (n – r) linearly independent solutions of this system, where r is the order of the basis minor of the main matrix of the system.

If we denote linearly independent solutions of a homogeneous SLAE as X (1) , X (2) , …, X (n-r) (X (1) , X (2) , …, X (n-r) are columnar matrices of dimension n by 1) , then the general solution of this homogeneous system is represented as a linear combination of vectors of the fundamental system of solutions with arbitrary constant coefficients C 1, C 2, ..., C (n-r), that is, .

What does the term general solution of a homogeneous system of linear algebraic equations (oroslau) mean?

The meaning is simple: the formula specifies all possible solutions of the original SLAE, in other words, taking any set of values ​​of arbitrary constants C 1, C 2, ..., C (n-r), using the formula we will obtain one of the solutions of the original homogeneous SLAE.

Thus, if we find a fundamental system of solutions, then we can define all solutions of this homogeneous SLAE as .

Let us show the process of constructing a fundamental system of solutions to a homogeneous SLAE.

We select the basis minor of the original system of linear equations, exclude all other equations from the system and transfer all terms containing free unknown variables to the right-hand sides of the system equations with opposite signs. Let's give the free unknown variables the values ​​1,0,0,...,0 and calculate the main unknowns by solving the resulting elementary system of linear equations in any way, for example, using the Cramer method. This will result in X (1) - the first solution of the fundamental system. If we give the free unknowns the values ​​0,1,0,0,…,0 and calculate the main unknowns, we get X (2) . And so on. If we assign the values ​​0.0,…,0.1 to the free unknown variables and calculate the main unknowns, we obtain X (n-r) . In this way, a fundamental system of solutions to a homogeneous SLAE will be constructed and its general solution can be written in the form .

For inhomogeneous systems of linear algebraic equations, the general solution is represented in the form , where is the general solution of the corresponding homogeneous system, and is the particular solution of the original inhomogeneous SLAE, which we obtain by giving the free unknowns the values ​​0,0,...,0 and calculating the values ​​of the main unknowns.

Let's look at examples.

Example.

Find the fundamental system of solutions and the general solution of a homogeneous system of linear algebraic equations .

Solution.

The rank of the main matrix of homogeneous systems of linear equations is always equal to the rank of the extended matrix. Let's find the rank of the main matrix using the method of bordering minors. As a non-zero minor of the first order, we take element a 1 1 = 9 of the main matrix of the system. Let's find the bordering non-zero minor of the second order:

A minor of the second order, different from zero, has been found. Let's go through the third-order minors bordering it in search of a non-zero one:

All third-order bordering minors are equal to zero, therefore, the rank of the main and extended matrix is ​​equal to two. Let's take . For clarity, let us note the elements of the system that form it:

The third equation of the original SLAE does not participate in the formation of the basis minor, therefore, it can be excluded:

We leave the terms containing the main unknowns on the right sides of the equations, and transfer the terms with free unknowns to the right sides:

Let us construct a fundamental system of solutions to the original homogeneous system of linear equations. The fundamental system of solutions of this SLAE consists of two solutions, since the original SLAE contains four unknown variables, and the order of its basis minor is equal to two. To find X (1), we give the free unknown variables the values ​​x 2 = 1, x 4 = 0, then we find the main unknowns from the system of equations
.

In this lesson we will look at methods for solving a system of linear equations. In a course of higher mathematics, systems of linear equations are required to be solved both in the form of separate tasks, for example, “Solve the system using Cramer’s formulas,” and in the course of solving other problems. Systems of linear equations have to be dealt with in almost all branches of higher mathematics.

First, a little theory. What does the mathematical word “linear” mean in this case? This means that the equations of the system All variables included in the first degree: without any fancy stuff like etc., which only participants in mathematical Olympiads are delighted with.

In higher mathematics, not only letters familiar from childhood are used to denote variables.
A fairly popular option is variables with indexes: .
Or the initial letters of the Latin alphabet, small and large:
It is not so rare to find Greek letters: – known to many as “alpha, beta, gamma”. And also a set with indices, say, with the letter “mu”:

The use of one or another set of letters depends on the section of higher mathematics in which we are faced with a system of linear equations. So, for example, in systems of linear equations encountered when solving integrals and differential equations, it is traditional to use the notation

But no matter how the variables are designated, the principles, methods and methods for solving a system of linear equations do not change. Thus, if you come across something scary like , do not rush to close the problem book in fear, after all, you can draw the sun instead, a bird instead, and a face (the teacher) instead. And, funny as it may seem, a system of linear equations with these notations can also be solved.

I have a feeling that the article will turn out to be quite long, so a small table of contents. So, the sequential “debriefing” will be like this:

– Solving a system of linear equations using the substitution method (“school method”);
– Solving the system by term-by-term addition (subtraction) of the system equations;
– Solution of the system using Cramer’s formulas;
– Solving the system using an inverse matrix;
– Solving the system using the Gaussian method.

Everyone is familiar with systems of linear equations from school mathematics courses. Essentially, we start with repetition.

Solving a system of linear equations using the substitution method

This method can also be called the “school method” or the method of eliminating unknowns. Figuratively speaking, it can also be called “an unfinished Gaussian method.”

Example 1

Here we are given a system of two equations with two unknowns. Note that the free terms (numbers 5 and 7) are located on the left side of the equation. Generally speaking, it doesn’t matter where they are, on the left or on the right, it’s just that in problems in higher mathematics they are often located that way. And such a recording should not lead to confusion; if necessary, the system can always be written “as usual”: . Don’t forget that when moving a term from part to part, it needs to change its sign.

What does it mean to solve a system of linear equations? Solving a system of equations means finding many of its solutions. The solution of a system is a set of values ​​of all variables included in it, which turns EVERY equation of the system into a true equality. In addition, the system can be non-joint (have no solutions).Don’t be shy, this is a general definition =) We will have only one “x” value and one “y” value, which satisfy each c-we equation.

There is a graphical method for solving the system, which you can familiarize yourself with in class. The simplest problems with a line. There I talked about geometric sense systems of two linear equations with two unknowns. But now this is the era of algebra, and numbers-numbers, actions-actions.

Let's decide: from the first equation we express:
We substitute the resulting expression into the second equation:

We open the brackets, add similar terms and find the value:

Next, we remember what we danced for:
We already know the value, all that remains is to find:

Answer:

After ANY system of equations has been solved in ANY way, I strongly recommend checking (orally, on a draft or on a calculator). Fortunately, this is done easily and quickly.

1) Substitute the found answer into the first equation:

– the correct equality is obtained.

2) Substitute the found answer into the second equation:

– the correct equality is obtained.

Or, to put it more simply, “everything came together”

The considered method of solution is not the only one; from the first equation it was possible to express , and not .
You can do the opposite - express something from the second equation and substitute it into the first equation. By the way, note that the most disadvantageous of the four methods is to express from the second equation:

The result is fractions, but why? There is a more rational solution.

However, in some cases you still can’t do without fractions. In this regard, I would like to draw your attention to HOW I wrote down the expression. Not like this: and in no case like this: .

If in higher mathematics you are dealing with fractional numbers, then try to carry out all calculations in ordinary improper fractions.

Exactly, and not or!

A comma can be used only sometimes, in particular, if it is the final answer to some problem, and no further actions need to be performed with this number.

Many readers probably thought “why such a detailed explanation as for a correction class, everything is clear.” Nothing of the kind, it seems like such a simple school example, but there are so many VERY important conclusions! Here's another one:

You should strive to complete any task in the most rational way. If only because it saves time and nerves, and also reduces the likelihood of making a mistake.

If in a problem in higher mathematics you come across a system of two linear equations with two unknowns, then you can always use the substitution method (unless it is indicated that the system needs to be solved by another method). Not a single teacher will think that you are a sucker and will reduce your grade for using the “school method” "
Moreover, in some cases it is advisable to use the substitution method with a larger number of variables.

Example 2

Solve a system of linear equations with three unknowns

A similar system of equations often arises when using the so-called method of indefinite coefficients, when we find the integral of a fractional rational function. The system in question was taken from there by me.

When finding the integral, the goal is fast find the values ​​of the coefficients, rather than using Cramer’s formulas, the inverse matrix method, etc. Therefore, in this case, the substitution method is appropriate.

When any system of equations is given, first of all it is desirable to find out whether it is possible to somehow simplify it IMMEDIATELY? Analyzing the equations of the system, we notice that the second equation of the system can be divided by 2, which is what we do:

Reference: the mathematical sign means “from this follows that” and is often used in problem solving.

Now let's analyze the equations; we need to express some variable in terms of the others. Which equation should I choose? You probably already guessed that the easiest way for this purpose is to take the first equation of the system:

Here, no matter what variable to express, one could just as easily express or .

Next, we substitute the expression for into the second and third equations of the system:

We open the brackets and present similar terms:

Divide the third equation by 2:

From the second equation we express and substitute into the third equation:

Almost everything is ready, from the third equation we find:
From the second equation:
From the first equation:

Check: Substitute the found values ​​of the variables into the left side of each equation of the system:

1)
2)
3)

The corresponding right-hand sides of the equations are obtained, thus the solution is found correctly.

Example 3

Solve a system of linear equations with 4 unknowns

This is an example for you to solve on your own (answer at the end of the lesson).

Solving the system by term-by-term addition (subtraction) of the system equations

When solving systems of linear equations, you should try to use not the “school method”, but the method of term-by-term addition (subtraction) of the equations of the system. Why? This saves time and simplifies calculations, however, now everything will become clearer.

Example 4

Solve a system of linear equations:

I took the same system as in the first example.
Analyzing the system of equations, we notice that the coefficients of the variable are identical in magnitude and opposite in sign (–1 and 1). In such a situation, the equations can be added term by term:

Actions circled in red are performed MENTALLY.
As you can see, as a result of term-by-term addition, we lost the variable. This, in fact, is what the essence of the method is to get rid of one of the variables.

With this video I begin a series of lessons dedicated to systems of equations. Today we will talk about solving systems of linear equations addition method- This is one of the simplest methods, but at the same time one of the most effective.

The addition method consists of three simple steps:

1. Look at the system and choose a variable that has the same (or opposite) coefficients in each equation;
2. Perform algebraic subtraction (for opposite numbers - addition) of equations from each other, and then bring similar terms;
3. Solve the new equation obtained after the second step.

If everything is done correctly, then at the output we will get a single equation with one variable- it won’t be difficult to solve it. Then all that remains is to substitute the found root into the original system and get the final answer.

However, in practice everything is not so simple. There are several reasons for this:

• Solving equations using the addition method implies that all lines must contain variables with equal/opposite coefficients. What to do if this requirement is not met?
• Not always, after adding/subtracting equations in the indicated way, we get a beautiful construction that can be easily solved. Is it possible to somehow simplify the calculations and speed up the calculations?

To get the answer to these questions, and at the same time understand a few additional subtleties that many students fail at, watch my video lesson:

With this lesson we begin a series of lectures devoted to systems of equations. And we will start from the simplest of them, namely those that contain two equations and two variables. Each of them will be linear.

Systems is 7th grade material, but this lesson will also be useful for high school students who want to brush up on their knowledge of this topic.

In general, there are two methods for solving such systems:

1. Addition method;
2. A method of expressing one variable in terms of another.

Today we will deal with the first method - we will use the method of subtraction and addition. But to do this, you need to understand the following fact: once you have two or more equations, you can take any two of them and add them to each other. They are added member by member, i.e. “X’s” are added to “X’s” and similar ones are given, “Y’s” with “Y’s” are similar again, and what is to the right of the equal sign is also added to each other, and similar ones are also given there.

The results of such machinations will be a new equation, which, if it has roots, they will certainly be among the roots of the original equation. Therefore, our task is to do the subtraction or addition in such a way that either $x$ or $y$ disappears.

How to achieve this and what tool to use for this - we’ll talk about this now.

Solving easy problems using addition

So, we learn to use the addition method using the example of two simple expressions.

Task No. 1

\[\left\( \begin(align)& 5x-4y=22 \\& 7x+4y=2 \\\end(align) \right.\]

Note that $y$ has a coefficient of $-4$ in the first equation, and $+4$ in the second. They are mutually opposite, so it is logical to assume that if we add them up, then in the resulting sum the “games” will be mutually destroyed. Add it up and get:

Let's solve the simplest construction:

Great, we found the "x". What should we do with it now? We have the right to substitute it into any of the equations. Let's substitute in the first:

\[-4y=12\left| :\left(-4 \right) \right.\]

Answer: $\left(2;-3 \right)$.

Problem No. 2

\[\left\( \begin(align)& -6x+y=21 \\& 6x-11y=-51 \\\end(align) \right.\]

The situation here is completely similar, only with “X’s”. Let's add them up:

We have the simplest linear equation, let's solve it:

Now let's find $x$:

Answer: $\left(-3;3 \right)$.

Important points

So, we have just solved two simple systems of linear equations using the addition method. Key points again:

1. If there are opposite coefficients for one of the variables, then it is necessary to add all the variables in the equation. In this case, one of them will be destroyed.
2. We substitute the found variable into any of the system equations to find the second one.
3. The final response record can be presented in different ways. For example, like this - $x=...,y=...$, or in the form of coordinates of points - $\left(...;... \right)$. The second option is preferable. The main thing to remember is that the first coordinate is $x$, and the second is $y$.
4. The rule of writing the answer in the form of point coordinates is not always applicable. For example, it cannot be used when the variables are not $x$ and $y$, but, for example, $a$ and $b$.

In the following problems we will consider the technique of subtraction when the coefficients are not opposite.

Solving easy problems using the subtraction method

Task No. 1

\[\left\( \begin(align)& 10x-3y=5 \\& -6x-3y=-27 \\\end(align) \right.\]

Note that there are no opposite coefficients here, but there are identical ones. Therefore, we subtract the second from the first equation:

Now we substitute the value $x$ into any of the system equations. Let's go first:

Answer: $\left(2;5\right)$.

Problem No. 2

\[\left\( \begin(align)& 5x+4y=-22 \\& 5x-2y=-4 \\\end(align) \right.\]

We again see the same coefficient of $5$ for $x$ in the first and second equation. Therefore, it is logical to assume that you need to subtract the second from the first equation:

We have calculated one variable. Now let's find the second one, for example, by substituting the value $y$ into the second construction:

Answer: $\left(-3;-2 \right)$.

Nuances of the solution

So what do we see? Essentially, the scheme is no different from the solution of previous systems. The only difference is that we do not add equations, but subtract them. We are doing algebraic subtraction.

In other words, as soon as you see a system consisting of two equations in two unknowns, the first thing you need to look at is the coefficients. If they are the same anywhere, the equations are subtracted, and if they are opposite, the addition method is used. This is always done so that one of them disappears, and in the final equation, which remains after subtraction, only one variable remains.

Of course, that's not all. Now we will consider systems in which the equations are generally inconsistent. Those. There are no variables in them that are either the same or opposite. In this case, to solve such systems, an additional technique is used, namely, multiplying each of the equations by a special coefficient. How to find it and how to solve such systems in general, we’ll talk about this now.

Solving problems by multiplying by a coefficient

Example #1

\[\left\( \begin(align)& 5x-9y=38 \\& 3x+2y=8 \\\end(align) \right.\]

We see that neither for $x$ nor for $y$ the coefficients are not only mutually opposite, but also in no way correlated with the other equation. These coefficients will not disappear in any way, even if we add or subtract the equations from each other. Therefore, it is necessary to apply multiplication. Let's try to get rid of the $y$ variable. To do this, we multiply the first equation by the coefficient of $y$ from the second equation, and the second equation by the coefficient of $y$ from the first equation, without touching the sign. We multiply and get a new system:

\[\left\( \begin(align)& 10x-18y=76 \\& 27x+18y=72 \\\end(align) \right.\]

Let's look at it: at $y$ the coefficients are opposite. In such a situation, it is necessary to use the addition method. Let's add:

Now we need to find $y$. To do this, substitute $x$ into the first expression:

\[-9y=18\left| :\left(-9 \right) \right.\]

Answer: $\left(4;-2 \right)$.

Example No. 2

\[\left\( \begin(align)& 11x+4y=-18 \\& 13x-6y=-32 \\\end(align) \right.\]

Again, the coefficients for none of the variables are consistent. Let's multiply by the coefficients of $y$:

\[\left\( \begin(align)& 11x+4y=-18\left| 6 \right. \\& 13x-6y=-32\left| 4 \right. \\\end(align) \right .\]

\[\left\( \begin(align)& 66x+24y=-108 \\& 52x-24y=-128 \\\end(align) \right.\]

Our new system is equivalent to the previous one, but the coefficients of $y$ are mutually opposite, and therefore it is easy to apply the addition method here:

Now let's find $y$ by substituting $x$ into the first equation:

Answer: $\left(-2;1 \right)$.

Nuances of the solution

The key rule here is the following: we always multiply only by positive numbers - this will save you from stupid and offensive mistakes associated with changing signs. In general, the solution scheme is quite simple:

1. We look at the system and analyze each equation.
2. If we see that neither $y$ nor $x$ the coefficients are consistent, i.e. they are neither equal nor opposite, then we do the following: we select the variable that we need to get rid of, and then we look at the coefficients of these equations. If we multiply the first equation by the coefficient from the second, and the second, correspondingly, multiply by the coefficient from the first, then in the end we will get a system that is completely equivalent to the previous one, and the coefficients of $y$ will be consistent. All our actions or transformations are aimed only at getting one variable in one equation.
3. We find one variable.
4. We substitute the found variable into one of the two equations of the system and find the second.
5. We write the answer in the form of coordinates of points if we have variables $x$ and $y$.

But even such a simple algorithm has its own subtleties, for example, the coefficients of $x$ or $y$ can be fractions and other “ugly” numbers. We will now consider these cases separately, because in them you can act somewhat differently than according to the standard algorithm.

Solving problems with fractions

Example #1

\[\left\( \begin(align)& 4m-3n=32 \\& 0.8m+2.5n=-6 \\\end(align) \right.\]

First, notice that the second equation contains fractions. But note that you can divide $4$ by $0.8$. We will receive $5$. Let's multiply the second equation by $5$:

\[\left\( \begin(align)& 4m-3n=32 \\& 4m+12.5m=-30 \\\end(align) \right.\]

We subtract the equations from each other:

We found $n$, now let's count $m$:

Answer: $n=-4;m=5$

Example No. 2

\[\left\( \begin(align)& 2.5p+1.5k=-13\left| 4 \right. \\& 2p-5k=2\left| 5 \right. \\\end(align )\right.\]

Here, as in the previous system, there are fractional coefficients, but for none of the variables do the coefficients fit into each other an integer number of times. Therefore, we use the standard algorithm. Get rid of $p$:

\[\left\( \begin(align)& 5p+3k=-26 \\& 5p-12.5k=5 \\\end(align) \right.\]

We use the subtraction method:

Let's find $p$ by substituting $k$ into the second construction:

Answer: $p=-4;k=-2$.

Nuances of the solution

That's all optimization. In the first equation, we did not multiply by anything at all, but multiplied the second equation by $5$. As a result, we obtained a consistent and even identical equation for the first variable. In the second system we followed a standard algorithm.

But how do you find the numbers by which to multiply equations? After all, if we multiply by fractions, we get new fractions. Therefore, the fractions must be multiplied by a number that would give a new integer, and after that the variables must be multiplied by coefficients, following the standard algorithm.

In conclusion, I would like to draw your attention to the format for recording the response. As I already said, since here we have not $x$ and $y$, but other values, we use a non-standard notation of the form:

Solving complex systems of equations

As a final note to today's video tutorial, let's look at a couple of really complex systems. Their complexity will consist in the fact that they will have variables on both the left and right. Therefore, to solve them we will have to apply preprocessing.

System No. 1

\[\left\( \begin(align)& 3\left(2x-y \right)+5=-2\left(x+3y ​​\right)+4 \\& 6\left(y+1 \right )-1=5\left(2x-1 \right)+8 \\\end(align) \right.\]

Each equation carries a certain complexity. Therefore, let's treat each expression as with a regular linear construction.

In total, we get the final system, which is equivalent to the original one:

\[\left\( \begin(align)& 8x+3y=-1 \\& -10x+6y=-2 \\\end(align) \right.\]

Let's look at the coefficients of $y$: $3$ fits into $6$ twice, so let's multiply the first equation by $2$:

\[\left\( \begin(align)& 16x+6y=-2 \\& -10+6y=-2 \\\end(align) \right.\]

The coefficients of $y$ are now equal, so we subtract the second from the first equation: $$

Now let's find $y$:

Answer: $\left(0;-\frac(1)(3) \right)$

System No. 2

\[\left\( \begin(align)& 4\left(a-3b \right)-2a=3\left(b+4 \right)-11 \\& -3\left(b-2a \right )-12=2\left(a-5 \right)+b \\\end(align) \right.\]

Let's transform the first expression:

Let's deal with the second one:

\[-3\left(b-2a \right)-12=2\left(a-5 \right)+b\]

\[-3b+6a-12=2a-10+b\]

\[-3b+6a-2a-b=-10+12\]

In total, our initial system will take the following form:

\[\left\( \begin(align)& 2a-15b=1 \\& 4a-4b=2 \\\end(align) \right.\]

Looking at the coefficients of $a$, we see that the first equation needs to be multiplied by $2$:

\[\left\( \begin(align)& 4a-30b=2 \\& 4a-4b=2 \\\end(align) \right.\]

Subtract the second from the first construction:

Now let's find $a$:

Answer: $\left(a=\frac(1)(2);b=0 \right)$.

That's all. I hope this video tutorial will help you understand this difficult topic, namely solving systems of simple linear equations. There will be many more lessons on this topic in the future: we will look at more complex examples, where there will be more variables, and the equations themselves will be nonlinear. See you again!