Fundamental decision system (specific example). Fundamental set of solutions to a homogeneous system of linear equations

To understand what it is fundamental system solutions you can watch a video tutorial for the same example by clicking. Now let's move on to the description of the whole necessary work. This will help you understand the essence of this issue in more detail.

How to find the fundamental system of solutions to a linear equation?

Let's take this system as an example linear equations:

Let's find the solution to this linear system of equations. To begin with, we you need to write out the coefficient matrix of the system.

Let's transform this matrix to a triangular one. We rewrite the first line without changes. And all the elements that are under $a_(11)$ must be made zeros. To make a zero in place of the element $a_(21)$, you need to subtract the first from the second line, and write the difference in the second line. To make a zero in place of the element $a_(31)$, you need to subtract the first from the third line and write the difference in the third line. To make a zero in place of the element $a_(41)$, you need to subtract the first multiplied by 2 from the fourth line and write the difference in the fourth line. To make a zero in place of the element $a_(31)$, you need to subtract the first multiplied by 2 from the fifth line and write the difference in the fifth line.

We rewrite the first and second lines without changes. And all the elements that are under $a_(22)$ must be made zeros. To make a zero in place of the element $a_(32)$, you need to subtract the second one multiplied by 2 from the third line and write the difference in the third line. To make a zero in place of the element $a_(42)$, you need to subtract the second multiplied by 2 from the fourth line and write the difference in the fourth line. To make a zero in place of the element $a_(52)$, you need to subtract the second multiplied by 3 from the fifth line and write the difference in the fifth line.

We see that the last three lines are the same, so if you subtract the third from the fourth and fifth, they will become zero.

According to this matrix write down new system equations.

We see that we have only three linearly independent equations, and five unknowns, so the fundamental system of solutions will consist of two vectors. So we we need to move the last two unknowns to the right.

Now, we begin to express those unknowns that are on the left side through those that are on the right side. We start with the last equation, first we express $x_3$, then we substitute the resulting result into the second equation and express $x_2$, and then into the first equation and here we express $x_1$. Thus, we expressed all the unknowns that are on the left side through the unknowns that are on the right side.

Then, instead of $x_4$ and $x_5$, we can substitute any numbers and find $x_1$, $x_2$ and $x_3$. Each five of these numbers will be the roots of our original system of equations. To find the vectors that are included in FSR we need to substitute 1 instead of $x_4$, and substitute 0 instead of $x_5$, find $x_1$, $x_2$ and $x_3$, and then vice versa $x_4=0$ and $x_5=1$.

A system of linear equations in which all free terms are equal to zero is called homogeneous :

Any homogeneous system is always consistent, since it always has zero (trivial ) solution. The question arises under what conditions will a homogeneous system have a nontrivial solution.

Theorem 5.2.A homogeneous system has a nontrivial solution if and only if the rank of the main matrix less number her unknowns.

Consequence. A square homogeneous system has a nontrivial solution if and only if the determinant of the main matrix of the system is not equal to zero.

Example 5.6. Determine the values ​​of the parameter l at which the system has nontrivial solutions, and find these solutions:

Solution. This system will have a non-trivial solution when the determinant of the main matrix is ​​equal to zero:

Thus, the system is non-trivial when l=3 or l=2. For l=3, the rank of the main matrix of the system is 1. Then, leaving only one equation and assuming that y=a And z=b, we get x=b-a, i.e.

For l=2, the rank of the main matrix of the system is 2. Then, choosing the minor as the basis:

we get a simplified system

From here we find that x=z/4, y=z/2. Believing z=4a, we get

The set of all solutions of a homogeneous system has a very important linear property : if columns X 1 and X 2 - solutions to a homogeneous system AX = 0, then any linear combination of them a X 1 + b X 2 will also be a solution to this system. Indeed, since AX 1 = 0 And AX 2 = 0 , That A(a X 1 + b X 2) = a AX 1 + b AX 2 = a · 0 + b · 0 = 0. It is because of this property that if a linear system has more than one solution, then there will be an infinite number of these solutions.

Linearly independent columns E 1 , E 2 , Ek, which are solutions of a homogeneous system, are called fundamental system of solutions homogeneous system of linear equations if common decision this system can be written as a linear combination of these columns:

If a homogeneous system has n variables, and the rank of the main matrix of the system is equal to r, That k = n-r.

Example 5.7. Find the fundamental system of solutions to the following system of linear equations:

Solution. Let's find the rank of the main matrix of the system:

Thus, the set of solutions to this system of equations forms a linear subspace of dimension n-r= 5 - 2 = 3. Let’s choose minor as the base

.

Then, leaving only the basic equations (the rest will be a linear combination of these equations) and the basic variables (we move the rest, the so-called free variables to the right), we obtain a simplified system of equations:

Believing x 3 = a, x 4 = b, x 5 = c, we find

, .

Believing a= 1, b = c= 0, we obtain the first basic solution; believing b= 1, a = c= 0, we obtain the second basic solution; believing c= 1, a = b= 0, we obtain the third basic solution. As a result, the normal fundamental system of solutions will take the form

Using the fundamental system, the general solution of a homogeneous system can be written as

X = aE 1 + bE 2 + cE 3. a

Let us note some properties of solutions to an inhomogeneous system of linear equations AX=B and their relationship with the corresponding homogeneous system of equations AX = 0.

General solution of an inhomogeneous systemis equal to the sum of the general solution of the corresponding homogeneous system AX = 0 and an arbitrary particular solution of the inhomogeneous system. Indeed, let Y 0 is an arbitrary particular solution of an inhomogeneous system, i.e. AY 0 = B, And Y- general solution of a heterogeneous system, i.e. AY=B. Subtracting one equality from the other, we get
A(Y-Y 0) = 0, i.e. Y-Y 0 is the general solution of the corresponding homogeneous system AX=0. Hence, Y-Y 0 = X, or Y=Y 0 + X. Q.E.D.

Let the inhomogeneous system have the form AX = B 1 + B 2 . Then the general solution of such a system can be written as X = X 1 + X 2 , where AX 1 = B 1 and AX 2 = B 2. This property expresses the universal property of any linear systems(algebraic, differential, functional, etc.). In physics this property is called superposition principle, in electrical and radio engineering - principle of superposition. For example, in the theory of linear electrical circuits the current in any circuit can be obtained as algebraic sum currents caused by each energy source separately.

Purpose of the service. The online calculator is designed to find a non-trivial and fundamental solution to the SLAE. The resulting solution is saved in a Word file (see example solution).

Instructions. Select matrix dimension:

Properties of systems of linear homogeneous equations

Theorem. A system in the case m=n has a nontrivial solution if and only if the determinant of this system is equal to zero.

Theorem. Any linear combination of solutions to a system is also a solution to that system.
Definition. The set of solutions to a system of linear homogeneous equations is called fundamental system of solutions, if this set consists of linearly independent solutions and any solution to the system is a linear combination of these solutions.

Theorem. If the rank r of the system matrix is ​​less than the number n of unknowns, then there exists a fundamental system of solutions consisting of (n-r) solutions.

Algorithm for solving systems of linear homogeneous equations

1. Finding the rank of the matrix.
2. We select the basic minor. We distinguish dependent (basic) and free unknowns.
3. We cross out those equations of the system whose coefficients are not included in the basis minor, since they are consequences of the others (according to the theorem on the basis minor).
4. We move the terms of the equations containing free unknowns to the right side. As a result, we obtain a system of r equations with r unknowns, equivalent to the given one, the determinant of which is nonzero.
5. We solve the resulting system by eliminating unknowns. We find relationships expressing dependent variables through free ones.
6. If the rank of the matrix is ​​not equal to the number of variables, then we find the fundamental solution of the system.
7. In the case rang = n we have a trivial solution.

Example. Find the basis of the system of vectors (a 1, a 2,...,a m), rank and express the vectors based on the base. If a 1 =(0,0,1,-1), and 2 =(1,1,2,0), and 3 =(1,1,1,1), and 4 =(3,2,1 ,4), and 5 =(2,1,0,3).
Let's write down the main matrix of the system:

Let M 0 – set of solutions to a homogeneous system (4) of linear equations.

Definition 6.12. Vectors With 1 ,With 2 , …, with p, which are solutions of a homogeneous system of linear equations are called fundamental set of solutions(abbreviated FNR), if

1) vectors With 1 ,With 2 , …, with p linearly independent (i.e., none of them can be expressed in terms of the others);

2) any other solution to a homogeneous system of linear equations can be expressed in terms of solutions With 1 ,With 2 , …, with p.

Note that if With 1 ,With 2 , …, with p– any f.n.r., then the expression k 1× With 1 + k 2× With 2 + … + k p× with p you can describe the whole set M 0 solutions to system (4), so it is called general view of the system solution (4).

Theorem 6.6. Any indeterminate homogeneous system of linear equations has a fundamental set of solutions.

The way to find the fundamental set of solutions is as follows:

Find a general solution to a homogeneous system of linear equations;

Build ( n – r) partial solutions of this system, while the values ​​of the free unknowns must form an identity matrix;

Write down the general form of the solution included in M 0 .

Example 6.5. Find a fundamental set of solutions to the following system:

Solution. Let's find a general solution to this system.

~ ~ ~ ~ Þ Þ Þ There are five unknowns in this system ( n= 5), of which there are two main unknowns ( r= 2), there are three free unknowns ( n – r), that is, the fundamental solution set contains three solution vectors. Let's build them. We have x 1 and x 3 – main unknowns, x 2 , x 4 , x 5 – free unknowns

Values ​​of free unknowns x 2 , x 4 , x 5 form the identity matrix E third order. Got that vectors With 1 ,With 2 , With 3 form f.n.r. of this system. Then the set of solutions of this homogeneous system will be M 0 = {k 1× With 1 + k 2× With 2 + k 3× With 3 , k 1 , k 2 , k 3 О R).

Let us now find out the conditions for the existence of nonzero solutions of a homogeneous system of linear equations, in other words, the conditions for the existence of a fundamental set of solutions.

A homogeneous system of linear equations has non-zero solutions, that is, it is uncertain if

1) the rank of the main matrix of the system is less than the number of unknowns;

2) in a homogeneous system of linear equations, the number of equations is less than the number of unknowns;

3) if in a homogeneous system of linear equations the number of equations is equal to the number of unknowns, and the determinant of the main matrix is ​​equal to zero (i.e. | A| = 0).

Example 6.6. At what parameter value a homogeneous system of linear equations has non-zero solutions?

Solution. Let's compose the main matrix of this system and find its determinant: = = 1×(–1) 1+1 × = – A– 4. The determinant of this matrix is ​​equal to zero at a = –4.

Answer: –4.

7. Arithmetic n-dimensional vector space

Basic Concepts

IN previous sections We have already encountered the concept of a set of real numbers arranged in a certain order. This is a row matrix (or column matrix) and a solution to a system of linear equations with n unknown. This information can be summarized.

Definition 7.1. n-dimensional arithmetic vector called an ordered set of n real numbers.

Means A= (a 1 , a 2 , …, a n), where a iО R, i = 1, 2, …, n– general view of the vector. Number n called dimension vectors, and numbers a i are called his coordinates.

For example: A= (1, –8, 7, 4, ) – five-dimensional vector.

All set n-dimensional vectors are usually denoted as Rn.

Definition 7.2. Two vectors A= (a 1 , a 2 , …, a n) And b= (b 1 , b 2 , …, b n) of the same dimension equal if and only if their corresponding coordinates are equal, i.e. a 1 = b 1 , a 2 = b 2 , …, a n= b n.

Definition 7.3.Amount two n-dimensional vectors A= (a 1 , a 2 , …, a n) And b= (b 1 , b 2 , …, b n) is called a vector a + b= (a 1 + b 1, a 2 + b 2, …, a n+b n).

Definition 7.4. The work real number k to vector A= (a 1 , a 2 , …, a n) is called a vector k× A = (k×a 1, k×a 2 , …, k×a n)

Definition 7.5. Vector O= (0, 0, …, 0) is called zero(or null vector).

It is easy to verify that the actions (operations) of adding vectors and multiplying them by a real number have the following properties: " a, b, c Î Rn, " k, lО R:

1) a + b = b + a;

2) a + (b+ c) = (a + b) + c;

3) a + O = a;

4) a+ (–a) = O;

5) 1× a = a, 1 О R;

6) k×( l× a) = l×( k× a) = (l× k)× a;

7) (k + l)× a = k× a + l× a;

8) k×( a + b) = k× a + k× b.

Definition 7.6. A bunch of Rn with the operations of adding vectors and multiplying them by a real number given on it is called arithmetic n-dimensional vector space.

Back in school, each of us studied equations and, most likely, systems of equations. But not many people know that there are several ways to solve them. Today we will analyze in detail all the methods for solving a system of linear algebraic equations that consist of more than two equalities.

Story

Today it is known that the art of solving equations and their systems originated in Ancient Babylon and Egypt. However, equalities in their familiar form appeared after the appearance of the equal sign "=", which was introduced in 1556 by the English mathematician Record. By the way, this sign was chosen for a reason: it means two parallel equal segments. And it's true best example equality cannot be invented.

The founder of modern letter designations for unknowns and signs of degrees is a French mathematician. However, his designations were significantly different from those of today. For example, he denoted a square of an unknown number with the letter Q (lat. “quadratus”), and a cube with the letter C (lat. “cubus”). This notation seems awkward now, but at the time it was the most understandable way to write systems of linear algebraic equations.

However, a flaw in the solution methods of that time was that mathematicians only considered positive roots. Perhaps this is due to the fact that negative values didn't have any practical application. One way or another, it was the Italian mathematicians Niccolo Tartaglia, Gerolamo Cardano and Raphael Bombelli who were the first to count negative roots in the 16th century. A modern look, the main solution method (via the discriminant) was created only in the 17th century thanks to the work of Descartes and Newton.

In the mid-18th century, Swiss mathematician Gabriel Cramer found new way in order to make solving systems of linear equations easier. This method was later named after him and we still use it to this day. But we’ll talk about Cramer’s method a little later, but for now let’s discuss linear equations and methods for solving them separately from the system.

Linear equations

Linear equations are the simplest equations with a variable (variables). They are classified as algebraic. write to general view so: a 1 *x 1 +a 2* x 2 +...a n *x n =b. We will need to represent them in this form when compiling systems and matrices later.

Systems of linear algebraic equations

The definition of this term is: it is a set of equations that have common unknown quantities and a common solution. As a rule, at school everyone solved systems with two or even three equations. But there are systems with four or more components. Let's first figure out how to write them down so that it will be convenient to solve in the future. First, systems of linear algebraic equations will look better if all variables are written as x with the appropriate subscript: 1,2,3, and so on. Secondly, all equations should be brought to canonical form: a 1 *x 1 +a 2* x 2 +...a n *x n =b.

After all these steps, we can begin to talk about how to find solutions to systems of linear equations. Matrices will be very useful for this.

Matrices

A matrix is ​​a table that consists of rows and columns, and at their intersection are its elements. These can be either specific values ​​or variables. Most often, to indicate elements, subscripts are placed under them (for example, a 11 or a 23). The first index means the row number, and the second - the column number. Various operations can be performed on matrices, like on any other mathematical element. Thus, you can:

2) Multiply a matrix by any number or vector.

3) Transpose: turn matrix rows into columns, and columns into rows.

4) Multiply matrices if the number of rows of one of them is equal to the number of columns of the other.

Let's discuss all these techniques in more detail, as they will be useful to us in the future. Subtracting and adding matrices is very simple. Since we take matrices of the same size, each element of one table correlates with each element of the other. Thus, we add (subtract) these two elements (it is important that they stand in the same places in their matrices). When multiplying a matrix by a number or vector, you simply multiply each element of the matrix by that number (or vector). Transposition is a very interesting process. It's very interesting to see him sometimes real life, for example, when changing the orientation of a tablet or phone. The icons on the desktop represent a matrix, and when the position changes, it transposes and becomes wider, but decreases in height.

Let's look at another process like: Although we won't need it, it will still be useful to know it. You can multiply two matrices only if the number of columns in one table is equal to the number of rows in the other. Now let's take the elements of a row of one matrix and the elements of the corresponding column of another. Let's multiply them by each other and then add them (that is, for example, the product of the elements a 11 and a 12 by b 12 and b 22 will be equal to: a 11 * b 12 + a 12 * b 22). Thus, one element of the table is obtained, and it is filled in further using a similar method.

Now we can begin to consider how a system of linear equations is solved.

Gauss method

This topic begins to be covered in school. We know the concept of “a system of two linear equations” well and know how to solve them. But what if the number of equations is more than two? This will help us

Of course, this method is convenient to use if you make a matrix out of the system. But you don’t have to transform it and solve it in its pure form.

So, how does this method solve the system of linear Gaussian equations? By the way, although this method is named after him, it was discovered in ancient times. Gauss proposes the following: to carry out operations with equations in order to ultimately bring the entire set to stepped view. That is, it is necessary that from top to bottom (if arranged correctly) from the first equation to the last one unknown decreases. In other words, we need to make sure that we get, say, three equations: in the first there are three unknowns, in the second there are two, in the third there is one. Then from the last equation we find the first unknown, substitute its value into the second or first equation, and then find the remaining two variables.

Cramer method

To master this method, it is vital to have the skills of adding and subtracting matrices, and you also need to be able to find determinants. Therefore, if you do all this poorly or don’t know how at all, you will have to learn and practice.

What is the essence of this method, and how to make it so that a system of linear Cramer equations is obtained? Everything is very simple. We must construct a matrix of numerical (almost always) coefficients of a system of linear algebraic equations. To do this, we simply take the numbers in front of the unknowns and arrange them in a table in the order in which they are written in the system. If there is a “-” sign in front of the number, then we write down a negative coefficient. So, we have compiled the first matrix of coefficients for unknowns, not including the numbers after the equal signs (naturally, the equation should be reduced to canonical form, when only the number is on the right, and all the unknowns with coefficients are on the left). Then you need to create several more matrices - one for each variable. To do this, we replace each column with coefficients in the first matrix in turn with a column of numbers after the equal sign. Thus, we obtain several matrices and then find their determinants.

After we have found the determinants, it's a small matter. We have an initial matrix, and there are several resulting matrices that correspond to different variables. To obtain solutions to the system, we divide the determinant of the resulting table by the determinant of the initial table. The resulting number is the value of one of the variables. Similarly, we find all the unknowns.

Other methods

There are several other methods for obtaining solutions to systems of linear equations. For example, the so-called Gauss-Jordan method, which is used to find solutions to the system quadratic equations and is also associated with the use of matrices. There is also the Jacobi method for solving a system of linear algebraic equations. It is the easiest to adapt to a computer and is used in computing.

Complex cases

Complexity usually arises when the number of equations is less than the number of variables. Then we can say for sure that either the system is inconsistent (that is, has no roots), or the number of its solutions tends to infinity. If we have the second case, then we need to write down the general solution of the system of linear equations. It will contain at least one variable.

Conclusion

Here we come to the end. Let's summarize: we figured out what a system and a matrix are, and learned how to find a general solution to a system of linear equations. In addition, we considered other options. We found out how to solve a system of linear equations: the Gauss method and talked about difficult cases and other ways to find solutions.

In fact, this topic is much more extensive, and if you want to understand it better, we recommend reading more specialized literature.