Lecture 1. Study systems of linear equations
We continue to study systems of linear equations.
Using the concept introduced earlier, we formulate a criterion showing when the system is compatible, that is, it has a solution, and, secondly, we study the set of solutions of a homogeneous system of linear equations.
Let's write down the system of equations in a general form. We compare the matrix of its coefficients with each system. This matrix also contains a column made up of the free coefficients of this system. This is an extended matrix. If we remove the last column, that is, the column of free terms, then the resulting coefficient matrix is called the main matrix of this system (let us imagine a line separating the last column from the main matrix).
Let's assume that we have reduced the extended matrix to a stepwise form. There are two possible variants, let's analyze them.
Variant 1. In a step matrix, each leading element is located to the right of the previous one, that is, the leading elements form a step. The slide shows the case when the leading element in the last row is in the main matrix. Let's denote the leading elements with letters a, b, c, and d. The leading element is the first non-zero number in the row. In this case, the rank of both the main and the augmented matrix is equal to the same number. The rank of a matrix is the number of basis vectors in the system. If we reduce the matrix to a stepwise form, then its rank is equal to the number of rows. Thus, in Variant 1, the rank of the main and the rank of the extended matrix are the same.
Variant 2. Let's assume that in a step system, the last row looks like this: the non-zero element is located in the very last place, it is the c element. If we remove the column of free terms, we get the main matrix, and the last row in it is zero. We have to cross it out to get a step system. In this case, the step system of the main matrix has fewer rows than the step system of the extended matrix. Thus, the ranks of these matrices are different.
What is the difference between these cases in terms of solutions? In Variant 1, we do not get a contradictory equation, so the system is compatible, that is, it has either one solution or infinitely many. In Variant 2, the last line gives a contradictory equation, which means that the system is incompatible.
Thus, we have the following theorem, called the Kronecker-Capelli criterion. A system of linear equations is compatible if and only if the rank of its extended matrix is equal to the rank of the main matrix.
Let us have an example. We are going to consider a system made up of two equations and use the criterion to check whether the ranks of its matrix are equal or not. Let's make up the main matrix. Since the rows in it are proportional, then, finally, we get one row, so the rank of the matrix is 1. We can perform an elementary transformation for an extended matrix: we subtract the first 3 lines from the second one. In this case, we get a step system consisting of two independent rows. This means that the rank of the extended matrix is 2. The ranks are different, so the system is incompatible.
Now we introduce the concept of a homogeneous system of equations. Let us take an arbitrary system of equations. If at least one free coefficient of an equation is not equal to zero, the system is inhomogeneous. If we zero out all the free terms, that is, we consider a system in which every free coefficient is zero, then we call it a homogeneous system of equations.
Let us consider some points for a homogeneous system. First, the zero n is the solution of a given system, which means that a homogeneous system of equations always has a solution: either it is zero or there are infinitely many solutions. Therefore, by the theorem, the system is compatible, so the ranks of the main and extended matrix coincide, and this number is the rank of a homogeneous system of equations.
Let us consider an interesting connection between the solutions of two introduced systems: the initial system and the corresponding homogeneous system of equations. Let a compatible system of linear equations be given, and the set of its solutions is denoted by the letter M. Let's assume that the system is compatible, that is, this set is not empty. We take an arbitrary solution, that is, an arbitrary m of numbers, and denote this m with the letter c. c is some randomly selected and fixed solution. Now we consider the corresponding homogeneous system of equations for this system. Let's denote the set of its solutions by the letter V.
The following fact holds: the set of solutions of the original system can be obtained as the set of solutions of a homogeneous system of equations plus a specific solution c, that is, the set M can be written as a sum: V+c.
Let's illustrate this fact with an example. We are going to consider a system of equations. It is a step system. We dealt with it in one of the previous lectures; we solved it in detail, and got the following set of solutions. It has the following form: the system has infinitely many solutions. Note that the variable x3 is free; we use it to express the variables x1, x2. We write down the general solution, for example, in the form shown on the slide.
Let's consider some particular solution of this system. We take the simplest one, for example, a triple of numbers (0, 1, 0). If we substitute this value instead of variables, we understand that this triple is the solution. Let's try to represent the set M as the sum of this particular solution, as the sum of the vectors c and the general solution of a homogeneous system.
After performing these transformations, we get the following set M: M is the sum of a specific triple (0, 1, 0) and the vector proportional to the vector (1, -2, 1).
Make sure that the set of vectors proportional to the vector (1, -2, 1) is a general solution of a homogeneous system of equations.
Thus, the set M can be represented as the sum of the general solution of the homogeneous system V plus some specific solution of the original system of equations.
We are going to make a geometric illustration of this fact. Let me remind you that a linear equation with three variables sets a plane in space. Besides, if the equation is homogeneous, this plane passes through the zero of the coordinate system.
Thus, if we consider a homogeneous system for the initial system, then its solution can be a line of intersection of two planes that passes through the coordinate origin. The letter V indicates this straight line in the figure. This is the set of solutions to a homogeneous system of equations corresponding to the system presented on the slide.
Next we get the vector c with the coordinates (0, 1, 0) and we add the vector to every solution of the homogeneous system. Geometrically, this means that each point of the straight line V is transferred to the vector c. As a result, we get a straight line setting the solutions of the original system, i.e., the set M. The M line is a line obtained by the direct transfer of the V line to the fixed vector c.
Now we are going to consider another homogeneous linear equation and the properties of its set of solutions. Note that the conclusions we are about to draw are of a general nature. In other words, any linear equation having a 0 free term has these properties, that is, a homogeneous linear equation.
Let's denote the solution of the considered equation by the letter V and write out some solutions. Let's start with the simplest one – when it is zero. The zero m is always the solution, i.e. the zero vector. Next, we take some non-zero solution, for example, the triple (2, -1, 0). This solution is checked by direct substitution. Along with this solution, the solution space contains such a triple of numbers: (-2, 1, 0), which is opposite to this one.
Next, we re going to choose another solution, for example, (-1, -1, 1). We multiply this solution by a number, for example, 3 (you can multiply it by any number), and we again get the solution of this equation. Now we add up any two solutions from the found ones and get a triple of numbers, which is also the solution of this equation. Using these transformations (adding solutions, multiplying an arbitrary solution by a number) we can get the set of all solutions of this equation.
Let's summarize. First, the zero vector always is in the set of solutions V.
Second, whatever solution c we take, multiplying it by any number, we again get the solution of the original equation. In particular, if we multiply the vector c by -1, we get the opposite solution of the original equation.
Third, by adding two solutions to a homogeneous equation, we also get a solution to this equation.
We can draw the following conclusion. We are going to formulate it in the form of a theorem. The set of solutions to a homogeneous system of linear equations is a vector space.
Let's introduce one new concept – the concept of a fundamental system of solutions. We need to recall the concept of the basis. The set of solutions to a homogeneous system of equations is a vector space. We denote this space by the letter V.
What is the basis? The basis is any linearly independent part of the original system of vectors, through which an arbitrary vector of this system is expressed. Namely, if we have a vector space, then the number of basis vectors is its dimension.
The basis of the solution space of a homogeneous system of equations forms the fundamental solution system.
The following theorem holds. Let us take an arbitrary homogeneous system of equations. Let there be n variables. If the r rank of this system is less than the number of variables, then this system has the basis, that is, a fundamental solution system consisting of n-r vectors. If the parameter n and r are equal, then the system has only one zero solution, which means that the fundamental system of solutions does not exist here.