Practical lesson 1. Fundamental system of solutions
In this lesson, we continue to work with linear equations. Let's consider the solution of homogeneous equations and talk about the relationship between the variants of solutions of an inhomogeneous system of equations and a homogeneous system of linear equations. We are also going to discuss the concept of a fundamental system of solutions.
Here is the task. We have three vectors (see the video). Can they be a fundamental system of solutions to a homogeneous system of equations?
We need to revise the concept of a fundamental system of solutions. This is the basis for the set of all solutions of a homogeneous system of linear equations. You know that the set of all solutions of a homogeneous system is a vector space. If this set consists of a single zero vector, then there is no basis. If the system has infinitely many solutions, then we can find the basis.
Can the vectors a, b, and c form a basis? Let us make a matrix of their coordinates and find out whether the vectors are linearly independent, because the basis is a linearly independent subsystem. Let's make it stepwise. We subtract the first line from the second one and the first three lines from the third on. We get the following matrix: the first row does not change, the second row is 0, -3, 3, and the third row is 0, -3, -6+9=3. As we can see, the rows are proportional, that is, subtracting the second row from the third one, we get a zero row that can be crossed out. The rank of the resulting step system is 2. This means that the rank of the original system of vectors is also 2. Thus, the vectors are linearly dependent, and they cannot form the basis. The answer is no. These vectors cannot form a fundamental system of solutions.
Let's take only two vectors: a and b. Can these vectors form a fundamental system of solutions? These vectors are linearly independent, because their coordinates are not proportional, which means that the required system of equations exists. We'll try to find it. Note that the coordinates of the vectors (1, 2, -3) add up to zero. Similarly for the second vector: 1-1+0 is zero. Thus, the coordinates of these vectors satisfy this equation (see the video). That is, a and b are two solutions of the specified equation. Moreover, the solutions are linearly independent. What is the dimension of the space of solutions to this equation? We have one equation, so we can say that we have a system consisting of one equation. There are three variables, rank 1, so the solution space is two-dimensional. That is, the set of solutions to this equation has the basis of two vectors. Since the vectors a and b are linearly independent and therefore solutions, they form the basis. So, here is the equation for which the fundamental system of solutions has the specified form: the vectors a and b.
Now we are going to continue to study this system. Now we will find one vector. Let us try to find a system of equations for which this vector is the basis of the sets of its solutions. In the previous case, we used one equation for two vectors, but here we can't use just one equation. We see that the sum of coordinates is zero. Nevertheless, I propose to write down the general solution of the required system of equations. Let us consider an arbitrary n of numbers, namely, a triple of numbers (x1, x2, x3), as the solution of a certain system. Since the basis of this system consists of a single vector, it means that any solution is proportional to this vector. We write (k, 2k,- 3k). x1 is k, x2 is 2k, x3 is 3k, and k is an arbitrary number, an arbitrary scalar. This means that we can consider the variable x1 as a free variable in some system that takes arbitrary values. The variables x2 and x3 are expressed in its terms. Since k is x1, then x2 is 2x1, and x3 is 3x1, and x1 is a free variable. Let us move all the variables to one part (see the video). We get a homogeneous system consisting of two linear equations. You can check that vector a is the solution. If we substitute the coordinates, we get the correct equation. Thus, the vector a is a solution, and this is the basis of the set of all solutions of the resulting system. This means that the vector a is the fundamental system of solutions for the specified system of equations.
Now, let us answer the question using the result of this problem. We know that for the system we have obtained, the set of solutions is proportional to the vector a. Let us write down the general solution of such a system of equations (see the video), which is not homogeneous, however, the left parts are exactly the same as for homogeneous systems. Let's recall the connection between sets of solutions. If we take some inhomogeneous system, we denote the set of its solutions by the letter M, for this system we write the corresponding homogeneous system, and the space of its solutions is V. Then the set of solutions of the original system can be represented as the sum of the general solution of the homogeneous system and some specific solution of the original system. C is the solution of the original homogeneous system. V is the set of vectors proportional to the vector (1, 2, -3). To write the set M, we need to find the vector c. We can do it by selection. To avoid guessing, we can express x2 and x3 via x1. x2 is 2x1-4, and x3 is 5-3x1. Now you can substitute any number instead of x1 and find x2 and x3. The easiest option is to take a zero, although you can take1, 2, 3; it doesn't matter. Let x1=0, then x2=-4, x3=5. As the vector c, we can take the triple (0, -4, 5). This particular solution of an inhomogeneous system is written as the sum of a vector proportional to (1, 2, -3) and this particular solution (0, -4, 5). Thus, we get the set of all solutions to this system of equations.