Practical lesson 2
At this lesson, we will work out an algorithm for constructing a fundamental system of solutions for a homogeneous system of equations.
Let me remind you that not every homogeneous system of equations has a fundamental system of solutions. But in case when the solution space consists of one zero, the fundamental system of solutions cannot be constructed. In all other cases, the fundamental system of solutions is constructed. At the lecture, we took into consideration a number of examples of building a basis, and now we will work it out in practice.
Let’s consider the following system of homogeneous equations (see the video). It already has a stepped view. We will try to find out whether this system has a basis for the set of its solutions. Let’s look at the last equation z=0. Using the last equation, you can express the second one by substituting zero instead of z. It turns out that the variable x is also equal to 0. We substitute zero values in the first equation, express y, y is also 0. We get exactly the case when the system has only one solution, zero. So, the triple (0, 0, 0) is the only solution. This means that there is no basis on this set. Therefore, this system of equations does not have a fundamental system of solutions.
Let’s solve this problem. A system of linear equations, a homogeneous system, is given. Let’s build a basis for it on the set of solutions and, using this basis, write down the general solution of this system. At the same time, we will repeat the Gauss method. We need a matrix of coefficients, and it is enough to consider the main matrix, the column of free terms cannot be written. We write -1 0 2, and carefully write down the zero coefficients for those variables that are not present in the equation. The second line is 2 1 -1, 3 1 -3, and, finally, -4 -1 5. We add the first two lines to the second one, add the first three lines to the third one, and subtract the first four lines from the fourth line. We get the following lines (see the video). So, the last three lines are proportional, that is, after we subtract the second from the third line, and add the second line to the fourth one, we get two zero lines, which we will delete afterwards. I will not write them down any more, we have done it many times. As a result, we get a step matrix and return to the system (see the video).
The variable x3 can be made free, and x1 and x2 are the main variables. We first express the main x2 through the free -3x3, then x1=2x3. So, the general solution is obtained through the free variable, we need a basis. By the way, you can note that since we have the rank of the original system equal to 2, and three variables, it means that the basis of the solution consists of one vector. This vector is what we need to find. To do this, we must assign any non-zero value to the free variable x3, for example, 1, and find the values x1 and x2, x1=2, x2=-3. We get a basis made up of a single vector (2, -3, 1). Since this vector is non-zero, it forms the basis of the solution space of this system. So, the fundamental system of solutions is built. Using this vector, we can express any solution and write it as k(2, -3, 1). The set of such triples forms the set of all solutions of the original system. I remind you that k takes any values here. You can specify that k belongs to the set of all real numbers.
Let’s consider the following example: the system consists of two equations and has 4 unknowns. I suggest not writing the matrix here. The fact is that after one transformation, we get a step matrix, and since we have a coefficient in the second equation for x1 equal to one, we subtract the second two from the first equation. And after that, we write the second equation in the first place, we get (see the video). So, the step system, x1 and x2 are the main variables, x3 and x4 are free. The basis for the solution space consists of the two vectors. To find them, we express the main variables in terms of free ones, we put - 3x2 to the right, then it turns out (see the video). We substitute x2 and calculate. So, we have expressed the main variables in terms of free variables, and now we need to assign values to the free variables in order to get two linearly independent vectors. To do this, as was shown in the lecture, we will build a table. In the first two columns, we write the main variables, then put a line, and to the right of the line, write the free variables. We need to specify two sets of free variables so that we end up with linearly independent vectors. To do this, the easiest way is to take the unit vectors corresponding to two free variables 1 0 and 0 1, as shown in the lecture.
However, if we take it this way, x1 and x2 will be expressed as fractional numbers. It’s okay, you can take them. But for those who do not like to work with fractions, so as not to make unnecessary mistakes, you can take any other set instead of such a set, as long as the resulting rows are linearly independent. So instead of ones, I’ll take threes. The result is a step system, which means that we get linearly independent vectors, while due to the three we will remove the fractions when calculating the main variables. So, we calculate x1. If x3=3, x4=0, x1=1, x2 will be equal to -5. We take a set of 0 3 free variables, substitute and find x1=4, x2=-8. So, each line gives us a vector, a=(1, -5, 3, 0), b=(4, -8, 0, 3). Thus, the fundamental system of solutions is found. We express arbitrary solutions in terms of basic ones. As you know, since the basis consists of two vectors, it means that any solution vector is a linear combination of data, that is, k1a+k2b=k1(1, -5, 3, 0) + k2(4, -8, 0, 3), k1 b k2 are arbitrary numbers. So, the general solution of this homogeneous system looks like this. Note that the basis on the set of solutions may consist of other vectors. If we took other values instead of x3 and x4, we would get other vectors. The solution is obtained.
Once again, note that we could take other values instead of the sets of free variables we took, as long as they formed a step system for the independence of vectors. In this case, we could get a different basis. The type of solutions would be written differently, but a set of solutions, of course, would not change, but simply would have a different form.