Lecture 3. A system of linear equations
A system of linear equations
We continue to study systems of linear equations.
At the last lecture, we discussed thoroughly the examples of systems that either have no solutions or have a single solution. Let's start this lecture with the following example.
We are going to consider the following system. It is stepwise. Moreover, if we start expressing some variable from the last equation, for example x2, and then insert its value in the first equation, we get that all three variables can be expressed in terms of some arbitrary value c, and c takes an arbitrary value, that is, the system has infinitely many solutions.
We get some number triplet for each number c.
Omitting the intermediate transformation, we will leave only what is on the slide, and talk about various forms of writing down the solution. Note that the variable through which we expressed others is free, in our case it is the variable x3.
We expressed x2 and then x1 through x3. We can write down the solution, leaving this variable, that is, specifying how other variables are expressed through a free variable. We also can, as we have just done, represent the variable x3 with the letter c and express x1 and x2 through this value, indicating that c takes an arbitrary value.
Moreover, the numbers triplet can be represented in the form already known to us - in parentheses.
Since the system has infinitely many solutions, the set of solutions is presented in this way. We get the set of all triplets of the form (c; 1-2c; c), where c is an arbitrary real number.
Now let's recall some of the facts that we talked about during our previous classes. First, let me remind you that using elementary transformations, we get a system that is equivalent to the original system, that is, the set of solutions does not change.
Then there is the theorem that any system of linear equations can be reduced to a stepwise form using elementary transformations.
Then we mentioned that only three cases are possible for an arbitrary stepwise system. We discussed two of them during the last class. Now we are going to speak about the third option, when the system has infinitely many solutions.
So, let me remind you of the fact that if we have contradictory equations in a stepwise system, we immediately conclude that the system is not compatible. If there is no contradictory equation, then in this case there are two options: either there is one solution or there are infinitely many solutions.
So, let's talk in more detail about the last case, when the system has infinitely many solutions. Let’s consider the following example: there are three equations in a system, three unknowns.
Initially, we can not say anything about the number of solutions, let's bring this system to a stepwise form.
To do this, as we did it last time, we will write a matrix of coefficients and achieve gradual zeros so that our matrix has a stepwise form.
To do this, we subtract the first equation from the second one, and we subtract the first two equations from the third one. This will provide us with zeros under the leading element of the first line. After these transformations, we get this system. So, there are quite a lot of zeros, but so far the system is not stepwise, because under the leading element of the second line there is a two, and we need a zero there. It is quite easy to achieve a zero - we have to subtract the two second lines from the third one. We get this matrix.
Here we have another situation that did not occur last time - a null line appeared.
According to the definition, there should not be a null line in a stepwise system, but it is easy to change.
As you know, crossing out a null line does not change the set of solutions. We cross it out and get a stepwise matrix.
Let's look at the situation we have here.
First of all, here we again get a free variable, but those variables that correspond to the leading elements of the matrix (they are highlighted in red) are variables x1,x3, we will call them the main variables.
The variable x2 is free, and x1,x3 are the main variables.
x3 equals zero, which follows from the last equation, and the variable x1 is expressed in terms of a free variable and again we get a similar situation.
If we assign an arbitrary value to a free variable, then we get our own numbers triplet for each value.
Now, I suggest that the variable x2 be represented by C, i.e., give the variable x2 an arbitrary value of C, then x1 =1-C, x2=C, and x3=0. We get the set of specified triplets, where C is an arbitrary number.
Let's look at another example. Let's increase the number of variables and take four variables. We still have 3 equations, but there are four variables in the system. Please note that the first equation contains three variables, and the second and third equations contain four variables, so when we go to write the matrix, we need to correct the first equation a little.
We will not always do it in this way — this is done verbally, but here I gave you a slide in which the summand 0*x4 is added in the first equation. So, we put a zero in the corresponding matrix in the right place, and then we carry out the same operations - we bring the matrix to a stepwise form.
Thus, we have selected the leading element of the first line. We get zeros under this element, fill up the already familiar transformation and calculate. We subtract the first line from the second one, and the first two lines from the third one. After that, you can see that the second and third lines are proportional, and to get a stepwise system, first, we subtract the three second lines from the third line.
After that, we get the zero line again and cross it out.
We get a stepwise system.
The leading elements are highlighted; let's analyze which variables are free here, and which variables are the main ones. We return to the system, i.e. we build a system equivalent to the original one based on the obtained matrix.
I again highlight in red variables that correspond to the leading elements, we consider these variables to be the main ones, and the rest are free.
x1 and x4 are the main variables. X2 and x3 are free variables.
Next, the variable x4 is already expressed, it is equal to one. We have to express the variable x1. We get what it is equal to from the first equation, i.e. we get an expression in terms of free variables.
There are two free variables here, so each of them can take an arbitrary value. That is, they do not depend on each other, so the letter C can represent the first variable x2, and the letter D can represent the second variable.
In other words, we use different letters. The following solution is obtained; X2 is C, X3 is D, X4=1, X1=3-C-2D.
Thus, we get this general solution. Please note that C and D are arbitrary numbers. In other words, our system has infinitely many solutions that depend on two numbers - C and D. This means that the set of solutions can be written down in the given way. It is a set of fours of the specified numbers.
Going further, we will consider the so-called geometric interpretation for the cases where the system contains either linear equations with two variables, or linear equations with three variables.
Let's consider the simplest case when we have 2 equations with two variables in the system.
Let me remind you of the fact that every linear equation with two variables sets a straight line on the plane, provided that at least one of the coefficients for the variables is not equal to zero.
In this case, we can interpret the solutions as follows.
Let me remind you that any system of equations, any system, if it is not zero, I mentioned this earlier, and any non-zero system of equations can have only three options: either have one solution, or have no solutions, or have infinitely many solutions.
Let's see how these options are used for this case.
If two straight lines that correspond to our equation are parallel, then the system has no solutions. If the lines coincide, then the system will have infinitely many solutions, and, finally, if the lines intersect, then the system has exactly one solution.
And this solution is the common data point of the two lines.
Thus, all three cases are implemented here.
Consider the following case, when we have a system consisting of two equations with three unknowns. Again, let me remind you that a linear equation with three unknowns sets a plane in geometric space, provided that at least one of the coefficients for variables is not equal to zero.
Let's discuss what cases are possible here, since we have two equations, then we have two planes. Let us indicate them by α and α1. Let's revise how two planes can relate to each other. They can be parallel — this means that these planes do not have common points. Then the system does not have any solutions.
The planes can coincide - in this case, any point obtained in the plane will be a solution - the system will have infinitely many solutions, and we will get two free variables in the corresponding stepwise system.
Further, two planes can intersect, and, as you know, planes can only intersect in a straight line, that is, if different planes have at least one common point, then they have a common straight line.
Again, we get the case when the system has infinitely many solutions, but in this case, if we bring this system to a stepwise form, we will already get one free variable; there are no other options here.
That is, the system of two equations with three variables cannot have one solution.
Finally, a more complex case is when we have a system with three equations that depend on three variables.
We have three planes.
Let's consider what the situation is in this case.
All three planes can coincide. In this situation, all three equations are equivalent to each other, so the system has infinitely many solutions, and we get two free variables in the corresponding stepwise system.
Now let's assume that there are two parallel planes out of these three ones. In this case, no matter how the third plane behaves, there will be no common points. If two planes are parallel, then there are no common solutions. In other words, the system is not compatible. Now let's assume that there are two planes that intersect.
The values will intersect in a straight line, then the third plane can intersect the other two planes in two other straight lines, and these straight lines can all be parallel to each other.
The situation is as follows. Three planes intersect in pairs, forming three straight lines that are parallel to each other.
In this case, again, there will be no solutions, i.e. there is not a single point that lies in these three planes at the same time, it means our system is not compatible.
Also, three planes can intersect in a straight line.
In this case, you get the following situation. Any point of this line is a solution of our system, i.e. the system of equations has infinitely many solutions.
In this case, there is one free variable in the corresponding stepwise system.
Finally, another option, when three lines of intersection of planes intersect at one point. In this case, this common point will be the only solution for this system.
Thus, as you can see, all three options are implemented for this example, when the system has no solutions, has infinitely many solutions, and has exactly one solution.