Математика. Материалы курса

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This practical lesson dedicated systems of linear equations.

In this lesson, we will try to work out the Gauss method, ie a method of successive elimination of variables with the help of elementary transformations.

Consider the following system.

Let me remind you that the lecture had considered various examples of systems and simple and complicated.

Now in practice as soon as we take a complicated system

There are four variables and four equations.

So, we try to work out a common algorithm.

As I said, we must first method of elementary transformations to simplify the system, namely, to bring it to echelon form.

To do this, we first write the matrix of coefficients, to escape from the unknown symbols, and then we convert it.

Let's do it carefully, paying attention that some equations do not contain all the variables.

What does it mean?

This means that if other variables zero coefficient.

So go into mass number zero.

So, part of the first line of the expanded matrix 7, -2, I will sign the variables, make it clear why we recorded a toe, because before we factor x3 0 before x4 - 1 and the free term - 2.

Next, the second row - 0, 1, 0, 1, 1.

I remind you that the last column is a column composed of the free members.

Further, 0, 3, -2, 0, 6.

Finally, the fourth line - 2, 2, 0, 5, 1.

So, here's this turned out matrix, which is called the augmented matrix is ​​because we have allowed the column of free members.

Now we bring to echelon form.

We need to under the seven were some toe two zeros already have, but dvoechka to hinder us.

What is wrong?

The fact is that in order to get a toe, we need to subtract the first two-sevenths of the fourth line

And then we will have a minus 2 2/7 * 7, that is 2 minus 2 - this will be 0.

But there are fractions that is not very convenient.

The lecture was such an example, and there we made the following transformation.

We aligned the coefficients of these equations, one can multiply the first 2, the latest 7.

Turn here and there to 14.

However, you can apply a different reception.

Yedinichku can be obtained by subtracting three-fourths of the first row.

Here are 7 minus 6, ie the unit.

I propose to make this conversion. is subtracted from the first row of three quarters.

So, I'll sign it, we subtract the three quarters.

It is a matrix:
1 -2-6 = -8, 0, 1-15 = -14, 2-3 = -1;
0, 1, 0, 1, 1;
0, 3, -2,0, 6;
2, 2, 0, 5, 1

So here are necessary to us edinichka.

Well, now we subtract from the fourth row of the first two.

We get the equivalent matrix.

Rewriting the first row.

Them we had no change.

Now we begin subtract 0, 2 + 16 = 18, then 0, 5 + 28 = 33, 1 + 22 = 23.

So, here are the leading element in the first row - edinichka.

In the second, too edinichka.

Achieve zero.

Here, everything is easy.

Immediately a third row subtract three second (minus three second), and subtracting from the last eighteen second.

A large number, but in the second row are the prime numbers so do not worry.

And keep in mind for the future, we could check the last line inside.

Then have a bit simpler coefficients.

so
1, -8, 0, -14, -1;
0, 1, 0, 1, 1;

Now subtract
0, 0, -2, -3, 6-3 = 3;

and the last line:
0, 0, and then think - here, too, 0, 33-18 = 15, if I'm not mistaken, 3-18 = -15.

Wonderful.

Let's look at the system.

Here are the key elements of

Every next leading element is located to the right of the previous one ..

We have stepped matrix.

And the last line can be divided by 15.

I'm not going to rewrite the matrix now, and immediately write down the system of equations:
x1-8 * x2-14 * x4 (x3 because when standing toe) = -1

Thus, the second equation:
x2 + x4.

At first, when it is difficult, you can write variables above the columns, not to be mistaken.

The third equation:
-2 * x3 - this is our third column
-2 * x3 - 3 * x4 = 3.

And finally, the last equation.

And I devide by 15.

I receive the line 1, -1, i.e .:
x4 = -1.

So, we have completed the first step.

The original system transformation method was changed to echelon form.

Now the second stage.

Now we begin to exclude consistently the unknown from the last equation, gradually rising higher and higher.

x4 is already expressed. I’ll circle it.

So, now what are we doing?

Now we put x4 into the second equation and get
-2 * x3 + 3 = 3.

This shows that x3 is 0.

Next, we substitute the values ​​found in the second equation.

So, now we go to the second equation and express from it x2.

Substitutes x4 with -1. x2 -1 = 1, then x2 = 2.

So we are moving from the bottom up.

Well, now we substitute in the first equation.

I will record it below
x1-8*2 (т.е.-16) -14*(-1) (т.е. +14) = -1:

x1-16+14 = -1.

Hence we express x1.

So, we get on the left -2, move to the right side - +2.
-1 = 1 + 2.

All variables are found.

So, this system has a unique solution.

Write this solution as an ordered qadruple (1, 2, 0, -1).

So the system is solved.

Now I propose to consider a problem where there is again a parameter.

We do not need to solve it completely.

The main thing is to understand, what values ​​of parameter b the given system is consistent, that is, it has at least one solution.

In this system there are four equations and three unknowns.

We proceed according to the same algorithm.

First we bring the matrix to an echelon form and then remember a theorem, which we discussed at the lecture.

It will allow us to understand, in what case the system will be consistent, and when it is inconsistent.

So, we build an extended coefficient matrix.

There will be four rows and four columns, accounting for the column of constant terms.
1, 1, -3, -1

2, 1, -2, 1

1, 1, 1, 8

1, 2, -3, b

We bring it to the echelon form.

From the second we subtract the first two, from the third one - the first one, from the fourth one  - also the first one.

Here we already have one, it is very convenient.

We get an equivalent system, more clearly equivlent matrix.

Thus, we subtract, and obtain the following elements0, -1, -2+6= 4, 1+2=3.

The next line - 0, 0, 4, 9.

Finally, the fourth line - 0, 1, 0 (because 3 + -3), and here b + 1. b - is our parameter, some fixed number, but we do not know it.

Thus, the pivot of the first row is 1, the pivot of the second row is -1.

We try to get zeros under this element.

To do this, we add the second line to the fourth one.

Again, we obtain a matrix of the following form:
1, 1, -3, -1 (thus, the first three rows, we rewrite unchanged)
0, -1, 4, 3
0, 0, 4, 9

Now calculate:
0, 1-1 = 0, 0 + 4 = 4, and finally, b + 1 + 3 = b + 4.

So what’s next?

Now we have to do something to have 0 below number 4

And it is necessary to subtract the third line from the fourth line.

Well?

We almost received the echelon system

However, due to the parameter we will have to do some research.

So, lines we copied the first three without changes:
1, 1, -3, -1
0, -1, 4, 3
0, 0, 4, 9

The fourth one is changed, accounting for our transformation:
0, 0, 0, b + 4-5 = b-5.

Well, almost an echelon system.

Why almost?

Well, because if b is equal to 5, we have a  zero line, and we must delete it.

The rest of the matrix is ​​echelon.

And if b is not equal to 5, we get an echelon system.

Let us remember that we are looking for.

We are looking for value b, for the system to be cosistent.

But the system is consistent if and only if there is no contradictory equation in the relevant echelon system.

Obviously, the first three levels are consistent, but the last can be controversial if some condition is carried out.

Let us understand what this condition is.

To do this, the last line is written in the form of an equation.

So,
0 (I will write in detail):
0 * x1 + 0 * x2 + 0 * x3 = b-5.

When the equation is contradictory?

When b is equal to 5, to be exact, if b is equal to 5, then the equation will be zero, and if b is not equal to 5, then it will be contradictory.

Thus, we conclude that if b is equal to 5 (I will indicate this equation by an asterisk, through an asterisk) then an asterisk is zero

So, let us note, if b = 5, then an asterisk - zero.

It can be deleted.

The system will have a solution, and only one, but the main thing that it is consistent.

But if b is not equal to five, then the equation is contradictory.

So the system is inconsistent.

It has an empty set of solutions.

Thus, we write in the answer b = 5, with this value and only with this value the system is cosistent.