Математика. Материалы курса

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We’ll have another practical lesson on systems of linear equations.

At this session, we’ll look at an example - a system that has infinitely many solutions.

At the last session, we considered in detail the example of a system that has only one solution.

So, the following system of linear equations is given.

Note, there are three equations and four variables in this system.

Therefore, we can say that there are two possible cases.

Either  the system is inconsistent if we have an inconsistent equation in the echelon  system.

And if there is no contradictory equation, then the number of equations is certainly less than the variables, and therefore in this case there are infinitely many solutions.

That is, it is impossiblr to get the only solution here.

So, let’s remember what we are doing

First we write the coefficient matrix and reduce this matrix to echelon form.

I will immediately write the matrix, swapping the first and third equations.

What for?

In order to get one in the first place.

We have repeatedly seen that one helps us to perform the transformation easily to get zeros below the leading element.

So, in the first line, I'll write down coefficients of the last equation:
1 3 -2 1 7

The second equation is left without changes:
2 1 -2 1 4

And in the last place I write down an equation that was the first:
3 -1 -2 1 1

Now we subtract the first two lines from the second line.

the first three - from the third -.

We get the following matrix.

The first line does not change:
1 3 -2 1 7

We calculate the second line:
0 1-6 = -5 -2 + 4 = 2 1-2 = -1 4-14 = -10

The third line:
0 next -1-9 = -10 -2 + 6 = 4 1-3 = -2 1-21 = -20

You can, of course, simplify the last equation by dividing all coefficients by 2.

But we do not need to do that.

We can immediately subtract two second lines from the third one because I substract -5 * (- 2) and get + 10, and below the leadibg element -5 we can see zero.

If this does not work immediately as I have said, it is possible to simplify the line, reducing them by a common factor.

So, to the third line, or rather subtract two second lines from the third row.

Check once again to see if there is zero -10 to + 10 eqauls 0.

The first line does not change:
1 3 -2 1 7

The second line, either.

You can leave it.

I just thought that there are a lot of minuses.

You can multiply by -1, but this is too much, not much will be simplified.

So, leave the second line:
0 -5 2 -1 -10

And calculate the third one:
0 0 4-4 = 0 -2 + 2 = 0 -20 + 20 = 0

What do we see?

Someone may immediately notice that the last line is proportional to the second one.

Thus there arose a zero line, which can be deleted, because it corresponds to zero equation.

Thet’s all.

In the sense that the echelon system is without a zero line.

So, you can return to the system and write down its decision
x1 + 3 * x2-2 * x3 + x4 = 7
-5 * x2 + 2 * x3-x4 = -10

I’ll circle the variables that corresponds to the leading elements.

It is these variables that can be considered  the main

Then the remaining variables will be free.

That is, they can be used to express all of the main ones.

Let's do it.

So x3, x4 are free.

From the last equation we express x2.

To avoid mistakes, we’re going to do this in detail.

First, I take off -5 * x2.

Transfer the remaining part on the right side.
-5 * x2 = -2 * x3 + x4-10

Next divide by -5.

We get
x2 is equal to (2/5) by x3 (yes, there are fractions, but keep in mind that we do not always obtain integers somewhere you have to work with fractions) further, minus 1/5  by x4 plus 2.

x2 = (2/5) * x3- (1/5) * x4 + 2

Now, from the first equation we express variable x1:

7 minus 3 x2, and x2 is already expressed, I substituting (2/5) * x3- (1/5) * x4 + 2.

This was x2, we immediately expressed it from the previous equations, to be more precise we placed it.

Further plus 2 * x3, because we transferred it to the right side.

Plus 2 * x3 minus x4.

And do not forget about a seven, which is located on the right side.

And I have already written it out, it is not necessary, all right.

So seven have already been recorded.

Well, now we accurately convert.

So, not to be mistaken I’ll rewrite.

Multiply three by each summand  in the brackets:
7-6 / 5 * x3 + 3/5 * x4-6 + 2 * x3-x4

Well, now we deduce the similar .

Thus 7 and minus 6 is 1

Further x3: -6 / 5 + 2 is +4/5 - the coefficient of x3.

Finally, when x4: 3 / 5-1 = -2 / 5, i.e. -2/5 * x4.
So, x1 is found, or rather the main variables x1 and x2 we expressed by two free variables x3, x4.

Now, to write the answer, let us give free variables arbitrary values.

Let x3 be c, x4 - d. c and d are arbitrary numbers.

In this case the answer can be written as follows.

More precisely, we write the solution in the form of a quadruple :

The first coordinate - 1 + 4/5 * s-2/5 * d

The second coordinate - x2 - 2/5 * s-1/5 * d + 2

next x3 - this is c
x4 is d.
(1 + 4/5 * s-2/5 * d; 2/5 * s-1/5 * d + 2; c; d)

Thus, the solution is a quadruple of the  indicated form (1 + 4/5 * s-2/5 * d; 2/5 * s-1/5 * d + 2; c; d).

And again c and d are arbitrary numbers, we get an infinite number of solutions.

That is, many solutions are the set of all such quadruples, where c and d are arbitrary numbers.

I now propose to consider the task where the parameter is present.

And depending on this parameter, we will get different solutions.

For some parameter values there will be no solutions, for some there is only one, but in some cases there will be infinite number.

That is, one such system will give us different solution sets.

So, where do we start?

Let's start with the same.

Let’s bring the matrix to echelon form.

The only problem - we do not know what k is, that is, k is some fixed number, but it can be any.

Therefore, at some point, we’ll again have to consider different cases.

But the first step is standard.

Write a matrix of factors:
1 1 k 1
1 k 1 1
K 1 (2-k) 1

Thus, subtract the first line from the second one,  subtract first k from the third line, where k - is a number.

For example, if k is equal to three, we subtract the first three.

But even if k = 0, our transformation does not change the solution set.

Just if k = 0, this means that we do not subtract anything, save the equation.

So, following these transformations we obtain the following matrix:
1 1 k 1

I'm doing the  interval greater, because due to the parameter our expressions will be longer. The second line:
0 k-1 Jan.-k 0

The third line:
0 1 2-k-k-k ^ 2 (and the last element - 1-k, I’ll shift the last column to the right) 1-k

Please note, the main elements of the last lines are opposite.

So if we add the second line to the third line, we get the required zero.

Let us make this transformation:
1 1 k 1

Further
0 k-1 Jan.-k 0

We add:
0 0 (here we calculate carefully) 2 + 1 is 3, -k + (- k) = - 2 * k and minus k ^ 2, and finally 1-k + 0 = 1-k
0 0 3-2 * k-k ^-k 1

Well, almost a echelon matrix.

Why may nuances occur?

Due to the fact that if for some k some factor turns zero, we get perhaps a null string.

Let's examine this case.

We understand when factor (3-2 * k-k ^ 2) is zero.

Such a complex expression.

However, it is a square trinomial.

To find its roots, you can apply the discriminant formula.

I multiply both sides by minus one to be written in a beautiful form, to have a  quadratic equation.

Then you can calculate through discriminant, and if anyone knows how to use Vieta formula, you can select roots.

In general, there is a variety of ways.

Let us do it  through discriminant. 4 + 12 = 16.

I will not explain the discriminant.

Who has forgotten how to solve a quadratic equation - remember.

Then the roots are -2 plus / minus 4 in half.

With plus we get one, with  minus this is -3.

I wrote briefly, sketchy.

Thus, the roots are  1, -3.

Also, they could be selected by Vieta formula.

So, now I propose to begin by considering these values.

Let's look at the first option, where k is equal to one.

That is, analyze the case when this one element is leading, i.e., the first non-zero.

Although at some k, it of course will be zero when this element turns to zero.

Let's explore this option.

What are we doing now?

Just substitute yedinichku instead of k:
1 1 1 1

0 0 0 0 and see the second line zeroed out, and the third line is:

0 0 0 0 - everything is zeroed out, that is, our system is equivalent to a single equation - equation of the form x1 + x2 + x3 = 1.

x1 - the main term, x2, x3 are  free.

We express the  main term through free ones and get a set of solutions.

And the set is endless.

Again x2 - is t, x3 - is d, where c and d - are arbitrary numbers.

Then, in this case the solution set is of the following form - a set of triples (1-c-d, c, d) - all kinds of triples of this type.

So, k is equal to one, the system has infinitely many solutions of this type.

So the first line is this.

By the way, let's note that we have considered the first case of k = 1,.

Now we’ll study other values of k.

Therefore, the second line can be reduced by (k-1).

If we had done it at once, we would have simplified it.

Who will see it.

So, we have k-3.

See, it turns -4, and here - 4 .-4 and 4, the fourth is equal to zero.

Finally, the last line 0 0 0, and here 1 + 3 = 4.

So, what do we see?

The last line in the equations language gives us a contradictory equation, i.e.  the equation of the form 0 = 4.

Again there are conflicting equations,  system is not cosistent.

We considered the case where k = -3.

Thus, the system is not consistent, its solution set is empty.

And finally, the third option, where k is not equal to 1 and k is not equal to -3.

In this case, we divide the second line by (k-1), we get: 1 -1.

Thus, the first line is
1 1 k 1

The second line. So I’ill write in detail. Let's write in detail:
0 k-1 1-k 0

And finally, the third line:
0 0 3-2*k-k^2 1-k

And among the leading elements there are no  zero ones.

In this case, the system has only one solution. Whicht?

Let's write down the resulting echelon system.
x1 + x2 + k * x3 = 1

Yes, as I said, the second equation is divided by k-1, given that k is not equal to one. We get x2-x3 = 0

And the last equation
(3-2*k-k^2)*x3=1-k

Now we already have all the transformations, so the answer is at hand (a general one).

What do we do now?

Now we will express x3.

Then we substitute and express x2, and so on.

But in fact, here there are bulky transformations.

But it only seems so.

In fact, everything is simplified.

The fact is that the quadratic trinomial with known roots is easily decomposed into factors, which is known from school mathematics.

Senior factor -1, so we do not forget minus.

And then, given that roots 1 and 3, we get  (k-1) * (k + 3).

Remember school mathematics, how we got such a product.

Well, then again, a common factor, which can be reduced.

Thus we get  x3 - is (1-k) divided by (1-k) * (k + 3) (introduce minusin  the first factor).

Reduce  x3 = 1 / (k + 3) A beautiful expression.

We get x2 from the second equation

It is equal to x3, that is, the same value.

Well, finally, we express x1 from the first equation:
X1 = 1-x2-k * x3 =

substitute:
= 1-1 / (k + 3) -k / (k + 3).

We reduce to a common denominator and calculate.

Problems with parameters are always complicated by the fact that here we have to carry out transformations with the letters carefully. k is deleted. 3-1 = 2.
2 / (k + 3).

A

That’s all. The solution is found. You can write down the answer.

So, if k is equal to one, the system has infinitely many solutions (I'm not going to rewrite this set, I can only show it to you again. That's it).

If k is equal to -3, the system is not cosistent – there is an empty set of solutions.

And finally, with the rest of k, that is when the k, is unequal or 1 or -3, the system has only one solution of the form: the first coordinate: 2 / (k + 3) and the other two coordinates - 1 / (k + 3) 1 / (k + 3).

So, only one triple (2 / (k + 3), 1 / (k + 3), 1 / (k + 3))

Here's an example.

Quite complex, but it provides all the options that may occur with the system of linear equations.

In this series of sessions  on this topic is over.

Thank you for attention!