0:00

[Boş_ses] Hello,

Until now, our work will add a new method.

Now you might say that we have already learned the determinant calculation,

Why do I need it.

I'll say a little bit more towards the end summarize the rationale for it, but in the big matrix,

When you get to the cofactor account the size of the hit in a matrix,

n cube so requires a multiplication operation.

But if we have seen in the Gauss-Jordan Yöndem n squared

It can be calculated as the process can be done with a hit.

Thus, preferred for large matrices in computer programs

They are the Gauss-Jordan elimination method.

Of course you can find in small matrix anyhow.

But this theory is important because the determinant of properties

We need to know.

We have already worked with the Gauss-Jordan elimination,

You know there is a purpose if you wanted to calculate the determinant,

Even if you want to solve the equations to calculate,

bring 1 on the diagonal, it is trying to bring under the 0's.

We are doing this with the following three works: a line to be multiplied by a number,

the first to propose that define the determinants,

Based on this premise, we prove the two properties.

Hit with a line number, the determinant does not change if we add to the other.

Here, hit the ith line numbers t, we've added the jth row.

Oh, by the general formula for these i and j that any random lines,

determinant does not change.

We also change the order of the two lines,

ith row of the jth line in here when we change the order

We see a change in the sign of the determinant.

They show us where we are, first, seventh,

found that results with the eighth feature.

These are called basic matrix operations.

Our goal is to bring 1 by applying them on the diagonal,

but also at a time if necessary, if we multiply it by a number,

If you need to change as well, here we are on the hit

multiplying the number of repeat determinant case we need to fix.

So let's see two examples.

3 of 3 one of a matrix.

Of course, this is simply because the matrix, especially because it also found 0

We can easily calculate the determinant of the cofactor method.

This one ourselves comparison

Let's ability to provide.

For the first column are 0, no matter which angle relative to the first column,

0 times 0 is located at the rows and columns you

After we get the determinant of 2 by 2 matrix remaining backward.

But is 0,

not worth the trouble to calculate the determinant, because the result will be 0 anyway.

Our mark of plus or minus 2 when we he was going,

We stood with minus 2.

2 found that the column and row cast, backwards, 2, 4, 0, staying 2.

2, 4, 0, 2 the 2 of 2

determinant of the matrix that will calculate the determinant of sub-matrices.

The third item of -1,

in the third row points to

+ Coming third row, first column,

3 + 1 = 4 because it will not change sign, we write -1.

-1 To the column and row threw in 2, 4, 6, 0 remains.

16 We find when we calculate them.

It calculates 2 by 2 matrix, we find by multiplying the numbers in front.

Now we hold it in our hands us a number for a comparison,

To be able to provide.

The accounts will make the Gaussian elimination.

Now is the time we begin with Gaussian elimination Just change

We need to do because we have to make diagonal 1, diagonal to get from 0 to 1

Let's change the second line to the first line where we can not get to her somehow.

The second landed in the first row,

The second line first appeared, we have changed our sign to this process,

We take it out to offset the mark.

So we have hit -1'l.

He also -1'l balancing multiplying again.

That to continue, we need to bring 1 onto the diagonal.

1 to bring this first line we have 2 compartment.

Therefore, we provide the balance by multiplying by 2.

2. We split, we hit 2'yl value did not change.

2 divided by 1, 3, 0.

The second line already had this one diagonal 0

there is also a -1 on the number.

The first line here,

If we add a third line to the first line 0 can bring.

Our goal is to bring to 0 under the diagonal.

The challenge of this process came to 0.

3 0 instead came up with this, we have added over 2 0.

It has not changed.

Now we will continue to do so,

first diagonal why our business is done, we will start with the second diagonal.

Again, our goal is to bring 1 on the diagonal.

This is easy, the second line 2 to divide, we multiply by 2.

-4 Means that we have changed the number on the front.

Numbers 1 and 2 occurred in this line.

We reached our goal, we bring on one diagonal.

We want to 0 below the diagonal,

We will add a second line to bring hit the -3'l 0.

When we do this we have added a hit -3'l arrived 0.

We hit 2 -3'l -6 arrived we add 2 -4.

As you can see, we have brought to the diagonal.

This diagonal matrix to the determinant of this triangle,

The product of the numbers on the diagonal, 1 times -4, -4.

At the beginning it means that there is a -4 -4 16 -4 times goes on.

We have already found the 16 with the cofactor method.

Our goal was already solid.

As you can see no difficulty in the process.

Gauss elimination method we learned earlier, on the diagonal 1

0. brought it under to reduce matrix triangle.

Now the second sample consists of a slightly larger matrix 4 4.

Now if you wanted to do with it cofactor, you'll see how many accounts.

Suppose we open by the first column.

There are 4 items in the first column.

3 of 3 for each will calculate the determinant of a matrix.

3 of 3 for 3, one for each 2 by 2 matrix will be.

So there are 2 of 2 2 collisions.

3'e carpacagız it 3 times.

4 to 12 times as many as 3 times 4 There will be bumps,

We also collect them.

Again, this is not something be done, but larger

Although a matrix, it can move more in this direction.

We can do exceptionally fast, whereas the same process with Gaussian elimination.

Our aim is still to start first with the bottom diagonal 0

bring; I'll do it quickly.

1 already on the first line.

-2'yl Hit the first line, we add a second line, we return 0.

If we add a third line to hit -3'l we return 0,

If we add the fourth line hit the -4'l 0 returns

with this subject with the other counts, -2, -3,

You need to add to -4'l hit.

These figures are obtained.

Now we have finished the first diagonal process.

We need to bring when we come to the second diagonal one.

This second line for -1'l We need to multiply.

So 1, 2, 7 is coming.

But to fix it, so we multiply the value of this -1'l

We stood in -1'l To change the determinant.

0 Now we bring the same effort again under the diagonal.

To do this, hit the second row with the third row 2, hit by 7

If we add the fourth line, we bring in will reset under the second diagonal.

Numbers they obtained by the challenge of this process.

As you can see here a minus four occurred in the third diagonal.

Therefore we split this former quadruples

We have changed that line the value of Determinant minus four times.

We need to keep the same negative determinant quadruples shock.

Coming at the beginning is a plus to minus four.

There was a negative one, and here.

Now we continue our goal.

We need to bring in third diagonal below zero, three lines for her

If we add a fourth line hit minus four to zero under one came.

And when we hit minus four to minus one it was four,

attaching was thirty-six to forty.

So because we have reached our goal in on a diagonal,

but that is not important in this last forty.

If you want to divide you hit your forties forty one here, one, remains.

Or, as we call it, we have already reduced the triangular matrix.

The product of the numbers on the diagonal gives the determinant.

Once a times a times forty, forty means that from here on the determinants of income,

There are also four front; We found one hundred sixty.

As you can see these calculations can be done quite quickly in a simple way.

Said the number of accounts where I brew

What would happen if I did show him with cofactors.

The main problem in the larger matrix of Laplace expansion

that is proportional to the number of transactions made at the opening cofactors cube.

The crossed able to show the size of a matrix.

You will find yourself making.

However, proportional to the square of Gaussian elimination.

Whether you think that the bin.

A thousand thousand square matrix by Gaussian elimination for the determinant,

so should the order process in a million

you'll have a billion transactions in order to cofactor.

So you put the computer a thousand times take more time

To calculate the cofactor.

We found the same result but a convenient method for application

the initiative is not clear Gauss, cofactors of opening.

However, more favorable Gaussian expansion for small matrices.

Theoretically at many places with little matrix, some theorems with three three-pointers matrix

is proven, some features are used in practice.

It is useful to know both respects.

Gaussian elimination when it is convenient for large matrices

but you can not make them by hand, the determinant of a matrix thousand thousand hands

They can not but Gaussian elimination method are used in computers.

Now, this is the method we have now found that up to one

As we do, we started the compilation; We started with two and three three-pointers binary matrix.

These two equations with two unknowns,

matrices from the three equations with three unknowns.

Here we met with a number that determines the unknown.

We also determinant of this number.

Because of the equation for determining mean something to the determin

determinant, we meet in terms of defining the determinant.

Here we examine the processes that occur when a sub-matrix,

There is also a cofactor that we have seen the two of them and still

Inspired by this formula, we reached two three three-pointers matrix.

This is not proof, but it is more unknown

We have seen that when we apply the equation valid.

Here we demonstrate a number of features.

D1 of the demonstration, D2, D3,

What we take as a premise in the eighties showed the D4'L page,

So in a sense make clear from the start, a physics equation,

As experience out of there, observing events

We found the cardinal rule of using an inductive method of rule.

I told you earlier, you also gözlese two people playing chess

You do not know quite how chess stones that no progress observed in the eye,

how the horses were going to pawn how it went,

Go to how the tower that you can see how it went elephants.

You can have a general rule here.

There are also rules about after these are the rules of chess.

As such we have made the process as well.

It's like we have done in the linear space.

We saw the first plane vectors, a set of rules here and it was removed,

We have to take them to the definition of linear space as propositions.

After that we remove the whole thing from there.

We did the same thing here.

And we calculated both properties using these propositions as we have also done a formula.

Here we learn the structure of the determinants of permutations with this icon.

We will take all possible multiplications, we take the product of all elements.

No product can not locate elements of the same row or column.

It also gives us a symbol of this permutation.

This time, the symbol of permutations gives zero either,

if we take the product to the matrix elements in the same row and column,

plus or minus a theory to us that we take as a

as it is brought to a formula that would be very useful.

The practical application of a multiplication of this n'l

By separating them on the minus one

get product in total, the equivalent of the cofactor is happening already.

And in this way we have completed the chain theoretically.

This basic calculation also the features we found

The method of the determinants or change, although changes in either a

We see that changing the cross or by the number sign.

Especially if we are making three triangular matrix.

Triangle means that all zeros below the diagonal matrix

matrix and its determinant is that the numbers on the diagonal

We previously demonstrated that the number of multiplication.

This complements our calculation tools.

Here are the questions.

I encourage you to do so.

Some of these cofactors, for finding sub-matrix and this

Calculate the determinant of the matrix method.

Duality in two three three-pointers and easy, you do the four quatrains

You have to start the way, but if nothing else can be done.

You can see how long it would be difficult.

This time with the same matrix structure triangle Gaussian elemer

I recommend bringing the calculation.

So learn topics

I hope you will ensure that he internalized.

After a summary here.

As we give each section behind.

There's sort of privilege of square matrix.

But in another matrix can be defined in a square matrix, rectangular matrix,

We offer features that can not be defined operations.

Here are two binary matrices and determinants of three three-pointers.

Here calculation, Laplace's calculation chain,

defining a matrix array calculation algorithm with a premise

and identifying a formula, but which, three are equivalent.

Starting with any others you can prove you can prove it.

We need to learn how to do that of course.

Not easy to always use this permutations symbol in the formula.

Even in three people may be surprised, but I recommend that you show an effort because

This permutation symbol is an icon for work elsewhere.

We summarize the main features of the determinant here again.

And other features that prove these areas.

That's it for this episode.

After that we come to an important betting on square matrix.

The inverse matrix, we can find the inverse matrix using determinants.

Thus becomes a determinant than the basic concepts.

It is possible to calculate the inverse of the matrix as an application.

Bye now.