[BOŞ_SES] Hello.

In an earlier session, two and three three-pointers duality

We have achieved some rules of the matrix.

Of course this is a demonstration, not a proof.

But I wonder if these rules are also larger matrix calculations

Can we say we are questioning.

Science is already developing such.

Are you doing some simple observations then you go from there to the general rule.

An important physicist, Richard Feynman says in book

He says: how observing rules of nature,

how to make a generalization for sampling.

Suppose you do not know says chess, watch the chess game of two people.

Here you see how carefully advancing the pawn,

how advanced the horses, castles, the queen of elephants,

You will see the king, and here's how to proceed carefully

You will receive a general rule you look.

Then you've set these rules

After you have discovered how to play.

Here we are in fact doing it here.

We make observations on a simple matrix.

We have achieved a general rule of here.

Now we apply the matrix in this nyn size.

As the calculation chain is improving.

When given a matrix that we obtain lower matrix.

In the ith row and jth column of this matrix by removing Mij M

from a number of minor ones, to letters

As we put it in position j.

So we are taking before i and j Element, the pillars,

The columns and rows,

this item's location in the minus one, we calculate that the negative

We are writing after the determinant of matrix union.

These changes are making a mark.

This conjugate matrix, creating what we call cofactor matrix elements.

The total number of rows and columns of Mija

We do not change the mark of minus one double because you're getting a double force.

plus it is happening.

Aij'n sign with Mija stay the same.

It's the same value.

But the number of rows and columns, order of

If we change also gives the total number of single bi mark.

According to these rules by opening the line and column is as follows.

See the collection here, i fixed in the first collection,

I do not perform an operation on, we're doing on the jar for collection.

I will give the first indication of a matrix line.

So do not touch the line, we collect items within the line hit.

According to this line it is going expansion.

If we write here the same formula i i collection on given lines

We're collecting on the lines that we come to go onto columns.

j remains constant.

It gave j column that is going on by opening the jth column.

According ith row in the previous formula.

We know that they all give the same result.

Now before you go to that general rule show a sample.

Here matrix.

We took in opposing the matrix ij'y

We want to find the matrix and its determinants.

For this, we will take i'yinc line,

j'yinc we take the column and the remaining (n-1),

(n-1) we find the determinant of the matrix in size.

If you notice here in line i minus plus we missed a good one.

We took the ith line that meant.

Looking at the column j j minus one plus one we missed.

That column, we took a j'yinc column.

Now here is of course that it

Although we have begun a ten-matrix, here it will be nine to nine.

This eight-eight from nine nines,

With a range up to two binary finally end, we came up with a hierarchy.

We find the determinant of the matrix into two binary.

So this rule gives us such a calculation chain.

Now let's apply it onto a matrix verse four.

There are also many zeros as you can see in this matrix.

Because we were on the line, according to which according to the column if

get one of these numbers, we found that we hit the cofactor.

Determinant of a sub-matrix equal, changing the sign E,

If we do not have to reset them to the account of cofactors.

We will reset again because something hit zero.

I want to make here two sampling.

One, let's make a açııl according to the first line.

According to the third column.

In fact, according to the most favorable here would open the third column.

Because three does not mean that we should have zero calculation of these cofactors.

This should only one.

But let us get our hands compared to the first line she used.

Then we will do according to the third column.

We do according to the first line we are taking the first line.

That we are taking the first column.

Because it's the first item in the first column of the first row.

We are writing matrix remaining after three three-pointers.

[1,0,3] [1,0,1,], [1,0,3,],

[1,0,1] and [2,1,0], [2,1,0].

We are writing it.

M 1 we are taking the first row again for two.

Because it determines the first indicator line.

1 2 sub-matrix.

1 2 line first line and the second column where we put the second column.

We will write the remaining three three-pointers matrix.

Including three three-pointers matrix had left, the first row and second column

threw [0,0,3], [0,0,1], [0,0,3]

[0,0,1] and threw it back in the last row of the second column [1,1,0] is staying.

[1,1,0] 1 3 M in the same way we do.

We put the first line, we take the third column.

We write the rest.

And we're finally M 1 4.

So we wrote the first line after the fourth column.

[0,1,0], [0,1,0], [0,1,0], [0,1,0], [1,2,1], [1,2,1].

They will account their determinants.

Now let's get the first M 1 from 1.

There are two zero in the second column as you see here.

So it will be convenient for us to open by the second column.

According to the second column we open merger,

column is the column and row and no one threw the remaining two

binary matrix of the determinant of [1,1,1]

We once wrote that there are two minus negative three.

We need to identify it as the sign alone.

It was three in the third row and second column

The sum of the number of rows and columns that three plus two single

We need to find a replacement for point 1 1.

Here we find in 1 1.

M 1 2 in the first column again that there are two zeros,

first row, second column can be calculated by.

For example, we calculate according to which line also compute,

I think we led by example this third column.

See these two hit trilogy remaining duality.

Reset here one product,

Reset the product of one of the determinant because it has zero zero.

In opposing cofactor conjugate matrix elements to zero again.

Because we will change the zero mark, but will remain zero.

One, we look at three still see there are two zeros in the first column.

According to the favorable turn with the first column is full.

As you can see here, once one,

Three times three, minus one, minus two, we find.

We have also set the mark.

The third element of the first column to determine the mark,

thus pointing to change.

Axis, which will give our kofaktörü minus two.

M 1 in 4 have accounts here we see two reset again.

Please assume that we open according to the first column.

The first column zero future, the future of a zero cofactor future.

Cofactor zero to one.

We stood Birla zero, zero hits one of them.

That means this is zero.

So we turn to find the determinants compared to the first line,

vector in-first line,

I'm sorry that three of four elements [1 2 0 0] vector

We hit the cofactor vector opposing this line.

Including two zero two zero minus minus,

here are the results we have achieved in the time we hit them; reset

We hit minus one for that effect.

We stood reset the two.

Reset something we stood, we do not even think we will hit zero.

We hit something again reset.

You do not need both of us to think negative here but there will be zero.

Results We find in this way.

Yet another application than the

Let's open the third column because according to the third column

There are three zero.

But before that her account of what we are doing

Let's see how we do it a little more practical application.

According to the first line will open.

We write the numbers on the first line.

[1 0 3], [1 0 1],

[2 1 0] we write the matrix with him three three-pointers.

After two but it will change the sign of the two.

We are writing to minus two.

Minus the row and column where we take two.

Back [0 0 3], [0 0 1], [1 1 0] remains.

We are writing these things take.

After reset, the sub matrix are writing it.

And again in this last zero, we need to change in the fourth zero-mark.

Plus sign, minus, plus, minus comes.

What if the yok.çünk minus reset but still need to do this one account

You get cofactor corresponding falls to reset to multiply.

Therefore, these last two terms is reduced.

If we open again this genus from the first term of almost cofactors

According to conveniently open the second column.

The second column before [0 0] will come at the end of the first.

Let's look at one marker; the second column of the third row.

A total of five rows and columns.

Therefore, we change the sign because it is an odd number.

The remaining sub-matrix [1 3 1 1].

We are writing here after.

Minus two is also still plenty reset

which can turn relative to the first or second column.

Suppose we turn according to the first column.

Back [0.3] [0 1] is staying.

We write it after.

The other two are writing after them was zero.

Just now able to calculate the determinant here.

1 times 1 is equal to 1.

Minus three minus two at the beginning of a minus 1 plus two coming here.

The second determinant is zero.

Because times 1 minus 0 0 3 times.

Therefore, only two remained.

So minus the product of the merger DETEMINANTS take 1 minus 2 gives us 3.

This is a fast write everything together.

In these applications without having to write separate grains cofactors this

done so.

But it would be open to the most appropriate third column in this example.

When we opened it zero times compared to the third column of

We hit the bottom matrix, zero.

Once again zero sub-matrices.

Needless to zero sub-matrix calculation.

Sub-matrix corresponding to zero once again.

To reset it, you do not need to calculate because it will be zero.

Only one remains.

The third column we need to determine the sign of one

fourth row by row and column order of

As the only number we collect will change the sign of four plus three seven.

We are changing the sign of one of these.

When left at the third column [1 2 0] [0 1 3]

[1 2 0] [0 1 3] and [0 1 0] remains.

Sorry [0 1 1] is staying because we took the third column.

Now, we can now easily account here.

This first line, according to a first column in-ve

One of the first column by the time we opened [1

3 1 1] is calculated by multiplying the sub-matrices.

You have to look at the sign.

In this one, because it is the first line of the first column 1 plus 1, 2.

Row row row plus column.

So mark that will not change.

This minus one standing here again we hit it

as when one gives us two of minus 3.

So as you can see here, we have provided more than a few things.

We apply our rules once in a matrix verse four.

However, rules that we have achieved and we have two binary matrix of three three-pointers.

We see that here also apply.

Again, this is a proof to show that, but that's not the way to develop

to show for and technique.

If they were even larger matrices four quatrains

Even if we could calculate so many zeros, but not so easy

here also we see that zero, zero, or with plenty of milk

To select lines gives us easy calculation.

We can not calculate by hand, of course, very large matrices, determinants.

It is effective and suitable software for this purpose.

But you have to know what you did to be able to use this software.

You need to learn these theories.

In some cases, the computer may need to develop your own software.

It belongs to a private matter rather than a special general software

You may need to develop software.

Then the work of these rules even more,

It is obtained by applying the rules we developed thereafter.

Now we take a break.

In the meantime, we'll go to more generalizations.

Find the basic properties of determinants,

to find the same basic features such as a chess player from their eyes