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[BOŞ_SES] Hello.

In our preceding session, four properties have proved,

We have identified four traits related determinants using the formula.

We say we're taking basic features.

In short, let us remember them.

We show matrices as follows: one of the first line,

in the second row two, three in the third row, Ajman to j'nc line,

n'nc line so that the n-component vectors each one of them.

We've seen this formula, wherein the opening formula,

Laplace expansion formula, using sub-matrices,

We have proven this formula and using just these four characteristics.

One Uniformity, homogeneous property; A line t

We like going in was hit by t hit the determinant.

A vector of any one row is similarly

If we add the sum of the determinants it was happening.

If its zero line of a matrix

We can easily calculate the determinant is zero.

And we saw that a unit of the matrix determinant.

We especially need to remember,

because it features four main features.

Now we Acknowledging the basis of these four characteristics

We can prove many more features.

Eight of them will see the cash in advance.

You will recall means of a matrix transpose

Coming opposed to changing the column with the line.

Determinant of the inverse of a matrix

equal to the determinant itself.

We can understand it very easily,

As opposed to this line because it changes the column,

determinant, for example according to the first column of the transpose

According to the first row of the matrix to calculate

equivalent to calculate the determinant, invariant determinants for him.

Here again, a simple calculation without the hassle

To see an example of one, two, three, four Let the matrix.

Four minus six of its determinants; We know that minus two.

On a diagonal line as you change and you will have four columns of the same,

UC will replace two places.

We see that change in the determinants of value when we do this,

Of course we have done over a two binary matrix, but this is generally true.

In a subsequent feature; two lines are equal in a matrix,

So let's say in the AI i-th row is equal to the number Aida j'nc line,

It's easy to prove that it is zero determinant.

A row of this determinant zero

but that is not an arbitrary line in the line of j'nc,

i.e. ji'nc row of which is equal to the AI value,

is zero, let's assume that the zero vector.

Our two was saying this; a matrix

If you add a line to a vector sum of the two determinants it was happening.

As you can see on the left we have the determinants det.

Here is the i-th row when we collect them becomes twice,

but j'nc line remains as it is for the zero vector of the future.

Now almost two special here, let's use the feature.

If we multiply the determinant of any line with a number t

it means that we are beating the count.

A line like you see here, which is two to

twice unchanging determinant of a matrix.

But here in the matrix of the matrix appears we received initially,

therefore here it has a determinant.

Therefore, the determinant is equal to two times the determinant.

This is only possible with the determinant is zero.

If you take this remains zero or left side of the determinant if you like,

The right of a determinant than two determinants çıkarınca only

is equal to zero prior to the determinant is obtained as a determinant.

The first line of our example one or two,

The second line is the same as you see happening zero determinant.

This is a huge matrix though any two lines equal to each other

provide that a simple numerical determinant be zero.

As a feature,

follows the immediately preceding and similarly again these four basic

It is obtained using the feature.

J'nc the ith row to row in a matrix is changed if we

we also have a determinant here, where there are Aja,

ie in the i-th row, Ajda j'nc line.

If we change the order in which the i-th row vector Aja income,

j'nc the line in the income.

Now how we do it, how us show that?

Let's start with something like this, Ajman plus of a line,

I think that there are still a plus one other line of Ajman.

Due to a previous feature theorem might say

because they are getting proven features, the two lines are equal.

E is a matrix of two rows of the determinant is equal to zero

We saw immediately preceding condition.

Now here's two feature

Let's open the totals using here.

We added another vector to a row of a matrix

determinant of the first vector matrix plus added time.

Here likewise Ai'li of

Aj'li stays and stays.

So here when we opened the one to the one,

Ajman in the second to third,

Combining these four determinants we obtain Aj in öbürkü.

Both of them are not going away.

See two lines of a matrix determinant is equal to zero,

again proving that fell from the previous one.

There where there's Aja, the determinant A.

There is Aja here Ai'yl changed to j,

There are two lines here in Ajman ones.

This zero, which is now here

zero, zero.

They fall after a summer rest and Herzegovina

Ajman line to the other line,

where Aj is the i-th row,

j'nc in line with, so we've replaced the lines of i and j.

Determinant in this fashion are zero.

This theorem proves that initially,

If you put one on the other side it seems to be negative determinants.

Again, a numerical example let one, two,

three, four Let's change the location of the line matrix,

three or four line up, coming down the two lines.

Here we find that the determinant minus two,

here are six minus four; plus have two,

we are proving here that feature work,

we provide.

If we add a line to hit with a number of other determinants unchanged.

So our determinants, matrix is the ith and Ajman and get j'nc rows,

others did not do them separately open always the same,

because they do not perform an operation on.

If we add the i-th row j'nc t'yl hit a line we get it.

j'nc we hit the line with a t, we've added the i-th row.

Now let's open this determinant.

According to a second feature, namely another in a line

The first plus second determinant determinant when we added vector.

There will be times Aija t.

To use this feature we take out the first in the t'li

As you can see two lines here turned into a matrix The determinant is the same.

Of course, this is zero, we prove that the property first.

If two identical zero line.

So here is zero when it falls only staying here a determinant.

Thus, at the beginning of the same determinant jth

According to the state line to hit t ith line we added.

Example; again our one, two, three, four [1 2 3

4] hit trilogy which will add to our second line matrix.

The second line was three and four.

It was here we had a nine attaching them.

We hit it three to twelve.

We've added twelve of fourteen was also doubled.

See, when we calculate it at minus forty forty-two

minus two and we see that as a determinant of change.

[1 2 3 4] that was giving us the usual minus two determinants.

Now we are going to special matrix.

Consider the diagonal matrix.

A1 diagonal matrix to separate from the others, A2, A3 were saying.

To that line D1, D2, DJ, let Dn.

Any DJ as follows: There are a number djj only on the diagonal.

Zero numbers in other locations.

This vector can be written as follows: djj'y unit vector

You will recall that we said we hit -l unit vectors in the IJ.

In the first of four feature I1,

I2 I posed the unit matrix.

Any line in the last jth line in the lj

There are a number of other locations in the jth zero.

Now when we do it here DJ.

Now let's use just the first feature.

Each line has the same structure we hit the d11'l unit vector.

But the unit vector of the n-th unit vector in the second row with D22

unit vector in line with dnner we hit.

If we remember this first feature for each d11

If we take out it is multiplied by the determinant d11'l.

If we take out the d22'y with them all the time we made the determinants D22

non-zero numbers coming forward, the unit vector in the matrix, a unit vector,

unit vector, but that means the unit stays unit vector matrix.

We know that there is a determinant of it.

Therefore, the determinant of a diagonal matrix

it is calculated by multiplying the number on the diagonal.

As you can see this visually explain if we take any line of DJs,

jth line all the coefficients are zero.

There are others just djj zero on the diagonal.

This diagonal matrix.

Not just the numbers on the diagonal zero.

How many of them can also be a chance to zero.

but also it needs to be non-zero numbers.

After all, all would be zero if the zero matrix.

Here is the determinant, such a diagonal matrix

determinant of the product of the numbers on the diagonal.

If we repeat the visual evidence here again

Taking out a unit vector d11 stay here.

Taking out a unit vector d22'y stay again.

The same operation on the challenge for all of these diagonal lines

numbers will come out.

Of the matrix [1 0 0]

[0 1 0 0] [0 0 0 0 1] as

reveal the identity matrix.

In this way the volume of the determinant of a matrix

We know that the fourth of the four basic features.

Therefore, we would be proved.

Triangular matrices have a very similar situation.

Determinant of the triangular matrix

Coming again equal to the product of the numbers on the diagonal.

We can very easily visualize.

Let's open this determinant by the first column.

This sub-matrix remaining time t11.

Others will be nullified whatever other lower because there will be zero

which nullified the determinant of the matrix will fall for them.

As you can see back again we had a triangular matrix.

We stayed determinants.

If we repeat this for all the numbers on the diagonal,

We see that the triangle on the diagonal matrix multiplication.

Here that is not a contribution of non-zero number.

As an example let; [1 2 3

4 5 6] from the diagonal matrix consisting think.

This diagonal matrix determinate according to first column

Once we calculate backward [4 5 0 6] remains.

He calculated that by four times in six.

one multiplied four times to six times.

He is not nothing but a product of the numbers on the diagonal.

Here we give a property as unproved.

It's quite a bit more to prove to occupy him,

but also an important feature.

(A, B] matrix can take the hit for it determinate.

We get to be a determinant and B determinate separately.

Determinants of the determinants of this product

We see that product.

This let to provide a numerical.

The first matrix is [1 2 3 4] indissoluble,

the second matrix is [1 -1 1 1] Let the product obtained in a matrix consisting of the AB.

We'll take this first column as you see in EU product horizontal

We will turn [1 2] will multiply.

We collect three.

Likewise, the first column of the second horizontal turning gene

line and hit the bar when we gather in 3 plus 4, was 7.

Similarly, the [-1 1], the second column of the resulting horizontal turn

eksib hit the minus one to plus two bath.Location had a three plus four.

Now (A, B) we know the determinant minus four minus two equals six.

Here's a determinant, but it is having a negative

It changes the sign translates into a plus two.

Both the determinant of matrix multiplication

If we calculate three minus seven, minus four oluyor.gerçek

We also know that the determinants of minus two.

We know that there are two determinants of the B's.

When we multiply these two minus four as easily be proven.

Using a previous feature, we compute in the following feature.

The inverse of a matrix, we do not know how to calculate the matrix

but when multiplied by the inverse matrix that their unit

and minus the matrix in which we combine, we say to the contrary.

Also this means is that the symmetrically opposite one to the minus.

Now we move through it; Here

A and B can look like the product of a matrix.

B minus minus being the first force.

We know that the determinant of a matrix.

This was one of the four basic features as the fourth feature.

Here the previous theorem, feature let's apply.

The determinant of the product,

the product of the individual determinant,

Now that B wherein B is a random, not vice versa

minus one equals that it is time we put the unit matrix.

Here's the obvious.

Determinant of a matrix unit.

So, with the inverse of the determinant 1 determinants giving.

So, the two are opposite each other.

We see, what we find here.

1 reverse split that divided the determinants of the determinants inverse matrix.

You can prove this in many places.

Now we want to be a break here.

We calculate twelve feature.

Doing what?

We went up here

We prove this by using the expansion formula to these cofactors.

Based on this, we calculate 12 feature.

The latter is the matrix with the inverse of each other

shows that the inverse of the determinant.

Now then we could stop here.

But there is also a determinant definition with a proposition like this.

It took four basic features,

these hypotheses, the basic premise,

axiom, we can define the determinant as postulated by assumptions.

In our next session, if ever determinant

What we do not recognize us as mere possibility topic

whether it will actually assume as a given circle.

We found these cofactors expansion formula.

From there we demonstrate these features.

Using them as a Laplace formula, the formula will account determinants.

Any one of these is enough to calculate the others.

So which of these determinants by cofactors expansion

We calculate the propositions already, but we have proved that giving up,

we were able to forget all about it in the beginning as an abstract.

What is the determinant of a matrix we asked,

When we make the following definition.

A determinant is a function that is a function of the coefficient.

Current features a function that he could count on these four basic features.

Characterized in uniformity; a line means to multiply the means to multiply the determinant t.

It means to add a car to a line, then add the two determinants.

If a line is zero,

it determinant determinant of zero and unit matrices 1.

They önerlemer we accept the basic definition, all attributes are also removed.

Matrix calculated this definition

It proved to be the only determinant.

Here one can challenge the number of these operations.

Now we adjourn here.

Here we'll get a formula.

Determinant defined to be three different accounts, like I said.

First, we give the cofactor, the second just a little before this

by definition propose.

Now we're going to a third way.

We will calculate a formula.

We have to be mixed,

each of which is sufficient to define determinant itself.

Each of the others so that all three definitions is enough to identify determinants.

The other two can be removed using them.

Indeed, we have found a formula for the cofactors expansion.

It is also more open formula, we calculate

by opening a basis of four properties

hypothesis, axiom alternatively,

hypothesis is not true, axioms, and postulates the premise,

We define a function assuming basic premise.

So the determinant of a matrix element by a previous recognition

a function obtained.

Now we get a significant but using it to come to this formula

We get little opportunity to reflect on these issues.