Doğrusal cebir ikili dizinin ikincisi olan bu ders birinci derste verilen temel bilgilerin üzerine eklemeler yapılarak tamamen matris işlemleri ve uygulamalarını kapsamaktadır. Cebirsel denklem sistemleri, sonuçların tekilliği ve var olup olmadığı, determinantlar ve onların doğal olarak nasıl oluştuğu, öz değer problemleri ve onların matris fonksiyonlarına uygulanışı vb. konulara derste değinilmektedir. Ders gerçek yaşamdan gelen uygulamaları da tanıtmaya önem veren “içerikli yaklaşımla” tasarlanmıştır. Bölümler: Bölüm 1: Doğrusal Cebir I'in Özeti Bölüm 2: Kare Matrislerde Determinant Bölüm 3: Kare Matrislerin Tersi Bölüm 4: Kare Matrislerde Özdeğer Sorunu Bölüm 5: Matrislerin Köşegenleştirilmesi Bölüm 6: Matris Fonksiyonları Bölüm 7: Matrislerle Diferansiyel Denklem Takımları ----------- This second of the sequence of two courses builds on the fundamentals of the first course, is entirely on matrix algebra and applications. Specifically, the studies include systems of algebraic equations including the existence and uniqueness of solutions, determinants and how they arise naturally, eigenvalue problems with their applications to diagonalization and matrix functions. The course is designed in the same spirit as the first one with a “content based” emphasis, answering the “why” and “where“ of the topics, as much as the traditional “what” and “how” leading to “definitions” and “proofs”. Chapters: Chapter 1: Summary of Linear Algebra I Chapter 2: Determinant Chapter 3: Inverse of Square Matrices Chapter 4: Eigenvalue Problem in Square Matrices Chapter 5: Diagonalization of Matrices Chapter 6: Matrix Functions Chapter 7: Matrices and Systems of Differential Equations ----------- Kaynak: Attila Aşkar, “Doğrusal cebir”. Bu kitap dört ciltlik dizinin üçüncü cildidir. Dizinin diğer kitapları Cilt 1 “Tek değişkenli fonksiyonlarda türev ve entegral”, Cilt 2: "Çok değişkenli fonksiyonlarda türev ve entegral" ve Cilt 4: “Diferansiyel denklemler” dir. Source: Attila Aşkar, Linear Algebra, Volume 3 of the set of Vol1: Calculus of Single Variable Functions, Volume 2: Calculus of Multivariable Functions and Volume 4: Differential Equations.
2x2 ve 3x3 lü Matrislerden Öğrendiklerimiz / Observations in 2 x 2 and 3x3 Matrices
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Doğrusal Cebir II: Kare Matrisler, Hesaplama Yöntemleri ve Uygulamalar / Linear Algebra II: Square Matrices, Calculation Methods and Applications
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Kare Matrislerde Determinant / Determinant in Square Matrices
Conheça os instrutores
Attila AşkarProf. Dr.
Matematik Bölümü (Department of Mathematics)
BOŞ_KAYIT Hello.
In our previous chapters,
We talked about general issues in our little session.
This time we'll start from the very concrete issues.
Reviewed by these two equations in two unknowns and three equations with three unknowns
the matrix will make these observations from small-dimensional matrix
and how these observations as well as the concept of natural determinants of us
revealed that out of these two and three-dimensional square matrix and
how you get the public to show us the way.
Let two equations with two unknowns.
Coefficients a11 a12 given coefficients,
x1 x2 numbers b1 and b2 to be given in our unknown.
Two equations with two unknowns.
We know how to solve it since junior high school.
The solution method; from the equation
From these two equations such as x2 is not whether to reduce one unknown.
Screening method.
Multiply the second equation by a12 to do it.
First multiply with the a22.
When we get hit with it like we did here a22.
We get it when we hit the a12.
Coefficients occurring before the X2 As you can see we had the same interests.
They hit them with what's already thinking about them
We chose thinking it would destroy each other.
If we remove them together, see here's the common factor x1.
A11 a22 a12 a21 minus in front of coefficients
b2'l b1 and this is happening is that as the numbers accumulate as we saw in the right hand side.
If we do so in a way that will not x1.
Not that the way to multiply the first equation by a21,
remove a11'l hit the second equation.
When we see something interesting as we did.
the same coefficient formed in front of the X2 is happening in front of the x1 factor.
There are many similarities in the right side.
Because here in the first row a21
b1 a21 times we got hit.
See, here's.
The first line in the axis for which we have come out of the second.
After all, this is a simple calculation that since junior high school in all of us
a screening method he learned.
But important observation here is that X1 and X2 dual
the multiplication of the number that has to be the same as the difference.
So this number determines the nature of the solution.
And these unknown number of partners must be an important product.
In determinants it said.
Because the means to determine the determin, that number determines the nature of the solution.
French in the English language are first examined these issues as well,
It is going to determine as to define culture.
If these coefficients are not zero Det say more here
we have written a summary using this definition number.
If we see zero can not solve this equation.
But if we divide the nonzero time after this, Det x1 and the same
Det dividing, dividing the x2 we obtain the same determinants.
So determinants, of course, also a very humble environment where two binary
under the number said we do ezilmeyel but it also gives the essence of the subject.
We meet with a number defining attribute of the solution.
That means to identify him from decisive in determining,
It called determinants that determines the meaning.
Now the same job in general the size
We may want to generalize the square matrix
but of course we can not do in general if such a thing hundred percent.
Therefore, here's one thing we can do to get some impressions.
I wonder how he can generalize these theories, develop theories.
But we have learned something but I did not learn anything from two binary too much.
We have learned that there is a fundamental issue that determines the quality of the solution.
But we've also learned a little of his character structure, but we also learn something more.
Maybe we'll learn a little something more to say we do three triple.
Now let's write three triple that.
Transport information as it is in the Gaussian elimination method
x3 x1 extent this factor without writing a
Meanwhile write the number on the right side to the right side
We achieve this extended writing matrix.
More generalized Gaussian elimination with what we just did,
more systematic, also to the regular.
The aim is to bring hit by dividing 1 and zeros on the diagonal below
subtracting from each other.
Finally, we've done it the previous one.
When we do this we show numbers with the base,
The values generated at the end of these processes work,
very easy to bring one to the first element instead of a11.
If you divide the whole line a11 If this number is zero, one income.
In order to bring the number below zero
If you remove this line with a21 hit zero income.
If you take the a31 hit zero income.
We may do so in this way.
If you are unsure about this or a previous summary of the
You can learn details by looking at the course Linear Algebra 1 course.
Extremely simple.
There is always something interesting on the diagonal consists of the number of Det Det Det.
It is not zero, it also Det 1
Also bring upward move on the diagonal
Det which now consists of the same number that we say what it was.
You can do it with numbers, you can do throughout the country.
And here we can see that x1 Det dividing the number immediately formed this right.
Det x2 divided by the number formed in the second row of the right again.
That gene is divided into two as in a binary number common
When we find that two x1 x2 x3 The duality
a top size of the matrix than we find a similar feature.
Det If this is zero, the possibility that the solution will not be as you can see this division
again this will determine the nature Det.
It's a little forbearance what we do these operations, Det
We see that as a structure formed.
If we compare this duality there are two double duality
a11 a22 a12 minus multiplication had had a21.
That meant two are missing a binary product of a cross hair.
Here, too, are coming together for the triple product is a matrix of three three-pointers,
There are also three issued.
So there is quite a partnership between two duality three three-pointers.
If we look a little further in the first a11 a22
row and first column, second row and second
column so that both partners come from non-overlapping rows and columns.
It is also here.
That same line, no triple product and not indivisible elements are the same column.
This is an important observation.
Here is a binary pair matrsit products,
The trio consists of triple matrix multiplications.
Some terms plus minus some of the terms are coming.
So half of the missing half of a cross hair is coming.
And as we see it a little later
Write able to organize in various ways.
As we do, we can write here, Det again.
See the a11 we see here have in common.
Here, too, there is a11.
Gather the a11 factor.
After that we look again a11 a12 a13 are the first line item.
We take the factor of E12.
We see that it came with a minus sign.
We take the a13 factor.
In this way we are organized according to the components of the first row.
Similarly, we can also arrange according to the first column of the Elements.
First we get the a11, the a21 then you take the multiplier,
After we get the multiplier A31.
And here we see it in a very interesting way in the A1 1,
A1 A1 2 and 3 or if we consider it as the first milk,
1 a1, a2 and a3 1 1 multiplier, binary,
We found two in binary matrix determinants.
Thus see them this summer as determinant is olşuy structure; here
2 a2, a3 we write the diagonal 3,
two out of three three two of the diagonal of the determinate type occurs.
Similarly, in a 2 a1 a2 1,
a3 3, a2 3 a3
1 If you get them in the summer, and this
It has the opportunity to write in a very systematic way, as you see it.
Three three-pointers, but could not yet rule out the determinants of two
binary matrix with a formula that can be obtained from the determinant
We meet, we encounter a calculation chain.
As we do this, let's take this coat matrix numbers again.
A1 a1 1 1 multiplier times the numbers in this matrix
Once back at the row and column where
the remaining half is going determinants of the binary matrix, let him M1 1.
Coming for Me minor,
we call it a sub-matrix determinant.
It said lower die; We are taking the line that contains any element and column
we look to the remainder.
The first line duality and here we take the two remaining first column
We are writing to bring here and we calculate the determinant of a matrix.
a1 2, found that the first line
1 a2 and a3 again threw three second column, a3 1
a2 3, which still consists of a two binary matrix.
Similarly, we consider a1 3, see we're bringing it a1 3.
A1 because it is the first line 3, we are taking the first row in the third column
Shoot for the third column, a2 1, 2 a2, a3 1 a3 2 remains.
a2 1, 2 2, 3 1, 3 2.
We calculate these determinants and determinants of the number we found on this floor
It consists of a plus or minus multiplied but were plus or minus signs.
Based on these, the MA minor,
If we change the mark of minor matrix,
plus or minus, minus, plus, minus the one we give the i + jth force.
For example, M 1 plus 1 plus a j i come here a second force
plus minus one because the force of the second coming, but we take a look at two
a two plus two at a time; three, comes as the third power minus one minus.
Now we're going right here anticipate a rule requirement.
In a matrix of a sudden Laplace
formula not see a summer night dream.
By studying a simpler system of this kind,
"How can I have a more general?" he must have thought.
Science is already developing such.
We are witnessing a more simple things, we're trying to find there more general.
Get to know physics so dramatically around the Sun.
We are removing the law of gravity from the Earth's rotation,
Newton's law of gravity also would remove the law.
Force equals mass times the acceleration.
So there thinking the same, you know the world is a very simple system
The sun revolves around a very short time to understand that nothing has been achieved,
It learned thousands of years, but all the force of a simple system of classical mechanics,
motion forces that determine their relationship is equal, we passed to mass times acceleration.
Here, similar structures.
We start from a very simple simple, here we are trying to learn the rules,
We find the rules, and that the M matrix,
alt matrislerin determinantlarını eksi birin
We say we have achieved adjoint matrix by multiplying the force.
In English this is called the cofactor.
Using this definition we can write such determinants.
The elements of the first row A1 1, A1 2,
A1 3 again this conjugate matrix of elements opposed to the first line,
cofactor able to bring the product of matrix elements.
So here are valid at least three three-pointers in a matrix
We found that the determinant formula.
According to the same thing when we held the line this time, the second indexes
we can show it on the collection.
And we saw it last
When we made in open software for any line
We have found in general a good if we put in place,
We find the formula for the determinant of expansion based on any line.
Overall, not a system, but is available in three three-pointers
It can be expressed in general now able to provide this same formula.
So you're testing it after finding a rule quarter, as five.
Then there are some more general abstract propositions remove
You are trying to identify.
Again, determinant, we calculate two kinds you remember where.
One of the column, it was on the first column.
Here, too, there are still two binary sub-matrices.
Signing on determinants of sub-matrices.
If we write this, see here for the second index partners.
So the first element of the first column,
Located in the second row where the first column,
The first column second place in the third row.
a i, j gives the location of the line, giving the location of j column.
As you can see here that these values are for a fixed one,
According to a happening it opened this column.
Here again, according to this first column we removed this rule, but it usually
we can show that by making this product are available.
Now this observation matrix of three three-pointers us now
n times the determinant of what we have learned from,
so the column and the line
I say seems to have paved the way for generalization to the matrix.
To be a little more modest, but this down
He has shown how to calculate the determinant.
Let us apply this now as a matrix three three-pointers.
It must be square to be determinant.
Determinant of this sub-matrix
as follows; We are taking the first row and first column for one,
back minus five four, five minus four, minus four and six remain.
We calculate the determinant of this.
A couple say again in the first line item in the second column.
So we are taking the first row and the second column.
Then back at two, four, one, it remains substandard.
Two, four, one, six.
We calculate this determinant.
When we calculate the third line,
We put the first line to the third sub-matrix,
We are taking the third column, going back a trio of sub-matrices.
There are also two, minus five,
a minus four we see that we are writing after that.
It's also the product of diagonal determinant of second diagonal minus, minus three turns.
Similar thing in the second row, making for the third row of the elements.
Relations are already here if we have the first column, there's one that's M2,
M 2 2 that they are in themselves constitute a matrix.
M1 1, M1 2, M1 3 first line, second line M2 1 2 2
2 and 3 which make up the third row of the matrix as shown here.
We started with here, here minors,
We found lower matrix from minor word.
Instead placing them see the numbers here, minus fourteen eight,
minus three, minus fourteen, eight, minus three and the second and third rows.
Now you'll remember from this negative as a sequential,
He considered jumping.
Minus sign stays the same because the one i QJ plus minus
here's one plus one plus primary strength remains.
plus, minus, plus, plus, we change it, we remained the same.
We stayed.
This has changed, it stayed the same, changed, stayed the same, changed, remained the same.
Here we're getting the matrix cofactors and the one we see that,
On the first line to the first line of these cofactors
See if we hit, we do.
Once -14, -14,
It +3 +16 -2 to -8.
A, pardon me e was +2,
We crossed 1 -3 -3, 2.
Minus 3 -1 giving.
Again it first with the second line,
If we multiply the second row of the cofactor, we find the same number again.
If we multiply the third row of the cofactors to the third line, we find a still negative.
Then we hit the first column to the first column cofactors
I write separately, but the number can easily watch too.
We find again a minus.
For example, let's do it.
The second column to a second column where multiply.
-2 To -8, +16
5'le 7 + 35,
When we hit them, we find the same old one.
Now this is very interesting.
Which means that the number of equivalent cofactor matrix,
Multiplying I multiply what we find with the same number.
Columns, co-factor, unless hit with the column we find the same again.
They say that this cofactor will be useful to us and gave us a general
It produces the calculation chain.
The result is always the same tactics.
If we want to find the determinant,
of course we do not need all of them individually each time.
To learn this, just because a procedure performed in order to observe the rule.
After determining that the result of this rule which
If you come to our business, which is easier if, especially if some elements are zero,
We hesaplayabiliriy the determinants using them.
We make these accounts more easily.
For example, such that in each case all the openings according to the first row
Needless to write the cofactor matrix.
1, we get according to the first line, 1, -2,
1, 1, -2, 1, we get them.
We stood with them in opposing the cofactor.
We stood here with two binary 1 after the determinant of the matrix.
Here -30
We find that 16 plus.
Second, we change the sign, plus, minus, plus marks was here +2,
-2, wherein for the cofactors
We put the first line, we take the second column.
12 minus 4, we get out of here.
Similarly, for the last element,
1 We are taking the first row and the third column.
-12 -4 1 times
We find in these numbers.
We find here the -1.
Similarly, we find we do still -1, for example by opening the third column.
Now determinant calculation
It seems to have discovered a method.
This discovery is the fact that he gives directly to the rule if possible
but it emerged from this rule, imagine that someone not have seen it applied,
a calculation chain that naturally come from within an account,
We see that the calculation algorithm.
Here the different aspects of this algorithm,
We will achieve this by developing a formula calculation chain.
Buf, a propositional therefrom
We can get the chain.
This proposition here in chains
We will see that came from calculated determinant features.
So the fact that we've ever done
I search for an answer to these determinants from comming off.
The next part in this more systematic calculation
We will develop the chain and this calculation
Taking some steps from the chain,
a proposition using properties
We will do a determinant of the system definition.
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