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.
Yöntem 2: Matris Fonksiyonlarıyla Çözüm / Solved Problems by the Use of Cayley - Hamilton Theorem
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From the course by Koç University
Doğrusal Cebir II: Kare Matrisler, Hesaplama Yöntemleri ve Uygulamalar / Linear Algebra II: Square Matrices, Calculation Methods and Applications
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From the lesson
Matrislerle Diferansiyel Denklem Takımları / Matrices and Systems of Differential Equations
Meet the Instructors
Attila AşkarProf. Dr.
Matematik Bölümü (Department of Mathematics)
[BOŞ_SES] Hello!
This differential in our preceding session
We found a method for solving equations of the team.
The essence of this method y a
z was to make a conversion in converting to the diagonal.
Bring a diagonal structure of the unknown which one single, one in each
We solved the unknown and brought into the equation and then we come back to y.
Here we proceed in a different way, but we still do the same
The solution under an initial condition that the equation again.
I want to remind you to start with.
We did in the first subdivision of this section,
The only unknowns in the equation solving method recall once more.
This method of interpreting a little different direction,
We'll get here the method we developed.
Again, the more a function here.
To a constant.
This is the initial condition.
Solution consisted of the following steps.
We find lean solution.
It is not the usual solution to over a x.
Indeed, we see easily that this equation is situated, it provides.
This actually happens to remind you again but I think
and because we need to emphasize here that the matrix function
It will be inspired to capture the air.
Special solutions was the second step.
Here we go again with optimism in a special solution.
We say: since this is a fairly simple solution contains information, let's use it.
Lean solutions now we multiply the unknown unknown function.
If you go to this starting,
284 pages in even-numbered pages.
We are on the left side instead of taking the derivative of this y.
y derivative course this e derivative.
plus a once u e to the ax hit the second first time
If there is a time derivative of the year on the right side.
So once again a plus where the function h.
Again, with the first customer left, the right first term
taking each other and we are getting a supremely simple equation for u.
Differential equations are not.
But there are only derivatives of derivatives.
Thus, the problem is coming to an integration problem.
We also find that taking the integration within certain limits.
This means that custom solutions are able to achieve this structure.
e to u multiplied by AX.
U here.
We also can write equivalent structure, taking the ax to them inside.
As you can see where x minus x h with a term of üssül
Coming to the product structure.
It is a grace and an equal representation in its own benefit.
We also add a simple solution for this particular solution we find a general solution.
Lean solution to secure a higher multiplying AX.
This is a simple solution.
And wherein the terms,
The term special solutions.
To find the general solution for x instead of y zero
sheep given year should not equal to zero.
This z, showing the flexibility of its c.
We will not be able to provide an initial condition c not here.
x instead
Put the car is staying here only to zero.
We have seen a definite integrals Integral also getting benefit earlier.
In this way, but not necessarily be zero unless x
Put the x base means that this term goes to zero zero drops.
This is not something the integration and simplifies our work.
Inasmuch as the value of y was equal to c at zero.
He also provided.
y is zero.
We find in the c.
So y can be brought to this structure,
instead of using the car where y is zero.
How is that here were two equivalent structures for custom solutions.
Here, via e ax throwing into a solution can be achieved in this way.
This matrix equation is to find our inspiration now.
Now that we are over the moon with y h plus the sheer number,
a component with a number of functions,
We found the solution with digital function,
Let the vector equation for the same solution.
E to the vector equation ax indeed,
that's where you see the matrix function, the name of the method comes from here,
Spirit also comes from here, with as yet unknown
We see now that there is a simple solution that we multiply by the constant vector c.
See our equation moon base
simple equation y equals a times.
Once the base year minus a simple function
As you can see hit the base year for which a fixed again
prevent falls and where the lack of square matrices
Because a'yl AX product via e ai no mater where you will have.
We put them in front of where the ax to c occurs.
Once again there is a YHA.
As you can see them taking each other.
So said a solution to this matrix.
We're going to have generalized solution we found in this digital function matrix.
Since, in the same way as we found a simple solution,
Using the same here as we inspired yh'y
e to the special solution from a matrix ax times x number of times
We assume again that the optimism of the product,
Put in it really to end
again it coming down to a simple equation results in simplification.
Already the biggest feature of the square matrix
The sequence is not significant as plain numbers and in all processes
it can be done any time of the collection of the product.
Quite a list at the beginning of this section,
We saw quite a list giving the privilege of a matrix.
So here's just the same numerical functions
the only remains of an integral business.
The base where the same numeric functions
as is, that is to say in the same way as in numeric functions
like going to the integral of the vector matrix multiplication.
The only difference here to ax over a matrix, a matrix function, a vector h.
In detail, it is giving a vector.
I still remember the special solution.
Ax'l u e to the special resolution was the product here
we put the ax to them.
Also here.
So we find special solutions.
General solution of the same so there's no need to even go through it.
We could have written after nephew.
Because the matrix, square matrix acts like the same numbers all at this.
And where do we get the same solution.
We kept it a little more general.
Scratch scratch, albeit not in a x x comes here a number of zero counts.
Now the whole issue came to that.
How do we calculate via e ai.
EUR for the calculation of the above, given the resources right here.
We know that.
If a lamp in one of the eigenvalues,
two lambda lambda n, the eigenvalues of the e a e two,
The matrix diagonalization most au q
e, including those to each column
köşegenleştiriy in which he functions.
So it is a matrix function of it going backwards
diagonal matrix with q as health soll
We know it's equally collision,
where they were given in these resources.
That work will be extremely simplified, let's see an example.
Let's take the previous example again the same.
Two unknowns, starting conditions are the same,
A matrix of identical right and it refers to it being the same.
We have found the eigenvalues of this matrix had been minus one and three.
We find the eigenvalues.
To bring paint to the unit eigenvectors matrix is symmetric
We know that makes it easy to calculate the Q and vice versa.
So for example we do here is completely the same as the previous one.
Now we come to the critical and different directions.
e to the e to the lambda once on the diagonal means Axe x
e to the x Place the right lamp twice in Q
Q minus comes from the left with a shock.
So it comes down to a matrix multiplication.
See how easy it is.
So if you find the eigenvalues of matrix as a hundred computers,
You find the eigenvalues, you create the Q,
You create the Q minus 1, the business remains a matrix multiplication.
You close your eyes that you just needed no computer command, bong
seen already how we could do this with MATLAB
We will give you the commands.
If we make this product, do the right product before.
We found that, had left matrix.
This matrix is the challenge of the multiplication results we have achieved here.
E to where we know the simple solution
With this year will hit zero, the simple solution here.
We have found the formula for a private solution.
E to the base where special solutions Axe
the integral of the product of HX base, we call it the base.
Let's do this multiplication.
We do this multiplication.
These are all things that you know as a bit time-consuming but extremely simple operations,
You do longitudinal multiplication, we found here.
We take this to the integral.
This means that integration means you get here integral.
We find that the zero and the value of x x base.
We came here because a change as cons
In order to get rid of it negative.
And that we obtain it, we stood with the ax to them,
e to the minus x over Axa we calculate it.
e to the minus Axe we achieved by replacing the x minus x base.
This integration is obtained after such account.
We stood with Axe with Ax'l to take it to higher,
wherein a multiplication result obtained from the special solution.
So all of these actions can be done blindly at all.
This will give the code with MATLAB software already, you'll see.
Here we find an overall solution,
As you can see, this consists of a collection and comes into this shape.
This equation a bit of work simplification,
If you divide a split coming here and getting four
results we have going here.
That we found earlier
Compared same solutions that we see now.
We see that the same solution here.
You can note the page.
On page 300 the same, exact solutions
giving us.
Now I want to take a break here.
This process is progressing in a very systematic way.
After that, I want to give an example.
This is a second order differential equation and this is a significant
differential equations, turning it into equation
This top exponent of this equation we see the team
by calculating that function matrix method and more
We will calculate the diagonalization method we have seen before.
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