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 1: Matrisleri Köşegenleştirerek Çözüm / Solved Problems by Diagonalization
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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.
Our matrix of the kind we saw in the first line of the previous session,
We see differential equation.
Here it is composed of unknown vector y.
He has derivative.
Constant coefficient matrix and a function h given in now
we solve it, we will proceed in a very simple way.
Let y transformation,
Suppose y, z as the product of the Q material.
So we will try our variable instead of the variable y.
Q. Let us also know now.
y base will be derived from this side.
Q should be formed from the constant factor that we take the derivative of y
When you arrive at Q derivatives.
Here we see the necessity of Q being constant coefficients.
It also makes it mandatory that the constant coefficients A little later we will see now.
If the x-dependent variables if this method work.
But this is real life, many problems
problems we face in this as a constant coefficient.
Let's place.
Q left, we put z base.
A var.
Let us instead be Qz'y of the year right here.
h var.
This immediately reminds us; Q minus multiply if we combine
So here we multiply the left with Q Q minus reverse merger Q unit matrix
z base hit of the fall remains to be just our base.
Q minus one, on the left to multiply.
We stood the whole equation because it will be the subject h Q minus the share merger.
Q minus the merger is going to h multiplied.
So we did an extremely simple algebraic multiplication operation.
We see here that if I chose now to the essence of Q vector,
I get the gist of the column vector,
This product gives us a diagonal matrix.
This supremely important we now see soon.
This will be a diagonal matrix in which we said that Lambda, the de on the diagonal
a core value of lambda lambda two, the light will be.
Then we understand that there must be a constant coefficient, because if,
Although x-linked matrix Q also would be attached to x.
We would not be able to process what we do here.
Because here y, while the derivative,
Q derivative times z times z plus Q derivatives revenue.
When Q is constant, but this is the case.
Q Q is constant, but the structure would have diagonalization.Lines
To obtain.
Q wherein Q minus the merger will be subject to the product of a new vector h.
Let him in game.
This visualization of the structure obtained as follows; There vector z on the left.
Here diagonal matrix,
just rest on the diagonal of the matrix core values are zero.
Multiplied by this speed and still gun right there.
If we take a typical jth line equation Let us say here
just equals the derivative of our base will only be found lambda j j times.
That is related to what we had with each other, we have found separate mixed components.
So with this structure wherein each of the one we have reduced to a single variable equations.
It is fixed also in this equation,
lambda is a matrix, which are fixed to fixed items; because of
We reduced the first-order differential equations with constant coefficients.
We will also find the right solution for each of these varieties.
We see the solution of first order equations.
j lean times homogeneous solution, simple solution to the exponential function.
g in a special solution that shows the impact of this future.
This is how we find now already looking thankfully g
We learn to solve equations in one variable.
Or we'll get with the integral formula or uncertain factors.
This means that the program has been in a supremely simple.
n in each of the grains having an unknown unknown mixed with each other
The grain came into the equation.
Each of them can solve,
We know that the solution even coefficient is fixed.
After that is not our problem, but we unknowns y and z.
That's what we y, z Q Once you get.
That's y we get back here again for the rest we know z.
The problem is that simple.
Now let's make an example; problems going with these steps: you see,
There are two unknowns, such as where the team first order differential equations.
Given the initial conditions.
y is the value at zero, you get a value of zero at zero Y2.
There are no equations in the first h.
These are the years we completed the vector components which Y1 Y2,
If we look at h, zero in the first equation to the second equation above
x What your matrix coefficients to see if one, two, two, one.
With these components, with this definition, these two
differential equations with unknowns,
It comes down to the following matrix equation; Once the base year plus years,
h and an initial condition y of the first component in a second component of zero.
So we reduce the problem of a matrix team.
Better visualize and Herzegovina, following the structure of the; Where certain matrix.
Now what do I need to do this?
Q matrisine.
To find the eigenvalues and the eigenvector matrix Q.
We start by finding the eigenvalues of this to say.
Easy to find the eigenvalues.
We are writing to business.
We have seen it many times in our previous chapter.
Minus lambda diagonal're writing on it in terms of lambda
We find that the core function of secondary roots.
Here comes lamp in the frame.
Lambda lambda less than minus one, minus two lambda.
A minus gives three negative quarters.
We know the roots of this immediately.
The root of minus one and three exits.
Place Herzegovina; Put a minus of one, minus a sheep by
plus two, plus two minus one equals three zero.
We see that it allows the three.
To find eigenvectors when we put in a lambda
opposing eigenvectors, we put the other way when the opposite
but not necessarily get from units of self-vector length.
As you can see here symmetric matrix.
Thus, taken as a unit also provides convenience.
No need to take the inverse of a matrix Q.
We put the vector of the first column.
We locate the second column vector.
Both the common denominator is divided by a factor of root two.
We put him in.
Q On the contrary, you can calculate directly the opposite if you want
but even if you get it come to this inverse matrix transpose matrix of the beauty of it.
It also now possible to find, does not require a new account to get the turndown.
We keep hard numbers on just the diagonal.
We are changing the place except the diagonal.
The resulting Q and Q cons of using no need to make this shock.
Because theorem to us, show the source before I theorems diagonal
on a lamp, the lamp will guarantee the two.
The mean minus one and three will be released.
We stood opposite to find the h and Q g.
See the Q stands for h reverse here and we achieve this vector.
You need to start conditions.
Whereby the initial conditions y, z Q once was.
z in Q will be y minus one times.
Here we write Q minus one to find z.
We are writing to y.
This value at zero, zero value in [1 0] 'di.
We find here our terms of initial conditions.
When we made it down to simple problems so far; the köşegenleş
For this matrix, in a known equation for z1 to z2
Z2 is an equation that is happening to one.
Taking the derivative of this equation is equal to z1 respectively
z1 plus minus times minus 1 divided by the roots instead of 2 exe this right also comes the term.
z because in this way again, this time with an unknown 2 2.
because it comes down to it's core values is a simple equation.
The homogeneous as we have seen them before,
We start with simple solutions.
1. there a simple solution to the equation minus 1
e to the x minus for that.
Special solution will be for the term to be ex X, on the right.
We do not know how many will be ex, but we know the structure, at least,
to that of XO.
We put the unknown at the beginning of an uncertain factor, a.
When we put the equation z derivative thereof
z 1 derivative of e to the x is staying just a times.
E to the x minus a times right here in e to the X and terms.
We would determine the value we arrange a time.
Similarly, in the second equation wherein
e to the solution to the core value is 3 × 3,
if it works really settled derivative 3 times to take over 3x.
E to 3 x 3 times here.
when there is no known term that B'La terms, when it does not exist.
This gave simple solution.
We believe that the solution also in this special.
On the right there is a term for the family of exponential functions.
e to the x that this will be a term X, but via an e
We do not know at the beginning it will be a few points.
We put him in an unspecified number of b.
Situated, we find that equation.
take the left side to put here a derivative z b e to the x times to be found.
On the right side there are three times b.
A 1 divided by 2.
All the above e e to the x and we reduce to the X. multiplied.
Here again there is no guarantee of a to b is also not a coincidence that way.
Because everything here is so many things we take for symmetric output.
So we find that z 1 and z 2.
We use the initial conditions and the corresponding yet
z 1 and z 2 are the unknown coefficients c 1, c 2.
We found our first in what will be the value 0.
Zero value at the root x 1 split will be.
Root = 2.
z 2 in the value of 0, it will be the same thing.
This is due to the simplicity we had selected, more account to suffocation
so you can see if the beauty we found.
Using them we determine c 1, c 2.
We're here doing this operation.
Here's what the C 1 C 2 's that what we find now.
And thus no longer using them only definite solution for z 1 and z 2,
We obtain the solution does not contain uncertainty.
y, q Kere z'ydi.
Z We now no longer a tool we use as a tool.
Z is our basis is not known.
Next. Q z is the time to convert it to y
We're getting.
Here q matrix.
Here z matrix vector z, z 1 and z 2 components.
Here we get the product structure as a function of solution here.
We can try to provide.
If the x-y providing a zero data
0 we know will come, because it was at the beginning of these conditions.
Indeed, if we put x instead of 0
0 to 1 will be exponential e to the x is negative.
There are 3 here.
Again the top 0.
3 got here 6.
Minus 2 out of here, because we put e to the x instead of 0 would be x is 1.
This means that 3 plus 3 minus 4 gives two numbers here.
Of course, because there is a root 2 4 1 over here did this
There are two times in two of our roots.
From there, divided by 4 1 was formed.
This 4 to 1 over 4 gives received by 1.
When we look at the second side put exponents x 0
because it still plus minus 1 3 3
0 will and we also provide the initial conditions.
Now it seems appropriate again to take a break here.
After that we will see the second method.
The same equation even another equation in the matrix
by calculating functions by matrix functions
We will find the solution methods and applications will show.
Bye now.
See you.
As you can see the many problems in admixture with each other,
of course we have to show the essence of drowning here two accounts.
But if a large computer system that in essence vectors
and it is possible to find the same method to promote core values.
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