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.
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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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Matrislerle Diferansiyel Denklem Takımları / Matrices and Systems of Differential Equations
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Attila AşkarProf. Dr.
Matematik Bölümü (Department of Mathematics)
[BOŞ_SES] Hello.
This session is the last session of the course.
We will close the last session at an example.
In this example, a second-order differential equations as you can see.
It will be because it is a second order two integral constant.
And hence the need for two initial conditions.
We choose to zero the value of y is zero.
This is our arbitrary choice.
Other numbers also could issue but confusion on this account, to recover from,
There is also a sense in the physical sense as well, and is also equal zero base year.
This equation is an equation significantly.
This mechanical vibration
sinus x with omega strain or an electrical circuit
equivalent to a strain that comes from outside the sinus omega x electric current.
You can find simulations at the atomic system.
We will solve this equation three kinds.
From that angle, that is why it is completely a second
We will solve in order differential equations.
Because this solution is very similar to the first order with constant coefficients.
For those that have not seen her for differential equations
I hope even be easily monitored.
Then we solve the matrix method now we have ever seen.
The first step in this second order differential equation one matrix for it
equation to bring the first-order matrix differential equations.
After we saw our two way.
We have solved the same problem from different directions in the last two sessions.
Diagonalization with the one, the matrix functions in one.
A matrix will consist here
By calculating the base of the matrix or that the matrix,
By diagonalization of the matrix will be formed.
Now let's solve it directly.
Yet before we get a simple solution.
Y2 base plus y equals zero, we are looking at solutions.
The solution is very simple because the sinuses twice
plus minus derivative sinus sinus happens when we take a simple solution is equal to zero.
When we receive the derivative equal to minus two times the cosine of coming himself.
Plus cosine we see easily that provides this equation.
If we multiply them by a factor c1 and c2 same feature
you get.
That is still going around.
When we take the car for the second derivative is constant minus c1 times here SiNx,
cosx minus c2 times.
When we gather in these two equations
plus as the second derivative itself, we see that zero.
That means it's a simple solution.
If we're looking for a specific solution right there right sinus omega, omega x sinus there.
So we know that x is in a special solution consists of a sinus omega but
We do not know what the factor now.
The name of this method already uncertain coefficients,
determined coefficients.
We do not know now you're getting sinus omega x one time.
Because we take the time derivative twice again that omega x, omega sinus x.
Be able to compensate for the right side.
Let's take the second derivative.
Each derivative used an omega future.
Once we found itself in negative second derivative of omega times.
See, when we gather here this year comes with its own omega minus the square.
He comes from 1pm.
Both the common factor is the sine of omega x.
Right side sinus omega x is estimated,
It was not a guess, of course, knowing a random guess, took us to the conclusion.
Because sinus on the right side, are able to select appropriate balanced.
What is unclear is the fact.
1 times 1 minus the square should be the omega.
Therefore, refer to 1 divided by 1 minus the omega square When you have chosen one will come here.
Sine sine omega omega x is equal to x will be compensated.
Thus, here we find special solutions.
General solution, we are adding a special solution simple solution.
Lean solution c1, c2, cosine sine x plus x plus this is our custom solution.
Initial conditions will give us just what this c1 and c2 should be.
See also sine value y is zero,
rather x y where zero zero.
x, where y is zero zero base.
These starting conditions we select our own.
This arbitrary, we could choose as we want, but they elected this problem.
It would sinus x x equals zero to zero as you can see, falls c1'l term.
Cosine x 1 would only remain c2.
Where x equals zero again when sinus zero, it also falls.
Here bi additive does not need to be equal to zero.
We see immediately c2 is zero.
When we receive and will account for derivatives at zero.
Cosine x comes from the first term.
So times cosine c1 x x equals zero as a value in
c1 to just stay in the first term.
We take the derivative of the sine cosine of the second term,
c2'l terms fell to zero is zero.
Sinus x we take the derivative of omega omega times kosinüsl term output.
Kosinüsl term x equals zero because it is only here once in omega 1
Omega minus one off the square.
Here we have two more C2 and C1.
We solve these two equations.
c2 zero output in a simple manner.
c1 here again simply be divided by 1 minus minus omega omega square.
So here we have found c1.
Wherein c1.
c2 fell.
As you can see there's a common denominator as 1 minus omega square.
Get her out of here if omega sinus x this term
omega minus sinus x times here as we have found the solution and the only solution.
Furthermore, as no solution.
We can provide here.
When we zero sinus x x is zero,
y is zero showed sinus x zero is zero.
When we take the derivative future cosine omega times here.
I've already got the omega.
When will we put here simply omega x zero.
Where will it be when we get the cosine derivative.
1 coming from the cosine.
So in this second requirement of omega omega minus,
We see the provision of initial conditions.
This problem still has a special status.
Omega zero, I mean zero occurs in the denominator as you can see if omega-1 data.
Omega-1 data denominator sinus x minus sine x occurs.
Zero divided by zero occurs.
This is actually an important condition in the physical care.
This is a resonance.
Whether electric circuit, whether mechanical circuit in atomic systems,
but this kind of resonance for the same equation in a similar equation
encountered events and important events.
The uncertainty that now if you divide zero zero
We use L'Hopital's rule to remove.
According to L'Hopital's rule that the omega of zero,
We take that as omega variable.
In fact, it is given anything but omega omega 1 while taking the omega changed
Going 1.
We take the derivative with respect to the share of the omega.
Not according to x x out off we received according to omega.
Kosinüs omega x.
Here, too, we received only omega-derived omega by one,
sinus x staying.
Get derivative of the denominator of the omega omega remains -2.
We also organize them a little and omega 1
After placing limits because of the way we organize omega 1.
Omega also has a minus sign to have put a 2 in the denominator when we see one.
1 split, we reserve 2.
This cons of the above dengeleyin by sinus x
cosine x minus x times as we find again the only solution.
Now we're going to solve the same problem matrix method.
Of course the first step of this second-order equations with two equations
two unknown terms, but only order-order differential,
You have to be brought to the first differential equation, we have seen it before.
In the introduction to this chapter a second order, even in general,
that in a more general equation, how the matrix,
vector of unknown, we have seen brought to differential equations.
As we were doing, quite simply; We say y = Y1,
We call on the Y2 y derivatives.
If y2, y derivative, we take the second derivative of y,
he's also derivative Y2.
You think to take place the equation, see, we get the following.
Our first equation,
whereby y = was Y1; in fact this equation Y1
base equals y2, there definitions.
Wherein the vector component of the equation
Y2 base situated, the second derivative of y,
y2'nc equal to the components of the first derivative.
As you can see here it comes Y2 derivatives,
It equals y we are moving to the right,
mark changes; y = was Y1, it was minus Y1, plus sinus omega x.
Initial conditions can be expressed in this vector is welcome.
y = y by that y1 (0), y1 is zero; so
is the value at zero of the first component,
base year value of zero is zero in the second component.
As we say when we say them.
See, there is a equation between y1 and y2, y1 derivative,
Y2 itself and right sides of the X, with a non-function.
That first line is zero, it brings as one, right in going to zero.
Indeed it is the first line, you multiply this matrix vector
You produce the same equation again, as supply.
The base of the second derivative of the equation minus Y1 Y2,
and Y2 is not here; zero,
and, but the function of x in a right side,
There is also an independent term of the year, it comes as sinus x omega here.
So we where y vector, consists of Y1 and Y2
When we agreed to the matrix where the output vector y
gets hit, and the right of the function x
vector which also appeared in this way.
So we put our equation, y = A · base
We bring y + h construction.
Before the second method
As we see, we need e to the Axa.
e to Ax as we find it; We find the eigenvalues of matrix.
Matrix, zero, zero, one, consisted of less than one,
We put a negative light on the diagonal.
As you can see here, the lamp in the frame, plus one turns Eigenfunction,
eigenvalues that force function.
Here the roots of lambda i, i.e. virtual i,
square root, square root of minus one, which is good, and minus turns.
These numbers finally, to be complicated to be virtual too much Or,
method does not change; but there is a change in a small place.
This we find opposing eigenvectors.
Their finding the length e1 of them stood with e1 equivalents.
We see them all the time, so if you look at the Hermit matrices,
to process the complex coefficient matrix.
That, with this eigenvector conjugate
gets hit; (1, i) 's, we are bumped to a minus i'yl,
means that one plus one; two turns, this is root for him.
This likewise e2.
These vectors, and uses the columns, we get the QR matrix.
Q matrix, taking the complex conjugate, so I have,
Turn the minus and we take the transpose,
We find that if we get the transpose Q-1, we find the opposite.
We can really make it.
Q-1, Q, where we found here gets hit,
see I'm taking the first column, bumping into the second deposit; a,
i multiplied by minus gives a more; two.
There is also a square root in the denominator here, there is a square root here; two.
That leaves the two split a two, one needs to go to the matrix.
You can easily go to zero outside the diagonal.
For example, multiply again, this time to the second line of the first column.
As you can see good times once a plus minus, giving zero.
We find that this is really to provide accurate accounts.
We find the exponential function as follows; on the diagonal,
e to the lambda 1x, 2x, e to the lambda,
here is a lambda lambda two, i and where i minus minus ix ix.
We put the right hand side Q-1, Q, we put left.
We have seen in detail how to do it perfectly in the two previous sections.
On the second part of the matrix functions and diagonalization.Lines
on the second and third sections of them we have calculated the matrices functions,
that's also a very powerful application.
This product finally multiplication of matrices,
right in the middle matrix-matrix multiply, take a matrix,
Do you have a better result you hit it to left matrix.
The result is seen as following.
So if you reckon you'll find it üşenmeyip,
Or, you can scroll to the next one accepts this.
Although the result will manifest itself already incorrect.
We know Lean solutions to over Ax'l to y0.
We found them to Ax, kosinüsl of X, X, sine.
And one thing more powerful than the matrix, because you will remember this rotation matrix,
cosine, cosine, sine, minus sinus, which is a rotation matrix.
Sine squared plus cosine square is determinant,
This comes from a rotation opposite.
But here the variable x, so this return demonstrates standing event.
Already with a swing in turns, something which is relevant to the return.
But y is zero, zero vector of hit, we find that the simple solution is zero.
This is a good thing, it simplifies our account.
At the end of the math and physics that does not come even as a compromise.
We knew this formula as a special solution,
e to the minus Axe
h'yl the multiplication, we will account the integration.
e to the Axa herein, kosinüsl of sine.
Where x is negative unless you change the x base, we get e to the negative base of the Axe.
So instead of x minus x base, bong
cosine negative consequences for the change, markers remain the same.
However sinus, minus sine minus alpha out for the exit,
Let's say for any sinus alpha alpha alpha times.
Here was the way to change the signs of sinus x minus base.
This is not a tall task.
It will hit h'yl after, we announced.
We found it in advance,
here on the way to the matrix differential equations,
It emerged spontaneously what the h.
With this product we obtain h,
We stood here because it will be a vector to vector matrix,
Results will be released as a vector.
There multiplying the sinuses as you can see in this vector,
There are also multiplied by the cosine of the sinus in one.
We also know how it is,
a sinus a sinus sinuses be the difference here when we say,
cosine of the angle difference consists of the sum of the cosine of the angle.
Cosine with sinüsün Kinder similar thing emerges, it means that
finally we find using simple multiplication and Trigonometric identity.
It will take the integration, it is also easy integration.
Here are the sine we find there just now,
the integral of terms, kosinüsl from the integral of terms,
We will do it in reset between x, it turns out that conclusion.
I give these details, because you are dealing with these accounts are true worth
because wrong though, would not overlap than previous solutions we find ultimately.
This is a minus omega and wherein,
while the denominator integral to the future, plus omega-on
this one has a plus minus Omega and Omega future denominator for him.
It will be less omega-square we take a common denominator.
Here above is held for these terms.
So here it is important to capture the flow or account details of this idea more
It does not matter, that collection be made by the school üşenmeyin
but we need to understand the impact of the method,
This continued for intermediate calculations here we come by here.
It was a special solution to this product with a higher integration of this.
We also found above the first to know.
e to the Axa cosine sine term here.
We wrote here kosinüsl the sine term.
We found the result of this integration is here.
Here there is a common denominator and common frame of omega minus
and here are the vector multiplier.
This vector is multiplied by the challenge of finally being used trigonometric properties,
made simple in this way would be simplified because in the end you'll also see why,
because here there is such cosine squared cosine future
There's a similar sine sine squared term future.
These supremely elegant sadeleşerek the term in this way comes
and this solution because they all lean solution zero
total solution overall solution had been simple solution plus custom solutions.
But also the overall solution tailored solutions for total lean solution is zero.
As a refinement here.
Where both Y1 and Y2, we find both.
So we get more than one function and its derivatives.
We found before we found a solution y terms,
solution of differential equations obtained directly here
We have the second-order equation solving.
Here we find a little bit different interpretations of the same results.
y1 y was his solution of differential equations we found this just now.
türeviydi y'nin Y2,
When we account for the derivative of y we found, that an internal control.
Get here derivative with respect to x times cosine of omega omega times cosine x x see.
Where x is based on a derivative of omega taken by this it had not affected the cosine x.
See also Omaga is taken out by the same terms that we find the common denominator here.
But we found it from a different direction.
You think that the system of freedom of hundreds
If the equation did with a degree from the hundredth degree equation
If I do not like to work, but to deal with the first-order matrix.
Matrix is a very powerful tool and we see PC,
much more work to be done in large matrix system.
The purpose of all this already great system to prepare for you.
We have already done this with the MATLAB linear algebra
than you do here with small matrix of problems
How do the elders of the code we will pass on to you.
Bye for now, but we still have a second problem.
This is the same differential equation
solve the diagonalization method.
We solved it, that we found this method more precisely,
otherwise we solve the problem in this way.
The same problem just to see the differences and similarities
We will solve the diagonalization method.
However, our beginnings in this second order differential equation.
This is completely the same as we did in the second step, we are making matrix ago.
Matrices same matrices, the right side of the same matrix starting vector,
The initial value vector is zero vector again.
The core values needed to Diagonalize,
They needed to find them already in the two core vector but merely here
You need look back no solution in itself says enough to get them
We put all the details again here, we are already in our hands.
They say in this equation after making yyqz
As we found in there before we insert Qz base comes first,
Q minus the merger has hit that we find in the triple product.
Q, wherein Q when we choose the earlier
theorems guarantee us that we will see this Q diagonal minus the AQ.
There's no need to do so on this account, it will come naturally.
First eigenvectors, eigenvalues i minus the second core value,
There guarantee that they came.
We calculate the terms of the merger on the right side of Q minus h
this is finally a vector by a matrix multiplication, we get them.
Initial values we y equals
Y values at zero since we also qZ
Q'yl to z product, z we see that zero here.
And we held them all the time differential
It come to the following equation structure.
The beauty of this again in the way we intended, of course, not be spontaneously,
wherein the matrix in terms of two unknown girls because the diagonal
Two years initially mixed together in this way, but in terms of our equations,
We see that the decomposition of the equation, the two of them.
And this equation also actually see each other conjugate
plus i have here minus here,
minus i have here but there is nothing missing.
When we look at them before first order
We find a simple solution to the equation.
Lean solution here e to the ix, down to the minus ix.
It's right there in sinus x Omega, the first order
cosine sine may come to that variants may come as well.
A1 chance to get to know them both times sine times cosine omega omega x A2 x
happening this particular solution, A1, A2 uncertain coefficients.
Similarly for the second unknown
Determined appropriate solution for Z2,
There uncertain terms that right inasmuch as sine coefficients will be sine
but we put in when we take the derivative of the cosine to cosine can come.
As we embed them in place a little longer time
This account may seem off two equations for A1 and A2 are
Also it comes with two equations for B1 and B2.
How comes?
See, right there in the sinus omega x,
here are cosine and sine omega omega x X and terms.
We combine sinus X, omega terms, they will balance the right.
When we look at them because they have zero cosine of omega-x to the right
so that the equation will be zero coefficients between A1.
Similar thing for B turns out that when we solve them
coefficients, we find a general solution we find, we find Z1 Z2.
They see things that many symmetries.
coefficients c1 and c2 came from simple solution.
We find them from the initial conditions.
x equal to zero, we find z1.
She zero.
Z2 was zero to zero that we find in their right hand.
We find here the c1, c2, and we find all the solutions ultimately emerges.
The initial solution to this whole
We know the way to go to y our starting point because it is already a
Let's choose a diagonal matrix y times z Qz
The qualifications will make diagonal, you get quality.
Therefore, we know the Q To find the y,
We also found the z, z vector components of the Z1 and Z2.
See here the square root of two common terms
and a minus omega square, we use them as well.
Z1, Z2 properties where we have achieved here is when we put out de
going out and gets hit again we found them before,
Or, we find that we have found a direct matrix,
matrix function
we find the solutions we find that we get the same in the second step.
We see easily provide that, where Y1 turns out,
wherein Y2, but y2 y1 derivative.
Here is a derivative of the sine you get a cosine derivative,
cosine derivative omega
cosine once it does, look at the way we organize them
omega minus sign will be coming here this where the cosine times.
When this de omega omega minus the derivative times the cosine terms it here.
So should we make as a result.
So that we in our lesson this subject, we finish this lesson well.
We will see if you look again to this summary is extremely simple.
We want to solve differential equations of the following kind.
The first method we call Q y, z times.
Q from here to the eigenvectors
When we create a product that gives the diagonal.
This diagonal matrix equations, we are dedicate unknowns.
One in each of the unknown and it's the first order equations.
The solution is easy to find it after we come back.
We will do so in the solution with the matrix function
We can actually see these things come to a solution.
This simple solution.
Special solutions, this integration comes to calculating the number of hit-matrices.
And here's the general when we stand together
we get a solution, we have seen several examples.
I really enjoyed making this lesson.
I hope you've learned quite a few things.
I learned how I can announce them because something better
he thinks that people are getting some different approaches.
Here we have learned the essence of the business.
Usually two binary matrix to calculate with three three-pointers
but we did not do the manual methods because they are larger.
If you need to make larger matrices using this computer.
To guide you and to also encourage the lay computer
how easy it is to program a computer without matlab
We leave it to your study by adding the sample.
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