Bu ders doğrusal cebir ikili dizinin birincisidir. Doğrusal uzaylar kavramı, doğrusal işlemciler, matris gösterimleri ve denklem sistemlerinin hesaplanabilmesi için temel araçlar vb. konuları içermektedir. Ders gerçek yaşamdan gelen uygulamaları da tanıtmaya önem veren “içerikli yaklaşımla” tasarlanmıştır. Bölümler: Bölüm 1: Doğrusal Cebirin Matematikdeki Yeri ve Kapsamı Bölüm 2: Düzlemdeki Vektörlerin Öğrettikleri Bölüm 3: İki Bilinmeyenli Denklemlerin Öğrettikleri Bölüm 4: Doğrusal Uzaylar Bölüm 5: Fonksiyon Uzayları ve Fourier Serileri Bölüm 6: Doğrusal İşlemciler ve Dönüşümler Bölüm 7: Doğrusal İşlemcilerden Matrislere Geçiş Bölüm 8: Matris İşlemleri ----------- This is the first of the sequence of two courses. It develops the fundamental concepts in linear spaces, linear operators, matrix representations and basic tools for calculations with systems of equations. The course is designed 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: Place and Contents of Linear Algebra Cebirin Chapter 2: Learning From Vectors in the Plane Chapter 3: Learning From Equations For Two Unknowns Chapter 4: Linear Spaces Chapter 5: Function Spaces and Fourier Series Chapter 6: Linear Operators and Transformations Chapter 7: From Linear Operators to Matrices Chapter 8: Matrix Operations ----------- 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.
Denklemlerin Gauss Eleme Yöntemiyle Çözümü / Gaussian Elimination for Solving Systems of Equations
Loading...
Do curso por Koç University
Doğrusal Cebir I: Uzaylar ve İşlemciler / Linear Algebra I: Spaces and Operators
25 ratings
Koç University
25 ratings
Na lição
İki Bilinmeyenli Denklemlerin Öğrettikleri / Learning From Equations for Two Unknowns
Conheça os instrutores
Attila AşkarProf. Dr.
Matematik Bölümü (Department of Mathematics)
Hello.
Until now always two unknowns, we deal with two equations.
Now the information we found here,
requiring much more knowledge before
We will apply the three equations in three unknowns.
Then the unknowns in the equation
How are we to understand the overall solution we have,
We know that almost completely used for that since junior high school
a more systematic elimination of two unknowns, starting from the two equations
We will solve the Gaussian Elimination method 'is to be brought to the structure.
We're doing three basic operations here.
The three basic operations are: a number multiplied by an equation.
As we have seen already that before two unknowns in the two equations
We stood with an equation to eliminate a certain number of unknowns,
then we add an equation in this second step.
This is also, we could destroy an unknown bringing resets here.
This second screening step.
Sometimes it can be useful to change the order of the equation.
This follows; Or, if you get the first equation of the third equation third
you may receive information in the second equation equation equation finally there
says that change.
This is called the three basic matrix operations processing.
There are two methods, these two methods complement each other.
One Gauss Elimination Method, the other Gauss-Jordan elimination method,
these approaches are the same.
As you'll see a little more of them
methods that directly unknowns.
This Gauss-Jordan method, but the transaction exist on this Gauss.
We use three basic operations like this: We write the matrix of floor numbers.
We are adding the right side to bring him.
And we're working to bring on a diagonal length,
We strive to be reset below the diagonal.
Now may not seem obvious, we will see it on the right of challenge examples.
We will still have three examples.
This time we will work with three equations with three unknowns.
Our equation could also say xyz, as given here, but
to be more systematic x x two x three Bill mAyAnlArImIz, here are
the number of times; for example where there is a factor of five minus two a right.
As we write them again considering the same matrix multiplication rule
As we gather here in the x x x Two x three in a column.
The rest are writing where numerals.
two minus one, minus two two minus five four, the same numerals here.
minus four to six same numbers here.
This two x x x trilogy still know when we hit those three,
In the third dimension by adding the third dimension
a vector in three dimensions again that x x Two x
bring it when we hit three horizontal consists of an inner product,
this time in three dimensions but in essence the same in two dimensions.
The terms we collect in a column on the right side again.
Now where x x two x three no need for our transportation.
Gauss others have seen it but I saw absolutely.
We fail to make a move that x x two x three.
As these coefficients are writing here,
We bitiştiriy also bring him to the right side of it.
We will take the with such a line
that has a value from the aesthetic or completely this line
Do this right given in the rightmost koymasa
We can understand that vector.
Now our goal is as follows; that the equation
a diagonal on the team, to bring under the zeros.
How do we do that?
These three basic work with.
With a line to hit a certain number, obtained Thani
To add to the bottom, and choose to come here so almost zero.
We will repeat it for more bottom line.
So this creates a cycle.
We need to bring on a diagonal in the first cycle.
Already have an additional procedure we do not need him.
This will bring one below zero.
Instead of both of these two instead of this one.
In order to bring this zero instead of two,
This first line minus two
My multiply and add this second line.
Do not touch the first line remains as it was the first row.
It hit with a minus instead of attaching two zero future.
We hit it with a minus two, plus four, minus one came here.
We hit this place with minus two, minus've added two to four, we stayed two.
We diameter with five minus two; We added ten minus minus three; minus thirteen.
We have successfully completed the first step in this first cycle, you see.
One came to the bottom of it.
Now again we want to always below the diagonal zero.
Very easy way to get this one instead of zero.
If we add here the first row minus combine hit zero income.
Now let's write the equation again.
Add to remain the same when we hit minus combine the first and second rows.
When we add to combine hit minus was zero.
It had two strikes minus plus merger.
We added minus four, minus two.
We hit minus merger, we add a minus six, was five.
We hit minus merger; minus five,
We add to this line as we get the results you see.
Now we're done now in the first row in the second stage process.
Because we brought on a diagonal, we brought these zeros below the diagonal.
Now here first thing in the sub-matrix,
so two in a binary matrix bring on the diagonal.
Or moves on to another vision of the diagonal first
do we encounter an item.
E this is very easy, if you multiply this equation minus the merger, it would have one.
We see it.
Currently we have reached the second equation
We want to make a diagonal element.
As you can see that we have come minus the merger has hit the building.
Using this equation the second element,
We will bring to zero under the second element on the diagonal.
E this easy.
As we are writing this equation here.
This in order to bring this one below zero
here i hit add to that second line.
Then the second line stays the same, of course, remains the same in the first row.
As you can see this happening when we add a zero.
When we multiply by two minus two to minus happening here four.
We have added five, stayed one and the same for the right side
Thirteen of these two
hit attaching we reach the building we saw.
Now we want to have achieved the structure.
There's one on the diagonal, there are always zeros in the lower diagonal.
Now what does this equation we consider comming,
If we write the last equation now clearly see here,
This last equation zero plus zero times x times x is two plus x three.
So by zero times x and x two of a dream that,
We read here as a thirty-three equals x.
an upper enters the equation after equation x one finds,
ie a zero times x x no x two minus two
x times x three, but we got three, so we found thirty-one.
So here it's happening in an equation with one unknown.
So there was an unknown way upwards already here below,
When we come to this line, we come to the second line from the bottom
x x three unknown when two unknown but,
Because there is an equation with an unknown, we arrange this equation,
This equation x minus two equals two times x three terms on the right.
x we know three, one was thirty,
When we organize it by x, we obtain two is seventy-five.
So we're solving an equation with one variable length, structure, and that this triangle
In the bottom of the triangle structure that we always bring zero
an unknown equation in every step of the way to the bottom up,
We arrive at an unknown equation.
Indeed, the bottom of the third equation, ie the depth of the first equation
When we arrived at the reading of this equation x minus two
x times x two plus three equals five.
E here we know where x three x two to know that here
When we put them to a still unknowns equation.
That's when we put these numbers in x one hundred twenty four that
you get.
So our solution is the only solution that's it.
Now a bit of a different equation
We're just now getting there is very little difference than that,
The first line of the same, um,
here gone
A minor change in the same second line third line.
These six instead of five came here.
Now we are making it diagonal in the same process, namely through the qualifiers.
Yet in the first evolution in the first cycle,
bring to a bottom corner of the first line we want to zero zero.
To do this, the second line, the first line minus two plus
hit when we add, as you can see this happening is zero.
Minus four so I was hit with that,
This is already a minus five to minus attaching the same two lines were the same.
We find it hit minus two plus two four attaching,
When we add two plus five minus three minus ten hit, we find here.
We need to bring back here again zero,
As the first line means that we have left.
We have achieved the reset from the second row.
Minus the time we add a third line combined to hit zero,
We have added a combined minus hit minus two, minus've added combine hit,
four, and we get the right term,
When we add to the merger had hit minus five.
So in the first cycle
We got what we wanted on a diagonal, they have it reset under one.
Now we try to here two binary matrix.
This will bring on a diagonal, it's easy,
If we multiply a negative future merger.
We do this process, we stood minus merger.
Here again, we will zero below the diagonal.
Adding to that I hit zero we have to make,
When we add to that I hit zero two minus two, it already aims,
here i've crossed that of minus four plus four, it was zero.
Now here's an interesting case out.
If we look at this last one equation that zero times x
plus two plus zero equals zero times x times x three-thirty-one.
E they are all based on this so that zero
Thirty-one equation equals zero going on, to this contradiction.
Judging from another angle zero times x plus
two zero zero times x times x three equal,
to ensure that no x equals thirty equation does not have a two x three sets of x,
x one will occur zero here's what you choose, you choose what x two
Choose zero will occur here, choose what will occur where x three-zero,
and we can not get rid of zero is equal to thirty-one contradiction,
hence there is no one solution that will allow this equation.
Again, there were two unknowns the only solution if we remember the two equations,
There were no non-state solution, no solution here in the second example.
Now again we receive the third instance in two unknowns
equivalent to a little more complicated by the two equations,
Or, it will appear in a little more sophisticated manner.
We take this equation,
A previous coefficients are the same, but a slight change in the right-hand side.
Twenty-one instead of a minus.
We do the same operations, so we are removing this gene X, the coefficients
We write the matrix, we find extended matrix by adding the right-hand side.
Our goal is still on the diagonal, bring under resets.
We do the same operations in the same way as the coefficients for them to be the same
zero zero future.
Which is different from the one the right-hand side
We've added combine to hit minus here,
We have brought zero, here we've added combine to hit minus was minus twenty six.
So the only difference here.
Now, here again there is a minus on the diagonal,
Our goal is always to bring a below zero on the diagonal.
This why, because last we have the first example in the equation
As you can see by reading the unknowns of an equation,
We'll come if the solution is only one variable in an equation.
In the second example we saw that there was no solution,
We know all this because her party was zero, zero was not right.
If you remember Cramer's rule, you'll see that he was in a similar.
The situation we have now achieved
We brought a combined slamming on the diagonal minus three to minus ten, was thirteen.
Now we bring below zero, zero, zero bring the look we get
i've loved to hit with that, I've added to that hit,
I've added to that hit and we saw that was the last element to zero.
So this summer the last equation, we see zero equals zero.
Now we can think of zero is equal to zero does not carry as much information, but most
at least not contradictory at all, a consistent state of their own.
We can interpret this equation somewhat different.
Zero plus zero times x times x times x two plus three equals zero zero
We can write, that does not impose a restriction on them x x two x three.
What are your three x x x two choose not you provide this equation.
So it will be infinitely many solutions.
Referring to one of the only equation the other hand,
but also it included stops in other equations, so whatever
we can also find solutions so infinitely our freedom of unlimited
not because of some limitations in the future, while the other equations.
Now try to move away.
Last we had was the case, the last line of zero is equal to zero.
Thus, a zero is equal to zero does not carry information does not pose a contradiction,
but also it does not carry information.
If we take this equation, we get two equations to three unknown like this.
But we know that we usually have two equations to three unknown
underemployed no restrictions.
We can choose to enjoy one.
The most systematic arbitrary choice of x to select three x three t say,
See what will be the last equation, x two minus two equals thirteen times t.
I take it proceed, the equation x three t
When we put this
Let my right side t to be t minus five.
Where t is minus two plus two t Let the right be.
We have two equations in two unknowns, as you can see and moreover the diagonals
on, there is always a cost.
So we find here directly x two x two thirteen plus two t,
because once the software this equation x t two plus two equals thirteen.
When we come to the first equation from the bottom upwards
the way we know what x two.
The first reading of an equation x once,
minus five minus two equals two times x t.
We know two of x, therefore x equals x two of which we know
We wrote in her right let's five plus two x two,
t here a minus of course,
t t of minus minus where this equation comes the following structure of our sheep.
Therefore, x and t from one species were found.
If we arrange them, let's create a vector x of x,
x x x x a thirty-one plus two of the three components were three t.
x two thirteen plus two t, x three and t.
As we can see there is a fixed number and there is also t'li.
Fixed among themselves gather, see also has a x a a thirty,
x two in a thirteen, no fixed x three, zero.
But t'li also have x t in a three,
x and t are also two two t have one, so we bring solutions to these structures.
Now why are infinite solutions, because what we give t let this solution.
Why not release the other two because it is completely the equation
t is possible then, lets get all the numbers, although the appendix,
x two, and we see that the freedom of a random one x
and we see that a solution is in this building.
This vector we saw earlier
we see the right part of the equation in space
Assuming he really is a plus x equals x zero
once u t in a space as is the right of the equation.
So this is all dots on the right solution in space.
There is not infinite, but also the right solution when we come out of this.
So there is a solution, but a random endless eternity,
not just in eternity all possible on this right
the solution can be an infinite number of points.
This forgotten the correct part of the equation in three-dimensional space AIf the following page
You will recall that the now forty-two and forty-third page if you look.
Now that we're here again, I'm having a break
If we have three equations with three unknowns on this diagonal
Under bringing a reset bringing the diagonal
if non-zero coefficient is the last equation
We find an equation of one variable as the only solution and the bottom up
increasingly used to solve an equation to correct the unknowns at every level.
If we thought that overall,
The unknowns in the equation will come to a building like this.
There are over a diagonal, namely those on the diagonal coefficients future
not first paid, base it shows,
but there are also resets the bottom of this one.
There is still a diagonal second reset under it,
I'm a little duraklayal the last one we move in this way.
The last one will occur in a number here, the BN base on the right.
They are always well-behaved two longitudinal baseline b b b 1 three
Longitudinal results from these processes that are changing the coefficient values reached.
Right here we see that the software last equation
anna base once the base bn xn times.
If NN base that number on the diagonal from scratch
We can solve almost xn'y it is different here.
xn'y the dissolution of the bottom up gradually each time xn minus one,
it is now known that xn'yl now be multiplied,
the right ones already known numbers below it upwards
We're going to get all the x values, we call it the only solution,
but the right side is zero, but in contrast
The number zero on the diagonal,
but it's the right side is zero or not zero, if not zero,
If you look at the number above zero in this equation that equals zero diagonal will be the base bn.
We now have a structure like this,
BN will have a conflict if the base is not zero.
So if you look at the number two examples that are even two to three three-pointers while
Although it still came second example of duality.
There is no xn unknown to this end,
therefore no solution of the equation.
But the numbers on the diagonal as well as zero
If the right side of zero, zero divided by partitions that do not need to be zero,
uncertain, but when we look at this equation it is equal to zero zero
It does not give us the information would be consistent state.
We can provide you with any xn.
Therefore, we also xn'y t,
three unknowns in the third example we saw a little while ago, as we saw in the three equations
We can find solutions to the increasingly upward.
Two and three, as we see this duality
to generalize the methods we have achieved in threes equation possible.
To do this, you can be forced in the first step, but much
There are two target objectives systematically.
We will bring on a diagonal one, it will also reset the diagonal below.
We will also look towards the last line, if you reset equals zero,
It means that the solution becomes infinite or zero is equal to zero
There is no solution if we encounter a number.
But also on the diagonal on the right side of zero,
even if just on the corner by dividing different from zero right side,
If you want to get the right side of zero, then xn equals zero interest, zero any
In these three examples that number divided by zero out as we see them.
In the following Gauss-Jordan elimination method.
For now, let this be a thought on what we have learned our share.
In our next session, we will see the Gauss-Jordan elimination method,
this is very simple, by Gaussian elimination
we have this triangle equation
we'll go into a little more than a simple structure by moving the triangle structure.
Explore nosso catálogo
Registre-se gratuitamente e obtenha recomendações, atualizações e ofertas personalizadas.
O Coursera proporciona acesso universal à melhor educação do mundo fazendo parcerias com as melhores universidades e organizações para oferecer cursos on-line.
© 2018 Coursera Inc. Todos os direitos reservados.
