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.
Ters Matris: Genel İnceleme / Inverse Matrix: General Study
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Del curso dictado por Koç University
Doğrusal Cebir I: Uzaylar ve İşlemciler / Linear Algebra I: Spaces and Operators
26 calificaciones
Koç University
26 calificaciones
De la lección
Matris İşlemleri / Matrix Operations
Conoce a los instructores
Attila AşkarProf. Dr.
Matematik Bölümü (Department of Mathematics)
[BOŞ_KAYIT] Hello,
Now we come to another important issue.
Inverse of a matrix; We know these numbers 1, 2,
3, something which is also the number 5, for example, that until given us number 5
What we find is to say the opposite by picking minus 5.
How did this happen?
If we collect the two time 0 gives these two numbers it is the opposite of each other.
Minus 3, and we also know that the opposite because these two plus 3
0 gives we collect.
If we have the number zero, because the property.
The converse is also having a number equal to him, so he got.
He's got the opposite, that is to say not much of a problem in the collection based on the opposite flat
According to multiply in number, but also said the number 5
and if you multiply that by the reverse split 1 We know that 5.
But when we think 0,
0 multiply it by a number when we say that such a number equal to 1 to give
not between finite numbers; by means that do not always have the opposite impact.
Now let's apply it to the matrix.
How is that if we take the sum of the two numbers gives them 0
we were saying the opposite of each other.
If it has collected with a given BCE
If 0 is the inverse of the matrix We say give this b matrix to a matrix.
b as it is the opposite of the opposite una b.
Because a plus b, b plus or equivalent.
If you assume the first is given, a given, the opposite will be a minus.
If we write this in terms of the matrix elements
b matrix money b, i, j number;
minus minus aija this is supposed to be equal to the prescriptions of a matrix.
So be the same size matrix with a matrix,
until a count of the number of rows of b rows,
b is the number of columns as the number of a column.
Geldiğmiz stand alone in a little different situation,
Not quite a bit different because of the serious consequences of changing order in shock.
If the kareys two matrix, there are ab matrix.
b square matrix but also if we saw examples.
Even though the EU has matrix of the existence of five matrs
not always guaranteed.
It is essential for the multiplication of a matrix as a matrix which means if provided,
it may be that when we hit a reverse left.
We call this the left and vice versa.
when a given that if you hit it right, right opposite.
We define this as follows: a a a a as the same number L
çarpsa will give you the number of units where a number of
When you hit the left to give the identity matrix.
When we hit a right again to give the identity matrix,
If it is left to them if such matrices
We say the opposite, and right opposite.
Now let's see what we have done on this sample it.
If, according to gather vice versa if the matrix is given in a matrix
it's easy to change the signs.
b When we gather with a we get really 0.
b with au gather them again you get 0.
Now let's still a rectangular matrix,
Let's look to the left to reverse this matrix.
From left to say otherwise, we will hit with an L left,
so that the end product will be the identity matrix.
Now here I write ai,
We do not yet know what hit it left this matrix.
We are looking for him because it's unknown.
Just a compatibility restrictions
Coming; here we do not know the size of the matrix at the beginning but left the
This will take the columns of the matrix a right to be multiplied.
We get the inside line on the left with matrix multiplication.
Hence we see the necessity of having three columns of numbers here.
Although this product is 4 or 2 though is undefined.
Therefore, we will be looking left multiplied by 3 column as defined.
On the other hand it needs to be multiplied in volume.
Then we see the necessity of having two lines.
For example, a line, but as we here in the matrix
It would be a row but two columns.
Three line to say; three columns, two rows of the matrix
but it is imperative that the unit would be square matrices;
hence the matrix resulting from the multiplication here
L for the unit may also need to have two lines.
So we do not know what it was but now emerges Le least structure.
Let's do this multiplication.
Here are the elements of the left matrix,
We will take the inner product on the right column.
We take the inner product of the first line of the first column.
We get second and so on in this way, we obtain four equations.
First L 1 L 1 1 plus 2 plus 2 L 1 3
One should not have to be equal to the right matrix.
Again we hit the first column of the second line
Le 1 plus 2 plus 2 times L 2 L 2 2 3 0 You have got to be equal but it's coming.
Because the right matrix element opposing this position is 0.
Again, take the second right-hand column of the matrix
If we get hit by the other two equations.
If you pay attention here because it gives the first equation 0
in the position that zero element 0 and 1 opposed.
A team examine the following equation.
We know we have our hands four equation whereas six.
So when we think of it as a constraint equation it makes every equation.
There are six of us so freedom of unknowns,
There are four restrictions this provision.
Therefore, it remains quite serbetliği.
Two of our freedom of staying back.
These two equations easier to solve if we solve right now.
L 1, L 3, we find that it is equal to 1 2.
L 2 3 1 plus the dividing rather let 2.
1 split, we find that 2 plus 2 L 2.
If see them take first place in the equation where L 1 3 there.
L 1 2 that we're bringing in now.
L 2 3 where there.
Instead, we bring the equation L 2 2s.
When we compile and solve them,
When we say them, we see the following.
See here for only 1 L 1 and L 1 2 stays.
3, L 1, L 1, L 1 2 3 times 2 equals that which remains is equal to 1.
There are two unknown in the equation, L 1 and L 1 2 1.
See also here we look at this second equation L 2 3
situated, plus 1 L 2 L 2
3 2 2 La're strong as it also because it is denominated,
we get this equation have collected twice.
There are two unknowns in this equation.
One of these two equations and the equation does not interfere with each other.
Unknown around here do not have the right.
The right not unknown here.
These are the details of the calculation, but where in fact two
There are unknowns in a single equation.
Here, too, a single equation with two unknowns.
We can solve for any variable t saying here.
We can solve as a parameter.
Here, any variable such as L 2
2 parameters, it is clear that a separate parameter.
If we solve the SA 1 L 1, L 2 we solve one of s and t are denominated.
We instead place them after we found L 1 1.
L 1 2, t, we find him.
We found also found, L 1 L 1 2 3 tr.
When we look at the other side, we find L 2 1 here.
L 2 1 minus 1 minus 3 if we find it.
L 2 2 there.
L 2 2 that we have the right to choose p.
We used her choosing.
L 2 2: 2 3 s if we find LA.
this equation is revealed as you find.
As you can see, there are infinitely many solutions of the equation because t and s
If you find a solution and the value.
Eternity is not a random two parameters alone,
an infinity two free grade.
Now included in the L, T to reveal that,
Allocate just the t and just as the term for water.
Here we get this structure is indeed them
martin third hit of the second matrix t'yl
When you hit balls with s, we get the equation here.
Now whereby t and p random numbers,
We find these basic matrix by selecting the appropriately,
There are infinitely many solutions, but not random eternity,
A two-dimensional infinity, and also where t'siz s'siz term
therefore, it will be in the same vector as three independent vector or matrix.
For example, if you get t and s to zero you find this matrix, the first matrix.
t a s falls s'l term if you get zero, where only
This matrix remains, two tactics that, when we gather here in the matrix.
Similarly, when we also received a t h is zero,
nevertheless have collected the first matrix of the third matrix wherein the matrix is emerging.
So here we are L1, L2 and L3 have achieved.
To the left L1'L the L2 has hit the L3'L
unit, the matrix, we can see how to provide it.
I do also here to provide it.
So given that this matrix of matrices
There are three independent left inverse matrix.
But these are mainly independent infinite matrix in this eternity,
öbürki are interdependent,
has hit a s, with each able to obtain or t'yl has hit,
but these three are independent, we found three reverse left.
These questions may come to mind, from right to left that snapped wonder reverse sturdy?
If we do this right we multiply the unit does not come when you see the matrix.
What L1 nor L2, L3 is not right nor wrong.
Now a similar matrix more
Let us examine our previous matrix actually, let's call it the right way around.
We are writing the inverse matrix of the right to say again,
We are writing this time unknown matrix R to the right.
Now we know the initial size of the R,
but it should be two rows of the matrix R to be that you have this product
We see now, because otherwise the product would not be compatible.
Here, take this column inside the first line were three
We will receive the product, incompatible, because where there is no element opposed to the third line,
compatibility means that we find in the end that the two lines.
In addition the unit will also find here to be three columns of a matrix,
sorry, this product will be three lines here, a fourth E,
You can also find column if you receive three line again.
But in order to get the three R's column is supposed to be three columns.
The structure of the matrix here now turns out,
four columns so that even if you can not provide this equation,
You can not get the identity matrix, also achieved if a three-line
You can not, even if you can not get a line,
You can also get square if, but when this structure would be both consistent product,
It will be compatible, and can also remove the unit matrix.
Now we're faced with the following situation: Six
There are many unknown RA.
L'was said, but we did this time correspondingly multiplied nine values
Turns out, this matrix is a matrix equation opposing each unit.
Search these equations we write here, unfortunately here
yet the separation,
wherein R1 and R2 are seeing dependent in itself,
R1 and R2 2'yl own between two dependent R1 R2 3x three own
between dependent and independent each team and three,
Each also has three against two unknown equation.
There's a three equation against a R1 and R2.
Thus a R1 and R2, we find one of the first two
Have we put in consistently been inconsistent third will look at him.
R1 is, one turns out.
R1 R2 is when we put one here, one half interest,
Let's put here after an R1, it was a.
Two.
Here we found a split of two, two times two is also an RA,
that is inconsistent with the situation as we see three equal zero.
Similarly, two where R1, zero.
R1 is time we put one over two zero two minus two R1 R2 zero two here.
If two of R1 minus one half, so here is a minus,
minus one equals zero again comes as an inconsistent state,
wherein R means that none of the solution in itself,
wherein R and here too if you follow the three R 1,
If this second equation is zero zero three R2,
If you put three of R2, so a zero occurs here as a discrepancy it is equal to zero.
So there is no solution for the unknowns in this equation.
If we extend this, the matrix does not mean that elements or himself,
that means that the matrix corresponds to the left of the three vice-versa
There is no right opposite, this is an interesting observation.
We will then look at the square matrix.
In the square matrix we see that the situation is quite different from the overall situation,
In that regard, many times more efficient use of square matrices
used as is, off the more interesting results.
According to a rectangular matrix of square matrix
If there is a reverse from right to left in a matrix are counterproductive.
These two matrices are equal and there is always one on one.
If left or vice versa, there is no right and vice versa.
This is a very interesting building in terms of square matrices feature.
Now we will see them on the examples.
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