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.
Denklemlerden Matris Çarpımına / From Equations to Matrix Multiplication
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Doğrusal Cebir I: Uzaylar ve İşlemciler / Linear Algebra I: Spaces and Operators
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İki Bilinmeyenli Denklemlerin Öğrettikleri / Learning From Equations for Two Unknowns
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Attila AşkarProf. Dr.
Matematik Bölümü (Department of Mathematics)
[BOŞ_SES] Hello.
Now again we will proceed with the information we get from the equation with two unknowns.
Previously the two equations with two unknowns here
As we wrote, we found the solution to the overall structure.
We encountered the concept as an important determinant we see here.
These determinants are zero, or zero if the solution,
So there is no solution or infinite solutions.
But we see that the determinant is non-zero and that the only solution
We also examined the third example embodiment.
Now here we learn a little more topics
By examining an expansion to depth, we will progress to a new area.
As you know, this equation, together and as a coefficient
In that place, a one, a two of a two-one, a two x two to write one,
x wrote two of the unknown and a right side b,
b, we obtain a still compiles both two-component vector.
We call this Amp coefficient matrix.
x unknowns.
These column.
This is a fact that can be seen as a matrix, the matrix because one column.
As two columns and two rows of a matrix.
We see this as a matrix with one column.
Now we we wrote them: it was accepted,
We also went on the more we accept without stopping.
We will examine it a little more now.
We wrote that the coefficients.
x A x two of getiririp've written here as a product.
That is the product of a matrix with the vector x equals gives a vector b.
Now let's examine what we are doing with this a bit hit.
We're doing that; We take the first line in order to obtain the first equation.
We've written here.
We are writing because as a vector magnitude defined by two points.
x A x two of which we're writing as a vector.
This is not different from anything we have seen before domestic product.
We were doing it in the domestic product.
The first element and the second element of the vector,
We collect second element also hit each other.
Indeed, when we collect it the first line of the equation on the left
We're getting side.
When we wrote this, a equals b would remove the first equation.
We are writing here again for the second line as a vector equation.
We are writing as a vector x x two.
And when we get where we find inner product in the second equation.
Right side b eşitleyin two of two equations we have achieved in this way.
So a little more as a product of processes
the essence of our work looked.
Now let's do a little more ilerletsek a matrix.
Here are a factor, such as a one with a two-item two from four.
Whether the B matrix.
How can we do so that the product of these two matrices earlier rule
do not conflict and define a process that is compatible with it.
Now here we see the product of a column vector of a matrix.
That means water the first column of the second matrix
a b a a and b are two one let's say the first column.
Let the water two in the second column.
Here let's do the following operations; Once a product of a vector.
C In a say.
a time, the second column with the same product in again to get a car two times.
And they also brought the first column, and we obtained vectors
Let's write and say that that product is also A'yl matrix B as the second column.
So we have obtained is as follows: We are writing the first matrix,
We are writing the second matrix, which like the first column
While the process has been like no previous second column
We getiririy landscape of this column vector as we did,
We take the first domestic product this line, we are writing here after.
When we hit the first column to the second row to take the same again in the second
When we receive the product line, we are writing this second element of the interior.
And making a second column with horizontal take similar actions.
We take the inner product of the first-line or second-line to the inner product
We bring the vector we obtained from this're writing a second column.
Now let's apply it on an example.
One, two, three, four Let the matrix.
a minus five, minus two, let's get the three matrices and multiply them.
Now, of course you do not have to divide us, but every time this column
In order to do that we first column sensed the detail work
Let p be a vector, let's take the second column of the water as two vectors.
We are the product of a vector by a matrix, we know that.
We are doing that; We take the first line
we convert it to a vector, with one or two elements.
In this right, the second, we get the horizontal turning vector inner product.
We take the inner product of two vectors a minus.
Domestic product means the product of the means to collect and receive mutual components.
Minus four minus one equals three.
We do the same process for the second row.
We again achieved a vector from the second row.
Vector in the right-hand side, we obtain a vector to bring it horizontal.
Yet if you look from here to take the inner product of three minus eight: minus five.
This gave us the car we are also a column.
Similarly we repeat the same process with the second column.
We still get a column vector.
They also brought in last summer as the first and second columns of the matrix
When we say that the matrix multiplication of the two matrices obtained.
Thus, starting from a simple place we go a little further and
We encountered the concept of multiplication of matrices.
[BOŞ_SES] Now the product matrix
Get emerges as just two more possibilities.
One unit matrix inverse of the matrix in one of them.
Due to the following; If we had our one unknown one equation, ax equals b.
A scratch different though, so these are not the matrix,
we know the numerical scalar equations.
ax equals b, if not zero.
If we multiply this equation a is reverse that
The converse is also a means that: a negative one that hit the merger of a number.
As example, a reverse split of five is five.
Here a negative one will hit the left side of the combine.
Because a minus for a combined x such that x is because once a given
once the right side will be a minus b.
Now we think that generalization of matrices where only
There are also two binary matrix, but usually we do this.
We meet as a concept.
When we say that the matrix be the reverse of that of the matrix multiplier,
We also had the concept is a unit matrix instead of a number.
So we are faced with a sudden two new concept.
The opposite concept to the concept of an identity matrix.
See, we apply it to the matrix equation Ax equals b,
Minus merger left to multiply.
We hit the left here, minus once,
We hit minus combine to ensure equality in the right course.
Negative because it is combined with a matrix multiplication unit,
Here is a how to fell x're just.
From here, the same one the matrix equation
we are able to bring the structure of an unknown solution to the equations.
This is an important gain.
We know how to take action because the flat number.
We see the same process can be performed with matrices.
We met with two key here is, as I understand.
One unit matrix, it's there; We know a number of flat
It also reverses left or right with the number we find a çarpsak.
For example, a five though opposite a trailing five.
He left in a split feeder çarpsa your çarpsa right we find a bear.
And the number of units which feature the making of the hit X again you hit the XO.
This brings us right this: it's equivalent to the alpha 1 a'yl
Once brought a- 1 1 instead of 1 is the matrix multiplication.
So one of the right and left matrix multiplication
We came to the concept of matrix units.
If so, how is he hit the 1-up x x veriyor any of these units matrix
I would like one that also hit the x to be given by a vector x.
So we do not really know yet the identity, then again in Western languages
I come from the capitalization of the ID i.
To ensure that this agent and only then diagonals
provided that zeros in on a matrix with a diagonal from outside.
In fact, let's look to make it.
I'm with x multiply again before we develop rules
We make x1 x2 according to the horizontal, we stood JOIN with 0 to 1.
Only x1 staying here.
The second line horizontally, again with x1 x2
We take it to bring the domestic product, the X2 is staying here.
So then we provide that x times x equals.
In the identity matrix, the matrix unit English
or identity to, said the identity.
Telling them for tips,
Not everyone speaks English but must also look to the international literature
If those who want to easily he can understand those terms.
[BOŞ_SES], [NOISE] If we find then on,
so let us ask how we calculate the inverse of a wonder here.
We do not know yet subject to the reverse.
In which say unknown x.
We want to say a'yl x have hit the unit.
We want to be left from the left, but we have not seen yet unknown çarpsak unit.
His right to an unknown matrix diameter of the unit that we want to be.
We want viewed as given a coefficient
Similarly coefficients x, as we saw before,
We have seen the product a'yl to b, where b is a very special and so
b a'yl has hit a matrix that is the matrix unit.
For him, as for the unknown x is called.
Let's take it again the columns.
However, a column with a matrix multiplication, however, the second column of a matrix multiplication.
If we calculate the uid bring, let's apply the Cramer rule.
x 1 1 above, the denominator,
The right side of the first
obtained when we write the determinant column instead.
As you can see here, just a 2 2 remains.
They do this by dividing the determinant AA.
When we hit again by a second unknown x x 2
1 to find the gene that in a matrix
The right side this time we put in the second column.
The second column in a zero.
As you can see, when we calculate this determinant is zero - a 2 1 off.
We are writing here to bring that too.
[BOŞ_SES] and
so also a determinant, we find when we divide x 2 1.
Taking this second set, ie the second pillar of the unknown matrix with a
Cramer again multiplied in the Implementing Rules, the
To find the right before x 1 2 0 1 matrix,
0 1 vectors we put in the first column of the matrix.
As you can see 0 1 comes instead of the first column.
Other parts of the same.
Here we find the product of the diagonal coefficients of these determinants,
As you can see here, minus the product of the second diagonal.
It should be a division determinants.
Cramer's rule because he says so.
Obtained using up the right side of
divided determinant of the matrix, determinant of this coefficient matrix.
But in both cases because of the same coefficient matrix
that the determinant of the same.
It's starting to take shape.
Here we see when we found the 2 x 2's always a determinant
It appears in the denominator.
Again we encounter the determinant.
And it will be an important determinant to say but
If we can not do zero determinant in this process.
A- if not zero, we can find one of these factors.
After digital processing will do.
We have found right a- 1.
We observed that when we hit again left to the unit matrix.
[BOŞ_SES] Current calculation
We have made the chain: Given the matrix, now that every time
The first matrix to deal with the determinants vesaireyl
See arrived at the inverse matrix 2 2 here.
First diagonal was the first item.
A 1 1 is diagonal, the main diagonal elements of the underlying.
These are not the way to get close to the first one in the matrix,
1 1 column and row where I close by a 2 2 remains.
We are writing to bring it instead.
Similarly, the rows and columns of a 2 to 2-to-1 lifted
We are writing this instead of staying 1 2 2 brought in it.
The number of non-diagonal stays that way.
2 found that a 1 buy milk and lifted a 2 1 remains.
Here is staying in a 1 2, ie the second diagonal
Numbers also are changing place, going substituted in the first diagonal ones.
Something quite systematic.
We found this, we found in a previous row
We want to achieve the results we want to produce them again.
As you can see there are negative signs.
And here came a 1 2, here came a second one.
But we had a cross hair marks.
This increase their location changes, we change the sign.
We try to understand it like this: Here are the pros, cons latter.
On the diagonal plus, minus, except diagonal.
Then we are changing their place,
because it is on the reverse below a 2 1, 1 2, above.
We are changing the place of them.
This cofactor in the first matrix is called the matrix.
Although we are experiencing here in the times still 2 by 2 matrix N.
cofactor in the matrices of the same and do the same process.
A little differentiation to be made, but of course these accounts will remain intact chain.
This is the second diagonal those areas,
to change the place of trying to get the number transpose it.
Ousted in Turkish is also used, namely the matrix transpose.
You get it, you knock,
You mean like this reflect the returns on a diagonal.
The converse is also the matrix, the inverse of the matrix,
the matrix so that when it hits its own unit,
wherein the obtained cofactor transposed
to divide the determinant matrix transpose,
but here are a number of determinants of a matrix.
We divide this matrix, the number of these determinants.
As we saw in the previous solution,
determinant in the denominator, the denominator here, of course.
Just have the following questions, have always a- 1?
We see that there is not always a-1.
If there is a-one-zero determinant.
But the determinant is zero if no a- 1.
This is not surprising,
in solid number if you do not and vice versa If a zero,
You will not find a number such that you gave when you hit a reset itself.
Therefore it as a generalization of the determinant
Lack of zero turns out to be necessary and sufficient condition.
Sub-matrices have a variety of names,
It called sub-matrix determinants.
Do minor, the minor's coming term.
It can be defined as cofactor conjugate.
It also depends on the following: the a, i, j, ie diagonal
The one minus with a plus on an item in a
Multiplying that remains as one minus the second force is a plus.
When we come here, it's a plus for minus three is two.
It also means that if we go on this way, we generalize to larger matrix
The same rules will apply and when we find determinant in this way.
We can really make a.
a'yl multiply a negative one.
In general we see here.
A bit later, we'll see just numerical example.
Indeed it on a diagonal and a diagonal
on the determinants that are outside but when we ae in a split,
here is also a determinant of AA, we see that one coming.
Let's look at an example: The first example we've already done.
This factor one, two, three, was four.
There were two and zero in on the right side.
So here we were, we write the equation ax equals b and fix it.
Here we solve, we find x x two are one and we know that.
Let's do it by using the inverse matrix.
Once the determinant of a matrix four times a minus
Three different twice as zero.
According to our previous finding means that there is a negative one.
We find a negative one as well.
We take a bottom here we find the determinants of the matrix.
So we're covering the place where one is located, we remained four.
We are writing to bring it where there is one.
We do not have the row and column of the two is located.
Here are three places where it is to bring these two are writing.
Similarly, remove the columns and rows of the three is two stays.
Bring two of these three, and finally we write instead of
Four columns and rows that do not have to obtain a stay of the back.
We are writing here to bring in someone.
Lower than that of these minor matrix
Do we have obtained from the cofactor in passing.
This also a negative factor for a negative one,
As you can see we get this number by multiplying one here.
Transpose that to have overturned the requirement.
He comes from the same mold those on the diagonal to rotate around the diagonal.
Minus coming to where he finds two of three.
Minus two to three, three to minus comes from where.
the determinants of a division of a matrix inverse is also the cofactor matrix.
Determinant a eksi ikiydi.
There are primarily two minus one over.
A four minus two, minus three and one.
As you insight here we hit them with all the elements hit the minus two,
minus two divided by three divided by two and two turns.
So we found the opposite of a matrix.
Just have to make it.
a'yl a negative one should not hit the unit matrix.
Again a minus au slammed the merger should give the identity matrix.
If we really make this product,
We make the first right-hand column of the horizontal work again.
We take the inner product of the first line.
See plus or minus two
Three of them stand out because two plus two is divided by three, minus two plus three is the data.
We found one that means.
Again, this can multiply the first column of the second line to bring it horizontal.
Three times two minus six minus three over four times two, plus one in six.
So minus six, plus six zeros giving.
We found zero.
Again, multiply the first row of the second column of the deposit.
One minus one divided by the product of two times two minus one again so gave it a zero.
Again when we hit the right side and laid on the second line,
three times, four times a minus
one half that three minus two, he gave a.
As we saw a times a negative one, gave the unit matrix.
Similarly, the çarpsak left, now I leave you with this process,
it also gives the unit matrix.
Therefore, we are to have obtained the supply.
How can we solve this equation using a negative one?
Just as we did there.
A number of unknowns in this equation even though the ax equals b.
Also nonzero.
If we multiply a negative combine, left x stays
Or, as a negative right to be more up to date once we divide words,
wherein x remains, which remains a divide b.
It was written to say a little more abstract,
a negative one to use.
Because a negative one, one divides a.
When we do the same operation on the matrix,
We stood combine a minus, a negative product of a merger of a unit matrix.
X x matrix multiplication unit again.
Multiplying the right side with a reverse b.
We know b.
We found a negative one.
So a minus b to combine our çarpıtığı able to find the time x.
Indeed you can see that multiplication by challenge
as minus four plus zero, minus four stayed.
When we multiply two three here.
It was only three to zero again.
Whether it's with the Cramer rule, elimination of the traditional equation
The solution we found with the method, we see that the ritual.
If we do not provide it, so if we multiply x a'yl
indeed, we are writing a here.
Where a given.
we found that x is x.
This multiplication did, four times a minus, minus four.
Three times two is six plus six minus four, two.
Likewise, here minus three minus 12 four times.
Four times three, plus 12.
Minus 12 plus 12, it also gives zero.
So far two unknowns here always two
We tried the equation and have achieved important concepts.
When you have a solution?
There is only one solution, no solution has infinite solutions, or when?
We understand that.
This was also thanks to the determinant.
That reflects the meaning of the word determinants.
Determinants, determines the nature of the solution determines the meaning.
After this solution is somewhat more concretely
Cramer's rule we have developed in order to understand what happened.
The strikes we made a matrix with a vector of this process,
We now see that the multiplication product of the matrix here.
Inverse matrix and the matrix by multiplying the unit matrix, we get the concept.
We always have done in two and two, two equations with two unknowns
We did, but we'll see this much larger equation teams
For all these are largely available and is available with the same name.
Gauss elimination method that will now begin a little more practical solution.
For now, take a break here and
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