[BOŞ_SES] Hello!
Gauss elimination method we saw in the previous section.
That the equation with three unknowns,
We implemented as concrete examples of the unknowns and the general
The method is applicable in the equation and we saw how to do it.
In summary, at the end of Gaussian elimination process to bring on a diagonal,
We aim to be reset below the diagonal.
The final step in the diagonal on a number, n,
so we assume that the base number in the column n'yinc the n'yinc line.
Wherein the base also initially a 1 2,
a number, such as 1 3 of the right side of b1, b2,
b3 reached as a result of this the number of Gaussian elimination,
We believe that the numbers are.
If the bottom diagonal scratch in this number
This last equation is different
Of course there is equal bn xn base of mother,
From solving it
We could find and below
an increasingly upward unknowns in each row
From a bottom-line equation is happening because, for example, where it is known that xn
unknown to our right were able to get up.
Gauss-Jordan is a step beyond that.
Just under one and a diagonal zeros on the diagonal
It not aims to bring the zeros on the diagonal.
This again is a line that is the same basic process completely certain
adding a top hit with a number or not
where the main operations carried out by bringing on a diagonal.
If we can get this structure, it is now time
You do not need to find a new process for the unknown.
We read directly unknowns.
Indeed, these are the first line x1 equals the value B1.
x2 b2 have equal value.
i'yinc on line x equals the b each have the value i
There is an unknown and therefore no longer needs line b to a new account.
Is completely unknown are compiled in the right column.
Now let's see how we do this by applying to the previous example.
Our first example was as follows: start with where we have seen coefficient
and we eventually came to the following structure of Gaussian elimination.
We see that the only solution to that one on the diagonal
because x3 We find here.
x2 x3 to go to the top of a, we find x1 go to the top.
Now the diagonal of qualifying Jordan
We intend to bring to the position on zero.
Now this here in the second row minus
instead of the two that we want to zero below this one.
Instead, one on top of the one we want to zero in on this gene.
It will make the process very simple.
See if we add the last line I hit the second line with that place will be zero.
31 i will be attaching to that hit 13 to 75.
If you multiply that do not deal at all with zero because zero is what hits zero.
It does not change anything in the upper attaching.
For her this nice of Gauss and Gauss-Jordan elimination method
side purifies us from unnecessary work and saving you time
avoid making mistakes in the process can proceed even by.
These are computer software for very large systems.
The work is done with software that translate into computer language.
Now we get to zero as you can see here, but this is the third diagonal
One of the elements we need to be reset on the last.
It is also very easy.
Let's write this equation here again.
The last line of the first row hit minus merger
If we add this to see where the next one zero.
With that i've hit 31, 62
but we need to combine negative impact.
When we add in this, I said it wrong here,
This last line minus multiply merger will add here.
This occurred where one zero.
We hit minus 31 to minus 31 combined.
25, we have added five and 26, but with minus 26 occurred.
We finished this cycle.
After one of these things to be done to bring the zero on the second diagonal.
Let's write it here equation.
We have the following structure.
These two cons are adding a second line to hit two plus two is reset.
I. attaching coming here with that hit zero.
I've crossed that of 75, 150.
Attaching 124 minus 26 arrived.
I repeat again.
Others do not have to deal with because of the hit reset to zero if you multiply this bout,
Add here does not change anything.
This does not change what you hit zero this one and from there directly
We can read this unknown, and the same results
or only with direct elimination
We get the results we found by Gaussian elimination.
The advantage again
No need for a procedure to calculate the unknown: Direct
We identify the unknown to the reading the right numbers.
When we look at the second example was again slightly different coefficients
where the number was lower in the previous five it was here.
The challenge of this process, we have the right structure to structure what we see here.
In the last line, zero, zero, there are zeros.
These coefficients are unknown.
There are also known side 31.
So here are 31 such a discrepancy is equal to zero, we see contradictions.
Therefore, this equation will allow any x1, x2,
x3 can not unknown.
Therefore there is no solution of this equation.
Indeed, we have reached the same conclusion it with Gaussian elimination.
When we look at the third example again, there's a slight difference.
A previous example, the same coefficient but this coefficient matrix
The last element of the trailing line 21 minus the trailing line.
When we do this process, here at the end of Gaussian elimination
We arrived at the building we saw that all elements in the last row zero.
That equals zero zero.
This is not something new, but consistent results to us layer.
Therefore, there is a solution to the conflict but does not contain infinitely many solutions
Although initially it will have three unknowns because we see three equations
in essence they actually had three against two unknown equation.
Because the last equation equals zero, an equation that does not contain information such as zero.
Thus, if we take this equation it remains behind these two equations for three unknowns.
These equations are also, of course, is not enough to determine the three unknowns.
We can choose any unknown as a parameter.
X3 to select the most systematic.
x3'et right to say let us quote.
T minus five in the first equation will occur.
The second equation will consist of 13 plus two t.
They are known for getting the right column.
T is known for the breed.
We were reduced to a smaller two binary matrix.
Here bring our goal of zero on the diagonal.
We see this again writing the equation here.
Very easy to zero.
If we add to that the last equation I hit the first zero income.
We do not need this hassle again at zeros.
It will not change what you have to çarpsa attaching one by one.
Thus we arrive at the right structure.
Namely: the last line we added two plus hit first
when I hit 13 two plus 26 plus five, 31.
Four hit two and two t i t t attaching the minus three t.
No account here again, as you can see by reading the unknown
We know without having to do it,
x 31 plus one equals two t, x two plus two equals 13 t.
x does not appear even here we choose three that t x three x three and t.
We still say we like the hard numbers and t'l
If the numbers compiled in two bundles, 31,
13 and the first vector to zero, we obtain the constants vector.
As the t multiplier as t'li plus three, two, consists of a vector.
Again, this space shows what we see as a right that before
This means that an infinite number of points on the correct solutions of our equation.
Although the solution is not a random endless eternity,
It can only be taken as points on the solution that's right.
However, there are an infinite number of three-dimensional point in three-dimensional space,
all of them are not the solution, we can not random points.
But is this true on those solutions.
This solved the equation we make a deliberation as AmAsInI,
one in Jordan on the diagonal Gaussian elimination results in the first equation
It came after we saw the numbers on the right side had an X occurs just here,
x two, three, and we see that we are studying x single solution.
The second equation was the last line contains a contradiction
such as zero equals 31, so that will allow this equation x one,
x Two x is not three and we realize there is no final solution.
In the third one on the diagonal equation
making zero equals zero as the last line does not contain contradictions
but we arrive at an equation that does not contain a description.
Threw the remaining two binary equation in the equation
When we get zero obtained on the diagonal
where t is the solution and the solution for all terms
t t values for finite values can be accepted,
We get infinitely many solutions and still is not a random eternity,
the infinite point on the right,
We define an infinite number of points would be the solution.
Now here we are finished with this section.
So I put a summary end of each chapter, we've seen this in previous chapters.
I encourage you to pass through them.
So what you have learned well,
You will be able to see that what matters incomplete information,
If you go back and look at the text to complete the missing information,
You can easily identify missing.
This is a very good summary, I think.
Because you can see here the basic concepts tidy together.
After passing the linear spaces and linear processor section.
So far, we have created an infrastructure in the first three sections.
The second and third sections of concrete linearly
We have seen, and the foundations of the matrix space.
We accept them as an infrastructure but immediately
also includes basic information on all of them, so that more concrete information
I hope this will help you understand the abstract structures.
There are two subunits.
There is a linear spaces.
We will see within this linear space of finite size,
We will see their function in infinite dimension space.
Linear processors shows the relationship between the two linear spaces.
Indeed, in a space in an equation ax equals b type x unknown
matrix allowing the relation between x and b in the known space in space.
We will see and we will see them more as a general matrix operations.
After finishing this part of the second section with square matrix
We'll see about examples but they are just very basic detail
all these two equations in two unknowns
We have seen the essence of the concrete.
