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[BOŞ_SES] Hello,

the previous department has an infrastructure

preparation and generalizations

We saw the plane in terms of vectors used to.

We made transactions with these vectors.

So now again more concrete examples to strengthen infrastructure

two unknowns on the two equations We want to review what they teach.

In fact, this may seem like a simple problem.

Because middle school since we solve two equations with two unknowns,

but the unknown,

The equation, in very large numbers, even though the basic up and down

Topics There are two unknowns in this equation.

Now we are starting here three numerical examples for it

We want to remove that features what you like.

For example, something important determinants.

Something important to know whether the solutions of equations.

How is the transition to the matrix here will see us very concrete information.

Let's start with the two unknowns equation as follows: In general, a,

b, c, d, e and f are numbers given coefficients

Now we have two types, namely whether this equation.

Them as a preparation a, b, c of the two

we have set the indicator 1 1 1 2 2 1

A second factor of 2, and the term on the right side of b1, b2 define as.

Yet given these numbers.

But our unknown x1 and x2.

I.e., where x, y instead of x1, x2 are putting because these indicators,

we used index will be very easy to work with them to extend to n.

There is a point in this geometry.

We have seen before.

How do you get out of the equation, the intersection of lines or planes.

Their equation in each plane, an accurate equation of the plane.

When we get both, we are also examined the intersection of the two lines.

Of course we know that the two planes

When we get them right they can intersect at a point.

We have only one solution.

They are extremely useful convention parallel to never intersect.

We also have no solution, and two in a row

we can say that there are infinitely many colliding the solution.

That now, in this geometry of these concepts in algebra

We have seen what it means.

The only solution is no solution or an infinite number of such inability is not the solution.

This may seem very simple, but two equations with two unknowns matrix

many fundamental issues that we will examine relevant are included.

Now we want to start three numerical examples.

As you can see here, our first equation (x1 + 2

x2 = 2) (3 x1 + 4 x2 = 0).

Close down in the second and third equation coefficients but together

equation with minor changes.

But as we shall see will be very different after them.

Now we know also that we have used since ancient times, two equations with two unknowns

these two equations for the coefficients of the equation when it upfront

When we collect or remove disappears when one of the unknowns.

For example: If we multiply this equation the first three to

see here consists of three x1, 6 x2 occurs here.

Here, 6 is formed.

We remove them from each other x1 disappear.

He had been here 6 × 2, 4 by 2 x2 x2 interests remain.

There remains six.

These equations, we find that 3 x2.

x2, 3 was brought here when we put 12

3 split comes at a minus sign.

We can see that x1 minus four.

We found the only solution to that is to say x1 and x2,

They intersect in the geometric mean of the two lines that point.

If we take the second equation, see there is a slight difference,

came here three instead of two, the rest is the same.

I hit the second from that of the first equation

As you can see subtracting 2 x1, it will give 0.

It will also be 4 * 2 0 here.

There will be 4 will be -4 to our interests.

See equation came the following structure: 0 x1 x1 were staying,

0 x2 = -4 x1 and x2 so that what you receive will be zero left.

As we came to an inconsistency.

Minus four equals zero should be to provide the solution to this equation.

Of course, such a contradiction is not providing this equation is not possible to define x1 and x2.

What you get here are taken x1,

If you can not provide what it takes to get this equation x2.

No solution of this equation that means.

Yet little different

an equation, again X2,

If we want to destroy x1 I multiply the first equation is doubled, the second from Let.

This time something interesting happens.

X1 and x2 each other in the left-hand drive

The numbers on the right side as taking each other.

Then we come to the equation as follows: 0 times x1 + 0

Once you have what it takes to get x2 = x2 or x1 0, 0 = 0 turns out.

So you take any x1 and x2 is that you can make the equation.

This means that there are infinitely many solutions.

See the example that even a simple equation

The only solution for the intersection of two right in this geometry,

or that there are no solutions at a point of intersection,

If you get two lines parallel but separate line,

x1 and x2 can not provide any of this equation and the equation as well.

Geometric them as there is no solution, they intersect.

In the third case there are infinitely many solutions.

If these two lines parallel to each other and overlap,

each of x1, x2 will equation thereof.

As we see, there are still a consistency between algebra and geometry.

Again, as we said before, the geometry to move in three dimensions in two dimensions

When we go beyond the three dimensions possible, but we can no longer benefit from geometry.

But the benefits of forming a front geometry's intuition.

If we draw this equation consists of the following varieties confirms, as I said just now.

L1, right opposite from the first equation, L2,

They intersect at a point right and their counterparts from the second equation.

He had been parallel in the second equation.

There was no common points.

Therefore, we say no solution.

Üçüncğ solution in the L1 and L2 are in conflict, even if different coefficients.

Indeed, if you look at this equation the second equation, the first equation is two-fold.

Therefore, it is true that here impr this equation,

The same is true of this equation is true that impr.

Both conflicts.

And the general solution of the equation in two unknowns let after seeing these examples.

Where a 1 1 1 2 coefficients of these two variables Two

Let's try to solve the equations in general.

Yet we are moving with the same approach.

So here we bring x1 or x2 is the only unknown by hand.

See first equation as a 2 2 with çarpsak wherein the right side of the off

Let separating said, we are right also been hit with a 2 2.

The second equation multiply by 2-to-1.

Here again we multiply by 2 and subtract A1 Or, minus multiply by 2-to-1.

It was the same in front of the x2 factor see when we get these two equations.

Our strategy was to destroy the already x2.

For this, the coefficients of the first equation of the second equation,

We hit the coefficients of the second equation in the first equation.

Remove them when we see a situation as we are experiencing.

X1 have a common factor here.

A 1 1 2 2 making sure that we organize a somewhat aesthetic one,

A two-two minus one or two a two-one, come as a factor of x.

x deuces was not, in a twice on the right side

b a and a minus two, b two occurs.

Likewise addressed if we take the second unknown equation,

namely to destroy the first one here this time x

a second equation of equations combine two

If we subtract a hit a merger,

As we encounter something interesting, though two of x consists of a factor,

but we see that this same factor in both equations.

Right side are different, but left those same tactics.

So the equation is the solution of equation

x x two coefficients a and the common factor is going.

This solution sets.

In determinants say.

See show with det number you see here.

Now the term determinant of where it comes from,

determinant term in Western languages to English and French to determin,

It means to determine, so determining the solution,

The decisive solution used here means determinants.

You are, if we look at them in a term on the right side

determinant in the structure because you bumped twice the number in the determinants,

You stood other two coefficients, you are removing from each other.

Here in twos and differences have multiplied,

in twos and differences have multiplied.

Cleansers see them as follows, note down the numbers here.

xa this deta'yl is multiplied.

x deta'yl the two still are multiplied, the determinants of these factors.

So we want to find x one term on the right-hand side

We will divide the determinant of the x two again, we want to find the right

We divide the determinants of the term in exactly the same way.

This is an extremely important determinant of mean things, determines the nature of the solution.

To make this division is already well on the zero determinant

We have to be different.

If a non-zero determinant, we see that the only solution.

Determinant to zero, we face the following situation at that time.

Let's see if the first chief determinant of zero, zero times x equals a

We can not find the right one x is not zero.

We are also in this case there is no solution,

similar to the determinant to zero as x two,

if no solution is right is not zero.

The right side also becomes zero, then any solution x,

any solution would be an x two.

Then take out the infinite solutions.

Here is a little more concrete in the left side of zero in the first equation

x for a right side is not zero, c is a non-zero coefficient,

Or, as in any place, if similar

One is not enough, we can find a solution here.

There is no solution.

As you can see if the two sides are equal is zero zero zero

We meet like-compliant result, again, we see that an infinite number of solutions.

So even if we encounter an important equation in two unknowns understand,

determinant.

Determinant determines the nature of the solution, here in Western languages

Determin to the French and English, to determine, deterministic say

We are faced with a coefficient used in the sense given to Turkish,

This coefficient determines that the quality of this solution.

Now this one is a determinant of

A We found two two jobs with this product.

We should organize this as coefficients.

In the first row there was a a a a a two,

I had a two one two two on the second line, we head to the equation

coefficients in a first line of a two, two two two.

We identify the determinants of the case.

See the above numbers,

As we held in the denominator of the numbers a, b a b two here,

we organized a two a two-two here, it's reached the same structure.

Determinant we see that we calculate as follows.

We stood on this diagonal terms.

A gives a one-two-two.

We stood in the second diagonal coefficients

A one two, one two dying, we remove from each other.

B coefficient once that we look at here on the first diagonal

A two-two, the second in a two be two diagonal pantry,

This way we find them çıkarınca share.

To say that this is also a determinant of the structures and x

it is also a determinant for a deyel I also combine related.

Similarly, the two determinants in the right side this time

b a and b two terms of the second column

We are writing and again we find the term that he calculated the determinant of the denominator.

Therefore, if the determinant is non-zero determinant of,

determinant of coefficients, b on the right side of the equation

two counts as the first or second column of a B

We find determinant of shares he calculated as column writing.

In the same denominator Determinants partners.

Thus, a finding that the x and x two coefficients

We have achieved is a very practical formula.

This is called Cramer's rule.

Cramer determining this rule, anyone who's name.

Dig into this with our previous solution Now Cramer's rule.

As you can see this was our first equation works.

There was a slight change in our second equation,

He is only two of the three numbers, right side are the same.

In our third and left sides of this equation is the same

While staying on the right side instead of the four we put zero,

that is quite close to each other equations.

If we write them by organizing,

The coefficients a and two of them are writing the first line,

The second factor of three and four, we are writing them in the second row.

Right side of two and zero,

here again a factor of two,

this time for three instead of two and four,

The numbers on the right two and zero.

Here the coefficients in the same manner except that the right two and four.

Let's calculate immediately before and determinants.

The first equation of the determinants number one two three four,

We were beating them napıyım four times those of the two diagonals

We are releasing the first and second three times and minus, minus, we find two.

Now our previous observation that this is nonzero

We decided that the only solution based because it comes determinant denominator.

This is a one and a one,

We've been doing as a determinant of one and a two.

We put in place the first column on the right terms.

Nevertheless right determinants for second

We are writing terms as the second column.

Here he calculated determinant of the numbers on the diagonal, we stood still,

We stood here in numbers, we remove from each other, as you can see eight

We find here, where we find zero minus six, minus six.

Coefficients generated from the time we receive one det rates

When we divide the determinant, we knew before qualifying

We find the methods we have achieved four number x minus one,

x two for the two det when we get this rate Deta,

When we get to the ratio, we find three numbers.

If they instead take them to settle in the equation x minus when you put four,

I'm starting from the second equation, minus 12.

we put two instead of three times x plus 12 is indeed zero.

If we go to the first equation, minus four plus

We provide the equation that equals six and two,

We reach the solution we found with other types of screening and also equation,

we see that we provide the solution.

We do have a second similar operation

As you can see in the equation coefficients, one, two, two, was four.

One, two, two, four on the right side when we place here

term two and zero determinant here

once we account when four minus two times two,

We know that zero of a determinant.

We just give here the result is not the only solution of this equation.

What can, or infinite solutions, there are many solutions or no solution.

a one and two, i.e. first column

right sides or placing first,

the second column we have achieved by placing the right party

We see that the determinant of the matrix is zero.

So we look to the equation x one eight zero division,

x two minus gold, such as divide by zero or zero

You need to find one to eight times x, x no such number.

That means there is no solution.

Similarly, we need to find the zero six times x minus two,

We do not have such a number x two, means that no solution of this equation.

Already here now dictated that the only solution is zero determinant,

but to no solution of two possibilities

We can see an infinite number of solutions to be using the right-hand side of this.

The third team was this equation,

coefficients again one, two, two, four.

Right side, there are two and four.

Here again, we calculate the determinant, times the number of determinants,

minus four four zero four turns.

instead of the one we're writing right side for a first column.

Eight minus zero eight turns as you can see.

For the second determinant, the right side for a two second

We are writing the column, see here again comes four minus four to zero.

So this equation is zero divided by zero.

Or we write it differently, so be zero by

denominator that we have no right pane,

Let's make this a up so old, taking a zero times x

You have to be zero, if it lets you choose what you choose an x,

There is therefore infinite solutions.

We start at zero to reset the divide Similarly x two,

uncertain at this time to give another interpretation zero

x two summers when we continued uncertainty in again,

but we interpret a little different, we choose what we choose, we see that this is the solution.

Therefore, we understand that the solution to the infinite.

But we also use the solution, but we benefit from Cramer Cramer solution.

Now, um,

I want to take a break here, we see that the equation.

We have seen that there are solutions to the equation.

Now we get comments like what the factor here,