[BOŞ_SES] Hello.

Before finding the inverse of a matrix

We found the substantially cofactors method.

You will recall the determinant calculation

cofactors was also an important method.

But as well as the elimination of Gauss-Jordan

as well as a second method was used in calculating determinants.

Here, too, we can estimate can be used as the determinant

seeing here is very important and one of the cofactor of the process,

One of the Gauss-Jordan elimination, the opposite of the matrix

essentially cofactors and is based on determinant.

Therefore, the inverse matrix of the Gauss-Jordan elimination could consider the calculation.

Now the situation is this; a matrix when given

multiplied by the reverse unit provide matrix.

Let's write the unknown and take it now.

So a given, known.

The inverse of an unknown matrix of x.

So we have a square matrix times the size of the still unknown

n times the size of a matrix unit hit the matrix to be obtained.

Now what do we do with that Gauss-Jordan elimination?

Unknown We do not write,

because they do not bring additional information during calculations.

x A x A remains.

x Two x two stays.

The important factor therein.

Here we find the extended generalized matrix.

We are writing coefficients.

We find expanded matrix by adding the right-hand side.

A matrix and the identity matrix right here.

We are writing to bring the unit matrix here.

The aim is that the expanded matrix

still basic matrix operations

Using the left side to get the identity.

It means to achieve the unit on a diagonal matrix future

on the diagonal and also resets the bottom of the future.

How will this be done?

With three basic process, we have three basic process; change the order of rows, a

line to multiply by a number and add a line to the other to hit a certain number.

Three of them.

Let's do it on the samples.

Again, our goal is to start with and under get a zero on the diagonal.

The following example, let's start with a very simple situation.

We are writing this matrix, one, two,

three, four matrix are writing the right side of the unit matrix.

It was expanded matrix.

This unit will try to the left of the expanded matrix matrix.

Way to do this already exists on a diagonal, bring below zero.

The first line of minus three to zero to bring additional hit,

We are adding to the bottom line.

This challenge comes as the first diagonal below zero, we add minus multiply by three.

We hit minus three to minus six, we've added four; minus two.

We've added three multiplied by minus zero; minus three.

We hit minus three to zero; zero has not changed, we've added one.

We continue now again, we have a fetch on the diagonal.

The number on the second diagonal here to make an

We need to make our minus divide.

We divide this whole line minus two; a.

Minus three minus two divided by three when we cut it,

When we divide into two minus one minus one half.

As you can see below, we get zero diagonal,

now we need to make zero on the left side of the unit, let's create a Matira.

On the right side, then the reverse will occur in the matrix.

Let's add a second line to the first line and hit the former two to achieve this.

There was zero.

Multiplied by minus

minus three minus was when we multiply it by two.

Here again when we added it to the first line

We find a number of anti-minus've added two hit.

Here we hit minus two; It was minus three, we add one; minus two outputs.

Again minus two

minus one hit by a slash output, and one we add to zero; output.

As you can see on the left side of the unit matrix it occurred.

Opposite to the right side of matrix occurred.

A divide here

We can get out number two as a partner,

A split would be even less if you take the two of you would also avoid doing.

This provides a more compact, indivisible manner matrix de

We can write but a superficial arrangement in the second degree.

But more importantly, we found that A'yl and vice versa

We argued that we need to find matrix matrix unit gets hit right,

hit the left still we need to find the identity matrix.

We are writing to the first product.

We are writing to one minus this right, we have found here a negative one.

Is it a product of the two is not that we need to observe.

Take the first column when we hit first deposit minus two

three divided by two plus the two he hit two of three,

minus two plus three; a, lets say that we have one of those.

If we again take this hit the first column to the second row minus six.

See here for three splits in two the denominator

He brought up two of the six plus two will remain.

So minus six plus six; zero.

Others likewise first takes the second column

When we hit a line, minus one; It gives zero.

And finally we hit the second column of the second line again three times; three,

four times minus one half; minus two, three, minus two; a.

Unit matrix is composed as you can see, we have found the means to provide.

You do not need to do so already the second, it gives it, but just as identical

habit hands thoroughly and let this product to be credible in the other.

Minus two, one, three divided by two,

So you're writing a minus minus one left, we are writing to the right.

Slamming tres above order.

See also a challenge of this multiplication,

we hit them on the first line of the first column three times two plus minus three; a,

and so again the first matrix is the identity matrix is obtained.

We also found the same sample before cofactors method

In the fifteenth hundred pages, four pages before that.

There we find out that the same inverse matrix.

So make it a double, maybe no need to provide all of these, but

our house thoroughly to digest visually

I believe it is beneficial to provide this.

Now a bit more complicated, let's large size matrix.

Three three-pointers.

Here we are writing again three three-pointers matrix.

We are writing to the right unit matrix, this time three three-pointers unit matrix.

The first column of the Gaussian elimination again

We need to bring resets the bottom of the first diagonal.

Therefore, we are adding a second line to the first line hit minus two,

here is zero.

We multiply by minus two; It was four, minus've added five; It was a minus.

We've added two hit with a minus; It was two.

We've added two hit with a minus; was a minus, they have not changed.

This process not only left matrix,

We're all in the extended matrix.

To bring again under the second diagonal below zero this one,

We are adding a negative hit, again we came to zero.

We continue the process on the right side, it came in this way.

Now that we're done it first diagonal.

As we begin to combine the second diagonal, we have a fetch.

We stood combine to bring a minus.

This is a plus or minus two, minus two, it was zero.

In this line, the second line is hit by two

We have brought below zero when we added the second diagonal.

The same process, we are doing the right tarafat.

For example, we hit on the right side with two two; four, minus've added one; three.

We also continue like this right.

Here we have provided the diagonal we write again.

First on a diagonal and below there are zeros.

But the third, this will reset the diagonal top

I have to bring two zero's already here.

To bring this minus zero where two of

We are adding two hit with the second line.

We also make sure that the right side.

Although we have achieved unit matrix,

Despite three three-pointers in the left side of the matrix,

We find one minus three three-pointers in the matrix on the right side of this one,

We have achieved here as we have seen happening.

We can do it again supply.

Already we face this inverse matrix in the sixteenth page

We did so with the Laplace cofactor expansion, we have reached this conclusion.

But the sly come to our house a little more, let 's understand the business so visually,

Minus combine to before after before after or as a minus minus

As we see that we have achieved really hit the unit matrices of.

Thus the supply,

It is also ensuring that found that our account is accurate.

Already have the same inverse matrix we find the cofactor

this inverse matrix we find the Gauss-Jordan elimination, it is already providing a.

Now here's a

It had better break.

Last issue we say this quite Cramer's rule

used less frequently in small size,

very few unknowns in the solution of equation

We will achieve this is a convenient method of Cramer's rule.