Sign In | Subscribe
Start learning today, and be successful in your academic & professional career. Start Today!
Loading video...
This is a quick preview of the lesson. For full access, please Log In or Sign up.
For more information, please see full course syllabus of Linear Algebra
  • Discussion

  • Download Lecture Slides

  • Table of Contents

  • Transcription

  • Related Books

Bookmark and Share
Lecture Comments (15)

1 answer

Last reply by: Professor Hovasapian
Wed Oct 30, 2013 3:11 AM

Post by Fawzy Hegab on October 30, 2013

I want to thank you professor for this great Lectures!

2 answers

Last reply by: Professor Hovasapian
Sun Sep 1, 2013 10:30 PM

Post by Christian Fischer on September 1, 2013

Hi Raffi,
Great video! Just a quick question: You state that in Cramers Rule A1 is formed by taking the n'th column of A and replacing it by vector b, but to me at 57 minutes it looks like you replace column 1 (and not "n" which i consider to be the last column in the matrix) with vector b. Is that an error, or have i misunderstood that part?

Best wishes,
Christian  

1 answer

Last reply by: Professor Hovasapian
Mon Jun 10, 2013 3:18 PM

Post by Manfred Berger on June 7, 2013

I know I'm jumping ahead a good bit, but I'll ask anyway. Are you going to come back and clarify what the adjont means in terms of finite dimensional linear operators? Most authors seem to be happy to skip that concept altogether.

1 answer

Last reply by: Jimmi Aastrom
Sat Jan 5, 2013 9:33 AM

Post by Jimmi Aastrom on January 5, 2013

Michael;

It probably doesn't matter at this point. But the problem is that Mr. Hovsapian wrote the last entry of the solutionvector b as 3 instead of -3.
If you calculate the determinants using 3 instead of -3, you get -10, -24 and -50 respectively, in stead of -4, -6, -8 which are indeed the correct values for the determinants.

1 answer

Last reply by: Professor Hovasapian
Sun Aug 19, 2012 8:55 PM

Post by michael bentt on August 19, 2012

Mr. Hovasapian:

Using Maple, I attempted to find the determinants you were seeking in "Cramer's Rule" section, part of the Cofactor Expansions topic, however, my values are different from yours.

I double checked and I derived the following values: det(A1) = -10, det(A2) = -24 & finally det(A3) = -50 whereas yours are -4,-6-8, respectively.

I've also included a screenshot of my work:
http://s1064.photobucket.com/albums/u365/mikebbc/?action=view&current=06303fa4.png

Thanks in advance

3 answers

Last reply by: Fawzy Hegab
Wed Oct 30, 2013 2:47 AM

Post by Romin Abdolahzadi on October 10, 2011

His remarks are awesome! "There is no reason in the world to believe this is true. There is nothing in intuition to lead you to actually to investigate that this would be true. But when you start fiddling with things, when you start following logical conclusions, and seeing where particular logical mathematical derivations lead you, you end up with something really extraordinarily beautiful. The inverse, its adjoint and determinant is actually related! That's very very strange."

Cofactor Expansions

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Cofactor Expansions and Their Application 0:42
    • Cofactor Expansions and Their Application
    • Example 1
    • Example 2
  • Evaluation of Determinants by Cofactor 9:38
    • Theorem
    • Example 1
  • Inverse of a Matrix by Cofactor 22:42
    • Inverse of a Matrix by Cofactor and Example
    • More Example
  • List of Non-Singular Equivalences 43:07
    • List of Non-Singular Equivalences
    • Example
  • Cramer's Rule 52:22
    • Introduction to Cramer's Rule and Example

Transcription: Cofactor Expansions

Welcome back to educator.com, thank you for joining us, this is linear algebra, and today we are going to continue by discussing co-factor expansions and using the co-factor expansion of a matrix to compute determinants, and to also compute the inverse of a matrix.0000

Now we did talk about the inverse of a matrix last time, and although this particular procedure that we are going to show you today with co-factor expansions with something that we define in a minute called the ad joint.0018

It is the radically, useful computationally it's not something that you want to use for more than a 3 by 3 or a 4 by 4, but again for theoretical reasons, it's always nice to be introduced to it, and to see how it functions.0026

Let's just dive in and write now...0038

Okay, let's start with a definition...0043

... Okay, that's always we will let A ...0051

... Be an N by N matrix...0059

... And we will let M, capital Mij be the N - 1 times N - 1.0063

Sub matrix, and again these definitions are more for formal purposes once we actually do examples, anything that seems a little strange and unusual here will make a lot more sense...0082

.. Of A obtained...0093

... By deleting, the ith row, and jth column...0102

Now let's say so let A be an N by N matrix, let Mij be the N - 1 by N -1 sub matrix of A obtained by deleting the ith row and the jth column.0121

Now the determinant of this Mij...0133

... Is called...0143

... The minor...0147

... Of the entry...0151

...Aij, so for example if we had A32, that would be the third column, second entry, we would knock out that row and that column.0158

We would take the determinant of what was left over, and then that's called the minor of that particular entry, and again we will do an example and it will make more sense.0169

One mo0re definition, the co-factor...0178

... Of Ai, oops let me put an i there of ij, which we denote...0188

... Capital Aij, so co-factor, we use the capital, minor we use the M...0201

... Is the following Aij equals -1, raised to the power of I + J times the determinent of Mij.0211

Okay, don't let all these subscripts and i's and j's and -1's scare you, let's do an example and it will make a lot of sense, so let's define our matrix A over here.0228

We will have (3, -1 and 2), we will have (4, 5, 6), and the third row will be oops, excuse me...0241

... We will do (7, 1, 2), now M12, so M12, this is the sub matrix we get from crossing out the first row, second column.0252

If I knock out first row second column, I am left with (4, 6, 7 and 2), so we have (4, 6, 7 and 2).0268

And that's exactly what it is, first row second column, go back to the original matrix, cross out, and the numbers that you have left over, those form sub-matrix.0282

In this case 2 by 2, because the original was 3 by 3, so we cut and remember up the definition N-1, by N - 1, 3 by 3 becomes a 2 by 2.0292

Now if we take the determinant...0302

... Of this M12 and again the determinant of a 2 by 2 matrix is just going to be, put some parenthesis, it's just that times that - that times that, so 2 times 4 is 8 - (-42).0307

What we end up with is -34, so let's put that aside for a second, so we have the minor, we have the determinant of the minor, and then we have that other thing that we defined which is the co-factor.0322

Well the co-factor of (1, 2) is -1, and the power that you raise it to is the sum of this (1, 2) right here, so it's 1 + 2 times the determinant that we got.0337

M12, now -1 to the third power is -1, so it's -1 times the determinant we found already, which is -4, so our co-factor is 34.0354

Lets go through this again, we have a matrix, we have a minor, we have the determinant of that minor, and we have something called the co-factor.0371

Our matrix, this is our original matrix right here, let me actually use, so this is our original matrix, we decided to take the minor, the M12, which means crossed out the first row, second column, so we crossed out the first row , the second column.0380

What we are left with was A 2 by 2, that's our minor, it is a matrix.0395

When we take the determinant of that minor, we actually end up getting this number right here, so this is a number when you take a determinant , remember a determinant actually gives you back a number.0400

And then what we do is we find the co-factor, the co-factor (1, 2) is the determinant that we got, multiplied by -1, raised to the power of the sum of the indices, ! + 2.0410

Let's do another example, go back to my, actually let me go back to blue ink here, so let's see this time let's do, let's calculate the M23 minor.0428

When we go back to our original matrix and we knock out the second row, and third column, what we end up with is another, the 2 by 2, which is (3, -1, 7 and 1).0444

When we take the determinant of M23, it is going to be 3 times 1 - (-1) times 7, it's going to equal 3 - (-7), which is 3 + 7, we get 10.0461

That's our determinant, and now our co-factor A23, that's equal to -1, raised to the power of 2 + 3, the row + the column, times our determinant.0478

I will just go ahead and put that here, okay -1 raised to the fifth power is -1, so you end up with -10.0494

Once again, we have our matrix, the original matrix, we knock out the second row and third column, because we are interested in the M23.0505

Anf we have a matrix, we take the determinant of that matrix, an then from that we derive something called the co-factor, so it's the co-factor that's actually going to be the real important thing that we continue to deal with.0514

Okay, so now what we have this thing called a co-factor, as it turns out we can use it to evaluate determinants, so before what we do this, we use the properties of the determinants to manipulate the matrix.0528

Find as many as 0's as we can, may be factor out some numbers, simplify things as much as possible, essentially put it into upper triangular form if you remember from the last lesson.0541

And then just multiply everything along the main diagonal, well this is another method of actually doing it, and again computationally it may not be as efficient, it may or it may not, it depends on the situation.0551

But theoretically it comes in very handy, and it will make more sense as we proceed with linear algebra, but for right now let's just go ahead and work on actually finding a determinant using this co-factor expansion....0563

Okay, we have a theorem, okay, A = Aij and remember this symbol is just a short hand symbol for the entire matrix, all of the entries, okay.0579

A is N by N, okay...0597

... Then for each I, which is less than or equal to N, greater than or equal to 1...0604

The determinant of the matrix A is actually Ai1 times the co-factor.0615

I1 + Ai2 times the co-factor, i2 + so on as many rows or columns Ain, oops, that should be a small A.0624

All this...0645

... in times Ain, now there is another one of these four of the columns and but again we have a lot of indices, we have a lot of A's, you know lowercase, uppercase.0649

Instead of sort of throwing at a bunch of symbolism, let's just go ahead and do an example, and it will make more sense, essentially what this says is that I can pick a particular row or column of my choice, and I can expand that matrix along that row or column.0663

We will just see what we meant in a minute, that's all this is saying, that you can sort of add that entry of the row and, let's say you pick the first row, you can take the first row entry times the co-factor for that entry.0679

The second entry in the row and the co-factor for that entry, add them all together, and as it turns out, you end up getting the value of the determinants.0692

Let's just jump into the example.0701

We will let A = (1, 2, -3 and 4) (-4, 2, 1, 3) (3, 0, 0, -3), (2, 0, -2 and 3).0704

This is our matrix, now we take a look at this matrix and we want to make things as easy as possible for us, so we want to pick the row or the column that has the most number of 0's, because that way those terms just drop out of this sum up here.0728

They actually don't show up at all, so it makes, makes our life a lot easier, so when I look at this, I see the third row, has two 0's.0744

I am going to go ahead and expand along this row, now I am going to give you a little bit of sum as little bit of check board pattern.0753

In this case we have a 4 by 4 determinants, and as it turns out, remember that -1 raised to the power of the I + J, well as it turns out when you, I + J is going to iterate, 1, 2, 3, 4, 5, 6, 7.0764

And what happens is that -1 becomes -1 + 1 - 1 + 1, so instead of keeping track of all the symbolism, as far as the definitions are concerned, and this is going to draw out the +, - pattern for a 4 by 4, and also a 3 by 3 for you so that you know.0779

+, -, +, -, basically it's just alternating +'s and - all the way through, -, +, and you can never have a + next to a - vertically or horizontally.0798

that's all it is, so you can do this for a 5 by 5, 6 by 6 if you need to free yourself, that's +, that's -, that's -, +, -, -, +, -, +.0809

And here we have +, -, +, and you will see what this means in just a minute, -, +, -, +, -, +, so these are the things that you want to keep in mind when you do your co-factor expansion, okay.0820

Now we have decided to actually expand this along the third row, so here is what the expansion look like, now I take, I am doing 1, 2, 3, 4, so I am going to have four terms.0834

Well my first term over here is a 3, and if I go over here and if I take a look, I see that this is a positive, so that means that whatever term is there, in the end I stick a positive sign in front of it.0849

In this case it's going to be a positive and I like to actually put my positives in front of my positive values, it's just to have it that I have, it's up to you, negatives of ‘course, you need positives, you don't necessarily need, but for me it helps keeps things consistent and balanced.0861

3, now we said that you are going to expand it along this and the co-factor is if you erase that row and erase that column, what you are left with is this, this,. this number, this, this, this number and this, this, this number, a 3 by 3.0876

Lets go ahead and...0894

... (2, -3, 4), (2, 1, 3), (0, -2 and 3) okay, now we move to the next one, I'll go ahead and write...0898

... What I actually know, I won't write up, because this is 0, it's just going to be a -...0913

... 0, this is 0, so it's going to be a + 0, now our last term -, minus is here we put that there.0921

This is -3...0935

... Now we knock out that row and that column, but that's in and we are left (1, 2, -3)...0940

... - (4, 2, 1)...0953

... (2, 0, -2), so now what we have to evaluate is this thing right here, okay.0965

We have done a co-factor expansion, we chose our third row, because it has a couple of 0's in it, we took this entry times...0976

... The co-factor of that, we took, if we were to actually write out this one, we would knock out that entry, and then that entry, and here this one we took this entry times its co-factor.0987

Now we have this, so now we have a 3 times the 3 by 3, couple of 0's, and a -(-3) times another 3 by 3, so now let's evaluate this, evaluate this and we will put everything together.0999

Go back to my black ink here, I am going to rewrite what it is that we had, so we had 3 times (2, -3, 4), (2, 1, 3, 0), -2 and 3.1013

We have -0, we have +0, we have -(-3) times (1, 2, -3).1030

Again with linear algebra, there is a lot of minus's, plus's and numbers floating around, I'd like to write everything, I don't like to do it in my head, and for example I don't turn this into a +3, as I am doing it.1041

I wait until the end, so that I make sure that as I am going down the list, every symbol that I have is completely explicit and clear, not something that I have forgotten (-4, 2, 1, 2, 0, -2).1054

Okay, so let's go ahead and expand this one, now this one, we can probably do, we can expand it aong this column or this row, because it has a 0 in it, it doesn't really matter which one.1070

Let's go, you know what I think I am going to go ahead and take this row, red, I think I am going to go ahead and expand along that row, so I have got...1083

... The 0 is gone, so I have a -2, now I go back to my pattern, let me rewrite my pattern here, so that I remember it for a 3 by 3, +, -, +, -, +, - and +, -, +.1100

Here it's going to end up being...1115

... -2 times, and when I knock out that row and that column, I get (2, 4, 2, 3).1122

And then I have this entry right here, the 3, well that's a + sign, so i put a + sign, I put the 3.1136

And now I illuminate the row and the column what that belongs to and I am left with (2, -3), (2, 1).1147

Okay, that is going to end up equaling...1156

... 2 times 3 is 6, 6 - 4 times 2, that's -2, well okay so this determinant is -2, -2 times -2 is 4 times -1 is -4.1163

That's why I want to keep track at everything, and I don't want to do it in my head early on, I want to see every negative sign and every positive sign.1180

Here we have 2 times 1 is 2, - of -3 times 2, so - of -6 is 8, 3 times 8 is 24, positive, 2, so this equals 20.1187

Okay, so this thing right here, this determinant is equal to 20, now we will do this one, okay that will be our, let's call this, so we will call this number 1, that's our first determinant number 1.1204

And we will call this number 2, so now we are going to do the second determinant, now we take a look in same thing, we have a 0, 1 here, so let's go ahead and expand it along this, so we look down here, it's positive.1221

It's going to be 2 times, well we eliminate the row and the column what that belongs to, (2, -3, 2 1), (2, -3, 2, 1), the 0 doesn't matter, but I'll go ahead and put it in anyway.1235

Actually you know what, we don't need it, it's not that important, and the -2 over here, it is positive, so...1247

... -2, I would like to write it that way, and when I eliminate that row and that column, I am left with (1, 2, -4, 2), (1, 2, -4...1257

... -4 , 2 okay, and when I do this multiplication, I end up with 16 - 20 = -4.1275

Now this is, these are just the square determinant, notice I haven't taken care of the 3 and the - (-3), so now I am going to do that, so I have my final answer as 3...1285

... Times the 20, okay, now it's...1297

... - (-3) times -4.1304

And when you do all of that you end up with 48...1313

... Again this last part, let's do it again, I got 20 for this determinants, I have got -4 for this determinant, but I have the 3 and the - (-3) from the 4.1319

I brought those down, 3 times the 20, - sign is here, -3 is here and the determinant that we solved for this one is -4.1331

Make sure you keep track of all these, a single - sign will make you go in an entirely different direction, that's why it's really important to write everything out and do it this way., it is the best way to do it.1341

B is clear and explicit is possible, don't do anything in your head along the way, okay.1352

Let’s see what else we have here...1361

... Let's talk about the inverse of a matrix by co-factor, so what we just did was introduce co-factors, we did a solved determinants by co-factors, and now we are going to see if there is a way to actually find an inverse by co factors, and there is.1365

So far the way that we have been doing inverses is actually very wonderful computational method, so if I have some matrix A, which is N by N, I actually end up forming the augmented matrix if you remember.1380

Okay, let's just put that there, I ended up forming this matrix, and then I subjected this matrix to Gauss Jordan elimination, in order to end up with the reduced row echelon form, and if it actually has a matrix.1396

If it has an inverse, that A turns into the identity matrix, and the identity matrix turns into some matrix B, well as it turns out, that B...1410

... Is the inverse, wonderful procedure to do it, always works, if it doesn't turn out to be an identity matrix, in other words if you get a row of 0's or a row of inconsistence.1423

Then that just means that the identity matrix, that means that the inverse doesn't exist, excuse me, we are drawing around all kinds of terms, identity matrix, inverse, determinants, try to keep them all straight, I will try to keep them all straight, okay.1433

Now let's introduce another way using co-factors, start with the definition...1449

... Okay, we will let A = be the matrix, IJ, okay and it is N by N of ‘course.1457

Well let's just write it out, N by N, then...1470

... Then N by N matrix, okay...1478

... ADJ of A, of the, called the ad joint...1486

... Of A...1500

...Is the matrix...1507

... Who is I, Jth entry.. is a co-factor...1512

... JI, in symbols, it looks like this, ad joint of A and sometimes you will see parenthesis around, sometimes not.1523

A11, A21, An1, A12, A22, A1n, all the way down to...1538

... Ann, okay, very careful here.1559

Let's go through this, very carefully, the first thing that I would like you to notice, so we have, if we start with some matrix, and we want to form the ad joint of that matrix, okay, what we do is...1566

Let's say for example the...1569

... For that entry, for the first row, first column entry of the ad joint, we actually take the co-factor of the entry for the original matrix.1582

But what's interesting is notice that the entry for the second, for the first row, second column, is the adjoint for the second row first column.1596

Everything is sort, is transposed, so what you end up is that, the order is reversed here, so when you are going along a matrix, and you are going along the rows and columns of that matrix.1609

When you form the ad joint, you are actually going to reverse the order, so we will give you an actual procedure, the method for doing this properly, so that you will always end up getting it right, okay....1622

... Best method...1639

... For forming the ad joint of A, very simple 1...1647

... Form, the co-factor matrix of A and then another words, the matrix of the co-factors of A, and again it will make sense when we do an example.1657

And then just take the transpose, that’s the best way to do it, take the transpose instead of trying to do each entry and try to remember where it goes.1673

Just do it for the matrix straight, and then just flip it along the main diagonal, okay let's do an example.1682

We have our standard matrix A, it is (3, -2, 1, 5, 6, 2, 1, 0, -3) okay.1690

And just to remind ourselves, I will go ahead and put +, -, +, -, +, -, +, -, +...1706

... Our pattern, okay, so now we are going to actually find the co-factor for this entry, this entry, this entry, for all of the entries, and then we are going to put them in into a new matrix, those entries, and then we are going to flip it.1716

Okay, so A11...1730

... = well, remember the co-factor, + then when you knock out this row and that column, you are left with (6, 2, 0, -3)...1736

(6, 2, 0, -3), when I take the determinant of that, it's 6 times -3, which is -18 - 0 +, so it's -18, that's that entry.1747

Now we will go to, where should I do it, I will do it, I will do it over here, okay, I know, it's okay I will do it over here.1762

A12, okay so this is the co-factor for the entry, first row...1778

... Second column, so it's a - sign, right, so - and then if I knock out this row, this column, first row second column, I get (5, 2, 1, -3) left over.1786

(5, 2, 1, -3) left over, 5 times -3, -15, -2 times 1 -17, - (-17), I end up with +17.1802

Again arithmetic is a real issue here, let me erase some of this here...1817

Okay, I want to actually be able to do it right underneath, so A13, so we are so on the first row third column, that entry is that, okay.1825

It's positive, when I knock out, the row and the column that, that entry is now left with (5, 6, 1, 0).1837

Okay, and the, 5 times 0 is 0, -6 times 1 + -6...1849

Okay, now let's move to the second row, so now we want to find A21...1858

21, well negative sign, when we come over here, the entry is a 5, we knock out that row, that column, we are left with (-2, 1, 0, -3).1866

That determinant is equal to -6...1886

We want to do A22...1893

...Positive, we are here, when we knock out that row, that row , that column, we are left with (3, 1, 1, -3)...1897

... (3, 1, 1, -3), this is -9 - that, you get a -10, okay.1909

Let's go to A23...1921

... A23, actually you know what; I am going to, so that we actually have something to refer to.1927

I am going to rewrite the original matrix here, and yeah, that's not a problem, so A, our original matrix was (3, -2, 1), (5, 6, 2), (1, 0, -3).1937

We want to be able to refer to it, and we said that the last thing that we are working on was the co-factor for the second row, third column.1954

Okay, so we go to second row third column, and row third column here, it's A negative, +, -, +, -, +, -, and then when we knock that out, and that out, we get (3, -2, 1, 0).1966

(3, -2, 1, 0), 3 times 0 is 0, -(-2) is + 2, -(-2), okay now we will work on the third row A31, positive, negative, positive, so it's positive, 3321 When I knock out that row, that column, I am left with (-2, 1, 6, 2), (-2, 1, 6, 2) when I solve that I get -10.1980

I have a (3, 2), so positive, negative, positive, positive, negative, negative, and I knock out that row, that column I am left with (3, 1, 5, 2).2014

Okay, I end up with -1, and my last one, the co-factor for the third row, third column, +, -, +, +, -, +, is +.2028

When I knock out that column, that row, I am left with (3, -2, 5, 6), (3, -2, 5, 6), and I should end up with 28 if I had done my arithmetic right, okay so now we can put all of these numbers into a matrix.2044

We can form...2064

... We start with, let's call it the, well we don't have to call it anything, the co-factor...2067

... Matrix, of A, okay that's our first step, we form the co-factor, then we take the transpose, so we put all of the numbers that we got in (-18, 17, -6).2078

(-6, -10, -2, -10, -1), so this one, the last one we did, and 28, and now after we do this, we want to take the transpose of this, okay.2095

let's go ahead and take the transpose, and we end up with the final matrix, which looks like this, so it's going to be -18, -6, -10, 17, -10, -1...2115

-6, -2 and 28, so I started with a matrix, I found the co-factors for each entry, in which numbers essentially.2137

And I take those numbers and I put them into, and I form a new matrix, okay, this is my co-factor matrix, and then that matrix that i got, I take the transpose of that.2152

And I end up with...2163

... This is my ad joint...2168

... It's a bit of a process, but this is how you do it, okay, that's the, here, now let's use our ad joint to come up with, the actual inverse.2174

We see, we will let A, be n by N...2192

... Then...2203

... Yeah let me write out the theorem and then we will settle with it, A times the ad joint of A...2208

... Okay, equals the ad joint of A, times A is equal to the determinant of A times the identity matrix, and the only reason the identity matrix shows up here.2217

Is that here we have two matrices multiplied by each other, that gives you matrix, remember the determinant is a number, so and fairer, i want to take turn a number into a matrix, and multiply it by the identity matrix.2231

It just means multiplying everything on the main diagonal, so it's a way it is converting a number to a matrix, that's what the only reason that it shows up here.2241

this theorem says that if I take the original matrix, and I multiply by its ad joint, or if I do it the other way, if I take the ad joint and multiply by the original matrix, they are both defined, because it's just N by N either way.2250

I end up actually with the determinant times the identity matrix, now again since this is an equality, anything that I do to one side, I can do to the other and maintain the equality, so let's just fit with this a little bit.2262

Let's take... let's take this one and let's take that one, we can, I just take randomly in order, so let's just go ahead and knock this out, the equality is still retained, but it doesn't really matter.2276

I am going to multiply both of these by inverse ion the left.2290

When I do that, the equality is retained, so the A inverse, times A times ad joint of A, equals A inverse times determinant of A times the identity matrix, okay.2298

Now, well A inverse times A , and remember associativity, A inverse times A is just the identity matrix, so I end up with the identity matrix times the ad joint of A is equal to...2309

... A inverse times the determinant of A times the identity matrix, and again the identity matrix as far as the matrices are concerned.2329

It is the identity, in other words, just acts like a 1, so when we do 5 times 1, we don't necessarily put that 1 there, we just say its 5, so, for all practical purposes, we can ignore these, and now since determinant is a number, I can divide both sides by that number.2341

Remember I can't divide by a matrix, but I can divide by a number, the determinant of A, so we can ignore this and this, they are just, they are there.2360

Now, the determinant, those cancel and I am left with the A inverse is equal to the ad joint of A, divided by the determinant of A.2371

This is my formula, so now if I am given a matrix and if I want to find the inverse, I can do it one of two ways, I can go ahead and set up that augmented matrix, and convert the matrix to the identity if it is such and the identity matrix will turn into the inverse if it exists.2383

If not then, the inverse doesn't exist, and here, or I can find the ad joint matrix, which again I can do very simply with mathematical software, it's not a problem.2400

And I can divide it, oops, I forgot to put the A, I can divide it by a determinant, well notice something really interesting here that determinant is a number.2409

Well, if the determinant is 0, we know that division by 0 is not defined, so in this particular case, if you know the determinant of a matrix is 0, that automatically tells you that the inverse doesn't exist., so now we have, we are going to actually write that as a theorem.2420

Okay, A is non-singular, and we will remember non-singular meant invertible, it has an inverse...2440

... Well, if and only if, which means is equivalent to the determinant of A, not being 0, excuse me, so if I am given a matrix, and if I take the determinant of that matrix.2455

Again with math software, however i want to do it, if the determinant ends up not being 0, I know the matrix, I know the inverse exists.2469

If the determinant is 0, this formula that establishes a relationship between the inverse of determinant of the ad joint tells me I can't divide by 0 that tells me that the inverse doesn't exist.2476

Okay...2490

Now, in the beginning of the lesson I mentioned that computationally this is probably not the best way to go about finding an inverse, but it is kind of interesting from a theoretical point of view to see that if you start with some inverse, and if you end up sort of fiddling with the numbers, the square array of numbers.2495

And you end up with something called the ad joint, or if you divide by it's determinant, it ends up being related to its inverse, this is what we look for in mathematics.2511

There is no reason in the world to believe that this is true, there is nothing in intuitions to lead you to actually investigate that this would be true, but when you start fiddling with things, and when you start sort of following logical conclusions and seeing where particular mathematical derivations lead you.2519

You end up with something really extraordinarily beautiful, the ad joint of matrix, its inverse, it's ad joint, and it's determinant are actually related.2536

that's very strange, and again there is no reason it should be that way, but there it is, we will lose dated a fundamental fact about nature, a fundamental fact not about mathematics, but about how numbers behave, how collections of numbers behave in this case.2546

Okay, let's go ahead and add to our list of no-singular equivalences, you remember last time we have a list where we said if A is non-singular meaning invertible, we can draw other conclusions that are equivalent to it.2562

like we said before we are going to continue to add to that list, and the list is going to get rather long, but it's going to be very powerful list...2574

... Okay, so we have, A is non-singular invertible, okay...2587

... that's the same as saying that if I take A and multiply by some variable matrix that the system, the linear system, the homogeneous system AX = 0, has only the trivial solution.2597

Okay, so if its invertible, if it has a matrix, the AX = 0, the homogeneous solution, so just by knowing that something is actually invertible, taking a determinant and realizing that the determinant is not 0.2609

I can tell you something about the homogeneous solution, I have made a qualitative statement about it, I don't have to worry about trying to finding it, because it tells me it's only trivial, it's not worth finding.2623

A is row equivalent to the identity matrix meaning, with those manipulations of exchanging rows, exchanging columns, multiplying one by the other, aI can convert it to the identity matrix.2633

AX = B, has a unique solutions of every b, so if I have a matrix, an N by N matrix A, and I know that the inverse, I know that the determinant is not 0.2644

that means the inverse exists, I know that the, if I have, that the non-homogeneous solution, that there is actually an unique solution, not an infinite number of them, and not no solution.2655

And of ‘course the last one, which is the one we did today, if it is non-singular invertible. that's the same as determinant not being equal to 0.2667

Okay, let's do an example, okay we want to compute the inverse of the following matrix if it exists.2677

Let's try, let's try blue ink, how's that?, so we have our matrix (4, 2, 2), (0, 1, 2), (1, 0, 3), so we have a 3 by 3 matrix.2688

First thing we want to do is find it's determinant, okay, so the determinant of A, I am going to do a co-factor expansion, and I think I am going to go expand it along this column.2705

I could do this this row, because a 0 in it, but I am going to go ahead and expand along this column this column, so again let me over here, write +, -, +, -, +, -, +, -, +.2722

When I go according to this column, the first term is going to be a +, second is going to be a -, which it doesn't matter it's a 0, and it's going to be a + over here.2733

The determinant is equal to 4 times; I knock out 1, 2, 0, 3...2742

... - 0 + 1 times (2, 2, 1, 2).2751

1 times is 3, -0, 4 times 3 is 12...2761

... And this one of ‘course doesn't matter, 2 times 2 is 4, -2 is 2 times 2 so it is +2.2768

Our determinant is 14, it’s not equal to 0, so that the inverse exist, now let's actually find it using our formula, okay.2779

We want to form, our first step is we want to form the matrix, the co-factor of A, we want to take A, which let me rewrite it again because just to make a little more clear what matrix we are dealing with.2791

(4, 2, 2), (0, 1, 2), (1, 0, 3), if we want to form the co-factor, then we want to take the transpose of that okay, so let's go ahead and do our co-factors.2804

A11, so i go up here, A11, that means I knock out this group, this row, this column and I am left with (1, 3, 2, 0).2819

The determinant of that is 1 times 3, is 3 - 0 is 3, its positive, A11 is 3...2831

... A12, I go up here first row second column, I knock that out, I knock that out, I am left with (0, 2, 1, 3), okay.2843

And since this is going to be a negative, 0 - of -, this is -2 is the determinant, but I have stick a negative sign in front of it, so it ends up being +2.2855

A13, I knock out that column, that row, I am left with (0, 0, 1, 1), so it''s 0 -1 is -1, and it's positive, so that's equal to -1.2870

A21, I will do one more, and then I will just write down what the others are, so now we are on the A21, which is second row first column.2887

I knock out this column, knock out that row, I am left with (2, 2, 0, 3), 2 times 3 is 6 - 2 times 0 is 6, so 6 - 0 is 6.2898

However we are expanding, we are using this co-factor, which is a negative, so it's -6.2911

And then when we continue along this fashion, we get A22 will go +10, we get A23 is equal to +2.2919

A31 is equal to a +2, A32 is equal to a -8, and A33, I'll actually do this one, 33 we are down here.2932

Knock out this one, knock out that one, I am left with (4, 2, 0, 1), 4 times 1 is 4, -0 is 4.2947

Positive, positive 4, okay, so now I am going to actually put this, I did like this, because these are the numbers that you are going to arrange, your matrix is actually going to be (3, 2, -1, -6, 10, +2, 2, -8 and 4) a co-factor matrix.2958

Lets form that, so our co-factor of A is equal to (3, 2, -1, -6, 10, 2) and we have (2, -8, and 4).2978

Now we want to subject that to transposition, and what we end up with is just flip the rows and the columns.2998

(3, -6, 2, 2, 10, -8, -1, 2 and 4).3007

This is our ad joint...3019

... Okay and now we said that our determinant of the original matrix is equal to 14, we know that the A inverse oops, well that was interesting...3023

... We know that A inverse is equal to the ad joint of A, divided by the determinant of A, well we just take this ad joint which we just got.3046

Divide every entry in there by 14, so we get three 14's - 6 14's, 2 14's.3061

Now I wanted to talk to you about reduction, yes, 2 14's is 1 se7th, 6 14's is 3 7th, it's up to you, i personally, I don't like, I like to leave my numbers the way that I found them.3072

Reducing is fine, if you want to, if you don't want to it's perfectly acceptable too, I like the degree of consistency, i don't like 7's and 14's floating around in my numbers.3086

It just helps me, thus so i actually don't reduce my numbers, I know most high school math's teachers will probably kill me for that, but there it is.3093

Okay, so two 14's, ten 14's - eight 14's - 1 over 14, 2 over 14 and 4 over 14.3103

This final matrix is our inverse that we were looking for, again it the ad joint of A over the determinant of A, determinant exists, we are good.3119

If the determinant is 0, we don't have to bother trying to find an inverse, so very, nice theorem to be able to use that.3135

Okay, we are going to close off this particular lesson with a really beautiful theorem called Cramer's rule, you probably remember it from your high school course, i, perhaps you didn't see it.3142

I imagine most of you have probably seen it though, it offers a way to find a solution to a linear system that is the same number of unknowns as variables, same number of unknowns as equations.3154

Let us write out, let me go back to my black ink actually, let...3170

A11X1 + A12X2 + A1nXn.3178

A21X1 + A22X22 + A2nXn and then we will work all way down, An1X1 + all the way down to AnnXn.3190

This is a system of, oh let me put my solutions here, B1, B2, all the way to Bn, so this is a linear system.3212

N equations and unknowns, M by N system okay, the theorem says if the determinant of A and the determinant of A is just the coefficient matrix, in other words I take all of these coefficients...3225

... And then I put them in matrix form...3245

... If the determinant of A does not equals 0, then...3249

... The system has the unique solution...3259

... X1 is equal to determinant A1 over determinant of A, X2 = determinant of A2 over the determinant of A etc.3273

X3, X4, okay and let's finish this off, we are...3293

... An is obtained...3306

... From A...3314

... By replacing the N'th column...3319

... Of A, with the vector B, which is the solutions, and again it will make sense when we actually do an example here.3330

We have the following system -2X1 + 3X2 -X3 = 1.3341

X1 + 2X2 - X3 = 4.3356

And -2X1 - X2 + X3 = -3, okay.3365

Let's do our matrix A, it is just the coefficients, let me do it in red here, so we have (-2, 3, -1, 1, 2, -1)....3375

... (-2, -1 and +1) good, and let's do our B, vector m, which is just spread here, (1, 4, 3) okay.3392

My X1 is going to equal, now when I take the determinant of this, let's actually go ahead and write out what we have.3407

I am not going to go through the process of it, I use mathematical software and the determinant is -2, okay, so now when I take A, let me erase this here...3416

... A1, I get A1, by replacing this first column, 1, with that number, so(1, 4, 3) with this whole vector I mean.3434

3 and then (3, 2, -1, -1, -1, 1) okay, when I take the determinant of that I end up with -4.3447

therefore X1 is equal to the determinant of A1, which is -4, over the determinant of A - 2 is 2.3461

This is my final answer for X, A2, well it's equal to this with that second column replaced by that vector, so I end up with (-2, 1, -2, 1, 4, 3, -1, -1, 1).3472

I subject that to determinant, mathematical software, I end up with -6, therefore X2 = -6 over -2, which is equal to 3.3493

And my final A3 is equal to (-2, 1, -2, 3, 2, -1) and the third column is replaced with that.3507

(1, 4, 3)...3522

... I subject that to the determinant function and I end up with -8, therefore X3 = -8 over -2, my final answer is 4, X1 is 2, X2 is 3, X3 is 4.3528

For my coefficient matrix, and I take the determinant of the A1, A2, A3 matrix that I obtain by sticking the solution vector in the particular column for the, for X1, X2. X3 and I solve it that way.3547

Thank you for joining us at educator.com, we will see you next time.3568