Birne Binegar: The average.
Birne Binegar: score on this exam with lower than the first one, I think, personally, I think the average was 89 meant a lot of people were looking at doing really well and, of course, on this exam the second exam the average was 84.
Birne Binegar: years old, it was a bit of a drop, but everyone is still doing relatively well I do want to give you one statistic that it's really striking.
Birne Binegar: I told you that that definitions would appear on the exam I told you exactly the four definitions that you'd be required to know.
Birne Binegar: As it turned out the students, the students who actually got all the definitions right that succeeded in memorizing them at least their average score was 97% so I mean and that's part of the reason why i've always insisted on definitions, because.
Birne Binegar: it's not just knowing the words but it's kind of hard to memorize without having a little bit of understanding as well, and it really helps you when you're asked to find a basis for subspace to know what you're looking for and so.
Birne Binegar: Those definitions really are a crucial pedagogical tool that I use, and I promise you on the final exam those four definitions are going to reappear again.
Birne Binegar: Until if you haven't got the definition quite memorized yet I still recommend you do so, because he are going to appear again on the final exam.
Birne Binegar: Okay, so that said, let me pass back the exams.
Birne Binegar: Abdullah not here, Mohammed.
Birne Binegar: Joseph.
Birne Binegar: De Leon.
Birne Binegar: Matthew Griffin.
Birne Binegar: Cameron killinger.
Birne Binegar: Oh sorry.
Birne Binegar: i'm looking at the wrong exam.
Birne Binegar: Actually, that was the first exam.
Birne Binegar: And he he's gonna have to pick it up and sell i'm not allowed to distributed against the friends and like.
Birne Binegar: All right, let me try again.
Birne Binegar: Not here.
Birne Binegar: David Calvin.
Birne Binegar: Johnson Hello.
Birne Binegar: there.
Birne Binegar: i'm still got the wrong i'm sorry I must have left them correctly again to my office.
Birne Binegar: i'll forward.
Birne Binegar: Mohammed Al.
Birne Binegar: Joseph.
Birne Binegar: The same.
Birne Binegar: All right, here we go Kristen Fisher.
Birne Binegar: carly.
Birne Binegar: Griffin.
Birne Binegar: Who.
Birne Binegar: Cameron calendar.
Birne Binegar: josh Lindholm.
Birne Binegar: Because.
Birne Binegar: Robert triplet.
Birne Binegar: Nelson Walker.
Birne Binegar: starts with.
Birne Binegar: I think starts with a.
Birne Binegar: With a J.
Birne Binegar: O D.
Birne Binegar: yeah I got a score for you.
Birne Binegar: Maybe got tangled up with other exams.
Birne Binegar: seem to have misplaced your exam but.
Birne Binegar: i'll bring it it's probably sitting on my desk, but if you want to see your score you got.
Birne Binegar: yeah.
Birne Binegar: You did very well.
Birne Binegar: i'll bring it next time.
Birne Binegar: Okay, so um.
Birne Binegar: I posted the solution to the exam there up on canvas, and so what I suggest you do if you have questions about how things were graded first look at the at the solutions and now.
Birne Binegar: You see how I sort of answer I was expected, but sometimes I make mistakes in grading, particularly if you reduce something to rational on form there's so many different versions of a matrix in.
Birne Binegar: One form that I may not recognize your solution is being correct so, but if it is if it's a bonafide roshan law enforcement, they given matrix i'm happy to correct that, but most of the mistakes actually.
Birne Binegar: were fairly simple you calculate a matrix wrong because you.
Birne Binegar: You don't remember to write things vertically, rather than horizontally, or you calculate the row space of a matrix instead of the column space of a matrix etc, but.
Birne Binegar: If you do have questions on how the examiner graded i'll be happy to address them.
Birne Binegar: But do have a look at the at the solutions, first, so we can.
Birne Binegar: Hopefully, the solutions will answer most most of your questions and, if not all, I can make adjustments.
alright.
Birne Binegar: yeah OK.
Birne Binegar: OK, so what I want to do today is, I want to make a little bit more progress on the term.
Birne Binegar: Oh i'll start off by reviewing what we did covered last time what determinants do for us and some.
Birne Binegar: Two different formulas for computing the determinants of an n by n matrix but.
Birne Binegar: As we've seen on Wednesdays lecture a while determinants are fairly easy for small matrix these if matrix easy to get moderately large like four by four beyond they get.
Birne Binegar: Really complicated really, really fast they're typically a polynomial in the entries of the of the matrix, but if you have like a.
Birne Binegar: For example, a six by six matrix if you actually want the determinant of a general six by six matrix it's going to be a polynomial degree six.
Birne Binegar: With 720 separate terms, so the determinants and some points are simple polynomial function, but in practice they can be really strenuous to compute until.
Birne Binegar: i'll go over the definitions of what the determinant is, but the main point of today's lecture is to give you a simpler way of calculating determinants that doesn't require you to work directly from the definition.
Birne Binegar: Rather, what we'll learn how to do today is how to reduce the matrix and get the determinant of the matrix very quickly that way.
Birne Binegar: Okay, so what's the determinant if you have a and make a square matrix own same number of rotors you have called.
Birne Binegar: The terminus supposed to be a function of the entries of that matrix such that, if that function is equal to zero upon evaluation, then you can conclude many things.
Birne Binegar: So i'll state this in the negative if the determinant is equal to zero, that means the matrix does not have an inverse and that also means that.
Birne Binegar: You have multiple solutions to any linear system where you use the given matrix as a coefficient matrix.
Birne Binegar: And it also is going to imply that if you have this particular linear system or the coefficient matrix times of variables equals zero.
Birne Binegar: That you're going to have more than just that trivial solution where you said X equals zero this will be an important consideration next week, when we start considering eigenvectors and eigenvalues.
Birne Binegar: It also means that he is not reducible to the identity matrix or that the dimension of the road space, the dimension of the column today's or equivalent Lee the rank of a matrix that's less than.
Birne Binegar: than the number of rows and the number of columns of the matrix now I stated these things, all in sort of a negative way in burst it's not exist or we we have more than one solution.
Birne Binegar: The fundamental theorems of linear algebra it's actually a con the negation of each statement what we proved just after the.
Birne Binegar: First exam was is that if a matrix has an inverse, then you only have one solution of such linear systems, in fact, this linear system has only the solution X equals zero.
Birne Binegar: And eight is reducible to the identity matrix if the inverse exists and the rank of a is equal to end if the inverse exists so.
Birne Binegar: The fundamental serum of linear algebra that we stated the fundamental theorem in vertical matrices each of these statements was is negated but they're all equivalent to one another that's just so I can take this one condition determinant a not equal to zero and directly draw the.
Birne Binegar: Conclusion but on the other hand, if the determinant a is not equal to zero, then the negation of all these statements.
Birne Binegar: A curse but we're actually going to be interested in this case, when, for the most part, what the determinant of a is equal to zero because that's going to tell us something about solutions have equations like this.
Birne Binegar: So it's going to be a polynomial function so yeah as soon as you have concrete entries and a then you'll be able to compute this and decide whether the number that results by evaluating that polynomial is zero or non zero.
Birne Binegar: Alright, so let me give you the formula for small matrices if you have a matrix with one row in one column then it's determinant it's just it's one entries.
Birne Binegar: notice that this is a linear function back to just the variable and it occurs to the first power and there's only one term there, we have a two by two matrix and while we have two rows and columns the determinant turns out to be the different to the diagonal products.
Birne Binegar: You multiply down this diagonal and then you multiply again down the Cross diagonal switching the sign, and this is a polynomial see each term has tube got two variables and factors and it's.
Birne Binegar: Just a polynomial in the entries of the matrix.
Birne Binegar: More generally, let me give you a.
Birne Binegar: sort of visual formula that you can also apply for a three by three matrix these I only did this.
Birne Binegar: For two by two matrix because these two formulas are pretty easy to remember, but if you had a three by three matrix ABC E, F G H I there's a sort of visual way of getting the formula for the determinant of this matrix sort of like this mimics this calculation what you do is you write.
Birne Binegar: The first two columns you write them again off to.
Birne Binegar: The side and then what you do is you take these dive on a product of this is a determined a is going to equal eight times eight times I.
Birne Binegar: And then B times X times g.
Birne Binegar: Plus C times D times, he.
Birne Binegar: And then.
Birne Binegar: we're also going to have some.
Birne Binegar: Just list here we have these.
Birne Binegar: diagonals from the upper right the lower left, those are going to come in with a minus sign or have be D.
Birne Binegar: minus a.
Birne Binegar: Sorry yeah as a eight and then CD G C, so there is a sort of visual formula for the determinant of a three by three matrix as well, you get that by writing the second first two columns again off to the right of the original matrix and just just taking diagonal products.
Birne Binegar: But for.
Birne Binegar: Four by four matrices and beyond, there is no such simple visual rule for calculating that eternity what we rely on instead, then, is.
Birne Binegar: A recursive formula for the determinant.
Birne Binegar: So, more generally, if you have a larger matrix and N by n matrix the way you calculate the determinant is by.
Birne Binegar: calculating determinants in terms of the miners of the matrix now what these miners are this is going to be, this is an n by n matrix this miner attached to a it's actually.
Birne Binegar: indexed by a row index and a column index of the original matrix a and what it is it's the matrix reform from a by removing from a it's a throw in jail call, so we take away one wrong, we take away one column we get an N minus one like N minus one matrix.
Birne Binegar: If we.
Birne Binegar: But this gives us a rule that allows us to calculate the determinant of an n by n matrix as a sum.
Birne Binegar: Of determinants of smaller matrices going to be basically one left row dimension in one life calm dimension than the original matrix.
Birne Binegar: Now, if you don't know precisely what that this determine if you could apply this formula again take this N minus one by N minus one matrix.
Birne Binegar: apply a similar formula and reduce the calculation of this thing into the sum of N minus two by N minus two and then keep applying this formula over and over again until eventually you get down to a determinant of, say, a two by two or you have an explicit formula for doing the computation.
Birne Binegar: Let me show you an example that here's a general three by three matrix applying that co factor expansion on this first row.
Birne Binegar: What i'll do is all march across the first row and i'll grab a term for each entry yet row those terms consist of.
Birne Binegar: The entry in the slot, then you have a sign factor that you get by adding the row and column index of the particular entry.
Birne Binegar: If that some is an even number, and this will be a factor one, but if you have if I plus J adds up to an odd number will have negative one to an odd power, which will be minus one.
Birne Binegar: So here's how this will work out.
Birne Binegar: by March cross the first row I have the first entry.
Birne Binegar: And then, let me put that formula up.
Birne Binegar: Basically, some and then you have the entry.
Birne Binegar: And then you have a sign factor.
Birne Binegar: And then you have a determinant.
Birne Binegar: Of a matrix I get from the original matrix by removing it I throw in jade column, and then here, you can some.
well.
Birne Binegar: In this case, let's sum over J K equals one to end and then here I is fixed.
Birne Binegar: Fixed realm index.
Birne Binegar: And that gives us actually I have a different formula depending on which which row I want to use, but if I use the first row.
Birne Binegar: Then I grab an element, the row there it is, I get the sign by adding up adding the row index and column index and right racing minus one to that power so for this guy one plus one is two minus one.
Birne Binegar: squared is equal to one, so I don't get any switching sign here, and then I have determinant of a minor.
Birne Binegar: which I get by here's the entry we're looking at remove the row and calm, it contains it and we're left with this to buy two minor a 22823832823.
Birne Binegar: All right, not just the first term, then I go to the second term in that row I pick up a minus nine because one plus two adds up to three, which is an odd number.
Birne Binegar: Until minus one to the third power is going to be negative, one I get entry assign factor, and then the determinant of the matrix I get by eliminating the rolling call that contains the one to entry.
Birne Binegar: And then I have the contribution for the third element in the row.
Birne Binegar: and similar construction there's the entry there's the sign factor of worked out to be one because one plus three adds up to an even number four and then here's the corresponding minor and then.
Birne Binegar: i've reduced the computation of the determinant of a three by three to adding up a bunch of determinants of two by two matrix ease.
Birne Binegar: But here we'll just use the Cross diagonal product formula to complete those determinants you get this, and now you see explicitly the polynomial in the entries of the original matrix that that corresponds to the determine notice that you have 12345 you have a total of six terms.
Birne Binegar: Six happens to equal three factorial.
Birne Binegar: So, and this turns out to be a general property of determinant when you take the determinant of a n by n matrix the number you'll get a polynomial but the number of terms in that polynomial is going to be n factorial So if you had a four by four matrix you'd have 24 terms.
Birne Binegar: If you had a five by five matrix you would end up with 120 terms and you see already this is getting quickly outrageous even for matrices of moderate size, if you had a six by six matrix you would have 720 terms in a polynomial expression for the determine.
Birne Binegar: So.
Birne Binegar: The upshot of this is that.
Birne Binegar: wow for two by two matrix sees a really simple to compute even for three by three you only have six terms, but when you start getting beyond.
Birne Binegar: With looking at major seeds that are four by four or five by five or six by six These determinants become.
Birne Binegar: Complicated functions to compute I mean it just it just have a tremendous number of terms and you probably don't have the stamina to compute the.
Birne Binegar: General determinant of a six by six matrix rather what you'll do it's not like you be able to avoid six by six matrices for the rest of your life rather what you will do is if it, you have to take determines of a six by six matrix you'll go out and you'll be you'll.
Birne Binegar: Then, far less time going to the library if you're going and what computer program will actually compute This, for me, and in fact that's one of the standard things every every mathematical software program, you will have a way of computing determines fairly rapidly.
Birne Binegar: All right, but we want to, we want to understand determinants a little bit better so.
Birne Binegar: The other thing I did on Wednesdays Besides giving you this recursive formula that allows you compute determinants by breaking it down into small determinants of smaller and smaller matrices.
Birne Binegar: there's another way of writing down an explicit expression for the determinant of the matrix that equally hard to compute but at least it visually it tells you a little bit of more about what the determine is here's the formula.
Birne Binegar: It involves the the idea of permutations of the numbers between one and four an n by n matrix now what a permutation is it just you take the numbers between one and and and you write them in a different orders and.
Birne Binegar: How many possibilities are there.
Birne Binegar: Those I take the numbers 123 up to the end and I now i'm going to consider other things like maybe 213.
Birne Binegar: So, this would be another ordering at the same list of numbers, the number of choices, you have children possible number of permutations you have of this standard list one to n is well for that first slot.
Birne Binegar: You have the first lot could be any one of these numbers, so you have no choices.
Birne Binegar: In the second slot and this law here you've used up one of your choices when you filled in the first slot so the numbers that you have left is going to be a total of N minus one choices for that second slot.
Birne Binegar: And then, when you move to the third element you've used up two numbers out of this list so you'll have.
Birne Binegar: A minus two choices left.
Birne Binegar: And you keep doing that, until you get to the last slot but that last slot there.
Birne Binegar: you've used up N minus one of these numbers so there'll be a unique number left from the original list that you can put in that last slot so that will be one, and now the count the total number of possibilities you just count up the number of choices.
Birne Binegar: You have no choices here and minus one choices here in mind your two choices here, these multiply together to give you the total number of choices, so, in the end, you have n factorial you just the product of all the integers between one and in one choices total.
Birne Binegar: So.
Birne Binegar: In this formula here when I sum over permutations I could be something over a large number of permutations it's going to be exactly n factorial that's how I was getting the numbers like polynomial for.
Birne Binegar: For a six by six matrix the number of terms is going to be six factorial which is 720 that's where it's actually from this formula that I was pulling identifying the number of terms The other thing you'll notice about this one.
Birne Binegar: Is.
Birne Binegar: These are the entries of the matrix the way i've written this this term here is the row indexes are incurring in their natural order one to the next factor would be a three something until I reach a doctor, so there are exactly in factors here, that means.
Birne Binegar: This term is a degree and term because they're in variables in that terms but Besides that, I just have a sign factor, this is going to be plus or minus one.
Birne Binegar: You look at the factors here the way I constructed, as I have.
Birne Binegar: Each each factor here lives in a different row but also each factor lives in a different column because.
Birne Binegar: These numbers signal one through six and are just a rearrangement of the numbers one two n so each of these numbers is distinct you don't have.
Birne Binegar: to choose, or one two threes or anything each number here is distinct and so effectively what these factors here represent is a product of entries of the matrix.
Birne Binegar: where the product is constructed by take from each row and from each column.
Birne Binegar: A unique element until you you take a unique element, the first row and unique element in the second row, we need help and third row, but they also have to live in separate columns as well.
Birne Binegar: And what you end up something over is all the possibilities all the ways, you can create factors like this, by choosing elements that live in different rows and in different columns.
Birne Binegar: All right, and then from that form of that this polynomial.
Birne Binegar: Oh yeah get the sign.
Birne Binegar: that's a bit tricky.
Birne Binegar: What you have to do for given ordering up the numbers between one and you have to count the number of times.
Birne Binegar: These numbers appear out of order, let me give you an example that.
Birne Binegar: calculate the sign of.
Birne Binegar: The permutation 312.
Birne Binegar: All right now the way I get this the sign is I count the number of times pairs of entries in this list pure out of order so three should not be appearing before one because one is less than three so that's out of order.
Birne Binegar: This is in order, so this is not in order, this pair is in order, one, two, but three and two are out of orders so.
Birne Binegar: So the number of times in this list number of times the entries are out of order are one and two and so that's going to be minus one to the two equals one if i'd had.
Birne Binegar: Fine.
Birne Binegar: 321.
Birne Binegar: that's the ordering now every parish out of Oregon this is how.
Birne Binegar: to proceed one well that that's out of order and then three, it does not come before one so that's out of order to, and so in this case we get minus one to the third.
Birne Binegar: Alright, so i'm just did a couple of the sign computation So you see that it's not something you can.
Birne Binegar: You actually have to write out of permutation and then stare at a little bit to figure out exactly what the sign of this is, but you can you can kind of.
Birne Binegar: work it out if it's the way I did here, you can work out what that shyness and unfortunately it is very little information it's just a plus one or minus one but it's kind of tedious to figure it out for any particular.
Birne Binegar: For any particular permutation, but if you have figured out all the signs, then this is these just turned out to be factors plus or minus one.
Birne Binegar: And then, what the formula then shows you is that the determinant of a n by n matrix is a polynomial degree and because you'll have n factors in the variables.
Birne Binegar: And each term has exactly yeah so that there's has exactly in factors and that the number of terms that occurs going to equal, the number of permutations which is going to be n factorial.
Birne Binegar: Okay, so.
Birne Binegar: The upshot of this is i'm not giving you.
Birne Binegar: A formula.
Birne Binegar: That any if any more efficient for computing determined, but at least it's revealing sort of a functional structure it makes it pretty explicit, except for the science and except for explicitly figuring out how many different ways, you can order the numbers one one to end.
Birne Binegar: All right, but we're not going to rely on that.
Birne Binegar: Rather, what I want to do.
Birne Binegar: Is I want to show you a simpler way of computing determinants that doesn't require you To sum up, for example, 720 separate terms in a polynomial.
Birne Binegar: All right now that's going to come about, because there are some if a matrix has have a relatively simple form, then, if determinant is going to be really trivial to compute.
Birne Binegar: The form i'm want to describe to you now is upper triangular matrix.
Birne Binegar: So these are matrices.
Birne Binegar: These are only.
Birne Binegar: Let me say it this way entries.
Birne Binegar: are only non zero.
Birne Binegar: When.
Birne Binegar: The row index.
Birne Binegar: Is.
Birne Binegar: greater than the column index.
Birne Binegar: What that's going to mean along this diagonal here from the upper corner, where you have a one one lower corner, you have a and in the second slot here you in a Tutu.
Birne Binegar: along this diagonal the row index and the column index or exactly the same to be upper triangular we're simply demanding that low this main diagonal we only have zeros.
Birne Binegar: that's kind of like real Echelon form but it doesn't actually have to be in row Echelon form what i'm really demanding is I.
Birne Binegar: scan the matrix I look along the from the upper left corner to lower right corner and I just asked or the entry below that diagonal all zero If so, we say the matrix is upper triangular.
Birne Binegar: Let me give you.
Birne Binegar: Know here's the condition when the column index is less than the row index I want the matrix to be zero.
Birne Binegar: here's a matrix so look at this guy it's equal to zero, this is an upper triangular here's the main diagonal and below it, you have nothing but zeros.
Birne Binegar: But the reason why this guy is zero is because the row index is too, but the column index is one since one is less than two, we have to put a zero there, and similarly for these other guys, but I think visually just.
Birne Binegar: Look at the matrix look that main diagonal and if you after you put a rule or something over those diagonal entry, if you only have zeros off to the lower left, then you have an upper triangular matrix.
Birne Binegar: notice this matrix is upper triangular but it's not rational on forums, I could hear, I have a pivot sitting directly below a pivot.
Birne Binegar: So it is not true that upper triangular matrix each are always in row Echelon form, but the opposite is true anytime you have a matrix enroll Echelon form it's automatically and it's automatically upper triangle.
Birne Binegar: What that tells you is if I start off with a matrix that's not upper triangular I can perform real operations that will transform it into a matrix and rational and porn.
Birne Binegar: And then i'll have a matrix that upper triangular so, even if I started off of the matrix that was not.
Birne Binegar: upper triangular I could make it upper triangular I just using our usual reduction algorithm I just have to take it to row Echelon form and i'm guaranteed that it's automatically upper triangular.
Birne Binegar: But the calculate the determinant this guy it's going to be really, really simple.
Birne Binegar: i'll show you why it's going to turn out that the determinant of this matrix it's a four by four matrix I should expect 24 different terms to calculate the determine, but if that I can tell you right now i'm just looking at determine if this matrix is six.
Birne Binegar: What i'm doing here is i'm oh it's athlete zero i'm multiplying along this main diagonal three times zero times two times one equals zero because they have a factor of zero.
Birne Binegar: Alright, so let me.
Birne Binegar: restate that as a theorem if you have an upper triangular matrix then if determining it's just the product of the element along that main diagonal where the row index of each factor always matches the column index.
Birne Binegar: Alright, so.
Birne Binegar: i'll show you our previous four four by four matrix and then you'll in the course of the proof you'll see why it's true in general suppose we start off with this general, this is the most general thing I can write down that's four by four and it's also upper triangle.
Birne Binegar: So the calculate determinant I can do a Co factor expansion along this bottom row.
Birne Binegar: So what i'll have according to my formula Keep it up here.
Birne Binegar: determine if a it's going to equals some over say.
Birne Binegar: guess how some of the columns will have entries I da sein factor, a one inch one I plus J and then I have determinant of a minor that corresponds to that entries, and so this is for.
Birne Binegar: A fixed.
Birne Binegar: row index.
Birne Binegar: What i'm gonna do is i'm going to use the fourth row with this matrix.
Birne Binegar: To start off this co-factor expansions all that I do in this page so like here's the entry in the first entry net fourth row was zero so actually the first three terms are all going to be zero because.
Birne Binegar: i'm zero have a term like this with this factor of zero, so I will have to compute the determine the second term in that row.
Birne Binegar: The entry net matrix is also zero, so I will have to compute that guy the third entry is zero, but when I reach last entry that's the first time, the actual entry in the matrix of the actual entry in a row, is not zero.
Birne Binegar: So that leads to this one.
Birne Binegar: First entry in the row is zero so I don't even have to compute this determinant here, the second entry net row is zero I don't have to compete that matrix determine if that matrix.
Birne Binegar: Third entry is zero but that fourth entry that's where we have the non zero entry and now I have something I need to compute because if this number is not zero this might not be.
Birne Binegar: That will have a positive contribution to the determinant.
Birne Binegar: I tell I write that out, and now I have a determinant of another upper triangular matrix will do the same thing we'll do a Co factor expansion along the bottom row to calculate the determinant of this matrix.
Birne Binegar: So i'll ignore these first three zeros I have an overall factor of a four four and then I i'll do a Co factor expansion of this three by three matrix so I have zero times.
Birne Binegar: minus one to the one plus three, so this is a no one three slot of this matrix so the flying factor is going to be minus one to the one plus three or simply a factor of one.
Birne Binegar: And then the corresponding minor for this guy I eliminate the row and call that contain it i'm left with this two by two matrix a 12813.
Birne Binegar: A 22833, then I go to the next guy zero the sign factor here to see this is in the third row second call three plus to add up the five.
Birne Binegar: So that's minus one to an odd number that's where that minus i'm from there's the entry itself, and then you have to determine if the corresponding mind I don't have to compute this determinant because it's being multiplied by zero and i'm left with simply this.
Birne Binegar: And then.
Birne Binegar: The only, this is not going to survive that's not going to survive, so it will have a four four times 833 times the determinant of this two by two matrix.
Birne Binegar: And there i've written out the determinant that two by two matrix by doing another co factor expansion and see determines Of these, one by one, or just the entry and when you work it out, this is just a product of the diagonal elements.
Birne Binegar: Now that's going to happen in general, this happened because.
Birne Binegar: I did a Co factor expansion I got this factor, but then I had to compute the turn of the corresponding minor it's also upper triangular so pick up this factor times the determinant.
Birne Binegar: Of a smaller minor but it's still upper triangular so i'll get this factor and i'll get the tournament of a smaller thing, but now we just a one by one matrix like just get that entry.
Birne Binegar: Well, that that pattern of calculation works, no matter how large the original matrix was you just steadily keep reducing it, and you find that the only thing that survives the calculations are these diagonal elements.
Birne Binegar: Alright, so that's great.
Birne Binegar: That means that if you had a six by six matrix and you want the and if you knew what was up or triangular then it's determinant would consist of only one term.
Birne Binegar: One term out of four more general six by six matrix you would have 720 different terms, so this is making upper triangular matrix these make life really easy because, instead of having n factorial terms to worry about you only have to worry about those product along the main diagonal.
Birne Binegar: Alright, so what I want to show you now is.
Birne Binegar: For most most cases the matrix you're going to be looking at it's not going to be upper triangular, and so the technique i've described for compound computing determines upper triangular matrices will not be applicable in general.
Birne Binegar: But we have a way of converting matrices into rotational on for matrices and those are examples of upper triangular matrix sees what i'm going to show you is that.
Birne Binegar: It is not true.
Birne Binegar: So.
Birne Binegar: hey can go to which row Echelon form.
Birne Binegar: Via.
Birne Binegar: Elementary roll off.
Birne Binegar: and
Birne Binegar: On the other hand.
Birne Binegar: If a prime equals the reduce row Echelon form.
Birne Binegar: A.
Birne Binegar: Then the determinant of a prime is just going to be equal well, you have to use the entries that this matrix all put PRIMES over my entries, but it just going to be the product of the diagonal elements of.
Birne Binegar: The determinant of a rule of thumb the rational on form of a matrix is going to be easy to compute, but the problem we have yes determinant a does not equal, the determinant terminate of its role Echelon form in general.
Birne Binegar: Even though this determination is easy compute it just a product of diagnose element.
Birne Binegar: it's not directly well, at least at this stage it's not directly related to the determinant of the original matrix.
Birne Binegar: However well the next theorem says is that when you do the real operations, most of the times.
Birne Binegar: The determinant of the matrix does not change, and the times when it does change are easy to understand.
Birne Binegar: I got four minutes i'll just be able to state what happened to you.
Birne Binegar: What happens is when you don't waffle with a matrix can you perform a single elementary row operation well the determinant does change if you interchange rose the determinant flipped sign if you.
Birne Binegar: replace a row via scale or multiple of itself well that fear factor is carried over into the new determinant.
Birne Binegar: And, but here's the nice thing here's the row operation you perform all the time you have a pivot below a pivot so you use an upper row to cancel out the pivot in the lower realms you do this, all the time, but this row operation does not changing the determinant.
Birne Binegar: So the way we actually calculate determinative a is we reduce it to relational inform.
Birne Binegar: Most of the time, the determinant not going to change, but if we interchange two rows I just have to remember that there's a sign flipping ball, or if I do a row rescaling I just have to remember that my determine is changed by the same factor.
Birne Binegar: So you can really do some matrix to relational on form, all you have to do to get the corresponding determinant is remember how many times you.
Birne Binegar: interchange rose then you'll have a as many factors a minus one from the determine as the original matrix or if you've.
Birne Binegar: done any row rescaling, then you have to take into account those factors as well, but for the vast majority of your elementary row operations that you use as you reduce your given matrix it's not going to have any effect at all on on the determinant.
Birne Binegar: And so here's the formula suppose a row reduces to a matrix a prime in in row Echelon for.
Birne Binegar: then determine if the original matrix can be calculated in terms of the determinant of this matrix in relational and form.
Birne Binegar: here's the that determine since it's in relational on form its upper triangular and so which determine it's just going to be the product of its diagonal elements.
Birne Binegar: easily computer now you have some corrections factors these factors of Lambda well, you have to remember how many times you've done row rescaling and if you.
Birne Binegar: If you put a factor in doing a row rescaling you're going to have to remove it, so you divide by it and expression for the determinant.
Birne Binegar: And then here this minus one to the arts that's number of times, she flipped rose in order to do the row reduction but as long as you keep track of the sort of errors that were introduced as you did the row reduction, you can take them into account and you get a nice.
Birne Binegar: A nice simple formula that when you once you get past the reduction, all you have to do is multiply together.
Birne Binegar: The diagonal elements of your matrix in Russia and form and then recall how many times you interchange rose and how many times you reach scale grows, you can you can get it very quickly that way.
Birne Binegar: Okay, so let's see one minute left I don't even have time for an example today so i'll let you go and what we'll do on Monday is will i'll do with lots of examples of how you use rubber reduction to calculate determines.
Birne Binegar: Good weekend.