Birne Binegar: Okay, so your guns are graded guys a little worse as time the average was 84 as opposed to, I think the last time you every like 89 or something, let me tell you something interesting now.
Birne Binegar: The students who did it like said definitions are going to be on the exam the students who memorize the definitions and simply wrote them down the average score they got on the exam was 97.
Birne Binegar: So this is why i've strived stress the definition so much it's not just to have the definitions memorized but it puts everything in context.
Birne Binegar: The bases and notion of linear independence, most of the things we're doing you don't really grasp fully until you.
Birne Binegar: take hold of the definitions and i'm i'm happy if you simply memorize them and have them have those definitions at your disposal.
Birne Binegar: But it really does make a difference on on the exams, if you take the trouble to memorize things verbatim that it helps you it helps you a lot, and it really should only take you about 15 minutes to memorize.
Birne Binegar: A definition, but you have to you have to memorize it exactly.
Birne Binegar: Otherwise, I saw lots of definition for people, setting a vector equal to a set I mean or setting subspace equal to a single vector yeah those kind of mistakes wouldn't happen if you if you if you just wrote things down verbatim.
Birne Binegar: Okay, well, let me, let me pass these exams back let's see here.
Birne Binegar: Jacob here.
Birne Binegar: Anthony.
Birne Binegar: robbins Roger Bennett.
Birne Binegar: David calvert.
Birne Binegar: Jackson pistol.
Birne Binegar: peyton there.
Birne Binegar: Lisa Fischer.
Birne Binegar: Derek Holloway.
Birne Binegar: Eric.
Birne Binegar: Lucas.
Birne Binegar: Colin meyers.
Birne Binegar: Jacob o'hare.
Birne Binegar: jaden patel.
Birne Binegar: let's see.
Birne Binegar: I can't read the first name last name read.
Birne Binegar: Taylor.
Birne Binegar: Jordan wheeler.
Birne Binegar: Alexander wilburn.
Birne Binegar: Madison Wilson.
Birne Binegar: Okay, no I butchered somebody's people's name or they've arrived late to anyone not get an exam yeah.
Birne Binegar: Others yeah.
Birne Binegar: Okay, so um I posted solution to those.
Birne Binegar: Any questions on the on the exam.
Birne Binegar: questions about how great things are working.
Birne Binegar: I can answer the questions like that, after class, so we don't have to.
Birne Binegar: disclose too much of what went on on individual exams, but any general questions on the exams.
Birne Binegar: Okay, so what I suggest you do is look at the Posted solutions and discover first how I taught the exam should have been completed and then.
Birne Binegar: If you if you do have questions.
Birne Binegar: Next week is not such a great week I visitors from France, next week and so i'm going to be doing my office hours via zoom again.
Birne Binegar: Monday and Wednesday of next week, but after next week, I intend to be on campus more regularly and i'm back on hoping to get rid of my mask after next week as well, so.
Birne Binegar: If you want to talk to me more extensively about the exam we can talk after class immediately or you could just hold off a week and then I should be my office regularly after after next week.
Birne Binegar: Okay um so questions.
Birne Binegar: Oh, you need an exam what's your name.
Birne Binegar: Sorry, I didn't hear.
Birne Binegar: Oh what's your.
Birne Binegar: what's your what's your last name read oh Okay, here we go.
Birne Binegar: There alpha bye good night.
Birne Binegar: Okay.
Birne Binegar: Alright, so.
Birne Binegar: By the way, those definitions that.
Birne Binegar: That I put on this exam I guarantee that they're also going to the same definitions are going to appear on the final exam so do take the trouble at some point before the final exam to really memorize those definitions goes for definitions are actually the core of the course it's a.
Birne Binegar: real substance of the language that you're learning in this course yeah.
Birne Binegar: Okay, so.
Birne Binegar: Okay, so what we're gonna do today is we're going to.
Birne Binegar: i'm going to give you a much more expedient way of calculating determinants but i'll start by reviewing what I went over on Wednesday.
Birne Binegar: What determinants are and what they're used for and and some simple formulas general formulas that will apply to any n by n matrix However, as we saw on Wednesday.
Birne Binegar: These formulas become very complicated very quickly, even for relatively small matrices the determinant turns out to be a polynomial in the entries of the matrix.
Birne Binegar: Even for a relatively small matrix one with six rows and six columns the number of terms.
Birne Binegar: That you would have in the polynomial that corresponds to the determinant that six by six matrix would be 120 separate terms.
Birne Binegar: So these polynomial CC determinant as functions of the entries become pretty complicated polynomials even for relatively small matrices.
Birne Binegar: But that doesn't be little there that their utility, and so the main thing I want to accomplish today is give you a simpler way of.
Birne Binegar: calculating determinants one that doesn't know that oftentimes can reduce the computation of up a determinant just to.
Birne Binegar: A single term in an otherwise polynomial that could have n factorial separate terms and that's going to be done will use real reduction in order to calculate to determine it.
Birne Binegar: Okay, so let me start.
Birne Binegar: Okay, so we introduced the determinant last time it's a certain function of the entries of the matrix and what it tells us is if the determine is equal to zero it tells us that.
Birne Binegar: an inverse does not exist or that.
Birne Binegar: Simple linear equations could have more than one solution so i'm stating it's a little bit more positively.
Birne Binegar: Actually, the way I described on Wednesday, a terminal not equal to zero means that at the inverse exists, but now i'm going to state it.
Birne Binegar: Take the positive condition determine your age equal to zero, that means that a does not have an inverse it also is going to mean that if you have a linear system or use a as a coefficient matrix you're going to have more than one solution and.
Birne Binegar: And in fact even this equation X equals zero is going to have solution other than the trivial solution it's also going to mean that oh sorry, it should be not didn't quite negate everything.
Birne Binegar: Is not reducible to the identity matrix.
Birne Binegar: Because if it were road lucille to the identity matrix a the miners first would exist and then.
Birne Binegar: This actually should be the rank is lesson in sorry got some typos here i'll correct those when I get back to my office and the reason why rank it's going to be strictly less than the number of.
Birne Binegar: columns is because well it's not reducible to the identity matrix.
Birne Binegar: When that happens, it means that when you reduce a you end up with a zero row at the bottom of your matrix somehow until the zero does not have a pivot.
Birne Binegar: And so you have one column at least one column that doesn't have a pivot the number of pivots in a row Echelon form determines the.
Birne Binegar: dimension of the row space and you mentioned up the call space one column of your incomes, does not have the pivot That means the rank is going to be less than in.
Birne Binegar: Alright, so this is typically white will be using the determinant later on in the course in fact probably starting on Mondays lecture this will use the condition that a determinant of the matrix equals zero to draw these kind of conclusions about a matrix.
Birne Binegar: Alright, so.
Birne Binegar: that's what he terminators going to do for us.
Birne Binegar: What is it exactly well it's a polynomial.
Birne Binegar: It the degree of the polynomial depends on the size of the matrix the basically the degree of the polynomial is going to be equal to the number of rows and columns in this square matrix so for one by one matrix since I only have one entry on one call it's basically.
Birne Binegar: A linear function, and in fact it's just the entry of the matrix.
Birne Binegar: Man as equal to two you know, two by two matrix then a polynomial that correspond to the determinant is up degree to which is natural for two by two.
Birne Binegar: matrix and it's just the difference of the diagonal products you multiply this way, and then you subtract the products along the opposite direction.
Birne Binegar: Now, unfortunately for larger matrix these.
Birne Binegar: formulas are not this simple and so.
Birne Binegar: The first way I introduced to calculate the determinant of a larger matrix was by a recursive algorithm that will allow you to do steadily reduce the calculation of determining to determine of.
Birne Binegar: one or the other of these two base cases I mean usually you'll stop here, because this this the determine a two by two matrix eat that formula is easy to memorize and it's a good.
Birne Binegar: When you actually have to calculate a determinant, this is an easy one to apply.
Birne Binegar: But, more generally, when you have a larger matrix say n is greater than to.
Birne Binegar: What you should rely on then, as far as general form that goes is this co factor expansion formula of which you have.
Birne Binegar: A couple versions.
Birne Binegar: You can fix a row index of your matrix and then you can mark across a given row, and when you calculate the determinant is you take an entry in the row.
Birne Binegar: You get a sign factor which can depend on the position of that entry inside the matrix you take it row index and column index some knows if that's an even number.
Birne Binegar: Then this factor sort of disappears it'll just be minus one to an even power, but if the some of the row index in the column indexes odd, this will contribute a factor of minus one, but it just a factor, plus or minus one.
Birne Binegar: And then here what this term is its determinant of a smaller matrix that you construct from a.
Birne Binegar: it's this m subscript I J it's the matrix you obtained from a by removing it's I throw and jade column that makes a matrix have one left with one less calm and one less row, and so a smaller matrix that.
Birne Binegar: This formula to compute the determinant up an n by n matrix as a some a bunch of determinants of smaller major cities, but this formula could be applied again if this matrix turned out to be.
Birne Binegar: Four by four, then you could apply this formula again to reduce the computation of this determinant into the son of a bunch of determines a three by three matrix see the miners of this matrix and if.
Birne Binegar: you've done those determinants of three by three matrix these could be used by applying the same formula just using.
Birne Binegar: A different value of end here, you get the same formula, and it can reduce it to a bunch of determines up eventually say two by two matrix ease and then you can go back and use a formula like this to get the explicit result.
Birne Binegar: So.
Birne Binegar: So that's how these formulas work, and you have a choice you can either fix a row index and just walk across a row grabbing each of these terms and summing them up.
Birne Binegar: Or you can fix the column index and move down a call the reason why those choices are advantageous is because you could have a matrix a 1232 to 01102 where.
Birne Binegar: The most expedient way to calculate this deep determine might be well, you would take advantage of lots of zeros in this column.
Birne Binegar: And so the determinant a.
Birne Binegar: With the let's see if I go down the second column the entry in that to one slot is to I have a sign factor this century it's in the first row.
Birne Binegar: And second column, and then you have the determinant of the minor okay that's the miners would I the matrix I get from the original matrix by removing the row and column that contain the entry question this one to entry and so that's going to be 2112.
Birne Binegar: And then you have another term plus zero times minus one to the two plus two.
Birne Binegar: determinant of lsc this is going to be now i'm looking at this entry tell me one three I just said to three up there.
Birne Binegar: One three and one two.
Birne Binegar: And plus you have.
Birne Binegar: Third entry in that column minus one it's in the third row and second call and the corresponding minor i'm looking at this guy so it's going to be 1321.
Birne Binegar: But the reason for choosing.
Birne Binegar: To expand the determinant of a along the second column is because these last two terms here on have zero as a factor.
Birne Binegar: So I don't have to compute that determinant, nor do I have to compute the bottom determined because it's being multiplied by zero, and all I have to do is calculate one lower the.
Birne Binegar: Lower dimensional determinant in order to get the result for this three by three matrix so what this is going to be it's going to be two.
Birne Binegar: Times minus one.
Birne Binegar: And then determine if this matrix is.
Birne Binegar: Two times two minus one times, one I guess we're going to get this will be a factor of three six there'll be minus six I guess so.
Birne Binegar: These co factor expansions.
Birne Binegar: they're they're tedious because, if I had started off of a four by four matrix like.
Birne Binegar: First, reduced to a bunch of three by three matrix see and then the three by three matrix each probably have to be computed in terms of some of a bunch of.
Birne Binegar: determinants of two by two matrix see eventually you get down to something calculator what it, but it tends to get messy pretty quickly and I don't think I would.
Birne Binegar: ever asked you the computer determinant of a four by five by five or six by six matrix using.
Birne Binegar: These co factors formula, but hopefully you see that in at least this gives up recursive definition of what the determinant is and it's not only a definition it's a it's a recipe for how to calculate it.
Birne Binegar: Even though it might be messy because you might have to apply that definition.
Birne Binegar: Multiple times, if you reduce the size of the determines that you actually need to calculate.
Birne Binegar: Okay, so.
Birne Binegar: let's do an example of a.
Birne Binegar: determinant calculation here i'll do it with arbitrary entries for preferred doing it just to show you again how the.
Birne Binegar: How the definition works, how you apply it, and then the show you explicitly how a polynomial POPs out of this calculation.
Birne Binegar: So here what i'm doing looks like i'm going to expand along the first column, so I have a one one times the determinant this minor, which is a two 283 81 but that's this term, and I.
Birne Binegar: The sign is going to be equal to plus one right because one plus one is an even number so i'll just have minus one to an even power just one.
Birne Binegar: The second term in that column it's the entry there's a two one now, I have a minus sign here that's because two plus one is an odd number and so i'll be looking at minus one to an odd power now pick up this minus sign, but then I, I have the.
Birne Binegar: Minor corresponding to this entry I eliminate the column and row that contain that entry and i'm left with a one to 813832833 exactly that.
Birne Binegar: And then I get to this last entry in the column here's the entry itself the sign factor is minus one race, the power three plus one or minus one to the fourth power just going to be a factor of one.
Birne Binegar: Oh, I have a plus sign there and then the corresponding minor eliminate the row and column containing that last entry and it's going to be 812-813-8228.
Birne Binegar: And then.
Birne Binegar: Here I reduce the computation of the three by three to have some of determinants of two by two matrix season those I can calculate explicitly using that difference of the diagonal products formula.
Birne Binegar: And Okay, and so there you see explicit formula it's a polynomial in the entries of the matrix it's polynomials degree three because this is a three by three matrix and they're exactly six terms 123456 the number terms is equal to the factorial.
Birne Binegar: corresponding to the size of the matrix so three factorial is six and that's why we end up with six terms for this, the termination of a general three by three matrix.
Birne Binegar: Okay, so yeah.
Birne Binegar: middle.
Birne Binegar: Know i'm going down the column, so I mean the second row.
Birne Binegar: First column.
Birne Binegar: Inside oh yeah oh yeah yeah yeah here this this is wrong anyone to.
Birne Binegar: rose, because if you go to the.
Birne Binegar: Maybe I did, maybe I did.
Birne Binegar: yeah you're right, this is where I I I was going cross the first row knock down the first call so here's where the mistake with with the.
Birne Binegar: One two here yeah and now now at right.
Birne Binegar: yeah good.
Birne Binegar: Okay, so.
Birne Binegar: Anyway, the formula.
Birne Binegar: No signs are coming from here, the way you get the sign factors you add that row index and column index if it's an even number, then this will be a factor of just one, the only one, I nj add up to an odd number, do you get.
Birne Binegar: That sign flip.
Birne Binegar: Okay, so.
Birne Binegar: that's one way of calculating the tournament I gave you another way that was more common tutorial.
Birne Binegar: What a second way of computing determinant uses the notion of permutations of the numbers one between one and so what a permutation is it some.
Birne Binegar: listing of the numbers one to end, but in some arbitrary order until each one of these numbers Sigma one through six Sigma m is one of these numbers and, in fact, these numbers all are distinct they're exactly these numbers, they just put in a in a different order.
Birne Binegar: What the notion of permutations colossus do is actually see the polynomial.
Birne Binegar: corresponding to determine it explicitly it's basically.
Birne Binegar: A product of the entries of the matrices and you can look at this product, the way it's formed is.
Birne Binegar: These row indices i'm just steadily increasing one goes to two and the third factor would have a row index of three the fourth factor would have a row index of four, but what i'm.
Birne Binegar: fooling around with here is the.
Birne Binegar: The column indices, and once I take an order.
Birne Binegar: signal, one could be any number between one and two and so.
Birne Binegar: When I sum over the permutations i'm actually constructing all possible products of the entries of the given matrix, but these products are constructed in a way that I have exactly each factor lives in a unique row and a unique call and so here, you see.
Birne Binegar: way i've ordered the factors I ordered them by their row index, then there the column index are out of order but it's some ordering up the numbers one two N, now the hard part about this formula is we're still going to have those plus or minus signs that occur in the determinant.
Birne Binegar: They actually they can be determined by looking at the permutation that corresponds to a given term.
Birne Binegar: What that is once you have a listing of numbers, between one and N in some order.
Birne Binegar: The fine at that permutation is the number of times that, as you scan across so as I increase this index i'm just.
Birne Binegar: moving across the order of these terms and then here i'm looking at the value of terms, so is representing which slot of this expression we're looking at and Sigma I it's going to be the battle, the value.
Birne Binegar: Of the integer that athlete occurs in the slot and what you're what you're looking for here is a number of times these guys are out of order so, for example, if I had permutation 321 well three and two are out of order because the natural order is.
Birne Binegar: To is less than three but three is less than two and actually these guys, this is out of order, three and one higher numbers proceeding, the lower number so that's how to order, and this is out of order as well.
Birne Binegar: To is greater than one but it's proceeding it in this list so in this case, what this number would be would be three so the sign.
Birne Binegar: Of this permutation 321 it's going to equal minus one to the three because there were three different pairings of the entries inside this list that are out of orders.
Birne Binegar: Okay, so.
Birne Binegar: This is.
Birne Binegar: This takes a little bit of fiddling around the count up the count up the number of times it's out of things are out of order but.
Birne Binegar: Once you see how to do that, then you can write down this formula, but it gets really complicated because the number of permutations you have for.
Birne Binegar: Number of permutations that you have and it's actually equal to n factorial and that number gets really big for even moderately large in.
Birne Binegar: For factorial is 24 five factorial is 126 factorial is 720 so just a six by six matrix when you construct this polynomial.
Birne Binegar: there's a recipe for it, but when you start to write it down, you have so many permutations to consider you end up with having to write down polynomial degree six with 720 terms, if you want to write down the general determinant.
Birne Binegar: Now so you're never going to do this in practice.
Birne Binegar: No one is maybe they did in the 18th century they actually wrote down several took a book filled the book up with calculations of the Germans, but nowadays, in fact.
Birne Binegar: What generally happens is you, you have a matrix it's larger than a four by four you'll probably use a computer to computer determinant and computers are not you know nowadays they can handle billions of calculation so.
Birne Binegar: Calculation up and determine up a 10 by 10 matrix which is going to have the number of terms in that polynomial is going to be like what 10 factorial so huge number.
Birne Binegar: I think it's like.
Birne Binegar: yeah well over a million, so you a 10 by 10 matrix you may have over a million terms, but a computer can can sum up a million terms fairly quickly, and so, while they may not be accessible to you by direct calculation this formula is worked pretty well if you can program it into a computer.
Birne Binegar: All right, but what I mean the main purpose of today, though, is to give you a much more expedient way of calculating determines one where you're not going to have.
Birne Binegar: Number of terms to compute it's not going to go up into the hundreds it's going to stay, you know relatively simple and that all you really going to have to do is row reduce the matrix to relational on form and then taking a take into account some factors.
Birne Binegar: So.
Birne Binegar: Let me just point out that, from this.
Birne Binegar: person permutation formula.
Birne Binegar: well.
Birne Binegar: fairly simple to state the see ya see the format, the determinant as a polynomial and and in fact the coefficients So these are the variables, the coefficient or just plus or minus plus one or minus one, so the sign is not very complicated just.
Birne Binegar: You basically summing up over every possible.
Birne Binegar: arrangement of the entries of a a like taking one entry from each row one entry from each column and multiply the things together.
Birne Binegar: Then you get a sign that you that you compute based on the permutation and you sum all those things up so that's an easy way to understand the determinant but it's.
Birne Binegar: Because there's so many permutations it's hard to compute in a in a practical sense.
Birne Binegar: Yes, it does show you that the determinant is Paul normal that each term has exactly in factors, so we would call this a homogeneous polynomial degree N, because not all the terms in the polynomial have degree in in the variables.
Birne Binegar: and
Birne Binegar: Because we have so many permutations in general, this polynomial is going to have n factorial terms.
Birne Binegar: All right, but that's, this is a formula keep in your back in your mind as to how complicated and determinant can be yet it's complicated because the signs are hard to compute and.
Birne Binegar: Writing down all possible permutations is hard to do as well, but it's a fairly simple formula, even though it's because the number of terms it's hard to compute.
Birne Binegar: Alright, so what i'm going to do today is give you a much more expedient way of calculating determinants.
Birne Binegar: So the first thing i'm going to do is i'm going to look at a special family of matrices so called upper triangular matrix Steve and i'll show you that the determinant of an upper triangular matrix is very simple to compute.
Birne Binegar: So.
Birne Binegar: let's begin with a definition a square matrix is called upper triangular if.
Birne Binegar: Whenever the column index is less than the row index, then the corresponding entry is zero.
Birne Binegar: What that condition is doing.
Birne Binegar: Is.
Birne Binegar: On this main diagonal.
Birne Binegar: This is where if you're on this main diagonal euro index is exactly the same as your column index.
Birne Binegar: This condition here is telling you that whenever the column index is less than the row index, then you have to have zeros So the first instance instance there would be this to one entry right because the column indexes one, but the row index is too.
Birne Binegar: And when the column index is less than the row index on both to have zero, and in fact what this condition does.
Birne Binegar: It just means that the low this main diagonal you only have zeros.
Birne Binegar: Let me.
Birne Binegar: give you another example here's an example.
Birne Binegar: Of a matrix that upper triangular so here, you see here's the main diagonal 3021 below that diagonal to the lower left, I have nothing but zeros.
Birne Binegar: It looks kind of like it's in row Echelon form but it's really not right here's here's a pivot and there's an entry below that pivot so, even though this matrix is not.
Birne Binegar: Quite in row Echelon form it's pretty close to it, and in fact the way you should understand upper triangular it's.
Birne Binegar: Slightly weaker condition than being in row Echelon form with relational on form we concentrate on how the pivots heart range, then we just want the pivots do occur in this matrix following a sort of descending care staircase for upper triangular.
Birne Binegar: What you do is you look at this main diagonal and you simply demand that all the entries below the main diagonal are all equal to zero.
Birne Binegar: So.
Birne Binegar: You have shot of what i'm saying here it's.
Birne Binegar: Being upper triangular is slightly weaker than being in real Echelon form, but on the other hand, if you're in row Echelon form you're guaranteed to be upper triangle and so.
Birne Binegar: Typically, what we'll do when we compute determinants is we'll take a matrix and we'll try to get it in a row Echelon form because we have a nice algorithm for doing that and.
Birne Binegar: It may be that we don't have to go all the way to relational on form if we simply get it to the upper triangular form we can actually compute to determine.
Birne Binegar: All right, so.
Birne Binegar: Why are we interested.
Birne Binegar: In matrices that are upper triangular.
Birne Binegar: Well that's because if you have an upper triangular matrix your determinant has only one term.
Birne Binegar: By had a.
Birne Binegar: I six by six matrix and I want to compute is determining what that permutation formula tells me is that I would have a 720 different terms to compute in order to arrive at determine.
Birne Binegar: If that six by six matrix, however, with upper triangular I only have to take the product at the diagonal entry to get to determine.
Birne Binegar: Let me show you why that's true it's going to come from the co-factor formula what i'm going to do here's a four by four matrix i'm going to apply the Co factor expansion on this fourth row to calculate determine of this four by four matrix.
Birne Binegar: Alright, so here's what we do well i'm sorry i'm gonna.
Birne Binegar: i'm gonna march along this fourth row first term is going to be zero times minus one to the four plus one times the determinant of.
Birne Binegar: The corresponding three by three minor and then we'll have a term corresponding that and that is well I work this out on the next page here's that first term.
Birne Binegar: corresponding to this guy second term its corresponding.
Birne Binegar: To the second element in the fourth row contribution of the third element in the fourth row and the contribution of the fourth element of fourth row.
Birne Binegar: This first three terms you look at the original matrix are all being multiplied by zero and so they're not going to contribute.
Birne Binegar: And so, in the end, all I get in the computation of this determinant of a four by four matrix is the determinant of a.
Birne Binegar: single three by three matrix but now i'll do another co factor expansion along this bottom row, and what will happen there alright, so I have a four four.
Birne Binegar: And now i'm going to do a Co factor expansion of this determinant on the bottom route to I have zero times minus one two it's in the first row.
Birne Binegar: Sorry, I send a third row and first column, so this is the one three entry minus one to the one plus three is going to be minus one to an even power, so let me insert those factors in here is, have you see how this was constructed.
Birne Binegar: At minus one to the cure i'm looking at the terminal this matrix so I guess it's minus one to the one plus three and that's why I don't get a sign this sign here comes about, by looking at the contribution of this factor that's a minus one.
Birne Binegar: let's see it i'm sorry it's this factor so that's in the third row and second column, and then I have the determinant for this entry.
Birne Binegar: I eliminate the row and column, it contains it it's 11130812 exactly that's the minor and then, I have a third term here and it's the only term it's not being multiplied by zero.
Birne Binegar: And so, if I simplify this expression I get a four four.
Birne Binegar: These two terms don't survive, I have a three three and then I have the terminal of this two by two matrix to compete well.
Birne Binegar: And then that two by two matrix it can be computed also using a Co factor expansion let's do it.
Birne Binegar: Say along that first call it'd be a one one times the determinant of this one one matrix a subscript to two and then you would have a one and two times the determinant of the corresponding minor, which in this case would be zero all right, and then, when we simplify that what happened.
Birne Binegar: The only thing it really survived because we had so many terms, that would be multiplied by zero, the only thing that survived with the product and with the entries of the original matrix along that main diagonal.
Birne Binegar: So what happened here.
Birne Binegar: Is.
Birne Binegar: When I computed The co factor expansion of this determinant of a I only got one term.
Birne Binegar: When I did the Co factor expansion along the bottom row and it picked up a factor.
Birne Binegar: A one four times the determinant of a small matrix but it's the same structure, here we go along the road will pick up this 833 and then we'll get a determinant of a smaller matrix.
Birne Binegar: But then we do a Co factor expansion of that when you only pick up one term there'll be a to two times to determine if that and so that co factor expansion, when you apply it repeatedly to an upper triangular matrix.
Birne Binegar: Effectively, it just picks out the product of the diagonal elements that happens anytime you have a matrix that upper triangular.
Birne Binegar: Okay well so.
Birne Binegar: that's going to make computation that determines, at least for upper triangular make your seat really simple.
Birne Binegar: Let me do a quick example suppose I had.
Birne Binegar: Equal 1-231-261-1213 oh let's make it upper triangular 0002120001 I see 1231234512345 the should be.
Birne Binegar: 03100002 suppose, this is a given matrix it's a five by five matrix, and so what I expect is that the general formula for the determinant of a five by five matrix is going to have 120 different terms.
Birne Binegar: But since this matrix is upper triangular it's determined and it's just the product of its diagonal elements and so.
Birne Binegar: I can just evaluate the terminal this matrix by inspection it's going to equal one times one times two times three times two equals 12 so if you're lucky enough to be given a matrix that upper triangular the calculation have to determine it's pretty trivial all you have to do is multiply.
Birne Binegar: Take the product of all the entries along that main diagonal.
Birne Binegar: But if you had if I had a non zero entry here that would screw things up and i'd have to go back and use the long formula in order to compute them the determinant well, except for one thing.
Birne Binegar: We know that every matrix.
Birne Binegar: can be brought into the form of a matrix in row Echelon forum, simply by applying elementary row operations in though in the way we've been doing all Semester and, in fact, these upper triangular matrix these are examples of matrices that are.
Birne Binegar: In row Echelon form or relational on matrix and relational and form is always upper triangle.
Birne Binegar: So the idea we're going to pursue here is well suppose we have a matrix it's not upper triangular not enroll Echelon form we can make it that way.
Birne Binegar: And once we get through the matrix in reducing one form, then we can easily calculate determine of that matrix the hard part it's going to be to relate that determine if the row H1 form for the determinant of the original makes group.
Birne Binegar: But it turns out that the difference between.
Birne Binegar: The determining the original matrix and the determinant of its role Ashkelon form is yeah it's pretty concrete and easy to keep track of the discrepancy there.
Birne Binegar: here's the theorem that all I won't prove and I will demonstrate it with two by two matrix So you see, essentially, how it works.
Birne Binegar: If I have an n by n matrix and.
Birne Binegar: let's suppose we apply an elementary row operation to that n by n matrix Well then, I get a new matrix and I can calculate it determine.
Birne Binegar: The first statement is if your row operation is just the interchange of rows, the only thing that that role operation did to the originally terminal is it ends up multiplying it by minus one so.
Birne Binegar: This is how a row operation changes to determine at least the inner row and changes.
Birne Binegar: Alternatively, if if.
Birne Binegar: You do a row operation, where you refill row well, what does that do to the determining it just we scale so determined by the same factor, so if you keep track of how many times you row interchanges and how many times you do row, you know do.
Birne Binegar: Real rescaling you can keep track of that as you compute determines and then here's the Nice part.
Birne Binegar: Most of the time when you read reducing a matrix you're trying to get rid of pivots using this operation you take on.
Birne Binegar: The first half when you're just trying to get the reports on form you taken you try to get rid of pivots that city below pivots and so you subtract from an upper row a multiple of another row from a lower row to get rid of that pivot there.
Birne Binegar: But this role operation doesn't change the determine at all so most of the time when you're doing row reduction that determinants not changing you want to think you have to worry about is when you.
Birne Binegar: When you flip two rows or when you do a row rescaling.
Birne Binegar: So let me show you this explicitly.
Birne Binegar: i'll just do it with a two by two matrix suppose a is equal to 1210.
Birne Binegar: So if I take this matrix and I entered the terminator a in this case.
Birne Binegar: it's just going to equal one times zero minus two times.
Birne Binegar: Two times, one is going to equal negative two.
Birne Binegar: All right, that's the difference of the diagonal product now suppose I.
Birne Binegar: said a prime equal to the matrix I get by.
Birne Binegar: interchanging real one in row two.
Birne Binegar: Well that's going to be.
Birne Binegar: The matrix 1012 just interchange those two rows and now the determinant of this new matrix.
Birne Binegar: it's going to be one times two minus zero times one equals two well exactly the negative of what we started off with.
Birne Binegar: Which agrees with this formula to do if you interchange two rows the consequence on the determinant is you get assigned flip.
Birne Binegar: If you have.
Birne Binegar: If you have let's see.
Birne Binegar: hey is still one to one zero and suppose we send.
Birne Binegar: row to goes to.
Birne Binegar: Two times for to.
Birne Binegar: Then.
Birne Binegar: New matrix it's going to be one to two zero and the determinant of that matrix.
Birne Binegar: it's going to be one time zero minus two times two.
Birne Binegar: equals negative four well.
Birne Binegar: Negative four is just two times the determinant of the original matrix and so that verifies this rule when you do a row rescaling what happens to determine if the determinant changes by exactly the same factor.
Birne Binegar: If I did to real rescaling than I would have two factors to adjust and then, finally, the most common elementary rule operation you apply the third rule says it it doesn't change the determinant all so those.
Birne Binegar: The vast majority of the real operation to perform on a matrix to get it in relational on form are not going to change to determine.
Birne Binegar: If you keep track of the number of times you flip rows and the number of times you reach scale a matrix, then you can take these adjustments into account.
Birne Binegar: And so what you end up with them is the following theorem.
Birne Binegar: If a matrix a is reducible to another matrix a prime enroll Echelon form, then you can use the deep.
Birne Binegar: You can calculate the determinant of a by calculating the determinant of this matrix in real life salon for this one is going to be easy, because if you enroll Echelon form you're also upper triangular.
Birne Binegar: So the calculation of the determinant of the matrix general actual on form, since its upper triangular all I have to do is multiply together the entries where the row index and the columns are the same, the entries along that main diagonal from the upper left to the lower right.
Birne Binegar: But then you have to take into account.
Birne Binegar: Since you reduce you've done operations that changed the determinant and so you have to account for factors that are introduced by those operations.
Birne Binegar: Here the determinant into be changed by a factor of Lambda if I want to get the originally terminal did you ever enough, the original matrix i'm gonna have to.
Birne Binegar: divide that determine if the new matrix the one I got by playing row operations i'm gonna have to divide it by Lambda so every time sorry we got it.
Birne Binegar: All i'm running over Okay, well, I guess i'll stop here, but these these factors that are coming from really scaling and this factor here is just a number of times you interchange rose.
Birne Binegar: But now it's to what you do to calculate the determinant issue really do so relational on form and you simply.
Birne Binegar: Take notes, as you do, you roll reduction, how many times, did I flip flip rose and how many times, did I perform a row rescaling those will give you the correction factors and you'll be able to simply calculate the determinant that way i'll do lots of examples of this on Monday.
Birne Binegar: Okay, so have a good weekend and I will see you on Monday.
Birne Binegar: Thank you.