Birne Binegar: Okay, so um I don't yet have your test graded they'll your test will come back on Friday i'm sure.
Birne Binegar: What i'm going to do today is i'm going to Fort ord with the course.
Birne Binegar: we're now having.
Birne Binegar: introduced a lot of abstract stuff so we began the course just doing.
Birne Binegar: Calculation mostly row reduction and and finding things to do with that and then since the first midterm we've been concentrating on abstracting like.
Birne Binegar: Sub spaces bases linear independence linear transformation now going back to things that you'll find our analytical directly in your applications.
Birne Binegar: Today, where we're going to talk about determinants which are certain functions, when you have a square matrix there's a certain functions.
Birne Binegar: Of the entry to that matrix that will allow you determine lots of properties of the matrix i'll review some of those in a in a second.
Birne Binegar: But the surprising thing is going to be is that once we have knowledge of the determinant.
Birne Binegar: For many purposes, we can avoid the road reduction algorithm we can use determinants to solve linear systems, we can use determinants to invert matrices etc what's nice about the formulas they're going to come out of that is.
Birne Binegar: The determinant or or an explicit function you just plug numbers into the function, you get a result.
Birne Binegar: that's different than the sort of algorithms we've been using thus far in our calculations, where.
Birne Binegar: You start reducing and matrix, but you have no idea what the end result, the relational on form is going to look like until you start working through the steps and see what happens as you go to the calculation.
Birne Binegar: So what's going to be nice when we get in this discussion of determining which we actually get a function that we can stare it and see some general properties of and will actually help us understand what's going on a little bit better.
Birne Binegar: Okay, so let me get started, let me first remind you.
Birne Binegar: What the textbook calls the fundamental theorem of convertible matrices like the fundamental therm of calculus that relates integration to differentiation, you know sort of.
Birne Binegar: You know, independently define things are actually turned out to be closely connected, the fundamental theorem so to for in linear algebra.
Birne Binegar: If we have any n square matrix, then the following five statements are all equivalent meaning if any one of them is true, then always true so, for example.
Birne Binegar: If you have a square matrix either the inverse exists or doesn't but if the inverse does exist that's implied in anytime you have a linear equation, where you use a as a coefficient matrix you can get a unique solution and.
Birne Binegar: Equivalent condition is that if you have well X equals zero is always a solution of this equation, but if a is equal equals if a's and vertical than in fact X equals zero is the only solution of this equation.
Birne Binegar: And another thing we've seen is that if he has an inverse then he has to be row reducible to the identity matrix that comes directly from our calculation, so when we try to compute.
Birne Binegar: At the miners first our algorithm.
Birne Binegar: starts with a joining to our given matrix the identity matrix doing row reduction.
Birne Binegar: Until we reach the reduce row Echelon for of this matrix.
Birne Binegar: I joined with a and what's going to happen is this reduce your actual on form is either can look like the identity matrix and something else.
Birne Binegar: And, in fact, if this is the case if the a part really reduces to the identity matrix that's what's going to happen to the identity matrix under the same row reductionist we're going to get a the minus first or if we get something else.
Birne Binegar: Then it doesn't matter what happens here, this means there's no inverse.
Birne Binegar: Alright, so I just reviewing that calculation just to tie in this fact that.
Birne Binegar: A matrix is going to have an inverse only if.
Birne Binegar: Reduce relational on form of that matrix is exactly the identity matrix.
Birne Binegar: But, in fact, the way we prove this theory is by analyzing the algorithm, but you will use to compute each step and, in fact, if you look at this.
Birne Binegar: Whether I competed at the miners first whether I sold this linear system or why I saw this linear system real reductions in ball and in fact the way these identities approving it.
Birne Binegar: Frequently and up counting the number of pivots in a row Echelon form and verifying that's the case oh yeah and the last one, is that the rank of a is equal to the number of calls this is just saying that.
Birne Binegar: When you.
Birne Binegar: The dimension of the column space of this matrix is equal to N and that's going to happen, because how would we get the dimension of the column space we would reduce a to a row Echelon form.
Birne Binegar: And for each column in the relational on form that have a pivot we get a basis factor well every enough for the identity matrix every column has a pivot and so all columns of the original matrix should be used, and so rank would be a and.
Birne Binegar: Anyway, so the on this, much is review we covered this before the first exam now what i'm going to do today is i'm going to add a new item to this list of equivalent situations.
Birne Binegar: i'm going to define a function, called the determinant when as soon as I evaluate that function, think of it is as simple as evaluating X squared except there will be more variables.
Birne Binegar: As soon as I evaluate that function, using the entries of a i'll get some number if the number is not equal to zero all know that the matrix it and vertical if the number is equal zero, I know that none of these statements it's true.
Birne Binegar: Though it's um it's a simple function that will tell us lot a lot about a matrix that working with.
Birne Binegar: OK, so the way i'm going to introduce it is i'm going to try to write down a function that.
Birne Binegar: One it's.
Birne Binegar: not equal to zero, I know a matrix is convertible when it is equal to zero, I know that matrix is nonverbal and i'll start with some small major see So you can see that how this can actually be carry out.
Birne Binegar: What i'll do.
Birne Binegar: yeah yeah i'm just saying i'm gonna have some function that will once I know what n matrix is like number should pop out of that like just plugging into a.
Birne Binegar: formula for calculating alright so here's the here's the basic example suppose I have a matrix one world and one column, so it just has a single entry.
Birne Binegar: And I want to look for conditions and one this matrix is in vertical.
Birne Binegar: Well, that means I want to find a matrix be so that when I multiply this matrix by be I get the identity matrix let's write down another one by one matrix.
Birne Binegar: We take the product, we get product of the entries, and we need that to equal one if, in order to match up with the identity matrix well that's going to mean that this auxiliary matrix D be has to be exactly one over a, otherwise the multiplication a time fee would not equal one so.
Birne Binegar: This matrix is going to have an inverse as long as I can find this matrix fee, which means I need to construct a matrix whose sole entry is one over a and I won't be able to do that if a is equal to zero right because you can't divide by zero.
Birne Binegar: So.
Birne Binegar: here's a simple function, the value of this determinant functions on the matrix with a single entry, a well, this is a number, so the value on this matrix is going to be the number a.
Birne Binegar: And then, this function defined that way, taking one by one matrix seats, the numbers.
Birne Binegar: will include that so long as this number is not equal to zero so as long as a is not equal to zero, so that we can define the inverse matrix.
Birne Binegar: is going to be unbelievable so that's your first example of the kind of function i'm looking for it's really simple it's just grabbing the entry of the matrix.
Birne Binegar: i'll get more complicated, as we looked at it larger matrices let's do the same thing or same kind of argument, but now we're looking at a two by two matrix alright so i'm going to look for a criterion on the entries ABC ED that if satisfied will tell me that this matrix isn't vertical.
Birne Binegar: Alright, so from the fundamental theorem that we started with.
Birne Binegar: We know that a matrix is square matrix just going to have an inverse only if they reduce relational on form that matrix is identical to the identity matrix because effectively that's how you calculate the inverse.
Birne Binegar: Well let's see what this requires of the original matrix so i'll start with a matrix now, even though i'll tweet easy a B, C D is variable not as.
Birne Binegar: precise numbers and i'll do the role reduction using sort of symbolic algebra rather than explicit numbers so first thing i'd want to do.
Birne Binegar: Is i'm heading towards reducing Echelon form just get my initial pivot equal to one so divide the first row by a.
Birne Binegar: Now i've got a pivot up here let's get rid of the pivot below so I subtract three times the first row from the second row replace use that to replace second i'll get this guy well.
Birne Binegar: let's get rid of this fraction here, and I can do that by simply multiplying the second row by a nano get a D minus DC now i'm at a place where.
Birne Binegar: I can start to draw some conclusions.
Birne Binegar: Suppose a D minus BC is equal to zero, then I ended up with a matrix it's going to look like.
Birne Binegar: One be divided by a 00, and this is, this is already in relational on form but it's it's not equal to the identity matrix and it, nor will it be.
Birne Binegar: You know won't be able to do any further row operation to get it to look like the identity matrix so if this entry in this corner here turns out to be equal to zero.
Birne Binegar: This matrix is not going to reduce to the identity matrix so what i'll use as my function that tells me whether or not a matrix convertible is i'll just use this entry here and so we'll define.
Birne Binegar: will define the determinant of a two by two matrix to be exactly this quantity, the difference of the diagonal products, a time to D minus B time seed.
Birne Binegar: And with that definition of the determinant.
Birne Binegar: We define it that way, then effectively the calculation on the preceding slide tells me that the determinant of this matrix not equal to zero is equivalent to.
Birne Binegar: This matrix having an inverse now i'm not computing in verse but i'm detecting whether or not it exists or not by evaluating this simple function.
Birne Binegar: And, of course, not only does it tell me that the inverse exists it tells me everything that was stated in that first.
Birne Binegar: First theorem is true that a X equals be will always have a unique solution, but the rank of the matrix of a two by two matrix is going to equal to if it's determined, if not equal to zero, so all that information will follow simply by plugging in whatever numbers, so if you had a matrix.
Birne Binegar: Suppose a is equal to 1234 then it's determined according to this formula would be.
Birne Binegar: One times four minus two times three and so that's going to equal what negative two since that number.
Birne Binegar: it's not equal to zero, I will know immediately that this matrix is not convertible because it's determinant is negative two rather than zero and, in fact.
Birne Binegar: You only kind of matrix only two by two matrix for which they make you matrix will not be convertible or those for which ad minus BC Z exactly equal to zero.
Birne Binegar: Okay, so i've done.
Birne Binegar: Thus far.
Birne Binegar: We know that the determinant.
Birne Binegar: of a one by one matrix is simply the it's only entry that eternal of a two by two matrix.
Birne Binegar: is equal to the difference of the diagonal products.
Birne Binegar: And i'll give you a formula now I don't think i'll write it on the on the whiteboard.
Birne Binegar: i'll give you another example, though, for three by three matrix you can do the same thing you can plug in.
Birne Binegar: Symbolic entries into a three by three matrix and then they tried to do the symbolic algebra that reduces this thing through identity matrix so you maybe divide the first row by a and then you use that.
Birne Binegar: That pivot one to clear out entries below and then you move into the smaller matrix and do the same thing what's going to happen is eventually you're going to get to.
Birne Binegar: A matrix where in that lower corner, you get something expression that ends up looking like this at the very end, and what you would need in order to.
Birne Binegar: be able to reduce this matrix to the identity matrix is it's going to turn out that this particular function of the entries of the matrix cannot equals zero.
Birne Binegar: Effectively there'll be it's going to appear in a denominator, somewhere, and so you can't divide by it and so that would prevent you from actually inverting that matrix.
Birne Binegar: And so you take this, then this function take that as a definition of what the determinant of the three by three matrix would be.
Birne Binegar: And then, as an with the proceeding examples it's just a function that we can evaluate as soon as we have explicit numbers for the entries of the matrix.
Birne Binegar: Evaluating that function will tell us whether or not the matrix is convertible whether or not you have unique solutions to linear systems where as the coefficient matrix and other stuff as well.
Birne Binegar: Alright, right now, now the hard thing about determinants, though, is that i've got up at sparse three by three and already starting get complicated it's a polynomial and the entries of the matrix where I have six terms if I tried to do a four by four matrix the same way.
Birne Binegar: To horrendous.
Birne Binegar: it's okay doing real reaction when you have numbers like 123 or four 200, but when you're doing everything symbolically, nothing is going to simplify two plus three if you could simplify that by five but a B is not going to simplify the anything simpler and so rather than try to.
Birne Binegar: develop determinant formulas for larger matrices using row reduction what i'm going to do instead is identify the pattern that we see in these first three cases and then i'll state a theorem that says that that pattern actually holds for larger matrices as well.
Birne Binegar: Okay, so um.
Birne Binegar: The state the pattern I need to make a.
Birne Binegar: sort of a simple digression.
Birne Binegar: suppose I have a square matrix so in rows and columns.
Birne Binegar: What i'm going to define is an auxiliary matrix that I can attach to a and get an exemplary matrix one for each row in each column of the matrix a.
Birne Binegar: here's the idea if you have a matrix a equals 123456789 i'm choosing these entries so that you'll recognize where everything came from in that matrix.
Birne Binegar: there's going to be a matrix that we call the one one minor a now what this is it's going to be, this is the one one entry right.
Birne Binegar: What i'm going to do to create this matrix is i'm going to eliminate the row and the column that contain the one entry what survives then.
Birne Binegar: it's going to be the bottom corner of the matrix 5689 so attached to this three by three matrix and this particular entry of the three by three matrix there's a two by two matrix and you can do that for all the entries of a matrix.
Birne Binegar: like this guy ABC ED.
Birne Binegar: yeah one one minor eliminate the row and calm that contains the one one entry and you're left with a two by two matrix E, F H Ai that it is.
Birne Binegar: If I look at this entry that's in the first row and second slot I eliminate the row in column that contains it see if I can do this you're left with a matrix E, F G I don't hear i'm.
Birne Binegar: Trying to cover up the things, but let me just display it D E, F G I those are the entry to a that don't live in the same call or row as the one to entry.
Birne Binegar: Similarly, one three we're looking at this entry we eliminate it trail and calm we're left with a two by two matrix de de de.
Birne Binegar: Alright, so.
Birne Binegar: here's my formula for three by three the determinant of a three by three matrix it involves six different terms, but now, what I want you to notice is, this is a.
Birne Binegar: One one entry and then this II II minus F H it's exactly determine of em one one, so this is the one one entry and then this is the determinant of the corresponding minor.
Birne Binegar: here's the one to entry and what this is di minus F G it's determinant of this two by two matrix.
Birne Binegar: and, similarly, the third term and the formula for the determinant is the third entry in the first row times that determine of the corresponding minor determine up the matrix we get by eliminating the rolling call that contain one three entry.
Birne Binegar: Until we get this formula for the determinant of a three by three matrix its first entry in the first row first column times the corresponding minor entry in the.
Birne Binegar: First row second column times to determine the corresponding minor entry in the first row third column i'm saying determine if the course my minor.
Birne Binegar: Well that's business it easy The other thing I want to point out that was there's a sign flip every other term first term comes in, with a plus sign.
Birne Binegar: And you get a minus sign now plus sign if I had a larger matrix what you would see if that there's a science flip every other term if you move across around.
Birne Binegar: All right, let's look at them, are you earlier examples as well and you'll see, they also follow this pattern here's a two by two matrix you know.
Birne Binegar: From our discussion of the.
Birne Binegar: ability to reduce this matrix to the identity matrix that this matrix is only convertible when a D minus BC does not equal zero well.
Birne Binegar: This is element in the one one slot and then, if I eliminate the rolling column that contain it i'm left with just a matrix D so.
Birne Binegar: That so this term here is a times the determinant of this minor one one minor, and then the second term here is this one to entry times the determinant of the one two minor matrix I get by eliminating the first row and second column what the original matrix and so.
Birne Binegar: On formula for the determine looks essentially the same as whether i'm looking at three by three matrix these are two by two matrix sees.
Birne Binegar: Each formula works out to be you move across the first row at you keep the entry of the matrix has a factor, and then you multiply it by the determinant of the matrix you get by eliminating.
Birne Binegar: The rolling column from the original matrix that contain to get an entry.
Birne Binegar: Alright, so my claim is that this um this generalize this.
Birne Binegar: If you have them n by n matrix.
Birne Binegar: Then you can compute the determinant of a in terms of the miners have a.
Birne Binegar: matrix each I create from a by eliminating a particular row in a particular column of the original matrix so here's what the formula generalized has to.
Birne Binegar: be signed factors.
Birne Binegar: minus one to the one plus and i'm just indicating that every other term you get assigned flip but, except for the science liked what you getting our entries of the matrix times determinants of the corresponding co factor now this formula doesn't give you.
Birne Binegar: A nice close expression if, as is a five by five matrix what this formula would do would reduce the computation of determine on a five by five matrix do the computation of.
Birne Binegar: determines a bunch of four by four matrices, but it could be reply, the cursor if this is these turn out to be four by four matrices I can apply this formula again and reduce.
Birne Binegar: The terminal have a four by four to the son of a bunch of determinants of three by three and the three by three matrix she's there determines it can be expressed in terms of two by two matrix ease.
Birne Binegar: and, eventually, you can get it down to a matrix that small enough that you have a formula for.
Birne Binegar: You get it down to a thing a one by one matrix, then you just use the single entry, if you get it down to two by two matrices you can use a D minus BC that cross diagonal product, so the idea of of this recursive definition of the determinant is to.
Birne Binegar: Basically, give you an algorithm for calculating that he terminate in terms of smaller matrices that if you keep applying the algorithm eventually you get this a concrete calculation that you can perform.
Birne Binegar: Right now, what um let me show you how this would work in a four by four matrix here's the matrix if we want to compute determine.
Birne Binegar: What I would do so I will march across say the first row of grab the first entry how multiply by the corresponding might determine up the corresponding then i'll read the second entry and all compute the determinant of the minor corresponding to that entry.
Birne Binegar: Let me see if I here are actually plugged in well, I have one times this matrix here is to make sure you get by eliminating the rolling column that contain the one one entry, the original matrix here's the.
Birne Binegar: One to minors is the one three minor and that's the one for minor.
Birne Binegar: Now of course i've got some more determined to compute if I want to get the determinant of the original matrix but at least i've simplified the problem of calculating the German for four by four into the calculating a bunch of smaller determinants.
Birne Binegar: OK now we'll finish that calculation in a minute, but I want to give you a more general expression.
Birne Binegar: What these kind of formulas they're called co-factor expansions and there's actually a number of co factor expansions that you can apply to a given matrix.
Birne Binegar: So let's see.
Birne Binegar: here's the statement, if you have any square matrix and you've defined the corresponding minor so for each entry and I wrote jake calm at the original matrix you can have a corresponding minor that you create by eliminating I throw in jail Tom at the matrix a.
Birne Binegar: And then, this will be a formula for the determinant of the original matrix expressed in terms of the determines of the smaller miners now what's new in this formula.
Birne Binegar: The proceeding formulas, I always.
Birne Binegar: chose the entries in the line in the first row of my matrix but, in fact, the same formula works, no matter which row you choose so here i'm leaving the index I as some fixed index might be one, two or three I can use any real I want.
Birne Binegar: But I keep the index I fixed and then I some over the columns yeah so what what this expansions isn't that it's referred to it as expansion along the I throw and what I pick up there are.
Birne Binegar: All the elements in the row and they're being multiplied by the determinant of the miners that correspond to those entries.
Birne Binegar: or there's an email yet another formula, the other alternative you have is to instead of working along a row, you can work down a column, so in this formula I keep the column index fixed.
Birne Binegar: or choose what some number between one and that will fix on fixed me on a column, and then I moved down individual rose by during this index I summing up the terms.
Birne Binegar: So my claim is is that, no matter which formula you use you always get the same result, no matter which row you start off with but which call me start off with once you do the final computation you get exactly the same terms, no matter how you you actually set up the sun.
Birne Binegar: Alright, so these numbers here the sign factors and determinants of the corresponding miners those things are called co factors and we'll see later that they have their own individual use as well, but for now that just.
Birne Binegar: Some nomenclature returns of the soul that terms that we get by getting a particular sign the way we get that sign is by we some.
Birne Binegar: row index in the column, and next be given entry and we just take a minus one to that power so when the row it index in a column index some to an even number that sign is one when they some to an odd number, then you get a minus zone.
Birne Binegar: And then.
Birne Binegar: That, then, we have the formula, so the first formula is called a Co factor expansion along I throw the shiancoe factors and keeping that.
Birne Binegar: index I fixed each word confining ourselves to looking at entries along the I throw in here, but we do is we keep the column index sticks and with some along the entries in in various roles so move up move down a column, you can keep this version.
Birne Binegar: But the important thing about the theorem is it doesn't matter how we which of these co factor expansion which we employ will still get the same result.
Birne Binegar: Right, so now let me return to the original.
Birne Binegar: Four by four matrix which I started the computation of determining what I did is I.
Birne Binegar: A couple slides back, we looked at this exact matrix and we did a Co factor expansion along the first row go ahead, one time to the determinant of this.
Birne Binegar: One one minor 0011100 to one that's exactly that, then we have zero times the determinant of a minor but, since this is going to be multiplied by zero.
Birne Binegar: I don't even have to write down that determine that it's being multiplied by zero so it's not going to contribute, then we get two times the determinant of this minor.
Birne Binegar: look like 101010101 exactly that guy and then we get zero times the determinant of the one for minor but since it's being multiplied by zero i'm not going to get any contribution.
Birne Binegar: Okay, so this is the computation we did three slides back, but where I reduced the determinant computation of the determinant of a four by four matrix to net, adding that up some contract.
Birne Binegar: express it in terms of determines of a smaller matrix 323 by three matrix ease and now I can use the same algorithm.
Birne Binegar: To compute this determinant let's again move across the first rub zero times the determinant of 102110 to one, and then I would have zero times the determinant of 1001 and then I will have one times the determinant of well when I eliminate that guy i'm left with the matrix 110.
Birne Binegar: And then.
Birne Binegar: Now i've got it now this determinative a three by three expresses it son up determinants of two by two matrix season these help just compute so determine if this matrix is one minus zero femina this matrix is one minus zero determine if this matrix is to minus zero.
Birne Binegar: And so we're explicitly able to reduce this determinant to just a simple number.
Birne Binegar: All right, what about this matrix so we had.
Birne Binegar: We had to do I just computed this determinant let's look at this guy next.
Birne Binegar: You can do the same thing, however, I want to do something a little clever now i'm going to exploit the fact that the second column has two zeros.
Birne Binegar: that's going to mean that i'm not going to have to compute the determine of this minor or this minor because they're going to be multiplied by zero.
Birne Binegar: And so that's where the advantage of this more general co factors expansion comes in, if you have a row or column that has lots of zeros it's going to make the computation much simpler, because there will be.
Birne Binegar: determinants of miners that you won't have to compute because they're being multiplied by zero.
Birne Binegar: So in this case, I think I moved down the second call, so this first contribution with this first term here is going to be zero because the determinant of the one two minor is going to be multiplied by this factor of zero, then I have this guy so I have one.
Birne Binegar: here's the correct sign the way you get the sign factor correct issue add the row indexing column index, so this is in the second row seven columns so I go minus one to the two plus two.
Birne Binegar: And then the minor I look at is the two by two matrix I get by eliminating the rolling column that this entry lives in so that's going to be 1111.
Birne Binegar: And so, and then i'm gonna get a contribution from last term which is going to be zero so the central term is the only thing it survives, and in fact it doesn't survive, either because the determinant of this two by two matrix is one times one minus one times one or zero.
Birne Binegar: Okay, so there's an example of a full computation we took a four by four matrix reduce this determinant to a combination of two.
Birne Binegar: Three by three matrix ease and then we evaluated determinants of the three by three matrix also by a Co factor expansion formula, but eventually we were able to reduce our determinants down to a one where we can apply a known formula.
Birne Binegar: Okay.
Birne Binegar: And so, in the end.
Birne Binegar: i've now computed these determines of the three by three I can put everything together to get to as the determinant of the original four by four matrix.
Birne Binegar: Okay, so.
Birne Binegar: What does she tell you is the terminator matrices it's going to be kind of tedious in general, because you're not going to get the answer right away, and you have to.
Birne Binegar: Keep reducing the matrices to smaller matrices until you get something that that.
Birne Binegar: You can actually know the formulas for rather than.
Birne Binegar: The co factor expansion for me.
Birne Binegar: Alright, so um but unfortunately that computation.
Birne Binegar: makes it hard to understand what kind of function is going to pop out of that algorithm you take a four by four you reduce it to a summer three by threes and then you reduce the summer three by three determinants do bunch of two by two determines and then finally.
Birne Binegar: Finally, you get an answer, but it's hard to predict what that answer it looks like, in particular, what kind of function is it is it assigned function is it a cosine exponential function of polynomial, how do we see the functional properties of the determinant.
Birne Binegar: Well, for that i'm going to give you a different way of computing to determine.
Birne Binegar: This way it's going to be.
Birne Binegar: starts off looking.
Birne Binegar: More like common at oryx and then linear algebra, but let me proceed i'm going to show you a different way of computing to determine if a three by three matrix but it's going to require me to make a slight digression.
Birne Binegar: will say that a permutation of n it's a listing of the numbers one through and so everything every number between one and then it's included in this list exactly once and what I what the permutation is it's just taking the numbers one through 10 and writing them in a particular order.
Birne Binegar: For the number when as equals three that i'm looking at numbers one, two and three, I have this ordering 123 or.
Birne Binegar: I can have one still be the initial number, but I can reverse the order of the last two, so another ordering the one to two.
Birne Binegar: Or the list could start with two and then I might have one three or 231 or the list could start with three and then I could have either one, two or followed by two one, but what i've listed here are the six possible ways writing the one numbers 123 in a particular order.
Birne Binegar: All right, and so and i'm going to refer to, not the standard right way of writing the numbers one two n in their usual order i'm going to call that the standard ordering.
Birne Binegar: Alright, so.
Birne Binegar: i'll show you the reason for these in a second, but before I do that I want to do one additional thing with the fermentation I want to.
Birne Binegar: associate to any one of these.
Birne Binegar: A particular sign either a plus sign or minus sign or you can think of, plus one or minus one what it's going to be is i'm going to.
Birne Binegar: Look at a permutation and i'm going to try to identify the number of times as I read across pairs of entry appear in order or appear out of order so in here.
Birne Binegar: We can think of three possible parents one with two one with three and two or three and this one one and two are in order, one in three are in order and two and three are in an order issue read across so this one.
Birne Binegar: This number S, which is kind of count the number of times things appear out of order would be equal to zero.
Birne Binegar: Now here one is before three that's Okay, one is before to that's Okay, but three if not before, to in the natural order, this has one sort of discrepancy in the ordering of pairs and so, in that case s would equal one, and this last one let's see.
Birne Binegar: Two is out of order with one three is out of order with two and let's see in one is out of order, with three as well, so, in fact.
Birne Binegar: This particular permutation has three different pairs 1213 and, two, three and they're all out of order, if you read across that list, and so the value of this number so counts the number of discrepancies in the natural order, it would be three for this.
Birne Binegar: Until the sign would be minus one, the third power or negative one.
Birne Binegar: Alright, so what i'm showing you is that.
Birne Binegar: You have a list of numbers, you can always write every possible way of ordering them down and what i'm giving you in this definition is for any particular permutation a way to assign a number minus one plus one to it.
Birne Binegar: All right, and so yeah so in here, I do a bunch of calculations have been signed yeah here's that last one I did each each of these pairs is out of order for this last permutation, and so we get.
Birne Binegar: Three contributions to F and the end result it's going to be minus one.
Birne Binegar: All right now.
Birne Binegar: All that was just to set up a formula for the determinant of an n by n matrix not in terms of determinants of other matrices but rather directly in terms of the entry to the original matrix.
Birne Binegar: So the way i'm constructing these terms is notice that the row index in this factor here is in natural order one to the next factor would be a three Sigma three and then you get to the last factor.
Birne Binegar: A row index is and now and then there's some number between one and end it will be a Sigma n so what's the signal is here it's just a listing of these numbers in some order and so don't think of signal, one is necessarily equal one, it could be the seventh element in this list.
Birne Binegar: But.
Birne Binegar: What you see here is that each term in the some it's going to depend on a particular permutation.
Birne Binegar: it's going to be a there'll be assigned factor, depending on how many times the entries in this permutation how many how many times pairs are out of order.
Birne Binegar: there'll be a plus sign or minus sign and then these factors here there's going to be a total of n factors, but these column indices are going to be distributed.
Birne Binegar: In some order then not necessarily going to be in the natural order 123 and so that's what this purpose of permutation is to change these column indices.
Birne Binegar: But it's term here is simply assign factor times a product of entries of the matrix of a one from each row in one factor from each column as well.
Birne Binegar: And when I some overall permutations i'm actually constructing all possible products at this form all possible products where you take an entry entries from distinct columns indistinct rows and you multiply them together.
Birne Binegar: Right now, I claim that that's that's another formula for the determine.
Birne Binegar: notice that what it does, though, is expresses the determinant as a polynomial and the entries right, this is just a product of variable so it's going to be a polynomial.
Birne Binegar: I thought i'll show you explicitly with this example here's a three by three matrix here's the possible permutations of the numbers one, two and three.
Birne Binegar: So if I wanted to compute this determinant i'm going to some over all these permutations i'll have to compute the sign of each permutation and then the corresponding product, so what you notice the way i'm getting these factors here as i'm.
Birne Binegar: writing the column indices of these factors in the same to be.
Birne Binegar: defined by this permutation so the role indices are always 123 in each of these term, but the column indices, and these factors are going to depend on what permutation on so here i'll have one, three and two.
Birne Binegar: And so.
Birne Binegar: This expansion is easy to write down once you know what all those permutations are.
Birne Binegar: And in fact.
Birne Binegar: If I play again explicitly for the a one one entry is a a to to entry is he a two three entry is I guess Oh, this is wrong, this should have been a.
Birne Binegar: shipment I there, but what happens when you actually take this data, you plug in and you also plug in for the signs you get exactly this and then.
Birne Binegar: Just regrouping things I can recover my earlier co-factor extension formula this.
Birne Binegar: Here, this first two terms, to have a common factor ah, well, I group them together and then II II minus F G is just the determinant of a minor corresponding that entry don't that I do reproduce the Co factor expansion, but it tells me some things about determining.
Birne Binegar: First of all, it shows explicitly that terminal is a polynomial second of all.
Birne Binegar: It gives us a way of counting how big that polynomial is.
Birne Binegar: How many how many terms is going to have it's going to have as many terms, as we have permutations Oh, how do we count permutations.
Birne Binegar: If you have the numbers.
Birne Binegar: One two N and you want to create a new list using just these numbers well for this first entry in this reordering of the numbers one through, and you have no choices right.
Birne Binegar: But then the second entry.
Birne Binegar: Of this permutation you've used up one of these numbers for this guy you're going to have N minus one choices.
Birne Binegar: Because you've already used one, and then the third one now you've used up two choices so for that third entry, you only get you're going to have N minus two choices.
Birne Binegar: And then finally you keep moving across when you finally reach the last guy well you've chosen every other number out of this list, except one until there'll be just one choice here.
Birne Binegar: So how many choices, do you have in total n times N minus one times N minus two etc down to one or you simply get n factorial choices.
Birne Binegar: So.
Birne Binegar: Actually number of permutations you have exactly n factorial and that should scare you.
Birne Binegar: Because well for three factorial six perfectly manageable for factorial is 24 kind of bad five factorial is 120 so I want you to use this formula.
Birne Binegar: to calculate the determinant of a five by five matrix i'd have to expect the polynomial with 120 different terms if I had a six by six matrix not very big 720 separate terms, so, even though this.
Birne Binegar: This is a pretty simple form it's really hard to write down when you get to a matrix it's any any bit much, much bigger than a four by four matrix.
Birne Binegar: yeah it's a nice formula, because you can see directly if this this function is a polynomial it's homogeneous meaning every term has the same number of factors and variables as exactly and factors, so the determinant is.
Birne Binegar: Paul no meal degree, and it has exactly n factorial terms.
Birne Binegar: So, because it's so complicated it's actually a very sensitive function, and it will be using determinants to detect things about matrices and.
Birne Binegar: This formula which is not a practical way of computing determinants It is nevertheless important because it shows you the functional forum what an eternal really is, from the point of view of algebra it's just a simple polynomial.
Birne Binegar: Alright, so i'll stop here today and what i'm going to do on on Friday is giving you a much.
Birne Binegar: more expedient way of computing determines one that doesn't require you to apply a recursive formula Nord a formula that requires you to list them.
Birne Binegar: In factorial different permutations and then use those permutations to compute the eternal what i'll give you as a way of finding determine really quickly using just row reduction.
Birne Binegar: Okay, so i'll stop here today and i'll see you on Friday.