Birne Binegar: Okay, so I don't have your test graded yet they'll come back on Friday.
Birne Binegar: And what i'm gonna do today is i'm going to move forward and to new material.
Birne Binegar: we're now.
Birne Binegar: Part of the course or we're.
Birne Binegar: going to return back to.
Jaxon Castillo: I would be.
Birne Binegar: Organizing.
Birne Binegar: principle.
Jaxon Castillo: I like I got on this course, like, I was trying to a little bit of modular machine on yeah.
Jaxon Castillo: Like a company, you know.
Jaxon Castillo: Just like this word.
Jaxon Castillo: For word veneer.
There we go.
Birne Binegar: Okay, so.
Birne Binegar: So what we're going to do today is discover a new calculation, one that that relates back to a fundamental theorem we had Oh, even before the first exam that related.
Birne Binegar: i'll put it up in a second but relate when you can actually find the inverse of the matrix and it related to other things, we discovered when can you solve a linear system and get a unique solution, etc.
Birne Binegar: And so i'll introduce the there's going to be a particular function that will tell us give us another equivalent condition and then we'll explore that function for small matrices.
Birne Binegar: and
Birne Binegar: The method which I used to introduce it sort of breaks down now, when the matrix starts to get even moderately large that you'll see how that happens, will develop a better formula that now can be applied to any square matrix and by n matrix but.
Birne Binegar: That in fact it's going to be a recursive formula and so it's more of an algorithm and an actual close formula for the determinant but then i'll finish today's lecture with giving you a nice clothes formula for the determinant and you'll see exactly.
Birne Binegar: What kind of function, the determinant of a matrix it, so I guess, let me see if I can get started here.
Birne Binegar: Alright, so.
Birne Binegar: You remember just prior to the first exam.
Birne Binegar: We were looking at.
Birne Binegar: Sea dragon this a little bit.
Birne Binegar: You know Eric and I can brag a little bit.
Birne Binegar: When we were talking about inverse of the matrix each we came up with the following theorems.
Birne Binegar: which was called the fundamental theorem because it ties together lots of different ideas we're working with up to that point so if you have any square matrix same number of roses, you have columns then there's what we had responded be equivalent statements one.
Birne Binegar: statement that this matrix has an inverse and then i'm know what's happening here.
Birne Binegar: Have a better copy.
Birne Binegar: Okay well i'm gonna have to flip this this lecture maybe.
Birne Binegar: Less.
Birne Binegar: Welcome to working I don't have a fancy pauses or anything in it Okay, so let me just remind you that.
Birne Binegar: We had a theorem baffling we were discussing the inverse of the matrix that related the existence of an inverse to other properties of the matrix.
Birne Binegar: For example, we do have an inverse then we'll always going to have a unique solution to any linear system that uses the matrix that's a coefficient matrix.
Birne Binegar: A has a has an inverse then this linear system is always going to have one and only one solution.
Birne Binegar: or another one another equivalent condition is that the equation eight X equals zero not some arbitrary vector be but.
Birne Binegar: Exactly the zero vector then that only has one solution corresponding X equals zero that X equals zero is a solution well you just substitute the zero vector here and matrix times any zero vector is going to produce the zero vector or another way for the know that the.
Birne Binegar: inverse of the matrix exists is by reducing the matrix that's how you calculate the inverse of the matrix if the matrix a row reduces to the identity matrix then we know the.
Birne Binegar: inverse exists and also turns out that the ranks of the matrix and rank is the.
Birne Binegar: Is the dimension of the column space, for example, well how would you find a dimension of the column space, you would reduce the matrix to rub Echelon arm and you would count the number of columns with tidbits.
Birne Binegar: Well, if the rotational on form is the identity matrix then every column has a pivot, and so the number of.
Birne Binegar: linearly independent column vectors is going to be exactly equal to the number of calls.
Birne Binegar: Alright, so what i'm going to do today is i'm going to introduce yet another condition.
Birne Binegar: So called determinant function which I haven't defined yet, but it's a function that.
Birne Binegar: that's actually going to be a polynomial in the entries of the matrix so unlike.
Birne Binegar: The reduction.
Birne Binegar: calculation that ties together these five conditions that we've studied previously, this time, these these conditions are linked by that rule reduction algorithm.
Birne Binegar: which, I stress, is an algorithm as opposed to a function and algorithm it's sort of this it's a sequence of structures and you carry out one step and you see.
Birne Binegar: results and then depending on what results you change, maybe use a different operation or different technique you proceed to the next step until eventually your lead to.
Birne Binegar: A final result, but the thing about working with algorithms So if you don't really see the answer until you reach the last step.
Birne Binegar: But on the other hand, if we had a polynomial in a that if simply we checked to make sure we evaluate that pole normally we don't get zero that would be.
Birne Binegar: We could just stare at the entries of the matrix and and perhaps the deuce these consequences, without having to do any role reaction and what i'll show you eventually.
Birne Binegar: Probably not this week, but early next week is that that's determined function can actually be used to avoid reduction.
Birne Binegar: We can solve linear equations without doing any reduction as long as we can readily compute determinants we can in fact find the inverse of the matrix simply by computing a bunch of determines so that's what's coming.
Birne Binegar: But way we're going to introduce this thing is i'm going to look for a function of the entries of a matrix that will tell me when the matrix is vertical or not.
Birne Binegar: So let me step through some and then the idea is that once we have this function on n by n matrix these define, then we can decide whether a given matrix in this set, we can decide whether it is convertible not simply by evaluating that function.
Birne Binegar: Alright, so here's here's.
Birne Binegar: An example that will clarify what i'm saying here suppose we have a one by one matrix so it just has a single entry.
Birne Binegar: And I want to know conditions for which this simple matrix is going to be convertible well, that means i'm looking for a matrix theory that when I multiply eight times that I get the identity matrix let's.
Birne Binegar: construct another one by one matrix we multiply these together the rule for matrix multiplication is.
Birne Binegar: single single column, and each so just multiply those together now the one by one identity matrix would be one by one matrix where the only entry is one, so what I need is a time to be to equal one or B is equal to one divided by a alright so.
Birne Binegar: here's a function i'm going to define on any one by one matrix it's just going to be the sole entry of that matrix and then this little calculation here shows is that.
Birne Binegar: This matrix is going to be in vertical only have one over a only it's a it's not equal to zero, because you can't divide you'd have to solve for be but if a is equal to zero this.
Birne Binegar: division by zero wouldn't make sense, so if this is our function, the fact that the determinants not equal to zero ne to a is not equal to zero, which means we will be able to find that in vertical matrix.
Birne Binegar: So this is the first example if the determinant function this function of the entries have a one by one matrix if it's equal to zero, then the matrix is not convertible because I won't be able to construct this jewelry nature be now, but if it's long as it's non zero then.
Birne Binegar: This can be calculated and we will have an inverse okay so that's simple that i'm going to declare now that determinant of a one by one matrix is defined as as it's only entry.
Birne Binegar: Right now that's a simple function but let's move on to more complicated months here we have a two by two matrix.
Birne Binegar: i'm going to apply the same logic i'm going to look for a function of the entries ABC and D that when I evaluate it for particular numbers ABC and D if that number is non Tyrol, I know the matrix is convertible if it's m equals zero than the matrix is not supposed to be not convertible.
Birne Binegar: Alright, so.
Birne Binegar: we'll look at this matrix and recall from fundamental theorem that an inverse of the matrix exists.
Birne Binegar: If, and only if the reducer national on form of that matrix is the identity matrix that's coming from what you calculate an inverse right, if you want to find a the minus first what you do is you.
Birne Binegar: form this double matrix the original matrix a and the identity matrix and you wrote reduce to reduce relational on form and then what happens is one of two things.
Birne Binegar: Either you get.
Birne Binegar: The side reduces to the identity matrix.
Birne Binegar: And you get at minus first here, or you get something else.
Birne Binegar: And on the other side well doesn't make any difference if you if you are in this when the left hand side that's not robbery.
Birne Binegar: To get into the matrix, then the inverse does not exist, and so this calculation shows you that you only have an inverse of the matrix when a row reduces to the identity matrix.
Birne Binegar: So that's what we're going to use we're going to try and reduce this matrix to the identity matrix and see if there are any anything that prevents us from doing that.
Birne Binegar: Alright, so here's the calculation yeah we started off with our general two by two matrix the ideas i'm going to try to rewrite it in reduce cholesterol on form, so I first thing I need to do is get rid of this pivot here so that's done by that real operation leads to this.
Birne Binegar: Well, if that's, the first thing I did is I read normalized pivot in the upper corner one by dividing the first row by a and then I clear out this pivot below the pivot in the first row I doing.
Birne Binegar: Multiple first row and subtracting it from the second row that leads up to this and then, when the next thing i'd want to do is get its enroll Echelon form now.
Birne Binegar: And there's no obstruction to these well except maybe dividing by a but um it's no obstruction to this calculation up to this point.
Birne Binegar: The last step would be well, we want to get this ticket in the second row, we want to get it equal to one so to do that well let's see I guess what I did that's I simplified.
Birne Binegar: Oh yeah I just multiply the second row to get rid of that denominator and now to get rid of to make that pivot in the second row equal to one i'd have to divide by a D minus BC but as long as I can divide by a D minus BC i'll be able to convert.
Birne Binegar: This entry into one, and then I can use the one there to clear out the entry there, so now what i'm trying to show you in this calculation is the reduction of a general two by two matrix to a matrix.
Birne Binegar: To an identity matrix in the form of identity matrix what a crucial step in the calculation, if this last step here, where you have to divide the second row by a D minus DC.
Birne Binegar: If that number is equal to zero, I won't be able to get rid of this guy I mean that that mean this guy Would he be equal to zero, and it would not be reproducible to the identity matrix so the main thing I need.
Birne Binegar: To complete this role reduction is for a that this product of entries a time T minus B time see I need that to not equal to zero so, but the reduction will proceed all the way to the identity matrix alright so.
Birne Binegar: What this calculation and shows is that a general two by two matrix can be reduced to the identity matrix only if this combination of the entries or.
Birne Binegar: The matrix entries does not equal to zero, and of course being reducible to the identity matrix this condition is equivalent to eight minus first exists.
Birne Binegar: So we have a function exactly.
Birne Binegar: A quadratic function right it's that degree to in the variables your degree one degree one add degrees you get a degree to term, this is degree to turn, so we have a function, the entries of a two by two matrix that.
Birne Binegar: When that function it's not equal to zero are in verses can exist when the function does equals zero the inverse is not going to exist and that's the that's the function on advertising is the one we're interested in.
Birne Binegar: function that simply upon evaluation will allow us to decide whether a matrix is convertible or not.
Birne Binegar: Okay, so we can continue like this, I did.
Birne Binegar: I did the one by one case.
Birne Binegar: One by one matrix, this is a convertible.
Birne Binegar: If, and only if.
Birne Binegar: The determinant of a.
Birne Binegar: determined that one by one matrix you find us the the sole entry of that matrix does not equal zero.
Birne Binegar: This this determinant that equals zero that's equivalent to.
Birne Binegar: Maybe say that that.
Birne Binegar: that's not equals zero and that's equivalent to.
Birne Binegar: This matrix having an inverse and what we've just done on this slide it look at two by two matrix a B, C D.
Birne Binegar: Is.
Birne Binegar: In verbal if, and only if.
Birne Binegar: The determinant of that matrix a B, C D and now I have a definition for it, I want it to be dysfunction a D minus BC.
Birne Binegar: If that does not equal zero, then this matrix will be convertible so, in fact, let me give you.
Birne Binegar: This example.
Birne Binegar: Does the inverse of this matrix exist well what we'll do is we'll calculator to determine it so we'd have it's a difference of the diagonal products will, I have one times for.
Birne Binegar: determinant 1234 it's going to equal one times four minus two times three.
Birne Binegar: And so that's going to be four minus six or negative two since that doesn't equal the number negative two does not equal, the number zero, I know that this this matrix has it inverse even though i'm not computing is determine.
Birne Binegar: Now.
Birne Binegar: That tells us not only that he does not have a determine.
Birne Binegar: If the determinant of a is not equal to zero, not only do we know that a minute first exists, but we also know that these other four statements are also true that a.
Birne Binegar: And so, this this one function it's actually giving us a lot of information about the matrix today.
Birne Binegar: telling us whether it has numbers what is like what is going to be it's going to happen if we try to solve a linear system using that matrix of the coefficient matrix that tells us that, if we try to reduce this matrix.
Birne Binegar: Take this matrix if I tried to Robert is it I know it's kind of reduced to the identity matrix and I did that not five to explicitly performing the reduction calculation.
Birne Binegar: But rather evaluating a particular function of the entries of that matrix and that function is not equal to zero, then.
Birne Binegar: All five of these conditions will be true for the for the given matrix.
Birne Binegar: Okay, so let's move forward a little bit um.
Birne Binegar: i've done it for one by one matrix these.
Birne Binegar: Two by two matrix ease I won't repeat the calculation for a three by three matrix.
Birne Binegar: But what happens for three by three matrix things you can again try to manually reduce this matrix to get it in reduce relational on form.
Birne Binegar: And look for obstructions that we keep that reduce your national unformed from being the identity matrix and what you'll find is that this particular function.
Birne Binegar: Of the entries of the given matrix so long as this function does not equal zero, then you will be able to reduce the given matrix to row Echelon form, so its inverse will exist, it will have unique solutions and all the other conditions in the fundamental theorem will follow.
Birne Binegar: Okay, now the problem I have in presenting this stuff, though, is is that.
Birne Binegar: about this stage and in fact I don't even do the calculation for three by three matrix it gets really complicated.
Birne Binegar: To try to figure out to reduce a matrix was symbolic entries where you have to keep track of all the variables, instead of adding integers together and simplify things stay, just as complicated as as the original presentation.
Birne Binegar: So what we're going to do next, then, is find a.
Birne Binegar: way of generalizing the results I stated so far in a way that will.
Birne Binegar: Allow me to define a determinant of them.
Birne Binegar: Arbitrary n by n matrix but without resorting explicitly to this role, reduce ability argument.
Birne Binegar: Well, instead of going to do it's going to look at the pattern that that sort of comes out of the three test cases we have been able to do and then we'll generalize that pattern.
So.
Birne Binegar: here's the here's the idea.
Birne Binegar: The key thing in the discussion now it's going to be we start off with a matrix of arbitrary size but still going to have the same number of roses, we have columns.
Birne Binegar: from any n by n matrix we can construct a bunch of smaller matrices the way we construct them is we take.
Birne Binegar: We take an entry of the matrix and we eliminate from the matrix.
Birne Binegar: The column, and the road that contains the entry so look at this entry a if I remove the column that contains it.
Birne Binegar: And the road that contains it unless with a smaller matrix its job I started with a three by three, but after removing a real mccollum i'm left with a two by two matrix.
Birne Binegar: We call this two by two matrix E, F H I the one to minor corresponding to the one one entry of the matrix the this entry here sitting in the first row and first call that's what these.
Birne Binegar: indices represent first row first column once I eliminate the first row, and the first column from this matrix i'm left with a smaller matrix E, F H Ai that's going to be called the.
Birne Binegar: One one minor of the original matrix this next matrix also two by two I get it by looking at the matrix.
Birne Binegar: The entry in the first row second column of the original matrix or this entry be here, I eliminate the call Nick contains it and the road that contains it and what survives is D E, F g.
Birne Binegar: So i've got a two by two matrix associated with this entry and, similarly, I can look at this entry.
Birne Binegar: eliminate it's rolling call and i'm left with a smaller matrix that's going to be D E G H So these are called miners.
Birne Binegar: You you create them by looking at particular entries in your given matrix we moving from the matrix the column and row that contains it you'll be left with the smaller matrix.
Birne Binegar: So here's the point of that this determinant of a three by three matrix which I said a reduction argument would tell you that, as long as this function of the entries of the matrix does not equal to zero.
Birne Binegar: As long as that determinant does not equal zero i'll know that this matrix is in variable well.
Birne Binegar: here's a replication of that formula, but now in terms of these so called minor matrices, this is a and what is he I minus F H well.
Birne Binegar: it's a difference of the diagonal product right he times I minus F times eight, so this is a times the determinant of this one one minor.
Birne Binegar: This next term here is the second entry times the determinant of the one two minor I need to be taught times I minus F time G exactly what you get here.
Birne Binegar: And then the third term here is the third entry in the in the first row times the determinant of the corresponding minor, so this is the one three entry.
Birne Binegar: I create the one three minor, and I take it determines the dh minus G EG so.
Birne Binegar: This gives us a way of understanding how we might generalize these.
Birne Binegar: These formulas for the current that we have previously done so, for example, even for two by two matrix I defined it as the difference of the diagonal products but.
Birne Binegar: I can also think of it as a times what what's the minor corresponding this guy it's only going to be a single entry it's going to be D.
Birne Binegar: So this term ad you can think of as the entry in the one one slot times the determinant of the corresponding minor determine out of a one by one matrix is just.
Birne Binegar: The entry D and, similarly, the next term would be its associated minor would be the matrix soul entry see, and so this expression for our determinant can also be thought as.
Birne Binegar: An expression, where we march across the first row with the matrix and we multiply the entries by the company determinants of the company minor.
Birne Binegar: Alright So what does this do for us.
Birne Binegar: If this formula holton general, then what what we can do is for a large matrix we can steadily reduce the size of the matrix for which we actually have to explicitly determine.
Birne Binegar: compute determines So if I start off with a matrix with in rows and columns what I can do is here's my formula for the determining what you do is you march across first row in this case and then for each term in that row you'll get.
Birne Binegar: you'll get the corresponding entry and then there'll be a determinant of a matrix but what i'm doing here is, this is actually the terminal of a smaller matrix if this is n by n.
Birne Binegar: After listening to me one row in one column i'm left with a matrix that is going to be N minus one by N minus one will have one less row in one left call, so this will be a smaller matrix and then.
Birne Binegar: If this formula holes in general, I can again reduce the determinant of this N minus one matrix by N minus one matrix.
Birne Binegar: I could expand along its first row, and I can reduce the computation of this determinant.
Birne Binegar: into the computation of a bunch of N minus two by N minus two matrices and you can steadily break things down until eventually hopefully.
Birne Binegar: you've got everything express it's determined enough maybe a two by two matrix or you want to go all the way you can bring it down to everything he's expressed in terms of the determinant of a one by one matrix.
Birne Binegar: Well, I mean, I think I do an example of such computation yeah here we go so here's an explicit matrix.
Birne Binegar: A four by four matrix and more of the way I proposed to calculate determine is by using this expansion formula, so your original matrix it's determinant can be.
Birne Binegar: computed in terms of the determines of it minors, so the formula is going to be march across the first row grabbing entry.
Birne Binegar: And then we'll create the corresponding minor so for this entry the minor 01001110 and zero to one that's what I got here, and then I had zero times the corresponding minor well eliminate the rolling column that contain this entry and you're left with let's see.
Birne Binegar: The rolling column so, but you left with 101.
Birne Binegar: and
Birne Binegar: Up sorry think Scott someone's flipped I get 101010 and one to one yeah so.
Birne Binegar: uh one thing to point out here is, as I march across these a given row, the first term comes in, with a plus sign second term comes in, with a minus sign third term comes in, plus sign and a minus sign so now.
Birne Binegar: Another feature of this formula is to recognize that the terms coming with alternating sign every every other term there's a sign slip.
Birne Binegar: But you get a nice clothes formula and now what I could do, I wanted to complete the determinant calculation of this determinant.
Birne Binegar: If I could look at this matrix and I could do a Co factor expansion, here it basically be zero times, the chairman of this matrix.
Birne Binegar: minus zero times the determinant of this matrix plus one times the determinant of this two by two matrix and then that would.
Birne Binegar: produce the determinant the three by three matrix and you'd have to similarly break this down into a bunch of two by two matrix determinants a two by two matrix ease and you can break this down both of those down into a some over determinants a two by two matrix ease.
Birne Binegar: But before I do that and I will do that.
Birne Binegar: I want to generalize that formula, a little bit because way i've stated.
Birne Binegar: That first row sounds kind of special when, in fact.
Birne Binegar: These kind of expansions you can you're not confined to just doing looking on the first row and multiplying corresponding entries.
Birne Binegar: Time to corresponding miners in fact you could do a similar expansion along the second row along the third row along the fourth row or up and down any column.
Birne Binegar: The only thing that's formula requires is that you multiply entries of the matrix by the determinant of the corresponding miners.
Birne Binegar: Alright, so here's the more general formula that will work, no matter which row you choose or, no matter which call me choose.
Birne Binegar: Mostly, the formula looks exactly the same, and the two cases what's different between these two lines is in the first line.
Birne Binegar: i'm keeping this index I, which tells you which row i'm looking at that's not changing when i'm something over here are the the column so here I fix a row index, and then I look at the entries.
Birne Binegar: In each column for that row and depending on.
Birne Binegar: Some of the row index and assign index I get a plus or minus sign here, but then I have the entry in the IPA slot times assigned that comes by looking at.
Birne Binegar: The parody of.
Birne Binegar: Some of the row index in the.
Birne Binegar: column index, and then you have the determinant of that minor that's associated with the iga entry.
Birne Binegar: Okay, so.
Birne Binegar: So when these formulas.
Birne Binegar: This combination here, where I take.
Birne Binegar: A minor corresponding that I throw in jail calm of a given matrix and multiply I take its determination and I multiply it by assigned factored it depends on the position.
Birne Binegar: Of the entry was in the matrix a that's called a Co factor, and we call these expressions for the determinant co factor expansions because each term.
Birne Binegar: In these formulas involved to co factor, it just depends on whether we want to.
Birne Binegar: add up all the Co factors along the a given row which case we use fix it we fix it a row index and we use this formula here are the Co factors and or we can.
Birne Binegar: fix the column index and move up and down a column computing factors like this here's The co factor, corresponding to the IPA entry of the matrix but the way we get to determine and as we march down a column multiplying entries by their corresponding co factors.
Birne Binegar: But the point of this more general formula is it doesn't matter which row or column with us, we always get the same result so whether I use this formula, and I use row one.
Birne Binegar: or I use this formula, and I use column 10 I would still get the number that's produced is always going to be the same that's not at all obvious but it turns out to be true, and.
Birne Binegar: i'll show you, if I have time i'll show you a little bit why that's true.
Birne Binegar: Well let's first do this computation So here we had a four by four matrix and what I had done so far is I.
Birne Binegar: Looked at the Co factor expansion along that first row, so I have the one one entry times the determinant of the corresponding PO factor.
Birne Binegar: And then I have the one to entry, which is zero well that's going to multiply a certain determinant but because this factor is going to equal zero it's not going to contribute, so we can ignore it, here we have a non zero entry so it's going to contribute to.
Birne Binegar: The computation determine it will be that entry times the determinant of the corresponding minor so for this entry we removed the first row, and the third column reflective 101010 and 101.
Birne Binegar: And then, this last entry here because it's equal to zero this zero would be multiplying a determinant because of because I have a factor of zero i'm only going to get contribution and zero.
Birne Binegar: All right, and then.
Birne Binegar: i've reduced the computation of the determinant of this four by four matrix and i've reduced it to the computing two three by three determinants.
Birne Binegar: Well here's one I can compute this determine if a three by three matrix by marching across the first row again zero times the determinant 110 to.
Birne Binegar: A minus zero times the determinant of I guess that will be 1001 and then one times the determinant Mel the corresponding minor is miss case 11021102.
Birne Binegar: And so I take a three by three matrix I can break it down into some of determinants of two by two matrices and then These determinants i'll just use the.
Birne Binegar: The cross diagonal product determinant of 1021 is one times one minus zero times two, which is going to be zero Well, this is multiplied by zero and then this term is going to be one times one times two minus one time zero and, at the end you get to.
Birne Binegar: and similarly for this guy here.
Birne Binegar: You can do.
Birne Binegar: You could do a Co factor expansion along the first row again but remember when you have a row or column that has lots of zeros in it that's going to me you're going to have.
Birne Binegar: determinants that you won't have to compute because they're being multiplied by zero so compute to determine is this matrix rather than do the Co factor expansion along the first row.
Birne Binegar: We can we can go down the second column because i'll have zero times the determinant of a minor but I don't have to compute that determinant i'll have one times the determinant of 1111 i'll have to compute that and then i'll have zero times the determinant of 1001.
Birne Binegar: But Tom in either case.
Birne Binegar: For this contribution of this term i'm going to get zero because the entry is zero here we have a non zero entries or going to have to compute that minor determine if the minor.
Birne Binegar: here's the factor one here's a sign correction that depends on which row and which call the entry this.
Birne Binegar: So when the rolling column indices add up to an even number, this is just a factor of one when they add up to an odd number I get a factor of minus one.
Birne Binegar: simple enough for rule and then we have a determinant of the corresponding minor, which is 1111.
Birne Binegar: Well, here we can compute this one times one minus one times, one is going to equal zero so it turns out, they turn in that matrix is going to equal zero which by the way, tells me that this matrix is not convertible.
Birne Binegar: I don't have to do any reduction soon as I know that if the determinant is equal to zero, I know I won't be able to reduce that matrix to the identity matrix.
Birne Binegar: OK, so the complete the calculation, you had a four by four matrix we express this determinant via a Co factor expansion on the first row.
Birne Binegar: that's this line here, and then I calculated using similar co-factor expansions that determinant of this three by three and determine of that three by three you know what worked out to be was one times.
Birne Binegar: The determinant was two and two times well, the chairman of that matrix was zero, and so I just get to so in general, when you're looking at determinants of three by three four by four or larger matrices.
Birne Binegar: There is a formula that you can apply that will allow you to calculate determinants in terms of smaller matrices, but if you have a six by six matrix.
Birne Binegar: first time you apply the formula you get a son of a bunch of five by five determines a five by five matrix as.
Birne Binegar: Well, that can help, so you apply the formula again, and then you reduce each determinant of a five by five to a four by four and then we boost those two determinants of.
Birne Binegar: Three by threes and then maybe when you get it down to determines of two by two you can use that cross diagonal product formula for the determines to actually get the an explicit number.
Birne Binegar: So what happens here is the calculation nothing determinant.
Birne Binegar: Eventually, you have to employ an algorithm and older computer, you have to steadily reduce the size of the determines that you need to compute until you get one that maybe two by two, and so, yet you.
Birne Binegar: just use the formula for two by two matrix well there's there's a close formula for a three by three but it's hard to remember the names will just reduce all the way to two by two.
Birne Binegar: Okay, so.
Birne Binegar: So the good good thing about it is determinants is once you know what that function is all you have to do is insert the values of the.
Birne Binegar: entries of your matrix into that function evaluated you get zero you know the matrix can be in it's not convertible if you get a number, other than zero the matrix isn't vertical.
Birne Binegar: What i'm going to show you next, though, is a close formula for the for the determinant done unfortunately in practice it's going to be just as hard to compute but its form is going to tell you a little bit more about what the what sort of function this determine it is.
Birne Binegar: So, to introduce that I need to introduce some jewelry concepts, the first is the idea of a permutation.
Birne Binegar: So if I take the numbers one, two and three, I can write them in various orders, I can keep one in the first slot but then I can reverse.
Birne Binegar: The positions of two and three, I get an a new a new ordering at the numbers one, three and two but i've listed here are the six possible ways of taking numbers one, two and three and writing them in a particular order.
Birne Binegar: So those ordering of a list of numbers, between one and it's called up the part of the permutations of in.
Birne Binegar: and
Birne Binegar: When I write a list of numbers in a strict order like this i'll call that the standard permutation so this would be the standard permutation, and these are non standard permutations where i've.
Birne Binegar: taken these things, these entry hours their natural order.
Birne Binegar: Okay, I need one more.
Birne Binegar: bit of information that are all attached to these permutations this be a so called some kind of or parody of a permutation what this is it's where you get this science function it's either going to be plus one or minus one.
Birne Binegar: way you assign a plus sign or minus sign to a given permutation you should look at the entries, and you count the number of times the entries are out of order so for this one, and this one, we have effectively.
Birne Binegar: three different kinds of pairs one to two pairs with three in one parish with three.
Birne Binegar: What i'm going to do is look at each of those pairings and decide whether they occur in the matrix in order or not.
Birne Binegar: So here one procedure to there in order to proceed three in one proceeds three so these numbers aren't perfect order, and so this number s which counts the number of times are out of order it's going to equal zero for this permutation one still proceed to and.
Birne Binegar: But three doesn't proceed to so the three and two are out of order here but.
Birne Binegar: So one in three are in order, three and two are out of order and and one and two are still in order so there's one sort of.
Birne Binegar: Order discrepancy for this permutation but this one, I guess yeah if I simply switch back I get these things in order so these things are out of order, one two.
Birne Binegar: Is should be after one it precedes it but one is before three and two is before three so the other other parents or in order so the way you would ascribe.
Birne Binegar: A number plus or minus one if you count you look at pairs of the of the entries, and you decide the number of times when you read across this list the entries appear out of order, so this is stated this way when so.
Birne Binegar: These hundred in this I am J are telling you the position that a given entry is occurring.
Birne Binegar: So here I am Jay can you can be in the first slot second slot or the third slot so those will be the loudest value for those indices, but then you have the numbers themselves as well, not the position in the in the in the list that their actual numbers and so, if the if the numbers.
Birne Binegar: are out of order but they're the row indices are in order, that's when you have the discrepancy and that's what we're trying to count.
Birne Binegar: Alright, so.
Birne Binegar: So, for example, for one, two and three here i've explicitly counted the number of times here's the three possible pairings we have and i've just skimmed across.
Birne Binegar: Each of these not count the number of times we're out of order to in one are out of order, two and three are in order and one and three are in order, so I get.
Birne Binegar: Plus sign for this out of order appear, but the other two pairings or in order so I just get zeros and then this way I can assign a scribe us a particular sign plus sign or minus sign to a given permutation.
Birne Binegar: Okay, so here's the formula, I wanted to present.
Birne Binegar: The determinant of an n by n matrix.
Birne Binegar: Is a polynomial see here, I just have think of these as your variables, this is just a product of n variables.
Birne Binegar: And in fact they're exactly this is the entry of the first row and the Sigma one column, though signal one it's going to be some number between one and N it's not necessarily equal to one.
Birne Binegar: So what what is this in this formula here each term I haven't I have an entry from each row i'm just changing the row indexes like move across factors, but these column indices column index I use on a given entry, it depends on which permutation i'm looking for.
Birne Binegar: But what this determinant formula tells you is that the determinant n by n matrix is a polynomial up degree in because I have.
Birne Binegar: These are all variables and I have exactly in factors so it's a polynomial degree in.
Birne Binegar: Where each term has an entry has a term from each row, and also a term from each column of the original matrix and.
Birne Binegar: So this is all the possible ways of gravity.
Birne Binegar: And entering the first row and pairing it with entries in.
Birne Binegar: In other roles, but but entries that live in different calls and then there's this simple sign factor that depends on which permutation we're looking at.
Birne Binegar: But what this tells us is something.
Birne Binegar: A little bit a little bit more function theoretic about this, not this determinant mention it displays it as a polynomial it also.
Birne Binegar: tells us how many how big that polynomial how many different terms you have.
Birne Binegar: To get that you have to count the number of possible permutations, but if you have the numbers one two N and you want to.
Birne Binegar: rearrange these guys and some list well for this first entry here you have any choices.
Birne Binegar: But then, once you go to the next entry Sigma to hear i've used up one choice i've already decided, my first entry is Sigma one so for the second entry here there's one choice it's missing because i've already used it it's all get N minus one choices here.
Birne Binegar: And when I get to the third entry i've used up two choices, or only have N minus three choices for that, and then you can up the possibilities, you have invoices time N minus one choices time and minus two choices, the total number of choices, you have for.
Birne Binegar: Setting up terms like this is going to be n times N minus one times N minus two are exactly n factorial so In fact this is pretty big polynomial if you think about.
Birne Binegar: A cdn for three by three we're looking at permutations of three there exactly six of those notice that six is three factorial if I had four I would be looking at a polynomial of degree for with for factorial terms or 24 terms.
Birne Binegar: let's go up a little bit higher and i'll let you go if we had a polynomial degrees are determine if a five by five matrix and going to have a determinant that has fine factorial turns.
Birne Binegar: Five factorial is 45 times for factorial it's going to equal 124 66 by six matrix there's going to be what.
Birne Binegar: six times 120 720 different terms, so these functions, while it's kind of simple to write down in a nice clothes formula is extremely complicated, even for moderately launched a matrix.
Birne Binegar: that's a polynomial it can have and really an incredible number of terms, even toward moderately large matrix like an eight by eight matrix.
Birne Binegar: OK so i'll stop here today i'm sorry I ran five minutes over my God, I apologize for that.
Birne Binegar: Maybe i'll let you guys go early on a Friday.
Birne Binegar: I didn't notice anything yeah.
Birne Binegar: there's.
Birne Binegar: Sometimes a treat this like a lost and found, but nothing today.