Birne Binegar: Okay, well, I guess we'll get started, so I have a number of exams that.
Birne Binegar: No one has picked up yet so if you haven't picked up the exam talk to me after class.
Birne Binegar: let's see so today today's plan is to.
Birne Binegar: You quickly.
Birne Binegar: results of last week, where we define determinants last week and we did some.
Birne Binegar: Some formulas for doing the computation so i'm going to do more examples of how those formulas are applied.
Birne Binegar: And then.
Birne Binegar: i'll start.
Birne Binegar: yeah I guess I end on an example that's going to.
Birne Binegar: come back to this Friday this.
Birne Binegar: week but.
Birne Binegar: At the end of this week, probably on Friday, we will discuss the eigenvector eigenvalues problem on one thing this week I need to talk about how one you can use determinants to.
Birne Binegar: do things that we normally have done using just row reduction solving systems of equations and or finding the inverse of the matrix.
Birne Binegar: You can actually do that using only determinants without doing any reduction, although as you'll see today, sometimes you use rubber ducky to compute the terminus but in principle.
Birne Binegar: The determinant determines lots of things about the matrix that you know, and in fact gives us a nice clothes polynomial that describes, we can use to describe solutions of linear systems are fine in versus a matrix eat so that's coming up on today, I just want to.
Birne Binegar: Work on how we actually compute determinants so that reminds you that determinant of a n by n matrix well it's going to be a function well here's your n by n matrix.
Birne Binegar: say a equals, they want one one and it's got to be a square matrix so same number of columns if you have rows.
Birne Binegar: And what the determinant is it's a function of these entries you think of these guys's variables it's going to be a polynomial in the entries of the matrix and what it's supposed to tell us is that when we actually substitute.
Birne Binegar: In actual numerical values for these entries and we compute that the value of that polynomial if it equals zero that tells us a bunch of theoretical stuff it tells us, for example, that that polynomial evaluates to zero that the matrix has no inverse.
Birne Binegar: Which and do you have a theorem that we covered, but the so called fundamental theorem of convertible matrices that we discussed just prior to the first exam.
Birne Binegar: If the inverse of the matrix doesn't exist, then that linear system like eight X equals D can have more than one solution in general you're going to have at least one dimensional solution space.
Birne Binegar: i'm a special case of this, the equation eight X is equal to zero well there's an obvious solution where you just said X equals zero and this equation will be satisfied, but if this condition hold that means there's more solutions and justice trivial solution.
Birne Binegar: If Ada minus first doesn't exist, that means a doesn't reduce to the identity matrix because the way we've calculated a the minus versus as far as we've written the matrix down or we formed the matrix a.
Birne Binegar: high and the computation of the inverse of a it's carried out by reducing.
Birne Binegar: To real epsilon form or, if you do select one form.
Birne Binegar: Of.
Birne Binegar: This matrix and, depending on what this reduce your lesson one form of this augmented matrix is, we can decide whether or not a has an inverse and in fact we end up computing it if it does, because what happens is that this.
Birne Binegar: When I form this matrix when I reduce it all the way to reduce relational on form you get one or two possibilities.
Birne Binegar: This left hand side.
Birne Binegar: either.
Birne Binegar: gets transformed into the identity matrix and if that happens.
Birne Binegar: The right hand side is going to end up being exactly the inverse of the original matrix or you get something else.
Birne Binegar: And in this case, if you get something else you don't get the identity matrix on the left, that means a to minus first does not exist.
Birne Binegar: So.
Birne Binegar: In fact, the purpose of this introducing guess determinant function, so we can answer this.
Birne Binegar: All these conditions, decide whether.
Birne Binegar: an inverse exists or whether a linear system associated with the matrix have solution or whether it's can be reduced the identity matrix or whether the dimension of the row spatial dimension calm today's lesson in these sort of.
Birne Binegar: theoretical conditions can all be determined simply by evaluating a particular polynomial that depends on the entries of the matrix.
Birne Binegar: So that's what we use it for it and become much that.
Birne Binegar: Principle application is going to come later this week, but what I want to do today is just to show you how you compute These determinants so.
Birne Binegar: The case when you have a one by one matrix matrix with a single entry it's determinant is exactly its entry, so that is a function of the entries of the matrix just just this so simple.
Birne Binegar: When you have a two by two matrix then here's the polynomial it this time this minus this time this was the difference of the diagonal products and.
Birne Binegar: it's a pretty simple formula it's turned out to be quadratic in the entries right because there's two factors in each term.
Birne Binegar: Now for larger major sees.
Birne Binegar: The formulas get much get complicated in fact they get complicated really rapidly when it gets greater than four or five or six.
Birne Binegar: But nevertheless, we do have a formula for computing determinant of the matrix of a square matrix of arbitrary size, the first formula, I presented last week with the so called co factor formula which.
Birne Binegar: there's some multiple versions of it.
Birne Binegar: One version, so, if you look at the inside the song this expression doesn't change the formula, what does changes here I something over the column indexes.
Birne Binegar: And I keep it in a row index fixed This is called a Co factor expansion along the I throw the index I stay fixed some number between one and N.
Birne Binegar: And then the terms we compute well we march across the entries in a given row picking up contributions from each column in that row.
Birne Binegar: So you'll get the entry itself science factor, and then you have a determinant of another matrix which.
Birne Binegar: is related to the original matrix simply by taking that matrix deleting it I throw in jade column you get a smaller matrix and the idea here is that.
Birne Binegar: You complete the determinant of a larger matrix you reduce it into the computation of the terminal a bunch of small matrices.
Birne Binegar: And if these major cities are not small enough to compute directly, then you apply a version of this formula again to.
Birne Binegar: reduce the computation of the actual determines to smaller and smaller matrices until eventually by making the bakery see small enough, you can eventually apply one of the more basic formulas.
Birne Binegar: The second line here similar keep a here I keep the column index fixed and I expand along the row I grab the.
Birne Binegar: API key entry, so I wrote a column there's a sign factor, and then the determinant of the matrix I get like deleting the ice road jake column from a.
Birne Binegar: Alright, so let me give you an example that here's a four by four matrix.
Birne Binegar: What i'm gonna do.
Birne Binegar: Is i'm going to take advantage of anytime this entry here if I either formula anytime is equal to zero, I won't have to compute this determine.
Birne Binegar: So when you apply these co factor expansion formulas, the thing to look for a rose or columns that have lots of zeros because that means I won't have to compute the associate term there, they are, they are here, I will have to compute something but let's write it out.
Birne Binegar: I have zero because the.
Birne Binegar: matrix entry here zero so whenever it multiple places equal to zero this zeros producing that that's zero producing that now, I have a non zero entry niche row so i'm gonna have to go back.
Birne Binegar: and apply this formula completely because this factor is now non 00 you do it, so I grabbed the entry, I get the sign factor, by looking where this entry lives and adding it row indexing column index together and I raised minus one to that power.
Birne Binegar: This example, this is going to be a factor of one because i'm raising negative one, to an even number and so minus one.
Birne Binegar: it's going to be minus one squared in some sense, so what you're going to just be a factor of one, and then I have.
Birne Binegar: That determined i'm from the minor when I eliminate the fourth row in the fourth column, those are the rows and columns that contain this for for entry i'm left with this matrix in the upper left hand corner 110020111.
Birne Binegar: And now.
Birne Binegar: I have another determinant compute, but I can compute it in a similar way I can look at the second row see that it only has one non zero entry and i'll do a Co factor expansion, this time along the second row.
Birne Binegar: first time I did a Co factor special in the fourth row this year as a co factor expansion along the second row second row of this matrix.
Birne Binegar: So it's zero plus two time okay so, then this contribution here is going to be to assign factor minus one, this is the two to entry, so I have minus one to the two plus two, and then I have the determinant of the corresponding minor.
Birne Binegar: Read off this matrix by eliminating the row and column that contained the entry and study so i'm left with a matrix 1011 that's what we have here all right now i've got the determinant of the original matrix right, this should be.
Birne Binegar: Determining day i've got it reduced to the computation of the determinant of a two by two matrix, and for that determinant computation i'll just take the difference of the diagonal products because you're just a two by two matrix.
Birne Binegar: And then, what we'll get is this factor of three these factors, the minus one end up just being factors have one, I have a factor of two and then this determine is one times one minus zero times one or simply one minus zero and you get six.
Birne Binegar: Alright, so, in practice it won't take that long to calculate determinant of the matrix like this, you just need to reply that recursive formula twice until you get an expression for the determinant the original matrix in terms of.
Birne Binegar: determines the smaller major cities that are convenient to compute directly.
Birne Binegar: Okay, so that's The co factor expansion formula and how it works.
Birne Binegar: We had a second formula for.
Birne Binegar: computing determinants it wasn't so much a practical value.
Birne Binegar: As far as direct.
Birne Binegar: explicitly doing the computation but I gave you this formula last week because it displays properties of the determinant.
Birne Binegar: When we think of it as a function of the entries it tell you it tells you that your determinant function is in fact a polynomial in the entries of your matrix.
Birne Binegar: This is a variable distance variable, this is a variable pitch we regard all the entries of a it's the variables and all it is is.
Birne Binegar: a bunch of signs plus or minus signs and I described, on Friday, how we actually compute this but in today's lecture the signs just think of them as not something that you, you look at a particular rearrangement of the numbers between one and and and you ascribe a plus or minus sign.
Birne Binegar: What that has to do with his the ordering of those numbers that they.
Birne Binegar: Do numbers in the list appear.
Birne Binegar: In.
Birne Binegar: The way they appear in the list as you go down the list if they if their size is out of order so it's three is coming before to then that.
Birne Binegar: We take take that into account, but if, when entries in that list or in the order like one is before three well.
Birne Binegar: We ignore those things, what we do is we count up the number of times the entries are out of order and we raise minus one to that power that's how this.
Birne Binegar: These plus or minus signs are computed but all I really wanted you to do is use this formula, not so much to compute because in fact it's not that practical.
Birne Binegar: The number of permutations you have an object it's going to be n factorial so the sun here becomes very huge very quickly if you have.
Birne Binegar: A six by six matrix, then you have to compute there'll be 100 700 and 720 separate permutations so that's.
Birne Binegar: Pretty laborious if you're going to have to look at 720 different first, you have to find 720 different permutations and then you've got to compute the sign of each one of them and then you've got to.
Birne Binegar: write out this polynomial that's going to have 720 different terms, so this is not a practical way of computing the determinant, but it does display that kind of function is it's a polynomial it's a polynomial degree and because each term has an factors in the underlying variables.
Birne Binegar: and, moreover, we can count the number of permutations and it turns out there's as many permutations as you.
Birne Binegar: The number of permutations of an object's is equal to n factorial so main purpose of this formula with just so that I can argue that the determinant of a.
Birne Binegar: matrix is always a polynomial function and that it's degree it's going to be the number, yet the degree of it as a polynomial on the variable it's exactly equal to the size of the matrix.
Birne Binegar: Assuming is an n by n matrix that polynomial is going to be a polynomial degree and and.
Birne Binegar: It shows that individual terms, here they consist of a plus or minus sign and then you get.
Birne Binegar: One entry from each row as a factor and you get one entry from each column is a factor and basically what these terms represent is all the possible ways, you can.
Birne Binegar: Take an element in distinct roles and different indistinct columns and multiply them together.
Birne Binegar: And then.
Birne Binegar: You count the number of permutations turns out the N factorial terms, but as i've stressed already this formula.
Birne Binegar: it's pretty simple to write down.
Birne Binegar: Some some signs and a common form for each factor.
Birne Binegar: it's simpler right now and to interpret as a as a function but it's not really used for computations.
Birne Binegar: Maybe maybe a computer might do it is the computer, can he can compute and keep track of all these permutations fairly easy, but if you're working out things out on a piece of paper this formula is kind of unwieldy, simply because you have so many terms to compute.
Birne Binegar: All right, so.
Birne Binegar: What I did it close of lecture on Friday, Dan was trying to give you a simpler method for computing determinants that are practical for determinants of moderate size.
Birne Binegar: This alternative way of computing determinants is based on the following theorem.
Birne Binegar: If you have a matrix that upper triangular but i'm going to state it in a way that's more familiar to us if a matrix is in row Echelon form.
Birne Binegar: Because we don't have a matrix in Russia on for, and we can always transform it into row H1 porn I doing elementary row operations, so this is a more familiar concept from what we've done in most.
Birne Binegar: In the course, thus far, this is it's just a slight weakling a matrix can be upper triangular meaning all its non zero entries.
Birne Binegar: yourself matrix the matrix is upper triangular It means, if you look below this main diagonal you would have nothing but zeros that the picture here.
Birne Binegar: The 000 all down here below this main diagonal you would only have zeros That would be an upper triangular matrix and what I showed you last time was that.
Birne Binegar: For if you have an n by n matrix you apply, for example, the permutation formula, you have n factorial terms to compute.
Birne Binegar: If a matrix is upper triangular though you only have one term to compute the determinant of an upper triangular matrix or as a specialization if a matrix is in row Echelon form computation of the determinant is really easy you just multiply.
Birne Binegar: The entries that occur along this main diagonal from the upper left to the lower right.
Birne Binegar: Alright, so the idea that I pursued on Friday was that well.
Birne Binegar: The determinant of an arbitrary matrix is really hard there's too many terms to keep track.
Birne Binegar: determinant of an upper triangular matrix in Russia and form it's really easy there's only one term to contribute so.
Birne Binegar: What happens if we try to start with an arbitrary matrix and we try to simplify it it's computation of determine by instead binding the determinant of a traditional on form which is going to be easy to compute.
Birne Binegar: Well, what happens is is that, unfortunately, as you roll reduce a matrix the determinant can change so one thing that happens is, if you interchange rose.
Birne Binegar: The determinant is certainly related up a new matrix is certainly related to the determinant of your matrix but this elementary row operation ends up.
Birne Binegar: Switching the sign of the determinant let me show you explicitly have a two by two you have a is equal to a B, C D sorry.
Birne Binegar: If you have a matrix to buy to a B, C D and I interchange the first and second rose.
Birne Binegar: I get a new matrix CE E a D now you're just just a two by two matrix so if we could computers determine.
Birne Binegar: What we're going to get is a times D minus the time see but for this matrix we have pain from a by doing this row interchange it's determined that.
Birne Binegar: it's going to be see times B minus eight times D it's still difference of the diagonal products, but now you see the sign is effectively reverse this is the negative of the determinant.
Birne Binegar: Okay, and so that's that's all this rule is telling you is how the determinant changes if you've done a row interchange.
Birne Binegar: But there's.
Birne Binegar: When you transform a matrix into relational on form if two other real operations that you generally need.
Birne Binegar: The second one is really scaling and that also affects to determine.
Birne Binegar: If you're.
Birne Binegar: In the course of real reduction, you need to simplify a row by rescaling it by a number well that's going to change to determine once you've done that once you've done this row operation then determine if new matrix is going to have an additional factor of land.
Birne Binegar: So these two operations are going to change the chairman, but what's the eggs, a day as the operation you do most often if you reduce as you try to.
Birne Binegar: arrange the pivots correctly to get in Russia line points if you're usually adding a multiple of one row to another and using that to replace a given.
Birne Binegar: And here the determinant of the new matrix it's going to be exactly equal to the determinant of the matrix you started off.
Birne Binegar: So, most of the time when you do a row reduction determine it's not going to change what you have to keep track of is how many times you interchange rose, and if you do any real rescaling she need to retain those factors.
Birne Binegar: But if you do that, then you arrived at the spoiling formula for the determinant.
Birne Binegar: Of the matrix is I don't really describe.
Birne Binegar: The computation to explicitly suppose i'm i'm road use a to a matrix a prime in row Echelon form, so it will be upper triangle.
Birne Binegar: The factor here you it's just the diagonal entries of this matrix and relational on form this these factors here are exactly the determinant of a prime that's what that is.
Birne Binegar: But um but what i'm arguing on the preceding slides is the determinant of a gets changed by the row operations, every time you do a row interchange you flip the sign so you'll have as many sign flips as you've done row interact now row interchanges as you've done this reduction.
Birne Binegar: rescaling a row also changes the value of the terminal it changes it by a factor of whatever you rescaling the row by So if you use the.
Birne Binegar: If we scale maybe the first row by a factor of Lambda one, the new determine with have that factor, so you, you have to pull it out, and so you divide by land.
Birne Binegar: And if you've done more rescaling in this process of reducing aid a prime well you're going to have to pull out all those factors as well, but once you've done that.
Birne Binegar: Once you've done the production and you've just kept track of how many times you interchange rose kept track of the number of times and the actual factors you used in row scaling, then you get a very simple computation.
Birne Binegar: These factors are all determined by your reduction process and then this is really easy once you get to Russia one form you just multiply the diagonal entries together.
Birne Binegar: Alright, so let me give you an explicit example that.
Birne Binegar: Lower this.
Birne Binegar: raise it a Lord I guess.
Birne Binegar: Okay, so let's look at this four by four matrix now, if I were going to do the permutation formula.
Birne Binegar: Using that to compute this matrix I would have what 24 terms because.
Birne Binegar: Now for factorial is 24 and so we're gen four by four matrix I should expect the polynomial degree for with 24 different terms that's really tedious to compute but.
Birne Binegar: Instead, what i'm gonna do is i'm going to reduce this matrix to row is one form keeping track of that reduction process and then i'll end up computing yes, the determinant.
Birne Binegar: Of the real Echelon form so let's do the reduction first year we have four by four matrix you have a pivot up in this corner, we need to get rid of that pivot and pivot, so I do some.
Birne Binegar: Elementary row operations and as advertised it these elementary raw operations are not going to affect them the value of the determinant.
Birne Binegar: They don't change the value of the determinant so the terminal this matrix is going to be exactly the same thing as the determinant of that matrix and notice i'm getting close to getting in relational on form, so this is almost easy to compute.
Birne Binegar: But to get it in relational on form, I need to do one of the couple other things, first of all these pivots are out of order, so I can.
Birne Binegar: interchange row to row for the end up with this nature, but now this pivot is off to the right pivot and the row below so i'll do another row interchange and now i've got a matrix that.
Birne Binegar: Not only in real life one form it's also upper triangular so what I know is the determinant of this matrix here.
Birne Binegar: is going to be one times two times two times two because its upper triangular and for upper triangular matrix seeds there's only a single term in the polynomial.
Birne Binegar: corresponding to which determiner it just the product of the diagonal elements but.
Birne Binegar: it's not the chairman of this matrix is not the same thing as the determinant of the original matrix because these two row operations change to the terminal.
Birne Binegar: This is going to introduce a factor of minus one, and this is also going to introduce a factor of minus one, but if I take those factors into account, then I can relate this matrix it's determinant to the determinant of that matrix and so that's what we do.
Birne Binegar: I had determine if the original matrix is going to be equal to the determinant of a row Echelon form, but because I had to use to.
Birne Binegar: row flips in order to carry out the role reaction i'm going to pick up this factor of minus one squared which happily enough just turns out to be a factor of one.
Birne Binegar: But then determine if this row Echelon form of a its upper triangular so to compute its determinant I just multiply down the main diagonal and I get.
Birne Binegar: A factor of one well one time two times two times two I just get eight alright so.
Birne Binegar: that's an example of how row reduction can be used to.
Birne Binegar: find the determinant of the matrix I will say that there's.
Birne Binegar: This one.
Birne Binegar: situation where this real reduction process is not that helpful it really works fast, as long as your matrix has numerical entries, but if my matrix actually has some variables and entry, well then it's it's harder to do the drill operations, because.
Birne Binegar: Well, instead of having integers calculable numbers, when you add rose together or when you multiply those together instead you have variables that you have to keep track up so.
Birne Binegar: Two times three is equal to six, but two times Lambda is equal to two times Lambda doesn't really simplify anymore.
Birne Binegar: All right, so.
Birne Binegar: So those are.
Birne Binegar: Effectively, three methods, but I think it really only counts as to you have the Co factor expansion form that applies to any n by n matrix, but the problem with it is you have to.
Birne Binegar: Keep breaking down the matrix until it small enough that you can actually computers determine it by hand but it works and you just got to be patient with it.
Birne Binegar: But a faster way of computing determinants is to use row reduction usually because, instead of in that last example that of having to calculate 724 different terms I just did the role reaction and then not take into account the discrepancies that arise when you.
Birne Binegar: reduce the matrix you have to take into account the number of row interchanges and possible rescaling but as long as you keep track of that stuff it makes the computation of a determinant really simple.
Birne Binegar: Alright, so what I want to do today to finish today's lecture is.
Birne Binegar: Do another example where we need to computer determine but here the emphasis is going to be on why i'm computing the determinant rather than how I actually compute that eternal.
Birne Binegar: So what i'm going to do.
Birne Binegar: This last example is figure out the numbers Lambda which I can satisfy an equation like this, or have a matrix times my unknown vector is equal to some scale or multiple of that vector.
Birne Binegar: When a condition like this whole matrix and a vector we call the vector at not the eigenvector and this number that works, we call it the eigenvalue i'm going to discuss the eigenvector eigenvalue problem a lot.
Birne Binegar: Actually, for probably the next two weeks or so, but the way I want to introduce That problem is simply well, but what numbers Lambda.
Birne Binegar: Can I actually find solutions of this equation, other than the one solution that's kind of easy.
Birne Binegar: When X is equal to the zero vector a times X is going to equal zero with X is equal to zero, whether I scale or multiply X by matrix or sorry, whether I matrix multiply X by matrix or I scale and multiply by squealer.
Birne Binegar: Zero vector times anything always produces zero vector it behaves essentially like the number zero and so.
Birne Binegar: What what I post here is when can I have alternative solutions of this of this equation, other than X equals zero.
Birne Binegar: well.
Birne Binegar: We will will determine and is i'll restate the problem little bit put it in.
Birne Binegar: You know, for a form that's at least formerly is more familiar.
Birne Binegar: Instead of having.
Birne Binegar: A X is equal to something that's proportional to X itself usually, when we talk about linear systems, I had eight X is equal to a fixed vector be.
Birne Binegar: But now the variable is appearing on both sides of the equation, and so, if you want to solve that variable you want to get the variable on the same side of the equation initial.
Birne Binegar: And so that's what i've done here i've taken this term and taking it over to the left hand side and then Lambda X i'm interpreting as Lambda times the identity matrix times X, so that.
Birne Binegar: That.
Birne Binegar: So that.
Birne Binegar: A X minus Lambda times X.
Birne Binegar: But.
Birne Binegar: i've got extra now isolated not quite isolated, but it certainly city only on the left hand side of the equation i'm just writing this term as a single matrix times X so.
Birne Binegar: This is just the single matrix so it basically it is a linear system it's just that it doesn't correspond directly to the system we started off with the original equation, with eight X is equal to Lambda X coefficient matrix.
Birne Binegar: And then some scale, but now I have a coefficient matrix times X is equal to zero, as long as I calculate this matrix I can solve this in your system and then potentially find other solutions of the original equation.
Birne Binegar: Okay, so.
Birne Binegar: If you write down the system.
Birne Binegar: This matrix here a minus Lambda time food, energy matrix it's itself is a matrix of the same size as the original matrix, the only thing that happened between this matrix and the original matrix a is such subtract the parameter Lambda from just the diagonal elements.
Birne Binegar: So it's effectively you take a and only along the diagonal elements you shifted by.
Birne Binegar: The variable Lambda shifted in a negative direction.
Birne Binegar: All right, but it's still.
Birne Binegar: it's still ends up being a three by three matrix or which.
Birne Binegar: We can interpret it as the coefficient matrix of a linear system.
Birne Binegar: Back it's exactly this linear system, so what we'll do.
Birne Binegar: Next.
Birne Binegar: Is.
Birne Binegar: i'll remind you that.
Birne Binegar: If the determinant of a matrix equals zero that tells us a bunch of stuff.
Birne Binegar: Here the effectively the negation of this fundamental theorem of convertible matrices if the determinant is equal to zero, then what are the consequences of that it's going to be that your.
Birne Binegar: equation eight X equals zero is going to have solution other than the trivial solution where X is equal to zero.
Birne Binegar: Now, in the case at hand we're looking at our original equation we're looking at, of course, was.
Birne Binegar: A X is equal to Lambda but this equation is equivalent to this, and now I can apply this condition it just that, instead of interpreting.
Birne Binegar: Well i'll use a different matrix here, instead of using the original matrix a here I use the shifted version, where I.
Birne Binegar: shift the diagonal entries that may a Maya factor of negative Lambda but.
Birne Binegar: If this condition is true it's an equation like this, and so, if I want to have solutions other than the trivial solution, what I need is the determinant of that coefficient matrix and equal zero.
Birne Binegar: So here's the condition we get if the original in your system is going to have a solution.
Birne Binegar: Other than.
Birne Binegar: The trivial solution.
Birne Binegar: occurs when you set that vector equal to zero, and what our question is is are there other solutions.
Birne Binegar: The argument as far as telling us, yes, we will have other solutions.
Birne Binegar: If the determinant of this matrix here is equal to zero so it's not determine of a itself if the determinant of the shifted matrix.
Birne Binegar: So let's um.
Birne Binegar: let's look at.
Birne Binegar: It.
Birne Binegar: For the matrix given where is it.
Birne Binegar: Here it is here's the original matrix but what i'm gonna do is i'm going to calculate the determinant not have a but up to that shifted matrix a where I subtracted Lambda from its diagonal elements.
Birne Binegar: Of that, here, you turn term a minus Lambda is the original matrix enjoying the Lambda and then the.
Birne Binegar: The shifting along that main diagonal and I compute the determinant that okay now how do we want to compute this determine we have at least three options we have.
Birne Binegar: A co factor expansion, we have a determined formula in terms of permutations and we also have role reaction.
Birne Binegar: Well, real reduction this kind of awkward when you have variables, instead of just pure numbers inside your matrix there's those variables, yet they don't simplify any throughout the calculation.
Birne Binegar: So maybe the most expedient thing would be is seeing, I have two zeros here let's do a Co factor expansion along this third call.
Birne Binegar: And so let's do that, we have zero times i'm going here's the one three entry, so I have.
Birne Binegar: One three entry points assigned factor minus one to the one plus three and determinant of the minor I get by eliminating the first row and third column from the original matrix.
Birne Binegar: matrix a minus Lambda time to die, but i'm not actually going to have to compute this determined because it's being multiplied by zero similarly for the second term in the third column.
Birne Binegar: it's being multiplied by zero like determine if the corresponding minor I won't have to compute and then I get the last time entry.
Birne Binegar: Time assign factor that.
Birne Binegar: determined by the position of that entry in the original matrix and then I have the determinant of the corresponding minor.
Birne Binegar: Alright, so what is that minor.
Birne Binegar: That minor you eliminate the third row and third call just to minus Lambda times one minus Lambda 112 minus Lambda.
Birne Binegar: But that's just a two by two matrix is determine it is going to be the different to the diagonal products or have to minus Lambda time to minus Lambda minus one times one that's when I computed there.
Birne Binegar: And so.
Birne Binegar: I write this out and what do you see the conditioning we get is that they're all has to equal the determinant this matrix.
Birne Binegar: If i'm going to have non trivial solutions other than sitting X equals zero if i'm going to have non zero vectors that satisfied an equation like this, then that determinant has to equal zero, which means this polynomial equals zero and that's what gives you an equation for Lambda.
Birne Binegar: You only like this polynomial is going to equal zero it's going to be if either one of these two factors is equal to zero so like either need.
Birne Binegar: Lambda equal to one, so that this factor is equal to zero until the total product is equal to zero or i'm going to lead Lambda equal to three, so the second factor of zero making the product equal to zero, so what this is going to imply.
Birne Binegar: Is that Lambda has to equal either one or three.
Birne Binegar: I think that's what I say, in the next line here yeah don't you either have to have them to equal one or three, and so the only non trivial vectors the only time, you can have a.
Birne Binegar: Non trivial vector at some vector other than the zero vector satisfies the equation like this well you can't solve it, in general, only for special values of Lambda and what we see is that determinant.
Birne Binegar: This a fling zero means that we have solutions other than the trivial solution X equals zero and what this equation shows that That can only happen for special values of this parameter Lambda.
Birne Binegar: So what we'll find on I think Friday of this week, if the vectors exit satisfying equation like this are called I can get through the bay.
Birne Binegar: And numbers Lambda.
Birne Binegar: These are called the eigenvalues of a.
Birne Binegar: and finding all possible eigenvectors an eigenvalues for a given matrix a that's called the eigenvector eigenvalue problem, and it has lots of applications.
Birne Binegar: Not just in mathematics and physics and engineering and in many fields you end up solving trying to solve and eigenvector eigenvalue problem.
Birne Binegar: Okay, so actually that's as far as I, I meant to go today, let me give you a preview of what we're going to do on Wednesday before we return to the eigenvector organizing problem.
Birne Binegar: i'm just going to put two formulas up on the blackboard and i'll drive them, or at least explain them on on Wednesday.
Birne Binegar: So.
Birne Binegar: There were two things we use.
Birne Binegar: Real reduction for one was for solving linear systems.
Birne Binegar: So if you had a X is equal to be remember what we did is, we would form an augmented matrix and we'd rather reduce to.
Birne Binegar: Reduce row Echelon form and then this reduce your national on form of this augmented matrix you would tell us me the augmented matrix of solution we convert back to equations and we expand the solution in terms of pre variable etc.
Birne Binegar: All I want you to remember about this line is that formerly up to this today's lecture the only way we had of solving linear systems which, by doing this role reaction this essential calculation of step, but now we have an alternative.
Birne Binegar: That i'll explain on Wednesday.
Birne Binegar: For n by n men matrices.
Birne Binegar: So let me make the statement if X is a solution.
Birne Binegar: A X is equal to be then.
Birne Binegar: Its components, because owners of the solution vector are going to be can be determined by evaluating some determinant bi guided by the terminus.
Birne Binegar: This is this is called kramer's rule and I guess I need to stay one more thing to make this.
Birne Binegar: comprehensible here the Ai is the.
Birne Binegar: n by n matrix.
Birne Binegar: You get from.
Birne Binegar: The original coefficient matrix a by replacing.
Birne Binegar: ice caused.
Birne Binegar: By.
Birne Binegar: This vector be.
Birne Binegar: And dumb so.
Birne Binegar: I have the matrix a I computed determinant and then from a I can compute a bunch of auxiliary matrices one for each one for each component of X.
Birne Binegar: So there'll be in such matrices but the value of the solution here, its components are just going to be determined not by using production, but now by computing the determinant.
Birne Binegar: Of the coefficient matrix and the determinant of the auxiliary matrices you get from a by replacing columns of a with that.
Birne Binegar: That through this on the right hand side of the original in your system.
Birne Binegar: Well, this is called kramer's rule.
Birne Binegar: The upshot of which, in order to solve linear system, I can do that just you see determinants.
Birne Binegar: And a second thing we use row reduction for was to compute.
Birne Binegar: In verses of matrices.
Birne Binegar: So you can do this.
Birne Binegar: direct reaction as I described earlier in today's lecture.
Birne Binegar: You form a different kind of augmented matrix where you would join the identity matrix and then you will reduce.
Birne Binegar: To reduce role, I shall inform again.
Birne Binegar: And then, yes.
Birne Binegar: In this reduce all Echelon form a prime.
Birne Binegar: Prime if this a prime happens to equal the identity matrix than the I prime it's going to end up being the inverse of day if the a prime is not equal to the identity matrix, then the original matrix a has no inverse but.
Birne Binegar: Our method of finding in verses of matrices was based on this rule reduction calculation.
Birne Binegar: I can replace that with an alternative formula.
Birne Binegar: In terms of determines.
Birne Binegar: It goes like this.
Birne Binegar: The inverse of a.
Birne Binegar: is equal to.
Birne Binegar: let's see it.
Birne Binegar: One over the determinant a time to transpose on a certain matrix.
Birne Binegar: Here see it's going to equal.
Birne Binegar: I J its entries are going to equal.
Birne Binegar: minus one to the I like gay times that the permanence of the iga minor.
Birne Binegar: Am I James.
Birne Binegar: So this is a minor some it's done matrix I get from the original matrix a by deleting is I throw in call.
Birne Binegar: Once I computed the determinant I just get a number, and so what i'm describing here is.
Birne Binegar: Effectively n squared numbers, what you have in choices for the index is in choices for the index J.
Birne Binegar: and
Birne Binegar: For each of these miners I get a matrix that I can compete with determinant get a number, and then I can multiply it by assign factor, these are cold co factors.
Birne Binegar: They of course occurred already in the co factor expansion of the determinant but now, what I do is I compute all these numbers.
Birne Binegar: And I arrange them in a matrix so this should be C one C and C and one C and and and my statement that will have described more completely on Wednesday is that.
Birne Binegar: The determinant sorry the inverse of the matrix can be found by computing this matrix using this these rules converting rosen the columns that that done by this trend transpose operation and then dividing by the determinant of the original matrix so I have the.
Birne Binegar: determinant of a n by n matrix and then I also have a bunch of determines N minus one by N minus one matrix used to compute but that inverse is completely determined by determining calculations, rather than row reduction.
Birne Binegar: That said, though of course oftentimes when you compute determinants huge reduction so maybe you don't completely escape or a reduction, but at least formally you can express determine if.
Birne Binegar: it's a.
Birne Binegar: Nice closed expression where what I know is that each of these guys is a polynomial in the in the entries of them, the miners at so it's also going to be a polynomial in the entries at a itself, and so this the inverse is a matrix where these entries are all.
Birne Binegar: polynomials in the entries of the original matrix but the inverse is not a polynomial function of the original matrix because these terms are in the denominator.
Birne Binegar: So basically it's going to be irrational function, rather than a polynomial function there'll be a bunch of polynomial entries, but each of which is divided by another polynomial.
Birne Binegar: Nevertheless, you can at least see the functional form the how the inverse of the matrix depends on the entries of the original matrix.
Birne Binegar: OK so i'll pick up a discussion of these two formulas on Wednesday.