Birne Binegar: words.
Birne Binegar: Okay, so.
Birne Binegar: here's a question on homework Joe this morning I assign problems that four and five to be due next Monday, let me step back a couple.
Birne Binegar: seconds to explain my rationale behind that first of all, we lost two lectures last week, which means my whole plan about having mid term the first midterm early next week is is.
Birne Binegar: Not possible anymore what we're gonna do is we're going to cover one last topic and we'll get through most of it today we'll talk about in verses of matrices but as far as homework goes homework said three is due today and.
Birne Binegar: You may have noticed that all the due dates for the homework on web assign have been set to the end of the Semester that's because.
Birne Binegar: Contrary the way things worked in the past, I can't just randomly change the due dates that people you know struggling getting something in on time extended a week doesn't seem to help very much.
Birne Binegar: What once it's got locked in there.
Birne Binegar: Either creates a new assignment or doesn't anyway, let you resubmit homework, so what i'm doing now is i'm just posting the assignments and making their due date on web assign the 17th week of the class or even after finals week, so the way you should.
Birne Binegar: know when US assignment should be done it's not looking at it's due date on whatever sign rather go to the canvas page.
Birne Binegar: And then you'll see towards the top of the our canvas page the due dates for the various homework assignments until right now, a homework set three is do today.
Birne Binegar: homework set four and five are going to be due next Monday that's because homework set for now that's concerned with solving linear systems and we've already covered that material.
Birne Binegar: i'm going to do another example first thing today to remind you of the process by which we fall systems of linear equations.
Birne Binegar: Holbrook set five is the last topic that covers in verses of matrix that's the last topic before exam.
Birne Binegar: I want to get both of those assignments done neither one is particularly long, so the fact that i'm doubling up two assignments, this week, rather than one that shouldn't be too much of an issue it shouldn't shouldn't take you much more than an hour to do both, but it is essential.
Birne Binegar: To get some practice of doing row reduction so certainly do homework set four and five, that will prepare you for the exam which I suspect, maybe.
Birne Binegar: could possibly be at the end of next week, but I think more likely, it will be two weeks from today would be the on the data first exam.
Birne Binegar: What i'm gonna do is i'm going to give you plenty of time to work on the homework problems, so they will the homework sets four and five are going to be due on Monday, if the test is the following on data gives you.
Birne Binegar: gives everyone a week to sort of consolidate consolidate the homework in theory and lecture and be prepared for an exam two weeks from today.
Birne Binegar: However, what i'm likely to do in today's lecture on Wednesday lecture is completely covered the material that's going to be on the exam except for the homework problems.
Birne Binegar: So what i'm likely to do on Friday and Monday of next week is probably.
Birne Binegar: go on to new material that will not be covered, on the first exam but the give you breathing space for the or the material that is going to be covered, on the first exam.
Birne Binegar: So what's likely to happen is we'll finish covering in verses matrices this week, then i'll be open to answering questions on the homework problems.
Birne Binegar: On Friday there'll be due on Monday, but we can I mean if you see the due date is now shifted all the way to the end of the Semester so it's not imperative that you get the homework done on the due date that's my suggested time so that you keep up with a lecture but.
Birne Binegar: Other than that there's not going to be any strict due date for the homework, so you effectively you could do all the homework, now the week before the night before the final exam and you would still get credit for it, although.
Birne Binegar: I am get it my web assign desk tell me when homework was turned in.
Birne Binegar: So if you get a D in the course and I noticed that you didn't turn in your homework until the night before the final exam i'm not going to be so sympathetic but that's about that's about the extent of like penalty for not doing homework on time.
Birne Binegar: only be the case if you were repeatedly turned in homework so late, would you find my sympathy kind of Dallas by that.
Birne Binegar: Okay, so.
Birne Binegar: If your questions okay good so.
Birne Binegar: To get started on homework set four and five, and if you run into the issue like how do I enter an answer you can raise those in class and we can do a DEMO of one of those problems.
Birne Binegar: But what i'm going to do today is, first of all, remind you of the topic we just left a week ago solving linear systems i'll step through one of those calculations and then we'll begin to talk about in verses that nature sees.
Birne Binegar: Okay, so.
Birne Binegar: All right, let me get this on screen.
Birne Binegar: help a little bit yeah okay so.
Birne Binegar: let's look at a system of three equations in foreign owned and i'm just going to quickly step through the process by which we not only solve the equations but we express our answers as a hyper planes.
Birne Binegar: Remember that a hyper plane is this geometric instruction that will use all the time and linear algebra hyper plane it's a set of vectors.
Birne Binegar: That you form by starting at a given point and then straying from that point on a fixed set of directions say we have a direction vectors V1 to decay and this is our starting point.
Birne Binegar: So these.
Birne Binegar: parameters range over the entire realign.
Birne Binegar: And then, what happens, for example, if this were a two dimensional hyper plane you pick a starting point zero relative to some coordinate axes and then you would allow yourself the strike in one or another direction.
Birne Binegar: Like arbitrary distances from that starting point, and as you vary the length of the factors by the scale or multiple you end up filling out the entire blackboard so you get this.
Birne Binegar: flat two dimensional surface in this case, and it would be if there was this were embedded in a three dimensional space, it would be a flat two dimensional plane embedded in a three dimensional space that's what we call a hyper plane.
Birne Binegar: And what we've seen yeah initially when we started talking about linear system that the system of.
Birne Binegar: One equation in foreign owned it's solution set is always a three dimensional hyper plane embedded in the four dimensional space of variables.
Birne Binegar: When I have multiple equations what we're looking at Dennis a solution is the intersection a bunch of hyper planes.
Birne Binegar: And intersections of hyper plane to always hyper planes as well it's just that there will be fewer spanning directions, but they'll still be able to write any hydroplane in this form.
Birne Binegar: Okay, so our goal in this example is to take the system of.
Noah Hightower: equations.
Birne Binegar: And somehow construct the hyper planes that corresponds to it solutions.
Birne Binegar: Okay, so um, how do we do that well, the first step is to.
Birne Binegar: Take the data that goes into specifying this linear system and encapsulated in this augmented matrix what we capture.
Birne Binegar: Are the coefficient of the variables and the numbers on the right hand side this this half of this augmented matrix contains all the coefficient 1102 and the first row to 215 and the second relic cetera.
Birne Binegar: And then the numbers on the right hand side we capture those as an additional column.
Birne Binegar: Then, the idea is that this augmented matrix is not it's data is equivalent to the data that it took to write down the original linear system.
Birne Binegar: What we'll do is we'll manipulate this data, until we get an augmented matrix that would correspond to the augmented matrix or the solution that's done like taking this augmented matrix and using the three elementary row operations.
Birne Binegar: We have grown interchanges.
Birne Binegar: That switching to wrote in the matrix we have.
Birne Binegar: The ability to replace a row by multiple itself, and we have the ability.
Birne Binegar: To take a row and replace it by at some with multiple another row.
Birne Binegar: And what we see is just using these three operations, we can systematically reduce an arbitrary matrix to unique for the so called reduce growth Echelon.
Birne Binegar: let's do that the first thing you need to do is get it in row Echelon for that means arranging doing real operations until you get a matrix where the pivots occurred this downward staircase from the upper left to lower right.
Birne Binegar: So this first operation reason why it was applied is because the low the pivot in the first row I have the pivot and second row I need to get rid of this if i'm to move towards.
Birne Binegar: row Echelon form and so like take a multiple of this row and I added to this role precisely so this leading entry cancels.
Birne Binegar: The recipe is like replace the second row is difference with two times the first row.
Birne Binegar: Alright, so that that produces that with our question.
Birne Binegar: All right, and then.
Birne Binegar: that's first step of row operation, but now we have a similar problem with this pivot in the third row line directly below the pivot in the second row so to get rid of this entry i'm going to take them.
Birne Binegar: Add to the third row some multiple at the second row just so I can cancel out that leading term and what we need to do is subtract the second rule from the third.
Birne Binegar: All right, and so we end up with this matrix all right now, let me point something out.
Birne Binegar: This zero row with the bottom.
Birne Binegar: If you convert back to equations it simply says zero is equal to zero well more is at zero times X one zero times X zero times X three.
Birne Binegar: Plus zero times X four is equal to zero, but after simplification, it just says zero is equal to zero, so what actually happened during this role reaction is three equations.
Birne Binegar: got reduced to really to effective equations that's because somewhere in this original linear system where some redundancies, the process of reduction not only simplifies the augmented matrix but it simplifies the corresponding equations as well.
Birne Binegar: And so what we actually just to production steps I not only reach row Echelon.
Birne Binegar: But because pivots are all equal to one and because low and above any pivot I only have zeros i've actually reached reduced row Echelon form and that's the form of the augmented matrix when you reach the last stage of reduction and you've got the augmented matrix of the solution.
Birne Binegar: So what we do.
Birne Binegar: Once we reach the matrix in reduce relational on form we convert back to equations.
Birne Binegar: But we use the augmented matrix in reducing national on form for one more thing.
Birne Binegar: you notice that first equation involves three variables what am I sold them course well.
Birne Binegar: What we do is we always choose the variable that corresponds to the pivot in a row, we think of that as a fixed variable.
Birne Binegar: And the other variables that occur in the equation correspond to ever given row, we regard those as the free variables, because that you can do from this first equation is you can isolate the fixed variable X one so that's the.
Birne Binegar: These columns two and four we don't have pits and that's why I mean i'm regarding X two in exports being free variables, but call them is one and three, they do have pivots and that's why i'm regarding them.
Birne Binegar: As a fixed variable I take the equations coming from the augmented matrix and we do show Echelon form, I write them down and then I moved the three variables over to the right hand side.
Birne Binegar: Now i've got the fixed variables expressed in terms of the free variables.
Birne Binegar: Alright, so these are what in high school, you might call these the equations of the solution, it tells you that, in fact, you have a two dimensional solution space, because you have to free parameters, but the other, the other variables are fixed in terms of the free parameters in the solution.
Birne Binegar: All right, and then.
Birne Binegar: we're aiming for a hyper plane expression of the solution, so the next step is to take our equations for the fixed variable the variables that correspond the columns with pivots and the renewal Echelon we take those equations.
Birne Binegar: We write down a vector that we want to interpret as the solution vector so first I just write down its components in generic form but then, where I can all substitute for the fixed variables X one and X three so i've replaced X one by each expression in terms of the free variables.
Birne Binegar: And the last step is then to look at this factor and rip it apart.
Birne Binegar: you'll have a constant terms and in some components 1020 I get this term that doesn't involve any free variables at all.
Birne Binegar: Then I have terms in each component that are proportional X two i'll grab those well I get a minus one, and the first component.
Birne Binegar: One times X two and the second component and next two deaths occurred in the last two components, so I just have zero there.
Birne Binegar: And then export appears in this sector as well we'll play by minus two in the first component doesn't occur in the second component minus one, and the third component in one in the four components and now what I got as a result.
Birne Binegar: is well, then that solution vectors explicitly displayed as a hyper plane here's your starting vector P zero and Here are two directions V1 and V2, for which I can.
Birne Binegar: Travel in away from this point by arbitrary distances X two and X four, and so this final expression on the right hand side that's what I mean by the hyper plane expression for the solution.
Birne Binegar: Okay, so that's the general procedure that you'll be applying a time and time again, as you work through problems set for.
Birne Binegar: You don't have to put it, you don't have to put it in the explicit set notation all you want to see is that starting point and the two directions and how the.
Birne Binegar: Free parameters coming to generate that entire space because X two X for our free parameters they're just as arbitrary if these numbers he wanted tk.
Birne Binegar: OK so without review, I hope you won't have any trouble with problem set for.
Birne Binegar: next step is to cover last topic that's going to be covered, on the exam and that's in verses of matrices which is going to give us an alternative way of solving linear systems.
Birne Binegar: In any and it's going to also involve role reaction, but.
Birne Binegar: The first part, the formal part it actually makes it look even simpler than it actually is.
Birne Binegar: Suppose we have a square linear system, meaning, we have this many equations, as we have a known.
Birne Binegar: or write it in matrix notation so we have a coefficient matrix i'm a column vector of variables is equal to call them director of numbers.
Birne Binegar: Suppose we had a matrix see such that when I multiply this coefficient matrix time see I get the identity matrix well if that's the case, then we can solve this matrix equation.
Birne Binegar: And essentially the same way that you use the solving linear equations involving one variable in high school.
Birne Binegar: If you had this equation.
Birne Binegar: To X is equal before well the way you would solve it, you would multiply both sides by one half and then you end up with X equals two right that's the standard way that we solve.
Birne Binegar: A single equation in one variables now just get rid of that coefficient well essentially that's what we're going to do here.
Birne Binegar: What the property of this factor one half is is it exactly cancels the factor of two there and allows us to isolate the variable X.
Birne Binegar: Well, this will also work as long as we have a matrix that when it hits the coefficient matrix it produces the identity matrix well we'll be able to write down the solution really quickly right so take this equation multiply both side by see.
Birne Binegar: Let me remark that we have to be careful and doing so, we have eight X is equal to be.
Birne Binegar: And I want to multiply both sides by see, I have to do it, a particular way.
Birne Binegar: Simply do that incorrect, the reason why is that matrix multiplication the order of factors makes a difference now.
Birne Binegar: i'm multiplying from the Left, on the on the left hand side that i'm multiplying the right hand side from the right by the same metrics.
Birne Binegar: i'm doing a different operation two different sides of the equation I can't expect the equality to be maintained Well, this is.
Birne Binegar: This is the wrong way to do it, you right see a and then, if you're going to multiply the left, the left hand side from the Left, I see make sure you also multiply the right hand side on the left by the same factor so.
Birne Binegar: This equation it's not going to be true in general, you can't derive it.
Birne Binegar: From this equation, it may not be true, but this equation will be true, so long as the original equation is true, because she did exactly the same thing to both sides.
Birne Binegar: Right so that's that brings that line, but then my assumption is that I have a matrix see such that when I multiply the coefficient matrix by it, I get the identity matrix.
Birne Binegar: Here I use that assumption, and then anything multiply by the identity matrix just reproduces when I started off with and so here, if I had this matrix.
Birne Binegar: I could fold this into your system by doing a simple matrix multiplication and that's the point it's basically we're doing a matrix algebra and log of this method of solution.
Birne Binegar: Alright, so let me give you.
Birne Binegar: formalize this a bit.
Birne Binegar: will say.
Birne Binegar: That matrix see that has that property that if you multiply.
Birne Binegar: By a given matrix a from the left by see you get the identity matrix and, furthermore, if you multiply a from a right bye see you still get the identity matrix.
Birne Binegar: We call the matrix see the matrix inverse of a if the analog of well one happening, the number which, when I multiply to I get to number one is the result.
Birne Binegar: Well, here we don't have numbers, we have matrices but see it's going to be the inverse of a because see times a produces the identity matrix just like one half times two is equal to one.
Birne Binegar: Alright, so let me give you listen to example your two matrices.
Birne Binegar: We use the.
Birne Binegar: Rules of matrix multiplication meaning to compute this matrix product I take dot products of the rows of the first factor with columns of The second factor.
Birne Binegar: I do it component by component and I get the identity matrix as a result, like do the multiplication in the opposite order, I still get the identity matrix.
Birne Binegar: notice.
Birne Binegar: In this example, though, that here, you see a kind of difference than what you would get with high school algebra you can't get rid of this to without introducing fractions into your calculations.
Birne Binegar: But notice in this matrix multiplication I found the inverse of the matrix a because see times a is equal to that again if you make sure.
Birne Binegar: I found this in verse about introducing any fractions at all so it's like most this problem from ordinary algebra.
Birne Binegar: Except it's a little bit different to finding the inverse matrix is not simply multiplying it's multiplicative inverse.
Birne Binegar: Because you're also doing easy addition to as well, you can actually find in verses of matrices that don't involve any multiplicative in verses of the numbers inside the matrix.
Birne Binegar: Because he is equal to the identity matrix and see a is equal to the identity matrix that makes see an inverse of the matrix a.
Birne Binegar: Okay, so.
Birne Binegar: Let me give you some to really quick properties of matrix in verses first of all.
Birne Binegar: They only they only work for square matrices when you have the same number of rows, as you have columns.
Birne Binegar: And second of all.
Birne Binegar: what's not obvious from the definition right said that step back.
Birne Binegar: Your i'm using an indefinite article a rather than, say, the matrix inverse of a it's just some make inverse of a but one of the first things you proved about this is that in fact there's only one in burst of a given matrix day.
Birne Binegar: So i'll show you that suppose you have two matrices matrix see and a matrix D that have that property.
Birne Binegar: That.
Birne Binegar: Effectively, is the definition of what it means to be a inverse of a you need see times at each one and see Tom K to equal one the matter which side you multiply a by similarly will assume the same per D.
Birne Binegar: And i'll show you that, in fact, if both these equation told so it's both C and D are in verses that they have to be identical to one another.
Birne Binegar: To do that, I take product of the three major sees a C and D and I compute it two different ways, first of all, I have three matrices multiplied together let's.
Birne Binegar: will play the second two factors first and then multiply by the first factory well, a time see is equal to the identity and the identity matrix on these equals a Dee da da see is equal to D.
Birne Binegar: or grouping things a little bit differently, I can calculate this product like first combining the first two factors and then multiplying by see if I do it that way, because D times a is equal to the identity matrix i'm going to get see out of this computation and so.
Birne Binegar: Both line to the right hand side or equal to da see and so he has to equal see So in fact my definition was a little bit loose I talked about a inverse of the.
Birne Binegar: matrix in the inverse of the matrix in the definition, but it turns out from this really short calculation that if we do have an inverse there's only one choice for it.
Birne Binegar: Okay, so that's one property and.
Birne Binegar: But it's really useful to us because, as soon as we find a matrix to sit see times a is equal to the identity matrix.
Birne Binegar: Because the matrix in verses are unique as soon as we see something multiplies a produces the identity matrix will be able to immediately identify it as the inverse of day because no other matrix could have that property.
Birne Binegar: The other thing I want to point out is that back in this definitions.
Birne Binegar: Here I put two conditions on the matrix see that whether I multiply a from the left or from the right, I still get the identity matrix.
Birne Binegar: Those look like two separate conditions, but in fact, if this is true, you can draw this this equation I don't do it because it's complicated it's.
Birne Binegar: Just it's not at all obvious you just try to compute these two products and compare it directly, you probably may not be able to actually prove that in fact this equation implies that.
Birne Binegar: But, in fact, it is true, and then, when I teach dance linear algebra soon as we come to this section I take a half a lecture to explain why this equation imply form.
Birne Binegar: But just take my word for it does make things easier That means that if you only if you could just show that see times as equal to the identity matrix, then you know that see in verse eight and that this equation would also also follow.
Birne Binegar: Okay, so.
Birne Binegar: we've got basis in versus the uniqueness here's something it's kind of cute.
Birne Binegar: suppose you have two matrices, each of which has an inverse will call them a the minus first and be the minus first.
Birne Binegar: These are going to be the unique matrices such that when this guy hit a, whether from the Right or the Left it's going to multiply and produce the identity matrix and be the minus first is going to be unique as well it's going to be the unique matrix for which, when I multiply.
Birne Binegar: The by a or B by beat the miners First, I get the identity matrix.
Birne Binegar: All right.
Birne Binegar: Now i'm going to show you how what the.
Birne Binegar: inverse of the product of A and B is going to be I claim this this matrix.
Birne Binegar: So the proven what i'll do is i'll simply take a B and multiply it by this suppose it in.
Birne Binegar: One when I do so I can regroup factors this the time being, the miners first will produce the identity matrix the identity matrix times either factors going to reproduce that factor than eight times 8am nine years first is equal to one so.
Birne Binegar: This matrix D minus first time to a to minors first has the property that when I multiply a B times it, I get the identity matrix that proves that be the miners first at the minus First is the inverse of this matrix because a B times that matrix produces the identity matrix.
Birne Binegar: So this uniqueness is useful it some it, it allows us to compute the inverse of a product really easily Now all we have to do is get the right ordering of these factors here, and then we see that we can produce the identity matrix.
Birne Binegar: In this fashion.
Birne Binegar: Okay, so notice, however it's not just replacing a eight minus first and be by be the miners first to get this inverse notice how the order of factors that switch.
Birne Binegar: same thing happened when we were talking about transposes the transpose of the product is also the product of the transpose but the product is written in the reverse order.
Birne Binegar: and similar phenomenon happens here, it is true that the inverse of a times B is made from the inverse of a versus B, but the order of the factors has been reversed and we actually need that to get these cancellations to occur properly.
Birne Binegar: Okay.
Birne Binegar: So.
Birne Binegar: Good so.
Birne Binegar: Now, what I want to describe.
Birne Binegar: pointed out that the reason why we're interested in matrix in verses is it gives us an alternative way of solving linear systems, we can solve it in a way that's more closely analogous the way we we solve systems with one variable and.
Birne Binegar: What I want to describe next, then, is how do I find the inverse of the matrix that's going to require a slight digression.
Birne Binegar: will tell you effectively how we're going to fund it.
Birne Binegar: The goal is to find some if i'm going to find the inverse of the matrix a The goal is to find some matrix such that, when some matrix see let's call it, so that when I multiply see either from the left or from the right, I multiply my given matrix guy that I should get the identity.
Birne Binegar: What i'm going to construct such a matrix for arbitrary matrix and.
Birne Binegar: These elementary makes your car i'm about to describe.
Birne Binegar: are going.
Birne Binegar: Well, let me just proceed and you'll see the utility of these elementary matrices.
Birne Binegar: Okay, so a definition.
Birne Binegar: Let our be any elementary row operation.
Birne Binegar: Oh, I just racing when I was going to write down.
Birne Binegar: We have three kinds of elementary operations.
Birne Binegar: We can.
Birne Binegar: interchange the order of two rows of the matrix we can.
Birne Binegar: replace a row by a multiple of itself.
Birne Binegar: or.
Birne Binegar: We can replace her row by at some with a multiple of another row.
Birne Binegar: Right Those are the three operations by which we convert an arbitrary matrix for example, into a real Echelon form or reduce real excellent.
Birne Binegar: Alright, so what an elementary matrix is is you pick up an elementary operation here i'm in and changing two rows.
Birne Binegar: And you do that operation to the corresponding identity matrix So if I switch roles one and two of the identity matrix i'll get this matrix that's going to be called the elementary matrix corresponding to this row operation, where I switched to rose.
Birne Binegar: If I.
Birne Binegar: consider an operation, where I replace road to bite three times row to well let's do that to the identity matrix and what happens is that.
Birne Binegar: The second row changes by a factor of three.
Birne Binegar: And Similarly, I can.
Birne Binegar: Do the real operation, where I agree place a row by some with a multiple to now that row I can also do that to the identity matrix and I get a particular matrix is result.
Birne Binegar: i'm here.
Birne Binegar: i'm using a to denote at elementary matrix the subscript here is telling you what row operation I performed on the identity matrix to get that and then here's result.
Birne Binegar: So this elementary matrix is exactly that matrix this elementary matrix is that this elementary that there are some natural correspondence between elementary row operation and their corresponding elementary matrices.
Birne Binegar: All right now, let me show you the utility of these gadgets.
Birne Binegar: If you have a.
Birne Binegar: Given elementary role operation call it script on then associated with that elementary row operation, well, we can apply it to the identity matrix and we get an elementary matrix.
Birne Binegar: here's the statement know why these guys are useful.
Birne Binegar: To implement this row operation on an arbitrary matrix I can do that I curse calculating this elementary matrix and then instead of operating your real operation directly on a I do matrix multiplication.
Birne Binegar: So here.
Birne Binegar: This real operation we're interchanging two rows way I would calculate the left hand side to side simply interchange two rows of a.
Birne Binegar: The way I would calculate the right hand side is I would interchange two rows of the identity matrix and then use that matrix the matrix multiplication I would multiply eight times.
Birne Binegar: theorem says that, no matter which way you calculate either side you get the same result.
Birne Binegar: Alright, so an elementary real operation, you can implement it two different ways, you can either apply it directly to a matrix or you can apply that row operation to the identity matrix and then do a matrix multiplication.
Birne Binegar: here's some examples of this here's what you see as an arbitrary two by two matrix let's see what happens when we interchange rose one, and you know to well, we simply flipped rose.
Birne Binegar: But we can also flip the rose up the identity matrix as well, and this would be the corresponding.
Birne Binegar: Elementary matrix now.
Birne Binegar: let's calculate what happens when I take this elementary matrix and I multiply from the left our original matrix a buy it right, so there it is there's the original matrix a calculate the matrix product what happened we got exactly the same matrix CD ab as we got by applying that.
Birne Binegar: That row interchange directly to the original matrix.
Birne Binegar: So i'm verified least in this example, this identity that interchanges of two rows of a matrix can be implemented, also by a certain matrix multiplication and in fact I know exactly how to make this matrix IE that will carry out that operation.
Birne Binegar: We can also do row rescaling by take a row, and I go to and I replace it by three times wrote to this, what would happen to my matrix a B, C D, but I can also multiply the second row with the identity matrix.
Birne Binegar: To get the corresponding elementary matrix and then I can.
Birne Binegar: Take my original matrix a and I can multiply it by that matrix and I get the same result same result as as if I did, that operation directly on a.
Birne Binegar: And last example here we're doing replacing a robot some with a multiples another row I can do that, directly or indirectly by.
Birne Binegar: figuring out what the corresponding elementary matrix is during the same operation to the identity matrix and then I can do the matrix multiplication and you see, I get the same result.
Birne Binegar: So what's The point of this well.
Birne Binegar: This is not making real reduction any easier, make it more complicated actually because, instead of being able to operate directly on a matrix.
Birne Binegar: I have to do the same operation to the identity matrix and then I have to follow up with our matrix multiplication so i'm not speeding up the calculation at all.
Birne Binegar: But what it is doing is it's showing that the effect of elementary rule operations can be understood as a kind of matrix multiplication.
Birne Binegar: And now i'll show you.
Birne Binegar: Why that's so important.
Birne Binegar: Suppose we have a matrix a that when we reduce it, we can actually get it.
Birne Binegar: By a sequence of elementary row operations, I can actually convert it to the identity matrix so on client one real operation and then another and another, and another and then, finally I reached the identity matrix got to doing all those role operations.
Birne Binegar: Effectively, the sequence of operations might be the row operation to perform in order to like this matrix in reduce role Echelon porn.
Birne Binegar: All right, and then Oh, by the way.
Birne Binegar: Remember that to reduce your left alone form of any matrix is unique to us.
Birne Binegar: If you look at the identity matrix 100100 it's always in reduce relational and form.
Birne Binegar: right because you'll only have non serial numbers hold on this main diagonal and so above the pivots are all equal to one and above and below pipits I have nothing but zeros So in fact.
Birne Binegar: If he can be reduced to the identity matrix the reduce role Echelon form of eight is exactly the again if you.
Birne Binegar: Alright, so i'm going to suppose that we have a matrix a that after I apply sequence of elementary row operations i'm able to convert it to a traditional Echelon form and that happens to be the identity matrix.
Birne Binegar: Well, because there's a different way of implementing these row operations, I can either act directly on a or I can multiply that doing this operation directly, I can multiply a by the corresponding elementary matrix and.
Birne Binegar: This.
Birne Binegar: term there's exactly the same thing is he subscript r1 times a and the second real operation would just be the second he subscript or two on this.
Birne Binegar: Until we see them is that the sequence of bro operations could be implemented by matrix multiplication but now we have.
Birne Binegar: A certain matrix product of these elementary matrices time today is equal to the identity matrix well good I found a particular matrix.
Birne Binegar: subset when I multiply eight times it, I get the identity matrix that makes this product of elementary matrices exactly the inverse of a because this guy himes a produces the identity matrix.
Birne Binegar: But we have a way now of calculating the inverse of the matrix we figure out how to reduce it to the identity matrix then we keep track of those operations will use each performance elementary matrix and then we multiply them all together, and that will be our numbers.
Birne Binegar: well.
Birne Binegar: Luckily that's not how we're going to calculate in verses this would be really tedious, not only would you have to do figure out how to reduce a to the identity matrix but then you'd have to for each of these.
Birne Binegar: Real operation to have to figure out what they do to the identity matrix and then you would have to multiply them all together so that's pretty strenuous calculation, if you have even just a few elementary row operations.
Birne Binegar: So.
Birne Binegar: Let me just put up our procedure and then i'll work towards making it more efficient.
Birne Binegar: So we have a sequence of row operations that convert a to the identity matrix.
Birne Binegar: The inverse of a it's just going to be a product of the corresponding elementary matrices or on the left, you have the first operation, the elementary matrix corresponding to the first row operation, you did.
Birne Binegar: Last factor on the left is going to be the last, the elementary matrix corresponding to the last row operation, and that will be renders.
Birne Binegar: Alright, so.
Birne Binegar: Now to make it go fast.
Birne Binegar: In the end.
Birne Binegar: These elementary matrices are you know meter object right when i'm i'm wanting to find the inverse of a I want this product, and if I can find a way of avoiding computing all these individual factors that will be a big plus.
Birne Binegar: So look this next slide does show you how you compute this product without computing the individual factors.
Birne Binegar: So here we go what i'm gonna do is i'm going to take my original matrix a and i'm going to join to it.
Birne Binegar: The identity matrix of the same size.
Birne Binegar: So this is n by N, and this is the N by n identity matrix.
Birne Binegar: Now what i'm going to do is take this matrix and i'm going to reduce it to reduce row Echelon porn.
Birne Binegar: Moreover, i'm going to assume that a can be reduced to the identity matrix going to have a sequence of row operations that if I apply it to a i'm going to get the identity matrix but let's see what happens to the left hand side here.
Birne Binegar: These operations are both to convert a to the identity matrix if I do the same sequence of bro operations, but instead of operating.
Birne Binegar: Only on the first half of this augmented matrix if I do it to the to the whole thing, well, it just means that when i'm adding rose you do.
Birne Binegar: You work column by column extending this matrix off to the left just means I have more columns to keep track of this I apply these row operations I don't have to change the role operation.
Birne Binegar: This this matrix has just as many roles as the original matrix so I can do the real operations on this augmented matrix as well.
Birne Binegar: So I do so and what's going to happen is that I can.
Birne Binegar: think a bit of acting on this double matrix I doing row operations on the first half and separately doing row operations on the second half, but it'll be the same sequence elementary roll off.
Birne Binegar: Now my assumption is that I can reduce a to the identity matrix so left hand side here, using this identity it's just going to simplify to the identity matrix What about the.
Birne Binegar: right hand side.
Birne Binegar: Well, here it is.
Birne Binegar: We know that.
Birne Binegar: He gets reduced to the identity matrix, then the same row operations are going to be applied to the left hand side.
Birne Binegar: And now, when I.
Birne Binegar: Take this right hand side and I computed.
Birne Binegar: By itself well I can implement these row operations, the multiplication by elementary matrices but so.
Birne Binegar: These row operations can be implemented by multiplying by this product of elementary matrices but it's just multiplying the identity matrix and so.
Birne Binegar: I can absorb that identity matrix saying to each one, and now I have exactly the same product of elementary matrices that I previously identified as a the minus first.
Birne Binegar: Okay So what does this tell us it tells me if a row reduces to the identity matrix.
Birne Binegar: Then.
Birne Binegar: This adjoin matrix a a join with the identity matrix it's going to reduce to I the identity matrix on the left hand side this thing here it's going to be exactly the inverse of.
Birne Binegar: itself what happens in is that when you reduce this matrix to reduce real Echelon form a is going to get converted I, and I gets converted to the inverse of a.
Birne Binegar: And so the way we calculate the inverse of a if we take this matrix reduce it to reduce your rational on form, and then we just grabbed this left that's right hand side, and that will be our interest.
Birne Binegar: Okay, so again i'm here i'm pointing out that that when you take a matrix like this and you write it in renewable Echelon form well if.
Birne Binegar: This this left hand factors it's going to contain all the pivot right you got a.
Birne Binegar: Non zero number and each of these roads and above and below pipits I have nothing but zeros and all the pivots or equal to one, so, in fact.
Birne Binegar: If a reduces reduces to the identity matrix the reduce real Echelon form of this matrix is going to be exactly this see you'll have Ai and then you'll have something else here, whatever it took to.
Birne Binegar: reduce the a part to the identity matrix it's going to take the I part of the original augmented matrix and it's going to convert it to a minus first.
Birne Binegar: Well here's the theorem that will end today's lecture with if you have an end my end matrix and it and it's reduced Echelon form has the identity matrix on the left and some other matrix on the right, then, that matrix on the right it's going to be exactly the inverse.
Birne Binegar: On the other hand, not every matrix is rho reducible to the identity matrix.
Birne Binegar: I had a matrix like this 111010 put it in put it in Russia and.
Birne Binegar: I had a matrix like this, I would not be able to use any real operations that would put a non zero number in this corner until I could not get the.
Birne Binegar: inverse matrix when that happens to you that just means he doesn't have an inverse yeah.
Birne Binegar: All systems of equations that have 20 solution also therefore that's right, I try and that's that's the big theorem that will prove on.
Birne Binegar: will prove on Wednesday.
Birne Binegar: yeah what we're going to find now and in the next lecture if there's gonna be lots of connections between solving linear system and finding in versus a matrix.
Birne Binegar: OK so i'll stop here and i'm will pick up.
Birne Binegar: More discussion of the matrix in verses on Wednesday.