Birne Binegar: Welcome back.
Birne Binegar: let's see.
Birne Binegar: So.
Birne Binegar: Having this two lectures last week is screwed up our schedule that I told you last last week that first exam was coming up, and I was expecting get through.
Birne Binegar: In verses of matrix seats and then begin preparing for the exam last week, but so this week i'm going to start off reviewing a topic that we haven't touched upon, since our last meeting.
Birne Binegar: How to solve linear system so i'll do give a quick example of that.
Birne Binegar: The real reduction part of the example is something that will use again today, but in a different fashion, who we won't use robe reduction so much to solve a system of equations directly rather will use rubbish option to figure out what the inverse of the matrix is.
But.
Birne Binegar: Let me first get some some administrative details.
Birne Binegar: homework set three.
Birne Binegar: Solutions actually the solutions to problems sets one through five have been posted on canvas I did that this morning that's because I want to, I want to help you get through the homework.
Birne Binegar: till i've done sample problems.
Birne Binegar: That what i've done is like replicated the assignment as best I can just remember that web assign it random nicest the numbers in the problem, so I can't do exactly the same problems that web assign would give you but i've done.
Birne Binegar: Similar problems where the numbers have been changed and that should help you work through.
Birne Binegar: What I really need you to do is get your hands dirty and do the row reduction or when you solve in your systems that that's a key calculation that.
Birne Binegar: Not that hard once you've done it a couple times, but the first that the first time you've done it is on the exam well all you reading is not going to help because you just just not going to recognize the situation you're in at a given step so i'll go through an example of that.
Birne Binegar: I think what will happen this week is all I will anyway get through.
Birne Binegar: The inverse of matrices but the homework that I can assign for that it's not going to be due until next Monday, and in fact what i'm assigning.
Birne Binegar: Today is homework set for which deals with solving linear systems and homework set five, which is a bit shorter, but it deals with calculating the inverse of the matrix.
Birne Binegar: When you once you finish homework six, four and five.
Birne Binegar: At least from a calculation point of view, you should be prepared for the.
Birne Binegar: The first exam but um.
Birne Binegar: I think the earliest that we could have the first exam might be Friday of next week, so that i'm not sure what the university policy is i'm supposed to have six week grades done.
Birne Binegar: By the.
Birne Binegar: Six weeks, but singing we last last effectively a week last week i'm not sure if they're going to revise that policy or not so.
Birne Binegar: My apologies if I don't get the grade Jacqueline done in time to give you, your six week great, but all strive to do that i'm actually not required because this is.
Birne Binegar: 3000 level classes and above are not required this submit six week grade but.
Birne Binegar: It is time that people should figure out how well they're doing in the course so we'll do an exam that will give you some feedback.
Birne Binegar: But that exam it's not likely to occur at the earliest maybe Friday of next week, more likely i'll give you the weekend to prepare for a Monday exam so maybe two weeks from today.
Birne Binegar: Okay, so any questions.
Birne Binegar: Alright, so what I want to do is like it's been at least in my mind such a long time since I talked about solving linear system i'm going to run through a quick examples to you know how that that procedure goes and then we'll take up a new topic in verses of matrices.
Birne Binegar: Alright, so let's see if my.
Birne Binegar: Goodness so here's a system of what looks like three equations and I count for unknown, so this is a three by four linear system, and what I want to do in this examples I want to express the solution of this system as a hyper plane.
Birne Binegar: Alright, so let's remember.
Birne Binegar: focus on our goal here by a hyper plane.
Birne Binegar: When I mean it's a set of vectors of the form, where you start off at a given point, and then you stray from that point along the fixed set of directions.
Birne Binegar: Okay, and then you let these parameters, do you want to tk.
Birne Binegar: range over the entire real line, and what that construction does is it takes a point.
Birne Binegar: and allows you to stray from that point in a bunch of fixed directions or linear combinations of those directions and you end up building out.
Birne Binegar: What looks to be like a smaller vector space embedded inside the larger space of variables.
Birne Binegar: So where this plane actually sits depends on what kind of vectors P zero P one and up through the chaos if these are 10 dimensional vectors this hyper plane is going to sit and attend dimensional space, but.
Birne Binegar: The number the dimension of that hyper plane is going to be lower than the 10 dimensional space it sits in its dimensions effectively determined by the number of independent directions, from which you can stray from your starting point.
Birne Binegar: Okay, so I do this, the first week of class and the goal part method of solving linear systems is actually to present the solution set as a bunch of vectors living on a on a hyper plane okay let's let's proceed.
Birne Binegar: The first step was to take the coefficient one one and 2221 and five arrange those, as you know what would be I guess.
Birne Binegar: Three rows and four columns or three by four matrix and then this.
Birne Binegar: last bit here be it's this column vector on the right hand side alright so here's the augmented matrix that corresponds to the original linear system.
Birne Binegar: And our method is to transform this augmented matrix to the augmented matrix of the solution, using row reductions so, for example, when you do reduction.
Birne Binegar: The first thing you look at in that.
Birne Binegar: World quite a while right now we're going to be concentrating on pivot oh.
Birne Binegar: Working in.
Birne Binegar: The pivots are the first non zero entries in a row there's a pivot there's a pivot and there's a pivot.
Birne Binegar: To get this first goal is to get this matrix in relational on form and so.
Birne Binegar: What i'll need to do is clear out this pivot that sitting below the pivot in the upper left hand corner, so I will subtract two times the first row, so this becomes we come to, and it could cancel that.
Birne Binegar: So we do the row operation we get a new matrix and now we've got it in row Echelon form least up to the second row.
Birne Binegar: We have a problem with that pivot line, the low a pivot in the row above so I knew another elementary real operation that as the effective.
Birne Binegar: Get making that row become intact to zero, because the second and third rope aren't academical So when I subtract the second row from the third row I get the zero RAL this, by the way.
Birne Binegar: indicates.
Birne Binegar: The original system.
Birne Binegar: Had redundancy.
Birne Binegar: Because what effectively happens is during this process of reduction that does not change the solution.
Birne Binegar: We found that our three original equations or equivalent to effectively to independent equations this last row it simply says zero equals zero and put no restrictions on the variables.
Birne Binegar: don't let that the original in your system is actually equivalent to a system of two equations in for unknown and what the row operations have done is they've removed the redundancy from the system of equations.
Birne Binegar: Alright, so i'm here i've gotten to real astronaut, for, but in fact.
Birne Binegar: we're kind of lucky in this case as soon as we reach row so one form we were also in reduce real Echelon.
Birne Binegar: I have the pivots occurring in a downward staircase that's real Echelon born the pivots are also equal to one and above and below pivots I have only zeros.
Birne Binegar: So when I have a pivot it's the only non zero entry in the columns and its value is one, so this is in fact a reduced row Echelon for and, as such, it's going to be the augmented matrix the solution.
Birne Binegar: So what I do, and the next step is I couldn't having reached the augmented matrix of the solution.
Birne Binegar: Having rishi augmented matrix of the solution or write down the corresponding equations and now we have to do a little bit of interpretation.
Birne Binegar: This augmented matrix and reduce your mind form.
Birne Binegar: Not only gives us the equations of the solution, but it gives us a particular interpretation of the variables.
Birne Binegar: Cross columns here let's attach them to their respective variables.
Birne Binegar: So on the coefficient i'd each column is related to a variable and what we do to make sense of these equations is, we take the variables that correspond the columns that have put it so that would be X one X three, these are going to be fixed.
Birne Binegar: i'll put it in quotes fixed variables.
Birne Binegar: And the variables that corresponds the columns to dump contain that would be X and X for those are going to be the free variables.
Birne Binegar: That so that we can take the single equation and thinking about expressing the free variable I sorry the fixed variable in terms of the free variable.
Birne Binegar: And this one this this equation can be used to express that fixed variable X three in terms of X for all you have to do to achieve that is moved the free variables, to the opposite side by subtracting them from both sides.
Birne Binegar: And now we have.
Birne Binegar: Effectively, the equations of the solution, where we can express the three variables in terms of sorry fixed variables in terms of the free variants.
Birne Binegar: All right now i'm looking for a hyper planes solution, so I want more than just the values of two variables in terms of other variables, I want to see this geometrically.
Birne Binegar: So the next step.
Birne Binegar: is to take.
Birne Binegar: The solution for the or the equations for the fixed variables that we just uncovered these two we are able to express X one in terms of X two X four X three x four.
Birne Binegar: We set up.
Birne Binegar: What to be a solution vector first we give its components and then wherever we can replace the fixed variable yeah will express it in terms of the free variables that's coming from these equations.
Birne Binegar: Which sorry came from the augmented matrix and reduce cholesterol and you follow that.
Birne Binegar: derivation there and then the last step once you've got your solution vector express in terms of the free variable you expand your solution vector in terms of the three variables so.
Birne Binegar: In one vector you peel off the constants one, and the first component know constant and the second component to in the third nothing in the fourth.
Birne Binegar: And then you have terms in various components that are proportional to X two I pull out X two is a common factor, and then I grabbed the minus one.
Birne Binegar: Here X, to the peers by itself so since like X two times one, and then the third and fourth component there's no X two zeros.
Birne Binegar: And then I have terms that are proportional to export and I grab those and it's a separate term in this expansion and now what i've got is an expression granny solution better.
Birne Binegar: i've expressed it as a point on a hyper plane, this is your starting point, p zero.
Birne Binegar: X two is a free parameter, so we could be anything but just re label it call it to one that parameter to you on his multiplying a particular vector V and then.
Birne Binegar: All export the second parameter and it's multiplied a second factor be too, and so now.
Birne Binegar: With this interpretation of the right hand side I displayed a solution vector is living on a particular hyper plane and we're done okay so that's the basic calculation and homework assignment for you'll you'll do a number of those but then it shouldn't be too too high.
Birne Binegar: And if if you run into any problems on the homework.
Birne Binegar: just raised them in class and we cannot, we can do some examples.
Birne Binegar: What I want to do, first let me ask the right questions yeah.
Over.
This entity was making sense less.
Birne Binegar: on the right side of the equation numbers alright so so here i'm just writing generic solution vector but you know I haven't imposed any conditions on my variables.
Birne Binegar: What i've learned from the last slide is from that augmented matrix up a solution that the fixed variable X one had to be related to the three variables X two X four X three is related to X four let's go back to stage, because this is really important.
Birne Binegar: The equations you write down directly from the augmented matrix and reduce us on form they don't tell you directly.
Birne Binegar: Which are free variables in which are fixed variables, the rule I would you make that determination is the.
Birne Binegar: Fixed variable so once you can solve for are going to correspond the columns with pivot so X one and extreme are going to be the variables, I can solve for.
Birne Binegar: The other variables corresponding second and fourth column they're going to be moved to the opposite side, and now I can replace two variables, with their expressions in terms of the free variables alright so that's where these two equations come.
Birne Binegar: And then, what I do is I start by writing down a generic factor, and the next step, I impose that if this vector is going to be a solution of my linear system, then X, one has to be related to X two X for.
Birne Binegar: Exactly by this equation, so I substitute in for a fixed variables, where I can X one an x Ray all right now i've got a vector that if I very free parameters X two X four it's going to.
Birne Binegar: vary over the entire solution space, all you want to do next is express that vector as a point on a hyper plane.
Birne Binegar: So I need a starting point, I need some parameters that I can bury times various fixed vectors.
Birne Binegar: Until that's what this expression represents I grabbed the constant vector that's going to be your starting point and then you have some parameter three parameter in the solution multiplying particular vector so this would be safe, be one, and this would be the two and then.
Birne Binegar: Standing across this equation, I see i've got an arbitrary solution vector expressed as a point on a two dimensional hyperplasia I have two directions, from which I can straight a starting point okay.
Birne Binegar: All right.
Birne Binegar: So.
Birne Binegar: yeah that's that's a basic calculation and.
Birne Binegar: What, in the end we get the geometric object that we saw a hyper plane expression for the entire solution set.
Birne Binegar: Okay, so what I want to do next, and then is introduce the last topic that's going to be covered, on the first exam and that's an notion of in verses of major cities.
Birne Binegar: And we should be able to get through most of this in a single lecture today.
Birne Binegar: here's the idea which is really simple.
Birne Binegar: suppose I have a n by n linear system, so I have the thing number of equations that I have variables and I write it in matrix notation so this would be an n by n matrix times and by one column vector is equal to an end by one column vector of numbers.
Birne Binegar: and
Birne Binegar: The idea here is that if we had a matrix or call it see such that see multiplying this coefficient matrix produces the invite an identity matrix I have another way of solving this linear system without using any reduction.
Birne Binegar: All I have to do is take this given matrix see and multiply both sides of this equation, I see.
Birne Binegar: one caveat, though, when you have a matrix equation, and you multiply both sides by a matrix you have to make sure that you multiply from the same side because.
Birne Binegar: A times B, if not equal to be times so if you multiply on one side of the equation you use see us as a left side factor and on the other side of the equation you use it as a right side factor.
Birne Binegar: That that's not allowed because you're doing a different operation two different sides of the equation.
Birne Binegar: You have to do exactly the same thing to both sides to the equations if you're going to maintain the quality, what I do if I multiply both sides of this equation.
Birne Binegar: From the left by this matrix see I don't put the sea it's easy to write it here and right here, but that would be incorrect, you have to write it on one side and then use the same side on the opposite side of the equation right, so we write that and then.
Birne Binegar: This multiplication this is so, she did so I can group this a little bit differently, instead of multiplying eight times X and then multiplying by see I can multiply two matrices together first.
Birne Binegar: Well, because i'm assuming that see times a is equal to one this product becomes the identity matrix.
Birne Binegar: And the identity matrix times any matrix or vector just reproduces the matrix or better you started off with and Lo and behold, we have we have found the solution to the original linear system.
Birne Binegar: Simply by doing a matrix multiplication by a particular matrix This, of course, is completely analogous to what you do in high school algebra.
Birne Binegar: If I want to default.
Birne Binegar: to accessing equal to six, what I would do is I would multiply both sides of the equation.
Birne Binegar: Maybe by one half and then I get an equation that says that X is equal to three right that's your usual way of solving a single equation is single unknown.
Birne Binegar: And so what this is telling us is that back, we can solve linear systems at least the matrix version of those in your systems, we can solve those.
Birne Binegar: effectively.
Birne Binegar: The same way that you you're you're accustomed to solving linear equations with only a single variable you just got to figure out.
Birne Binegar: What the inverse of a is analogous to one half being the inverse to you got to figure out what that is but, once you figured out what that is then you can.
Birne Binegar: quickly get to the solution, simply by identifying the inverse of a and then multiplying both sides of the equation by that inverse.
Birne Binegar: Okay, so that's that's the scheme we're going to pursue now and the principal problem we're going to face is well here I pulled this matrix out of a hat I just said, I had a matrix see with this property our goal for today is to figure out how to find such a matrix okay.
Birne Binegar: Alright, so.
Birne Binegar: let's start with a definition.
Birne Binegar: Let a be any square matrix with the same number of roses, we have called then anytime we have a matrix see such that only multiply.
Birne Binegar: A from the left by see we get the identity matrix and when we multiply a from the right by this matrix you can get the identity matrix we're gonna call this matrix see a matrix inverse of the matrix a.
Birne Binegar: Alright, so with this definition as regardless fish and see is going to be a matrix defined by these two properties.
Birne Binegar: So let me show you an example here some two by two matrix ease.
Birne Binegar: and
Birne Binegar: If I multiply.
Birne Binegar: A and multiplied by see from the ripe it multiple a by see from the right, I get this product and matrices memory, the way you compute matrix products, as you take dot products.
Birne Binegar: Have rows of the first factor with palms second factors that computation produces the two by two identities and, similarly, if I multiply the given matrix a.
Birne Binegar: From the left by this matrix see you do the calculation, you also get the identity matrix but notice here.
Birne Binegar: Unlike.
Birne Binegar: What happened here in this.
Birne Binegar: usual college algebra type calculation is that it get rid of that too, I had to introduce fractions notice the inverse of this matrix a doesn't involve any fractions now that's not always the case, but it does show you that fractions are not required to do matrix division yeah.
Birne Binegar: Well that's hard to prove I don't prove it in this course in an advanced linear algebra course I would prove that in fact this equation this equation implies this equation, and vice versa, so, in fact, these two.
Birne Binegar: You can prove that, in fact, these two are not separate conditions there are single condition, but it's hard to see at this stage of the game, well, I guess the equation implies this equation, but it is true and.
Birne Binegar: it's just that the yeah it's just that it's not the proof it's not that hard, but you basically have to understand how to find the inverse of the matrix and then then proving that this implies that are This implies that that's easy easily done.
Birne Binegar: Basically, so you have a unique solution and so yeah yeah it will work out but don't worry about so much that it will turn out that that Islam, if you can satisfy one of these conditions, then you'll have the inverse.
Birne Binegar: And if that we're going to use that.
Birne Binegar: But let me show you some quick properties easy properties and matrix in verses and then i'll show you explicitly how we actually calculate these indices.
Birne Binegar: Alright So the first thing is.
Birne Binegar: The way I look up the definition on the preceding slide I defined.
Birne Binegar: And inverse rather than the inverse of the matrix but what i'm showing you i'm telling you now is in fact there's only one in birth to a matrix if if if there was any at all.
Birne Binegar: So what I want to show you, you said, if you have two major C C and D that have that same property whether I multiply from the right or left.
Birne Binegar: multiplying a by see always produces the identity matrix and similar for this matrix D what i'm going to show you that, in fact, if these two equations are true, then it has to be that the matrix see is exactly equal to the matrix D.
Birne Binegar: Alright, so here's how you do it we've got three matrices ABC and D so let's look at the product of all three times eight times see.
Birne Binegar: Now we have a couple of different ways we can compute that first of all, we can multiply eight times see first and then followed up with multiplication ID what will happen.
Birne Binegar: Then, is that will effectively have the time to identity matrix just a times sees equal to the identity and the identity matrix times anything just producers, what you started off with, so the AC that can be reduced to the matrix D.
Birne Binegar: But we could have choose the initial grouping differently to compute this product, I could also compute.
Birne Binegar: This product first and then multiply by see but I multiply it times a I get the identity matrix the identity matrix time to see is equal to see.
Birne Binegar: And so now, if you just sort of extend the equal side gic is, of course, equal to DC and you taste this around see is equal to da C and D Ay Z is equal to D, in other words, he is equal to see so.
Birne Binegar: We have.
Birne Binegar: If we have an inverse to a matrix In fact we only have one choice there's only one matrix that will satisfy.
Birne Binegar: One unique matrix that will satisfy both these properties.
Birne Binegar: Okay, so that's going to be helpful, so that means that if we find any matrix such that, when we multiply a given matrix a buy it and we get the identity matrix you know, we have the inverse all we need do is just verify that the product of a in its inverse equals the identity matrix.
Birne Binegar: And then here's another simple property that actually we can prove.
Birne Binegar: If you have to make to seize that have in verses and you multiply them together, you get a new matrix what is its inverse well this theorem or.
Birne Binegar: The statement here tells you that the inverse of a product of to invertebrate matrices is the product of the end versus but notice that the order of the product is different.
Birne Binegar: than of having a time to be worth as the left hand factor at the minus first on the right hand side the right hand factor.
Birne Binegar: So what you have to do if you want to compute the inverse of a product of major cities, you have to take the inverse of the individual factors and then write that product in the reverse order.
Birne Binegar: Let me show you why this works.
Birne Binegar: Alright, so I claim that this matrix is going to be the inverse of a B so let's calculate what happens when I multiply a B times this particular matrix.
Birne Binegar: Well, you always have the freedom to carry out the multiplication in any particular order you want, so what i'll do is instead of calculating at first, then calculating this matrix.
Birne Binegar: i'll just grip the product differently now be time be the miners First is the identity matrix then this again if the matrix could be absorbed in either these factors and you're left with eight times eight and my first with the identity matrix what i've shown you is that the matrix.
Birne Binegar: That when you multiply.
Birne Binegar: The MIT a product matrix ad by this particular matrix you always get the identity matrix because identity matrix these are unique that's how we can include that this guy is actually the inverse of this product.
Birne Binegar: This proves it because a time this produces the identity matrix.
Birne Binegar: Okay, so frequently that's, how will verify that we've got the identity matrix We just have to make sure that the product, the appropriate product rule.
Birne Binegar: matrix on the inverse is equal to the identity matrix once you confirm that that's true, then you've got the inverse.
Birne Binegar: Okay, and then stressing there's only one inverse to a matrix until be might be the miners first at the miners First is the unique inverse of the matrix a B, but.
Birne Binegar: We could also say.
Birne Binegar: I could also say that.
Birne Binegar: The inverse of the day.
Birne Binegar: is equal to a the minus first be the miners first right you take the inverse of each factor, but you also reverse the order.
Birne Binegar: Okay.
Birne Binegar: Alright, so that's.
Birne Binegar: Nothing nuff of an introduction to what.
Birne Binegar: In verses of matrix needs are and part of what they'll do for so help us solve linear systems about doing role reaction Now I want to show you how they got to calculate them so sorry.
Birne Binegar: screen split.
Birne Binegar: So what i'm gonna do is i'm going to show you how you calculate the inverse of the matrix by sort of an indirect argument i'm gonna have to make a.
Birne Binegar: digression into some objects that we call elementary matrices so let me define those guys and then i'll show you their utility if you start with the an identity matrix with an rosen and calm and let the script RD any elementary row operation.
Birne Binegar: or elementary row operations are things like.
Birne Binegar: Switching two rows.
Birne Binegar: So that may be the first one, second one thing we could do we could send a row to multiple of itself.
Birne Binegar: or the third thing kind of thing we could do is, we can take a row, and then we could replace it by.
Birne Binegar: It some with a multiple of another row alright So these are the three elementary real operations that we use to reduce matrices that's what i'm talking about here this script art can be.
Birne Binegar: Anyone to be three kinds of operations.
Birne Binegar: And what i'm going to do is i'm going to do that real operation to the identity matrix to create a new matrix i'm going to call this matrix boldface he subscript are as the elementary matrix corresponding to the given row operation.
Birne Binegar: let's do some quick examples here's a two by two I identity matrix.
Birne Binegar: Is a row operation, where I switched row wanting row to if I do that to the identity matrix I get a new matrix it's not the identity matrix any longer it's rather it's called the identity, make the elementary matrix corresponding to this row operation on two by two matrix sees.
Birne Binegar: If I re scale a row I can do that operation to the identity matrix I get a new matrix it's the elementary matrix corresponding to this row operation.
Birne Binegar: and, similarly, for the third type of elementary row operation, where we place a row by it, some with a multiple of another row, you can also do that to the identity matrix get a new matrix the new matrix is called the elementary matrix corresponding to that role operation.
Birne Binegar: Alright, so I have a way of making a bunch of new matrices from the identity matrix here's the point about.
Birne Binegar: If you have an elementary matrix corresponding to a real operation, then you actually have two different ways of implementing that row operation on a given matrix a you can either operate directly on it, or you can do matrix multiplication I the corresponding elementary matrix.
Birne Binegar: So on the right hand side.
Birne Binegar: If i'm saying.
Birne Binegar: Applying this CERT operation, what I would do is I would take a row of a and I would add, replace it by some of a multiple have another rub.
Birne Binegar: or on this side, what I can do is I can do that same operation to the identity matrix get a new matrix but instead of offering directly on a i'm simply going to multiply it by this elementary matrix.
Birne Binegar: The claim that the theorem is no matter how you carry out that these operations are different on either side of the equation, but they have the same result.
Birne Binegar: So the upshot of this is that elementary real operation can either be up, you can either implemented directly on a matrix or like doing that real operation to the identity matrix and then doing a matrix multiplication.
Birne Binegar: We show you explicitly with some examples.
Birne Binegar: How that works out.
Birne Binegar: Alright, so here's a given two by two matrix for reference.
Birne Binegar: The two by two identity matrix is this guy.
Birne Binegar: What i'm gonna do is i'm going to.
Birne Binegar: Do this role operation first aid directly.
Birne Binegar: To layer it is but a second thing I can do is I can do this row operation why switched the two rows I can do that directly to the identity matrix So when I switched the first and second rows of the energy matrix I get this guy.
Birne Binegar: Right now, now now let's having compute this guy let's see what happens when I multiply hey from the left by this.
Birne Binegar: Elementary matrix corresponding to the original elementary role operation well you compute and notice the two by two matrix it's kicked out by this matrix multiplication coincides with the matrix I got from a by doing that row operation directly.
Birne Binegar: So I built least verify for one particular case.
Birne Binegar: I guess this row interchanges that in fact.
Birne Binegar: I have two different ways of implementing a row interchange either directly on a matrix or by multiplying the matrix by the corresponding elementary matrix.
Birne Binegar: Alright let's do a couple more example, like you, a row rescaling.
Birne Binegar: Is what it would do to that matrix a B, C D i'm we scale, the second row by three we scale, the second role at the identity matrix by three I get this matrix.
Birne Binegar: And then, if I take this matrix and I multiply the original matrix by it Lo and behold, I get the same result as before, maybe three see 3D and it'd be interesting to be so indeed.
Birne Binegar: Operating directly on the matrix a.
Birne Binegar: That calculation yield the same result as if I operated on this role operation on the identity matrix and then multiplied a from the left by that elementary matrix.
Birne Binegar: And there's one more example I can do this is what.
Birne Binegar: If I replace the second row with it, some with multiple the first row, this is what happens.
Birne Binegar: If I do that do the same operation to the identity matrix I get this guy and then.
Birne Binegar: You then multiply a by this elementary matrix you get it back exactly the same matrix as if you would apply that elementary rule operation directly gay.
Birne Binegar: Right, so what i'm doing here is i'm not making with the introduction of elementary matrices i'm not making the calculation faster i'm slowing it down right.
Birne Binegar: This is the easiest thing to do just operate directly on the matrix to find out what happens when you do this elementary row operation, the slower way to do it would be to do this role operation on the identity matrix and then do a matrix multiplication to see what happens today.
Birne Binegar: But speed is not my point here my point here is to.
Birne Binegar: replace this algorithm that sort of hides what's going on and replace it with something that looks a little bit more explicit that.
Birne Binegar: algorithm The thing that we do to a can be implemented the matrix multiplication and that's the essential idea and purpose of these elementary matrices not to simplify calculations, but to have a different way of interpreting these elementary row operations.
Birne Binegar: Okay, so.
Birne Binegar: That idea, hopefully in the back of your mind, let me show you now how you can calculate the inverse of a matrix.
Birne Binegar: All right move this again see if we get things on screen alright so.
Birne Binegar: My hypothesis on this slide is I my given matrix can be reduced to the identity matrix.
Birne Binegar: Right, so what that's going to mean that means that I start with a matrix a and I start doing elementary row operations some sequence of operations of those forms that eventually I get a doobie the identity matrix.
Birne Binegar: that's all tied up in this this hypothesis that there's going to be a sequence of row operations that transform a into the corresponding identity matrix.
Birne Binegar: Right, so you have something like this setup all right now here's where the elementary matrix she's coming.
Birne Binegar: When you look at this line, I have to perform one row operation then perform another row operation on the result until Finally I reach the identity matrix.
Birne Binegar: So this line is spoke to depict this process by which a sequence of elementary row operation changes, a given matrix to the identity matrix.
Birne Binegar: What we've seen in the previous preceding slide is that there's no alternative way of implementing these elementary rule operations, we could do it via a matrix multiplication as well, but now look at this.
Birne Binegar: We have a certain matrix just multiply these things together times a produces the identity matrix well, that means.
Birne Binegar: That.
Birne Binegar: Whatever matrix results from this product of elementary me sees.
Birne Binegar: It times a produces the identity matrix and so, in fact, what's enclosed inside the parentheses, there has to be the inverse of a.
Birne Binegar: Because anytime you have a matrix times another matrix producing the identity matrix, then the matrices heart in verses of one another.
Birne Binegar: Okay, so Oh, I guess it maybe it's not showing it the bottom of this screen there, but basically what we show them is that if you can reduce a matrix to the identity matrix then.
Birne Binegar: Take the corresponding sequence of row operations.
Birne Binegar: convert those.
Birne Binegar: reinterpret those rule operations of multiplication by the corresponding elementary matrix that we can once we know the intro operations.
Birne Binegar: We can do the same thing to the identity matrix and get a bunch of matrices and then what the shows is that back this product of elementary matrices has to be the inverse.
Birne Binegar: Of the original matrix so we have a at least now a procedure, a formal procedure at this point that tells us that how we could calculate a the miners first we figure out how to roadies at the identity matrix take those.
Birne Binegar: Elementary row operations in the same order.
Birne Binegar: Instead of applying them to a apply them to the identity matrix and then multiply everything together and that will produce the inverse of a.
Birne Binegar: Right and so.
Birne Binegar: What do we have so fly.
Birne Binegar: You have a sequence at elementary row operations, once you have that sequence you construct the corresponding elementary makes these, and you can construct that product.
Birne Binegar: All right, but that's.
Birne Binegar: that's not very efficient i'll show you how you can do it really quickly without having to figure out all these individual elementary matrices.
Birne Binegar: Alright, so what i'm going to do something similar to what we've done before i'm going to expand my coefficient matrix hey.
Birne Binegar: or this matrix a i'm going to augment it with some additional columns what i'm gonna do is i'm going to put the N by n identity matrix i'm going to just put it off to the right of the original matrix that I have.
Birne Binegar: This matrix here is going to be called an augmented matrix it's going to be, at the same number of roses a but the number of columns has been doubled.
Birne Binegar: But what's here is the identity matrix and what's here is the original matrix day.
Birne Binegar: Alright, so let's again assume that a can be reduced to the identity matrix.
Birne Binegar: Well let's do the same sequence of row operations and instead of operating on the matrix a let's use them on this make this augmented matrix a or I have joined with a.
Birne Binegar: Then, what will happen is these role operations remember those raw operations, the calculation to perform a column by column you're adding.
Birne Binegar: A row to another one, that means you're adding a component to it like component and if I add in more components it's still the same operation, so if I interchange two rows I can do that for a or I can interchange the same two rows of this augmented matrix.
Birne Binegar: So what will happen is when I apply the elementary row operation to this augmented matrix it's going to have the same effect on both sides.
Birne Binegar: The first row operation it's going to be for not only on this half but it's also going to be born on the right hand side and so will subsequent operations.
Birne Binegar: Our hypothesis is that we can reduce at identity matrix so this sequence of row operation supply today is equal, the identity matrix.
Birne Binegar: But we also have the same sequence of row operation now being applied to the identity matrix.
Birne Binegar: Well let's write that now i'm just taking a take this left hand right hand factor write it out and then each row operations, they could be implemented the matrix multiplication i'll just.
Birne Binegar: Take the operation and convert it to the corresponding elementary matrix and rather than doing these directly on I.
Birne Binegar: will carry out that these operations by multiplying by the corresponding elementary nature sees.
Birne Binegar: And what happens in when you look at that what's going to get kicked out it's exactly that product of elementary makes you sees that we already have identified is the inverse.
Birne Binegar: Of a the miners first these guys take a to the identity matrix and so their corresponding products it's going to have elementary matrices it's going to be the inverse of a.
Birne Binegar: Okay, so.
Birne Binegar: here's what we can conclude from this discussion, if you can reduce a matrix a to the identity matrix then.
Birne Binegar: When I joined the identity matrix a it's going to reduce to well as still going to be reduced to the identity matrix but what's going to happen to this slide it's going to become exactly at the minus first.
Birne Binegar: Okay, well, I have a minute left so.
Birne Binegar: here's the conclusion, if you can reduce at the identity matrix then by taking this matrix and row reduce to get to which reduce relational on form.
Birne Binegar: you're all what's going to be left hand side is going to reduce the identity matrix and the right hand side what's going to happen to that it's it's going to be.
Birne Binegar: Whatever whatever happens to this left hand side under the row reduction that takes eight I.
Birne Binegar: I hear it's going to get changed over to a the minus first so Whatever happened to.
Birne Binegar: The right hand side of the augmented matrix it's going to be the the inverse of a let me state this as a theorem and then i'll let you guys go we'll pick up with this on Wednesday.
Birne Binegar: If you have a square matrix and a can be reduced to the identity matrix, then the way you calculate this.
Birne Binegar: This inverse of Asia, reduce this matrix to which we do show Echelon form that will produce.
Birne Binegar: The identity matrix on the left hand side and what will be on the right hand side is now your inverse, on the other hand, if a cannot be reduced to the identity matrix then that matrix a is not going to have an inverse at all.
Birne Binegar: I thought stop here and we'll pick up with this discussion on Wednesday.