 So, as I know, last time, in case you didn't notice, the recitation moved from far away there to close up here on the subscript, so that's where we meet. So are there, are there, are people confused about how to learn things like, I don't know, find a distance between a point and a plane, or find a plane containing three points, or even those kinds of things? I don't move on. I'm happy to move on. So, does anybody have a point of confusion they would like me to clear up? How do you start the homework? How do you have trouble with the homework? Okay. You think, oh, it's good. So, is there something I can clear up? Okay, first it was like I, K, G, L, and you were confused. Okay, that's fine. I mentioned it briefly, but briefly only. So, if we have, say, in the plane, we call this guy I, we call this guy J. If we're in three-space, we'll call this guy I, this guy J, and this guy K. So, I is a one in the first place, and then as many zeros as you need to be the right dimension, J is a zero in the first place, and one in the second place, and as many zeros as you need to be the right dimension. Zero, this is not the same I as the square root of negative one. K is zero, zero, one. And if we're in a higher dimension, we're going to run out of letters pretty quickly. So, this is also called E1, and this is also called E2, and this one's called E3, and if I'm in a five-dimensional space, so in general, EK, well, okay, so this I is a different I from that I, how about EN? It's going to be all zeros, a one, and all zeros, where this is in the end place. So, the EIs are the standard basis vectors for R, M, where we put this is the I chord, the end chord. Yep, other questions or confusions? So, here's some play, let's say it is, I don't know, 3M plus 2Y minus C equals 2, does that work? Let's see, sure that one will work. And I have some point, I don't know which side of the plane it's on, I don't care. One, two, three, that better not be on the plane, let's see. 3 plus 4 is 7, minus good. 7 minus 3 is 2, okay, good. Okay, so I want to find the distance, so distance, here, let me go blanking it. So I want to find the distance between this point, so that's the problem I want to do. Okay, so really you know all the stuff you need to know to do this, you just need to realize that you know. So, usually when we're finding the distance between two things in this class, we're going to find a pair of vectors and project one onto the other. Now we just have to think about what are the right vectors that we want, and then we'll compute the projection of one vector onto the other. So there's sort of a vector naturally associated with this plane, I would ask her, but it's okay. Right, so the normal vector associated with this plane, and let me put it here just to draw something. There's the normal vector, let's call it n. And what vector is it? Do you see what vector it is having? If you don't. Okay, you don't think so. Okay, so, I mean some people do, and so the reason I'm asking hers, because she asked the question, so does somebody want to tell me what it is? Yeah. Go ahead. 3, 2, negative 1. Alright, so 3, 2, negative 1. Why is it 3, 2, negative 1? It's 3, 2, negative 1 because this plane is the set of all the points that look like some vector n dotted with, I don't know, p minus, what do I call it? p minus v equals 0. Where p is some point, let's call this guy p, p is some point in the plane, and it's a collection of all the vectors v here in the plane. So if I take the dot product of v minus p, well, really this is v, if I take v minus p and dot it with n, they have to be at a right angle. Because the plane is, I tell you which way it is and where to start it, and I want all the stuff that's perpendicular to it. So you can just read off from this this 3, 2, negative 1. That's what I'm dotting with my arbitrary vector v, p minus v to get it. Okay, and what's a point in the plane is anything that satisfies this equation. So since there's a 2 here and a 2 here, why not make this be a 1 and make those be 0s and then it's easy. So I can take 0, 1, 0. That satisfies the equation. I can also take anything else that satisfies the equation, but this won't work. So I can also take 0, 0, negative 2, or lots of choices. So now we have this vector p, this point p, this vector n here and this guy 1, 2, 3. Well 1, 2, 3, if I put the origin here, is this vector and that's not really of much use to me. I want a different vector. What vector do I want? I'm just looking for people whose names are there. What's your name? Sam. Sam. You can get the vector that's the difference from p to 1, 2, 3. Right. So if I start using colors, if I take this vector from p to 1, 2, 3, then what I want is this perpendicular distance here, which is the same distance if I make a rectangle there. And so now this is starting to look like a problem that you've done before. I have the red vector, let's make n a little shorter just to emphasize it, and I have the black vector n. And what I want is this green link, which is the projection of the red vector in the direction of the black vector. So that'll be my distance. So my distance is the projection of my red vector e is 0, 1, 0. So 1, 2, 3, minus 0, 1, 0 in the direction of 3, 2, maybe 1. Yeah. Wouldn't it be similar since you already know the normal vector? Yeah. Would you just make a line using the normal vector from the point you're given and then solve for where that line crosses the plane? Would that be simpler? I think that would be harder. Okay. Maybe it would be simpler. That's okay. I can certainly find a line that goes through 1, 2, 3 in the direction of n and then find out where that line crosses the plane. Which is a little tricky but still doable. And take the distance between those two. But this is actually a very easy calculation. Once you sort of conceptualize what's going on, right? I just subtract this minus that. That was easy. And now I'm going to take the dot product and dot it with, I didn't give him a name. I call it r. And dot it with r for red divided by the length of n. That's the length I want. So that's really easy because n dotted with this guy, that's r is 3, 2. Well, this subtraction is 1, so that's 2. And then negative 1 plus 3 is 2. And on the bottom, I have the length of n. The length of n is the square root of 9 plus 4 plus 1. So the distance is 2 over the square root of 14. So the distance is 2 over the square root of 14. Yeah? Okay. So we have to remember how we define a plane. One way to define a plane is we can do it parametrically where we define two vectors in the plane and we take any combination of them. The other way which is a little more familiar is to say it's everything that's perpendicular to some vector. And here it's just B-scoordinates in front of x, y, and z. Because this is really, so we do that over here, so a plane we can define as either some vector p which is a point in the plane plus t times v1, well, let's call it v, plus s times w where p is in the plane, v and w are non-linearly dependent. They're non-parallel in the plane. So this is not the useful formulation, although we can turn it into one. We're just saying start here at the origin, go to p, and then look at v and look at w and consider all the combinations of them. But the other way that is more useful to us in this context is we think about the normal vectors here n and we say it's all the guys who sit in the plane, let's call this guy, I don't know, s. It's no, I already used s. It's all the u so that n dotted with u minus p is 0. It's all the u's here so that if I look at the distance, I mean I look at the vector who lives in the plane between p and u, it's at a right angle and it's all those same guys. And in this case, these guys are of the form, these guys are always of the form, let me write it this way, x minus x naught, y minus y naught, z minus z naught, that's these guys. And n is something, a, b, c. So the dot product goes to 0. So that means it's something that looks like a, x minus x naught, plus b, y minus y naught, plus c, z minus z naught, is 0. And now if I gather all of the non-variables together, I get ax plus by plus, is this visible? Is it not really? Then that will fall apart, so that's what I can do. I'm going to give ax plus by plus c, z equals some number of k, where this k is negative, is ax naught plus by naught plus c, z. So just thinking about how planes work, that tells me my normal is a, b, c. So here I have the equation, 3x plus 2y minus c, that tells me that the normal vector is 3, 2, and 1. So that was a long trip to get somewhere, but I think it's important that you guys clear on stuff. In this one, this b is just from the origin to somebody in the plane, which did I call it v here? Here I called it v, just anywhere. So when you try and describe some object, it's all the collections of things that have a certain property. This is what we do a lot. It's all the vectors who, when I take a dot product with n, they're perpendicular. So it doesn't matter what you would have subtracted. So in this problem, I should get the same answer if instead of choosing the point 0, 1, 0, I came over here and chose the point, oh, I've lost the equation, 0, 0, negative 2. I should get the same answer. So if you do it with the, now of course, this red vector will be over here, so I'll be projecting a different vector onto the normal. But I should get the same answer because this plane is described independent of the base point. It doesn't care what point p I choose as long as it lives in the plane. So it has to satisfy this equation. There was nothing magic about the fact that I used 0, 1, 0, other than it was easy to see. 3 half is 0, 0, or it could be 1, 1, 3. Lots of choices. It's not very relevant to what I'm trying to write. Given the two points, and that's three-dimensional, how do you express them? Okay. So, parametrically, I mean, vector, vectorially? Sure. So here's two points. One, two, three, four, five, six. No, I mean, I mean, functions. Yeah. Express them. What do you mean functions? You want coordinates like x, y, z, just like x, y, z. So those are annoying, because you need three equations. And so usually people write something like x minus x naught over a equals y minus y naught over b equals z minus z naught over c. It should be because two dots can determine a line. Yeah, so this works. So pick a point x, y naught, and z naught. One point on the line. And a is actually x1 minus x naught, y1 minus y naught, z1 minus z naught. So, but this is really two equations, right? This is relating x and y and y and z. Pick two, and then I get two equations. So this is two equations and three unknowns. You can write it this way. And this is really, no, I think it's stupid. But people do this. And in some, I mean, when I took multivariable calculus back when there were dinosaurs and stuff, we used this form a lot. I think this form is basically stupid, but it's fine. Because really you're describing a plane and a plane. But I think the other way is sort of more natural. But it's okay? Yeah. When you have the vector that has ABCs in it, is that the vector to the point on the normal vector that intersects with? No, the ABC there is, I think, parallel. Is the vector. Did I get that right? I think this is right. What is the vector? The vector along the line. Along the normal vector? So, okay. So here we have a point. X1, Y1, Z1. And here I have a point. Well, I call it X0. X0, Y0. And I want to consider everything along this line. So, sort of a natural geometric way in my brain to do this is to think about this vector and just say I take X1, Y1, Z1. And to it I add any multiple of X1 minus X0, Y1 minus Y0, Z1 minus Z0. That's sort of the parameter. Do you have a question? Yeah. I was talking about the example before that. Oh, okay. Well then, I'll stop that. So what is the vector with ABC? Here. This ABC, that's the components of it. That's the vector to the point on the normal vector. That's the normal vector. The normal vector. It's not a point on the normal vector. It's the normal vector. Right? So for example, if I take the plane who has normal vector 1, 1, 1, passing through the origin, that will be the plane X plus Y. So the normal vector is 1, 1, 1. The plane X plus Y plus Z equals 0. 1, 1, 1 does not satisfy this equation. So it's not a point on the plane, nor is it a point. Okay, yeah. When you define a plane by everything perpendicular to the normal, when you're writing out mathematically, you're like the normal stops right where the plane starts. Right? It touches the point on the plane, like how do you define it on R1? A normal is a direction. Okay. A direction with a magnitude that we don't care about. Okay? It's a ratio. It's a pair of slopes, if you will. Written in a convenient form. So it's just saying, here's my plane, and that's the way you go to get away in the fastest way you can. Yeah. Now, that may or may not have a lot to do with where the origin is. If I'm talking about the origin, I just mean to think about everything perpendicular to the normal. Yeah. Right. That's why we have a point. Okay. That's why I have this other, this number k, sliding up and down all of those infinite series, that infinite collection of planes. So k is literally like the distance along the normal. Okay. Sure. I mean not really, but it's closely related to that. That's here. It's off by a second. Does it shift the plane up and down? The plane shifts the plane up and down, so what you should think of in terms of this number. If k is zero, this plane contains the origin. Yeah. If k increases, this plane moves away from the origin in the direction of the normal vector by some amount related to k. Okay. And if k decreases, then it moves towards, well, moves the other way. Yeah. Which decreasing negative moves away in the other direction. Right. And you should think of this, this part as describing a stack of pieces of paper plane things and k is choosing which one. All right. Which is the same as choosing these points. That's not why you're not seeing them. Okay. Yeah. After you long make the plane green and you find that if you want to find the distance between that point, yes? After you long make the plane green and you find that you're calling the point and you want to have these things. I don't know what that means. Normalize the plane and plot it in the order of the points. The AX minus B by plus three B. Yes. Normalize the use of A. Make sure that that's a unit vector. Yes. Okay. Sure. That's why. So if you do that, if you make sure that the length of your normal vector is one, then when we are calculating this dot product, the length of n is one. This is a one on the bottom. And so we're taking a unit vector dotted with some r. So sure. Because it's a unit vector. Because when you divide by one, not much changes. Okay. Yeah. Yeah. The number below the plane doesn't lie on the same side of the plane. Right. Yeah. Well, so what is above or below mean? So that's sort of the same as this question here about the k. If I plug in zero, where will it go? So the normal is going to define, if we consider, if you consider the normal vector as describing a line through the origin, it's zero, zero, zero around that. So consider the line in the direction of the normal with zero, zero, zero around that. Then, so now I'm going to consider, I don't remember if this is how it's phrased, this is p up. Because there's no sort of natural up. Right? Except for maybe think, well, did they actually find it as in the y-axis? Right? I forget. I don't know. Anyway, do you have to pick some notion of up because I don't know what above means. I do know what on the other side of the origin means. And so the question now is, is p here or is p here? Well, to decide whether p is here or here, well, should have something to do with looking at these vectors. Okay, so I'm going to move on because I think we've worked up about half the class already. No, not quite, only a third. A little more than a third. Is this not what I'm talking about today? And in fact, what I wrote in my notes is, remind people about the cross product. Yeah, I did an excellent job. So, you know, I know. So I'm going to actually, I think, skip over that. So get Albert to remind you about the cross product. I'm going to move on, but we can get too far. What? I think. Arthur. Yeah, Arthur. I'm sorry, Arthur. I just met him two weeks ago. I don't know what I'm talking about. I don't know what I'm talking about. I just met him two weeks ago. Get Arthur to remind you about the cross product. I think it shocks me. Yeah, I can use that name, too. That would make you remember, I guess. Anyone? Okay. Change gears a little bit, except it's not really changing gears, but it may not be clear that we're not changing gears. Does that make any sense? No. What I want to do is talk about systems and equations and solving equations, which, on the face of it, doesn't seem to have a lot to do with this. So if I have... Well, let's do one here. If I have, like, x plus y equals one and two x... I don't know why I'm using the one that I have. If I have this system of equations, probably you all know how to solve this. Totally. I hope. So what you do is you add them together, and I get three x equals six, and so x is two, and so now I take it. x is two, and I go back here, and I see that two plus y equals one, so that tells me that y is negative one. And I screw it up, and that seems to work. And... So that's it. That's a good thing. In terms of line, we're saying we have the line x plus y equals one. Do you have some line? Do that. Two x minus y equals five. This is just sort of a sketch. It's already a wrong picture, because y equals negative one. Two x minus y equals five. And we're finding this point. But you might ask yourself, what this has to do with this? It's clear what this has to do with this. Or in particular, what does the equation three x equals six have to do with this picture? Let's put that on hold for a minute. And then, similarly, if I go up a dimension, and I have, and since I had one worked out, it doesn't really matter, x plus y plus z is plus y plus z is two. Two x minus two y plus z is one. X plus y minus z is minus four. And again, my goal is to solve this system of equations. Is there anyone here who doesn't know how to do that? Anyone here who doesn't know how to do that? Anyone here whose hands actually... doesn't know how to do that? So let me not do that. Your hand doesn't know how to do that. So we use the same process. We use the same trick. We add together pairs of equations to make variables drop out. So let me just do it, I suppose. If I add, well, actually, if I take this one minus this one, that would be very useful to me. So here's a, b. So if I take a minus c, it gives me zero, zero, two z equals six. So z is three. So I found z very quickly. So z is three. I could take some other pairing here, or I could just plug this back in and turn it into an equation in two variables. So z is three. Then I know that, for example, x plus y equals negative one. Z is three. This one tells me that two x minus two y equals one minus three equals negative two. Did I screw up somewhere? And x plus y, this one is the same equation. Yeah, equals negative one. So this one is of no use to me anymore. Okay. And then I just do the same thing I did over there. Here I can divide through and write this guy as x minus y equals negative one. And then I can add them together and get two x equals negative two. So x equals negative one. Did I make a mistake? No. And then y must be zero. And so my solution is the point, well, there's my solution. And what I'm doing here, one way I can think about what I'm doing to try and relate this to what we've done before is these are three planes. The three planes, I don't know if it is like that, but in general if I take two planes, they cross in a line. And then I put some third plane in here. This will cross in some other line. It looks horrible. Cross in some line. So here's where's my third plane because I've already lost it. There's my third plane. That looks terrible. But anyway, these three planes cross in a single plane. And then here's one plane. Here's another plane. Here's another plane. And this adding together of solutions is giving me some different plane that is related to the line where the two planes cross. So now probably when you thought about this kind of thing before, outside of this class, you didn't think of these as planes. These were just collections of piles of things that you're doing some algebraic stuff. But it's exactly the same as doing this geometric recombining of the planes to find other planes and so on. If you think of these as planes, rather than as lines, well, or in the case, not as lines, as equations, or in the case of lines, if you think of these lines, so this line is something like, I don't know, x plus y equals something along the slope, something like that, these guys have different slopes, so they cross. But if I take x minus y equals one and x minus y equals negative one, they're parallel, so there's no intersection. So here I have no solution and here I have one unique solution and then I have the other stupid case where I took, say, x minus y equals one and two x minus two y equals two. These are parallel, these are the same line, and so here I have infinitely many solutions. So I have this situation where I have either a unique solution, no solution if the lines are parallel or if the lines are the same, then anything on the line is a solution, so I get lots of solutions. I have the same situation here, except that I maybe have a couple of different kinds of infinitely many. Right? I could have here a situation where I have a unique solution if the planes are crossing in sort of general position. I could have no solution if the planes are all parallel. Well, I guess if two of the planes are parallel, it doesn't matter what the other one does to get no solution, so it's even if, right, I can't be on all three planes if two of them are parallel. They don't have to be parallel. What? Don't they not have to be parallel for that unique solution? Yeah, or I could have a triangle, right? I could have, I mean, they don't have to be parallel. The lines of intersections have to be parallel. Okay. So if I reduce so that the two lines where they intersect don't meet, then, but they could also be like, as he said, making, I can't draw it. That's like, no, that's good. Well, it's even a triangle, right? But they could be, right, they could be, well, whatever. They could be, yeah. We could triangular prism. There's a triangular prism and extend them to make planes. Yeah, and now it looks terrible. Anyway, we could do that. I could have a line of solutions where they all come together in some line. Again, it's really hard to draw this jump. So again, I could have infinitely many solutions where they all meet in one line. And so I have some more variations of when I get infinitely many solutions. They could all be the same plane, in which case I get a whole plane. But again, I have the situation where I want, I have not, and I have infinitely many. And this is sort of a general property when you have systems of equations, you might have a unique solution, you might have no solution, or you might have lots of solutions. And so one of the sort of natural questions is to ask, when do I have one situation and when do I have another without actually having to solve the equations, right? So here, in the dimension two case, it's easy to see x plus b by equal c a2x plus b2y equals c2. Here, if we look at the slopes, if the slopes are different, we have a unique solution. So if a1 over b1 is not a2 over b2, assuming they are better yet, a1 b2 is not a2 b1. The reason this is a better form is because maybe one of them is 0, but b1 of the b's is 0, we have a problem. So this is a better condition, then that means that for sure they cross. Because the slopes are different, so two lines of different slopes have to cross in a way for them to get away from each other. So I'll have a unique solution. If these two numbers are equal, then either they were the same line or they were parallel lines in both cases, I get either no solutions or anything, right? But this doesn't generalize so well. Just out of curiosity, under what situations would you have more than one unique solution? Never. What does that mean? More than one unique solution. Unique means there's one. I mean, right? So those are sort of contradictory. But in these cases, for these linear equations, which correspond to lines and planes and flat things in the sense of not bending rather than flat in the sense of planar, we always have one, none, or infinitely many. We can't get three. On the other hand, if we have something bent, there's a quadratic term there or a cubic term or something like that, or even an x, y term, then we can't get more. But because we're talking about flat things, two points determine a line. So if they cross, then they cross. And they can't sort of cross and then come back and cross again. Because they're flat. Can't do something like that. So I think this is really the question we're asking. When might I get two distincts rather than unique? Two distinct solutions that I can't sort of push one and get the other. And in this case, because these are linear equations, never. If I got one, I got one. Maybe I get a lot. But I either get one, zero, or lots and lots and lots. How do you, like, distinguish between the different levels of magnitude and solutions? Did he say, did he swear or did he not? They're all the same. Does he say something like that? Well, yes, there is. There is a notion for that and it's called the co-dimension. But let's come back to that when we get something. So, let me say that. Let me say that a little. Let me go ahead and say that. So imagine, I mean, here in lines, we only have a choice. There's no solutions. There's a point where there's a line for solutions. But with planes, we have the, so, right, so if I have two variables, I get none, two equations. I get none, no solutions. I get a unique solution, or I get a line's worth. You can call that infinite. If I have planes, well, then again, I have the case of none. I have the case of one. And now I have sort of two cases. I have a plane worth and here I have a line worth. This is not a technical term. Right? So either my solution consists of a line of solutions, or maybe I have a whole plane full of solutions. That is, all three equations describe the same plane. In this case, if you think about three-dimensional space, meaning I have three independent coordinates, and a line meaning that I have one coordinate, then I have sort of one other, I have two directions I can move away from the line in that are independent. The total work is what's called co-dimension two. The total space minus how many dimensions of stuff I get. And this guy, if I have a plane worth, this is co-dimension one. Is that how you store this line? Depends on who you're talking to. And whether, you know, if you're in seventh dimension, if you have seven equations and seven unknowns, it's a little easier to say, well, I have three dimensions of freedom, then I have three and four. I have three that are in my solution set, and four that aren't. So now I don't know what to say. I have a space worth. I have a four space worth. I mean, yeah. So the co-dimensional number is directly affected by dimensions you're operating with. Because a line in the plane leaves you only one dimension to play with, but a line in space leaves you two dimensions to play with. So the co-dimension is the stuff that's left over. The co-dimension, plus the dimension, is the ambient space. And this is actually a theorem in linear algebra about... I don't want to tell you. I don't want to tell you. You've got to figure it out. It's about the dimension of the image and the dimension of the real space. The dimension of the image plus the dimension of the null space is the dimension of the space. Okay. That means nothing to you. You're so good. Okay. Now, where was I going with this? I don't know. Okay. We have these equations. Now, to solidify this process that we go through, really what we're doing is we are... we combine the equate... So how about this? If I have a system of linear equations, that means I have some number of variables and some number of equations. So, like in the case here, where I have two variables and two equations, I could have then... then if I take... So given this, if I take... So if I scale any of the equations by some non-zero number, the solutions don't change. In other words, if I have 3x plus 2y equals 5, that's a system of one equation in two unknowns, this has the same solutions... Well, let's put it another way. x equals 1. This has the same solutions as 3x plus 2y equals 5 and 2x equals 2 because I scale this one by multiplying it by 2. If I multiply it by zero, then I sort of screw this stuff up. But I can always undo this by multiplying by the inverse of that and not by the plus and zero. So that's one thing I can do. And then the other thing that I can do is that I can add together... Any two equations and replace one of those two y. In other words, if I want to solve 3x plus 2y... So if I want to solve x plus y plus z equals 5 and x minus y plus z equals 8, I can add them together and get the new equation 2x plus 2z equals 13 and use that one as well. So this collection of three has the same solutions. So I can replace this one with that one. Now the reason that's true is because if the choice x, y, z or however many variables I have satisfies this one and it also satisfies this one then it will certainly satisfy their size. So these together tend to be called elementary operations. So putting these two things together goes by the name elementary operations. And if I do so theorem which I proved by mouse rather than writing but the theorem is sort of obvious applying elementary operations and the system of equations doesn't change the solutions. So this is something you've learned in... I don't know when did you first do this? I don't know. In high school or junior high school or maybe elementary school, but a while ago, you learned this although you didn't say it this way. But if I just add together things it doesn't change and I just go through this process of trying to make variables go away until I figure out which ones are which and then I have everything I need. Okay? Why does this matter? Well, so that tells us that doing this kind of process like looking at A minus C doesn't change anything and the proof that that does is you just check that you always get the same solutions but we can also sort of codify this process a little bit by... suppose I have AX plus B, A1 X plus B1, Y plus C1, Z I don't know if that's enough equals K1 A2X plus B2 Let me just do it with the actual numbers. Sorry. Because I'm not proving anything. X plus Y plus Z equals 5 2X plus 3Y plus 4Z equals 9 I can rewrite this system let's make it square it doesn't need to be square I can forget about the XY's and Z's and just think about this array 1, 1, 1, 2, 3, 4 1, 1, 2 and play with this now somehow I'll have to use this 5, 9, 1 sort of play with that as well now behind the scenes I'm thinking of this as being the X's, the Y's, and the Z's and I can really remember that I have X's and Y's and Z's by trying to somehow think of this object as a collection of vectors 1, 1, 1 and then when I do something with X's, Y's, and Z's I get 5 this statement where I just peeled off the coefficients and wrote them in an array peeled off this and wrote it in a column is really the same statement as saying the vector 1, 1, 1 dotted with X, Y, Z is 5 this dotted with that is this and 2, 3, 4 dotted with X, Y, Z is 9 and 1, 1, 2 dotted with X, Y, Z is 1 these are exactly the same statement but there's a funny kind of multiplication going on here to combine the two and it's written by the name of matrix multiplication which many of you have seen before but we multiply a row times a column and the reason we do that is to preserve this equation business I'll come back to that in a bit but now we can sort of well actually let me also just for a minute change this let me change this to instead of this equation this equation 0, 0, 0 that is I'm going to forget about the 5 and the 9 and the 1 for a minute and I'm just going to think about this guy combined with that guy gives me all zeros now if we do that then we remove one of the possibilities from the solution so I'm going to put this on hold for a minute I'm going to think about planes and lines and hyperplanes and so on notice that if I have a system of equations and I'm thinking of them as either lines or as planes or higher dimensional planes or anything like that if I have a system of equations so equations in x, y, z whatever equals all zeros this is called a homogeneous doesn't mean that it's all mixed up like milk it means that it's all zero on one side but if you think about the situation I have ax plus by plus cz plus dw equals zero then always zero is the solution in this case again this is linear equations this means always the origin we always have one solution only the origin we rule out the case of the equations being inconsistent we might have infinitely many solutions we might have a plane of solutions or a line of solutions or whatever the solution case because we just made it we made them all pass through zero and that seems like I've lost something but I've gained something and what I've gained is I can just think about coefficients in terms of lines the question is are they the same slope or are they different slopes in higher dimensions so I can't use the same slope different slope stuff how many people already know this when I'm talking about so those of you that have linear algebra then probably know this better but even if I have maybe you get this in regular algebra I've lost track of where I am so if I pass into a homogeneous system and then I'll come back to the end homogenous case soon that means that with that equation I just have to look at one, one, one, two, three, four one, one, two and do stuff to it and what I'm going to do is something that goes by the name Gaussian Elimination I'm just going to try and get rid of columns to get ones in the front what I like in the end to get to something so it looks like one lots of zeros something like that I'd like to be able to do stuff to it to get to this situation because this is telling me if I do this again if I'm thinking in terms of X's and Y's and Z's I'm saying X plus Y plus Z 2X plus 3Y plus 4Z equals something and when I transform it to this I'm saying X equals something I know that's what 1, 0, 0 equals whatever will mean something I know so to solve the equation is to go through the process of transforming this leading matrix and keeping track of that 591 that goes with it and that will give us the solution if I'm doing the homogeneous case then I don't care it's just zeros they're always going to be zeros if I can get to the identity then there's a unique solution and less than the identity then I have at least a line's worth or maybe a plane's worth or maybe a higher dimension so if the point of this is lost on people they're all pushing at me very intently feel nervous okay so that process here let's just see if we can get from this guy to that guy I just start with I take this guy from this guy and that will give me I'll leave the 111 here I'll put the 5 here too I'll put the 5 there and I'm going to subtract twice this from that so that will mean I get a 0 and 1 and 2 and twice 5 subtract it from 9 is minus 1 and here this and subtract it from that I get immediately 0, 0, 1 equals negative 4 so that tells me right away z is negative 4 so now I know what z is and then I can do the same game to try and get rid of these ones and this too by repeating this process so I can take well I can kill off this entire column by using this one to get 1, 1, 0 I don't know what I'm going to get yet I'm going to take this line and subtract it from that one so this will be 9 because I take this minus that and I'll take twice this and subtract it from that one so that will tell me 0, 1, 0 twice this from that so now I'm done oh I'm not quite done so this has told me that z is negative 4 y is 7 and x plus y is 9 and now I know that y is 7 so if I subtract this from this I'll tell how I'll get that x is 2 y is 7 and z is negative 4 and that's part of it so computers are a lot better at this than I am but this process which I mean has everybody done this before so this is called something like Gaussian elimination what? it's the vein of your existence I don't like it much sorry okay so let's think about what I just did but it's good to say I'm confused because that makes me stop so we started with this is just another way of writing where did I start this guy is just another way of writing x plus y plus z equals 5 that's this first row this says 2x plus 3y plus 4z equals 9 and this one says x plus y plus 2z equals 1 but I just neglected to write the x, the y, and the z just because I felt like it ignore this this is too confusing now what I did here this first equation is 111 equals 5 there it is and I want to make this I could have chosen to get rid of the z's or whatever but let's get rid of the x's first so this line here this is twice that now I know it's okay to rewrite this as 2x plus 2y plus 2z equals 10 because it has exactly the same solutions as that and I also know it's okay to add two equations together and then replace one of them with the sum so I added this to this that kills the x and leaves me an equation in y and z which is what I did here so this is the equation no x's plus a y plus 2z's equals minus 1 and then I did the same game on this one except I didn't need to multiply by 2 I just subtracted 1 minus the other and that one has the good fortune for me since these are the same not only did I kill the x but I killed the y at the same time so I saved the step because that just fell away so now that already told me right here c equals minus 4 because I was lucky if I hadn't been lucky I would have got a 3 and a 5 here and I would have had to do more with layers so now I've reduced the problem I'm happy with this one and since I have a 1 here I can use it to kill off the entries in the other guys because these are all 0's that's the same as when over there I found out that z was 4 going back and substituting c equals 4 is just taking this one and using it to clear out this entire column so here I used this one to clear out both of those with 0's by saying well since c is negative 4 then y plus 2z is negative 1 is the same as saying y is 7 because c is negative 7 and so that told me a new equation with no z's and that leaves me x plus y is 9 which I think when I did it by hand over there I got when I did it before I had x plus y equals 9 so now I have the simpler thing x plus y is 9 and y is 7 that would mean that x is 2 but again I did it this way if you program it's very easy to write a computer program to do this because you just take the first, take the top thing take the first non-zero entry use it to kill off all the things below it ok now you got this and all 0's then move over take this guy who is the first non-zero entry in that use it to kill everybody in his column ok move over take the first non-zero guy use it to kill everybody in his column and you just keep annihilating the column 1 by 1 until you either wind up with 1 1 1 1 equals something or if I could change this a little bit I might wind up with something like 1 0 0 equals 2 0 1 0 equals 7 no other information so if I got a 0 there and that tells me I have a line of solutions because I've got a line of no information or if I got a number here like a 5 that would tell me there's no solution because it says 0 equals 5 yeah is the correct way to think about this that you're solving a general case first on the left 3 columns and then you're applying it to a specific case on the right I don't know what correct means that is a way to think of it another way to think of it is I'm changing coordinates until everything is easy kind of the way I I'll come back to that not today probably okay but again so here this is a way to solve sets of equations and to make the process extremely mechanical it may be the bane of your existence or a puzzle well it's only a puzzle if you don't like doing nasty arithmetic right the puzzle is how to avoid multiplying by 47 12 and things like that if you don't, if you're a computer multiplying by 2 and multiplying by 47 12 kind of the same but if you're a person you want to make sure that the arithmetic is nicer to do and then it becomes a puzzle because you have to say okay I can do this and my numbers will stay whole numbers and it's like oh that way oh no I hit terrible denominators so that's when that's when it gets in right so that's when it gets in but the concept is and I'm a pure mathematician so concept is fine for me actual you know doing I'll write a computer for you and this is exactly the application of this just saying do elementary operations nothing changes same solution we can also use this to sort of determine even though I don't care what's over here whether these equations are consistent or inconsistent right if I have equations like x plus y equals 1 well let's just make them be 0 so I'm going to do the homogenous case this is a very simple case this does not have a unique solution it's sort of obvious that it does not have a unique solution because this is 3 times that this means that if I write this in the matrix form I have 1, 1, 3, 3 they're better but I have that this is a multiple of that so if I do the game where I try and get rid of something I can get rid of these 3s quite easily if it becomes 1, 1, 0, 0 I have a row of 0s I'm not going to be able to solve this so I'm left with a whole line of solutions the same thing will happen no matter how many variables I have if when I do this I eventually get a row of all 0s something bad there that tells me that I have so I have so in the 2 by 2 case this is easy to see it's a little less apparent we have 12 equations and 12 unknowns but it's doable just start going so I have the system 1, 1, 1, 1 0 0, 1, 0, 1 0 0, 2 1, 2 0 this guy oh I guess I need one more 1, 2, 3, 4, 5 well it's already bad let's put some more in here 1, 1, 1, 1 1, 0, 1, 0 ok, fine none of those are the same it's pretty easy to see since this plus this plus this is this this is not going to have a unique solution plus this equals this so this will not have a unique solution because I can very easily get all 0s here which means that one of my variables is not going to be able to be found can you say that one more time so looking at this matrix see the third row is the first row plus the second row oh ok if I add this to this I get that which is how I made it up except that I made a mistake and I wrote 0 when I meant 1 then that tells me very quickly if this is the homogeneous case I don't have a unique solution I don't have enough equations to get a unique solution because this guy is really the same as those guys a pair of these describes the same plane as this guy and certainly you can see that if I take 2 equations and 3 unknowns or 2 equations and 3 unknowns you can see I won't expect to get a unique solution at best I can hope for a line because when I combine these I can get rid of one variable but I can't get rid of both so I get a line of solutions yeah yeah if you reduce the system the co-dimension will be the number of 0s you get exactly you have this situation where when you wrote it with the coefficients on the left side and the line of the solutions are on the right side if you had a situation where the left side added up to another there the right side did not that tells you it's inconsistent there can't be 2 solutions so if I augmented this guy I'm not thinking about the homogeneous case this plus this is this I have 1, 2, and 8 this has no solutions because when I add these together I will get something that is a contradiction this has to be a 3 to expect solutions but in general yeah in general we ran out of time in general we tend to work with the homogeneous situation reduce it to the trivial case and then go to the inhomogeneous case I didn't get nearly as far as I could hope today but that's because we got a little sidetracked with the earlier stuff but I hope I clarified earlier stuff for you what should I call it now alphons so alphons will tell you something there's some other name Armando Armando will tell you about alphons in grade 17 the recitation is to do the kind of stuff that I did at the beginning of the class that I don't have a lot of time to do but apparently before recitation look at all of the homework problems that are due on Wednesday and use recitation to clarify anything about any questions you have about alphons the homework that was assigned last week there is no two days there's homework every week homework every week is due on Wednesday