1 00:00:00 --> 00:00:06 2 00:00:00 --> 00:00:06 The last time I spent solving a system of equations dealing with 3 00:00:04 --> 00:00:10 the chilling of this hardboiled egg being put in an ice bath. 4 00:00:09 --> 00:00:15 We called T1 the temperature of the yoke and T2 the temperature 5 00:00:14 --> 00:00:20 of the white. What I am going to do is 6 00:00:17 --> 00:00:23 revisit that same system of equations, but basically the 7 00:00:21 --> 00:00:27 topic for today is to learn to solve that system of equations 8 00:00:26 --> 00:00:32 by a completely different method. 9 00:00:30 --> 00:00:36 It is the method that is normally used in practice. 10 00:00:34 --> 00:00:40 Elimination is used mostly by people who have forgotten how to 11 00:00:39 --> 00:00:45 do it any other way. Now, in order to make it a 12 00:00:43 --> 00:00:49 little more general, I am not going to use the 13 00:00:46 --> 00:00:52 dependent variables T1 and T2 because they suggest temperature 14 00:00:52 --> 00:00:58 a little too closely. Let's change them to neutral 15 00:00:56 --> 00:01:02 variables. I will use x equals T1, 16 00:01:00 --> 00:01:06 and for T2 I will just use y. 17 00:01:02 --> 00:01:08 I am not going to re-derive anything. 18 00:01:05 --> 00:01:11 I am not going to resolve anything. 19 00:01:08 --> 00:01:14 I am not going to repeat anything of what I did last 20 00:01:12 --> 00:01:18 time, except to write down to remind you what the system was 21 00:01:17 --> 00:01:23 in terms of these variables, the system we derived using the 22 00:01:21 --> 00:01:27 particular conductivity constants, two and three, 23 00:01:25 --> 00:01:31 respectively. The system was this one, 24 00:01:28 --> 00:01:34 minus 2x plus 2y. And the y prime was 25 00:01:34 --> 00:01:40 2x minus 5y. And so we solved this by 26 00:01:38 --> 00:01:44 elimination. We got a single second-order 27 00:01:42 --> 00:01:48 equation with constant coefficients, 28 00:01:46 --> 00:01:52 which we solved in the usual way. 29 00:01:49 --> 00:01:55 From that I derived what the x was, from that we derived what 30 00:01:54 --> 00:02:00 the y was, and then I put them all together. 31 00:02:00 --> 00:02:06 I will just remind you what the final solution was when written 32 00:02:05 --> 00:02:11 out in terms of arbitrary constants. 33 00:02:08 --> 00:02:14 It was c1 times e to the negative t plus c2 e to the 34 00:02:13 --> 00:02:19 negative 6t, and y was c1 over 2 e 35 00:02:18 --> 00:02:24 to the negative t minus 2c2 e to the negative 6t. 36 00:02:24 --> 00:02:30 That was the solution we got. 37 00:02:30 --> 00:02:36 And then I went on to put in initial conditions, 38 00:02:33 --> 00:02:39 but we are not going to explore that aspect of it today. 39 00:02:36 --> 00:02:42 We will in a week or so. This was the general solution 40 00:02:40 --> 00:02:46 because it had two arbitrary constants in it. 41 00:02:44 --> 00:02:50 42 00:02:49 --> 00:02:55 What I want to do now is revisit this and do it by a 43 00:02:54 --> 00:03:00 different method, which makes heavy use of 44 00:02:58 --> 00:03:04 matrices. That is a prerequisite for this 45 00:03:02 --> 00:03:08 course, so I am assuming that you reviewed a little bit about 46 00:03:06 --> 00:03:12 matrices. And it is in your book. 47 00:03:09 --> 00:03:15 Your book puts in a nice little review section. 48 00:03:12 --> 00:03:18 Two-by-two and three-by-three will be good enough for 18.03 49 00:03:17 --> 00:03:23 mostly because I don't want you to calculate all night on bigger 50 00:03:21 --> 00:03:27 matrices, bigger systems. So nothing serious, 51 00:03:25 --> 00:03:31 matrix multiplication, solving systems of linear 52 00:03:28 --> 00:03:34 equations, end-by-end systems. I will remind you at the 53 00:03:34 --> 00:03:40 appropriate places today of what it is you need to remember. 54 00:03:39 --> 00:03:45 The very first thing we are going to do is, 55 00:03:43 --> 00:03:49 let's see. I haven't figured out the color 56 00:03:46 --> 00:03:52 coding for this lecture yet, but let's make this system in 57 00:03:51 --> 00:03:57 green and the solution can be in purple. 58 00:03:55 --> 00:04:01 Invisible purple, but I have a lot of it. 59 00:04:00 --> 00:04:06 60 00:04:05 --> 00:04:11 Let's abbreviate, first of all, 61 00:04:07 --> 00:04:13 the system using matrices. I am going to make a column 62 00:04:11 --> 00:04:17 vector out of (x, y). 63 00:04:13 --> 00:04:19 Then you differentiate a column vector by differentiating each 64 00:04:18 --> 00:04:24 component. I can write the left-hand side 65 00:04:22 --> 00:04:28 of the system as (x, y) prime. 66 00:04:24 --> 00:04:30 How about the right-hand side? 67 00:04:29 --> 00:04:35 Well, I say I can just write the matrix of coefficients to 68 00:04:33 --> 00:04:39 negative 2, 2, 2, negative 5 times x,y. 69 00:04:36 --> 00:04:42 70 00:04:38 --> 00:04:44 And I say that this matrix equation says exactly the same 71 00:04:42 --> 00:04:48 thing as that green equation and, therefore, 72 00:04:45 --> 00:04:51 it is legitimate to put it up in green, too. 73 00:04:50 --> 00:04:56 74 00:04:55 --> 00:05:01 The top here is x prime. What is the top here? 75 00:05:00 --> 00:05:06 After I multiply these two I get a column vector. 76 00:05:08 --> 00:05:14 And what is its top entry? It is negative 2x plus 2y. 77 00:05:18 --> 00:05:24 There it is. 78 00:05:22 --> 00:05:28 And the bottom entry the same way is 2x minus 5y, 79 00:05:33 --> 00:05:39 just as it is down there. Now, what I want to do is, 80 00:05:42 --> 00:05:48 well, maybe I should translate the solution. 81 00:05:49 --> 00:05:55 What does the solution look like? 82 00:05:58 --> 00:06:04 We got that, too. 83 00:05:58 --> 00:06:04 How am I going to write this as a matrix equation? 84 00:05:58 --> 00:06:04 Actually, if I told you to use matrices, use vectors, 85 00:05:59 --> 00:06:05 the point at which you might be most hesitant is this one right 86 00:05:59 --> 00:06:05 here, the very next step. Because how you should write it 87 00:06:00 --> 00:06:06 is extremely well-concealed in this notation. 88 00:06:00 --> 00:06:06 But the point is, this is a column vector and I 89 00:06:01 --> 00:06:07 am adding together two column vectors. 90 00:06:01 --> 00:06:07 And what is in each one of the column vectors? 91 00:06:02 --> 00:06:08 Think of these two things as a column vector. 92 00:06:05 --> 00:06:11 Pull out all the scalars from them that you can. 93 00:06:09 --> 00:06:15 Well, you see that c1 is a common factor of both entries 94 00:06:14 --> 00:06:20 and so is e to the negative t, that function. 95 00:06:18 --> 00:06:24 Now, if I pull both of those out of the vector, 96 00:06:22 --> 00:06:28 what is left of the vector? Well, you cannot even see it. 97 00:06:26 --> 00:06:32 What is left is a 1 up here and a one-half there. 98 00:06:32 --> 00:06:38 So I am going to write that in the following form. 99 00:06:35 --> 00:06:41 I will put out the c1, it's the common factor in both, 100 00:06:38 --> 00:06:44 and put that out front. Then I will put in the guts of 101 00:06:42 --> 00:06:48 the vector, even though you cannot see it, 102 00:06:44 --> 00:06:50 the column vector 1, one-half. 103 00:06:46 --> 00:06:52 And then I will put the other 104 00:06:49 --> 00:06:55 scalar function in back. The only reason for putting one 105 00:06:52 --> 00:06:58 of these in front and one in back is visual so to make it 106 00:06:56 --> 00:07:02 easy to read. There is no other reason. 107 00:07:00 --> 00:07:06 You could put the c1 here, you could put it here, 108 00:07:03 --> 00:07:09 you could put the e negative t in front if you want 109 00:07:07 --> 00:07:13 to, but people will fire you. Don't do that. 110 00:07:10 --> 00:07:16 Write it the standard way because that is the way that it 111 00:07:13 --> 00:07:19 is easiest to read. The constants out front, 112 00:07:16 --> 00:07:22 the functions behind, and the column vector of 113 00:07:19 --> 00:07:25 numbers in the middle. And so the other one will be 114 00:07:22 --> 00:07:28 written how? Well, here, that one is a 115 00:07:25 --> 00:07:31 little more transparent. c2, 1, 2 and the other thing is 116 00:07:30 --> 00:07:36 e to the negative 6t. 117 00:07:33 --> 00:07:39 There is our solution. That is going to need a lot of 118 00:07:38 --> 00:07:44 purple, but I have it. 119 00:07:41 --> 00:07:47 120 00:07:47 --> 00:07:53 And now I want to talk about how the new method of solving 121 00:07:50 --> 00:07:56 the equation. It is based just on the same 122 00:07:52 --> 00:07:58 idea as the way we solve second-order equations. 123 00:07:54 --> 00:08:00 Yes, question. 124 00:07:56 --> 00:08:02 125 00:08:04 --> 00:08:10 Oh, here. Sorry. 126 00:08:04 --> 00:08:10 This should be negative two. Thanks very much. 127 00:08:07 --> 00:08:13 What I am going to use is a trial solution. 128 00:08:10 --> 00:08:16 Remember when we had a second-order equation with 129 00:08:13 --> 00:08:19 constant coefficients the very first thing I did was I said we 130 00:08:17 --> 00:08:23 are going to try a solution of the form e to the rt. 131 00:08:21 --> 00:08:27 Why that? Well, because Oiler thought of 132 00:08:23 --> 00:08:29 it and it has been known for or 300 years that that is the 133 00:08:27 --> 00:08:33 thing you should do. Well, this has not been known 134 00:08:32 --> 00:08:38 nearly as long because matrices were only invented around 135 00:08:36 --> 00:08:42 or so, and people did not really use them to solve systems of 136 00:08:41 --> 00:08:47 differential equations until the middle of the last century, 137 00:08:45 --> 00:08:51 1950-1960. If you look at books written in 138 00:08:48 --> 00:08:54 1950, they won't even talk about systems of differential 139 00:08:53 --> 00:08:59 equations, or talk very little anyway and they won't solve them 140 00:08:57 --> 00:09:03 using matrices. This is only 50 years old. 141 00:09:02 --> 00:09:08 I mean, my God, in mathematics that is very up 142 00:09:06 --> 00:09:12 to date, particularly elementary mathematics. 143 00:09:09 --> 00:09:15 Anyway, the method of solving is going to use as a trial 144 00:09:14 --> 00:09:20 solution. Now, if you were left to your 145 00:09:18 --> 00:09:24 own devices you might say, well, let's try x equals some 146 00:09:22 --> 00:09:28 constant times e to the lambda1 t and y 147 00:09:28 --> 00:09:34 equals some other constant times e to the lambda2 t. 148 00:09:33 --> 00:09:39 Now, if you try that, 149 00:09:37 --> 00:09:43 it is a sensible thing to try, but it will turn out not to 150 00:09:42 --> 00:09:48 work. And that is the reason I have 151 00:09:45 --> 00:09:51 written out this particular solution, so we can see what 152 00:09:49 --> 00:09:55 solutions look like. The essential point is here is 153 00:09:53 --> 00:09:59 the basic solution I am trying to find. 154 00:09:56 --> 00:10:02 Here is another one. Their form is a column vector 155 00:10:00 --> 00:10:06 of constants. But they both use the same 156 00:10:06 --> 00:10:12 exponential factor, which is the point. 157 00:10:10 --> 00:10:16 In other words, I should not use here, 158 00:10:14 --> 00:10:20 in my trial solution, two different lambdas, 159 00:10:19 --> 00:10:25 I should use the same lambda. And so the way to write the 160 00:10:26 --> 00:10:32 trial solution is (x, y) equals two unknown numbers, 161 00:10:32 --> 00:10:38 that or that or whatever, times e to a single unknown 162 00:10:38 --> 00:10:44 exponent factor. Let's call it lambda t. 163 00:10:43 --> 00:10:49 It is called lambda. It is called r. 164 00:10:46 --> 00:10:52 It is called m. I have never seen it called 165 00:10:49 --> 00:10:55 anything but one of those three things. 166 00:10:52 --> 00:10:58 I am using lambda. Your book uses lambda. 167 00:10:55 --> 00:11:01 It is a common choice. Let's stick with it. 168 00:11:00 --> 00:11:06 Now what is the next step? Well, we plug into the system. 169 00:11:03 --> 00:11:09 Substitute into the system. What are we going to get? 170 00:11:07 --> 00:11:13 Well, let's do it. First of all, 171 00:11:09 --> 00:11:15 I have to differentiate. The left-hand side asks me to 172 00:11:13 --> 00:11:19 differentiate this. How do I differentiate this? 173 00:11:16 --> 00:11:22 Column vector times a function. Well, the column vector acts as 174 00:11:20 --> 00:11:26 a constant. And I differentiate that. 175 00:11:23 --> 00:11:29 That is lambda e to the lambda t. 176 00:11:28 --> 00:11:34 So the (x, y) prime is (a1, a2) times e to the lambda t 177 00:11:32 --> 00:11:38 times lambda. 178 00:11:35 --> 00:11:41 Now, it is ugly to put the 179 00:11:38 --> 00:11:44 lambda afterwards because it is a number so you should put it in 180 00:11:42 --> 00:11:48 front, again, to make things easier to read. 181 00:11:46 --> 00:11:52 But this lambda comes from differentiating e to the lambda 182 00:11:50 --> 00:11:56 t and using the chain rule. 183 00:11:53 --> 00:11:59 This much is the left-hand side. 184 00:11:56 --> 00:12:02 That is the derivative (x, y) prime. 185 00:12:00 --> 00:12:06 I differentiate the x and I differentiated the y. 186 00:12:03 --> 00:12:09 How about the right-hand side. Well, the right-hand side is 187 00:12:07 --> 00:12:13 negative 2, 2, 2, negative 5 times what? 188 00:12:09 --> 00:12:15 Well, times (x, 189 00:12:12 --> 00:12:18 y), which is (a1, a2) e to the lambda t. 190 00:12:15 --> 00:12:21 Now, the same thing that 191 00:12:18 --> 00:12:24 happened a month or a month and a half ago happens now. 192 00:12:22 --> 00:12:28 The whole point of making that substitution is that the e to 193 00:12:26 --> 00:12:32 the lambda t, the function part of it drops 194 00:12:30 --> 00:12:36 out completely. And one is left with what? 195 00:12:35 --> 00:12:41 An algebraic equation to be solved for lambda a1 and a2. 196 00:12:39 --> 00:12:45 In other words, by means of that substitution, 197 00:12:43 --> 00:12:49 and it basically uses the fact that the coefficients are 198 00:12:48 --> 00:12:54 constant, what you have done is reduced the problem of calculus, 199 00:12:53 --> 00:12:59 of solving differential equations, to solving algebraic 200 00:12:58 --> 00:13:04 equations. In some sense that is the only 201 00:13:02 --> 00:13:08 method there is, unless you do numerical stuff. 202 00:13:05 --> 00:13:11 You reduce the calculus to algebra. 203 00:13:08 --> 00:13:14 The Laplace transform is exactly the same thing. 204 00:13:11 --> 00:13:17 All the work is algebra. You turn the original 205 00:13:14 --> 00:13:20 differential equation into an algebraic equation for Y of s, 206 00:13:19 --> 00:13:25 you solve it, and then you use more algebra 207 00:13:22 --> 00:13:28 to find out what the original little y of t was. 208 00:13:27 --> 00:13:33 It is not different here. So let's solve this system of 209 00:13:32 --> 00:13:38 equations. Now, the whole problem with 210 00:13:34 --> 00:13:40 solving this system, first of all, 211 00:13:37 --> 00:13:43 what is the system? Let's write it out explicitly. 212 00:13:41 --> 00:13:47 Well, it is really two equations, isn't it? 213 00:13:44 --> 00:13:50 The first one says lambda a1 is equal to negative 2 a1 214 00:13:49 --> 00:13:55 plus 2 a2. That is the first one. 215 00:13:52 --> 00:13:58 The other one says lambda a2 is equal to 2 a1 minus 5 a2. 216 00:13:56 --> 00:14:02 217 00:14:00 --> 00:14:06 218 00:14:06 --> 00:14:12 Now, purely, if you want to classify that, 219 00:14:09 --> 00:14:15 that is two equations and three variables, three unknowns. 220 00:14:13 --> 00:14:19 The a1, a2, and lambda are all unknown. 221 00:14:16 --> 00:14:22 And, unfortunately, if you want to classify them 222 00:14:19 --> 00:14:25 correctly, they are nonlinear equations because they are made 223 00:14:24 --> 00:14:30 nonlinear by the fact that you have multiplied two of the 224 00:14:28 --> 00:14:34 variables. Well, if you sit down and try 225 00:14:32 --> 00:14:38 to hack away at solving those without a plan, 226 00:14:35 --> 00:14:41 you are not going to get anywhere. 227 00:14:37 --> 00:14:43 It is going to be a mess. Also, two equations and three 228 00:14:41 --> 00:14:47 unknowns is indeterminate. You can solve three equations 229 00:14:46 --> 00:14:52 and three unknowns and get a definite answer, 230 00:14:49 --> 00:14:55 but two equations and three unknowns usually have an 231 00:14:53 --> 00:14:59 infinity of solutions. Well, at this point it is the 232 00:14:56 --> 00:15:02 only idea that is required. Well, this was a little idea, 233 00:15:02 --> 00:15:08 but I assume one would think of that. 234 00:15:05 --> 00:15:11 And the idea that is required here is, I think, 235 00:15:09 --> 00:15:15 not so unnatural, it is not to view these a1, 236 00:15:13 --> 00:15:19 a2, and lambda as equal. Not all variables are created 237 00:15:17 --> 00:15:23 equal. Some are more equal than 238 00:15:20 --> 00:15:26 others. a1 and a2 are definitely equal 239 00:15:23 --> 00:15:29 to each other, and let's relegate lambda to 240 00:15:27 --> 00:15:33 the background. In other words, 241 00:15:31 --> 00:15:37 I am going to think of lambda as just a parameter. 242 00:15:34 --> 00:15:40 I am going to demote it from the status of variable to 243 00:15:39 --> 00:15:45 parameter. If I demoted it further it 244 00:15:41 --> 00:15:47 would just be an unknown constant. 245 00:15:44 --> 00:15:50 That is as bad as you can be. I am going to focus my 246 00:15:48 --> 00:15:54 attention on the a1, a2 and sort of view the lambda 247 00:15:52 --> 00:15:58 as a nuisance. Now, as soon as I do that, 248 00:15:55 --> 00:16:01 I see that these equations are linear if I just look at them as 249 00:15:59 --> 00:16:05 equations in a1 and a2. And moreover, 250 00:16:04 --> 00:16:10 they are not just linear, they are homogenous. 251 00:16:07 --> 00:16:13 Because if I think of lambda just as a parameter, 252 00:16:11 --> 00:16:17 I should rewrite the equations this way. 253 00:16:14 --> 00:16:20 I am going to subtract this and move the left-hand side to the 254 00:16:19 --> 00:16:25 right side, and it is going to look like (minus 2 minus lambda) 255 00:16:23 --> 00:16:29 times a1 plus 2 a2 is equal to zero. 256 00:16:28 --> 00:16:34 And the same way for the other 257 00:16:32 --> 00:16:38 one. It is going to be 2a1 plus, 258 00:16:35 --> 00:16:41 what is the coefficient, (minus 5 minus lambda) a2 259 00:16:39 --> 00:16:45 equals zero. 260 00:16:43 --> 00:16:49 That is a pair of simultaneous linear equations for determining 261 00:16:48 --> 00:16:54 a1 and a2, and the coefficients involved are parameter lambda. 262 00:16:53 --> 00:16:59 Now, what is the point of doing that? 263 00:16:58 --> 00:17:04 Well, now the point is whatever you learned about linear 264 00:17:02 --> 00:17:08 equations, you should have learned the most fundamental 265 00:17:06 --> 00:17:12 theorem of linear equations. The main theorem is that you 266 00:17:10 --> 00:17:16 have a square system of homogeneous equations, 267 00:17:13 --> 00:17:19 this is a two-by-two system so it is square, 268 00:17:16 --> 00:17:22 it always has the trivial solution, of course, 269 00:17:20 --> 00:17:26 a1, a2 equals zero. Now, we don't want that trivial 270 00:17:24 --> 00:17:30 solution because if a1 and a2 are zero, then so are x and y 271 00:17:28 --> 00:17:34 zero. Now that is a solution. 272 00:17:32 --> 00:17:38 Unfortunately, it is of no interest. 273 00:17:35 --> 00:17:41 If the solution were x, y zero, it corresponds to the 274 00:17:39 --> 00:17:45 fact that this is an ice bath. The yoke is at zero, 275 00:17:44 --> 00:17:50 the white is at zero and it stays that way for all time 276 00:17:48 --> 00:17:54 until the ice melts. So that is the solution we 277 00:17:52 --> 00:17:58 don't want. We don't want the trivial 278 00:17:56 --> 00:18:02 solution. Well, when does it have a 279 00:17:59 --> 00:18:05 nontrivial solution? Nontrivial means non-zero, 280 00:18:03 --> 00:18:09 in other words. If and only if this determinant 281 00:18:09 --> 00:18:15 is zero. 282 00:18:10 --> 00:18:16 283 00:18:18 --> 00:18:24 In other words, by using that theorem on linear 284 00:18:22 --> 00:18:28 equations, what we find is there is a condition that lambda must 285 00:18:28 --> 00:18:34 satisfy, an equation in lambda in order that we would be able 286 00:18:33 --> 00:18:39 to find non-zero values for a1 and a2. 287 00:18:38 --> 00:18:44 Let's write it out. I will recopy it over here. 288 00:18:41 --> 00:18:47 What was it? Negative 2 minus lambda, 289 00:18:44 --> 00:18:50 two, here it was 2 and minus 5 minus lambda. 290 00:18:49 --> 00:18:55 All right. You have to expand the 291 00:18:51 --> 00:18:57 determinant. In other words, 292 00:18:54 --> 00:19:00 we are trying to find out for what values of lambda is this 293 00:18:58 --> 00:19:04 determinant zero. Those will be the good values 294 00:19:03 --> 00:19:09 which lead to nontrivial solutions for the a's. 295 00:19:06 --> 00:19:12 This is the equation lambda plus 2. 296 00:19:09 --> 00:19:15 See, this is minus that and minus that, the product of the 297 00:19:14 --> 00:19:20 two minus ones is plus one. So it is lambda plus 2 times 298 00:19:18 --> 00:19:24 lambda plus 5, 299 00:19:21 --> 00:19:27 which is the product of the two diagonal elements, 300 00:19:25 --> 00:19:31 minus the product of the two anti-diagonal elements, 301 00:19:29 --> 00:19:35 which is 4, is equal to zero. And if I write that out, 302 00:19:35 --> 00:19:41 what is that, that is the equation lambda 303 00:19:39 --> 00:19:45 squared plus 7 lambda, 304 00:19:43 --> 00:19:49 5 lambda plus 2 lambda, and then the constant term is 305 00:19:48 --> 00:19:54 10 minus 4 which is 6. How many of you have long 306 00:19:52 --> 00:19:58 enough memories, two-day memories that you 307 00:19:56 --> 00:20:02 remember that equation? When I did the method of 308 00:20:02 --> 00:20:08 elimination, it led to exactly the same equation except it had 309 00:20:10 --> 00:20:16 r's in it instead of lambda. And this equation, 310 00:20:15 --> 00:20:21 therefore, is given the same name and another color. 311 00:20:21 --> 00:20:27 Let's make it salmon. 312 00:20:25 --> 00:20:31 313 00:20:30 --> 00:20:36 And it is called the characteristic equation for this 314 00:20:34 --> 00:20:40 method. All right. 315 00:20:36 --> 00:20:42 Now I am going to use now the word from last time. 316 00:20:40 --> 00:20:46 You factor this. From the factorization we get 317 00:20:44 --> 00:20:50 its root easily enough. The roots are lambda equals 318 00:20:48 --> 00:20:54 negative 1 and lambda equals negative 6 319 00:20:54 --> 00:21:00 by factoring the equation. 320 00:20:57 --> 00:21:03 Now what I am supposed to do? You have to keep the different 321 00:21:03 --> 00:21:09 parts of the method together. Now I have found the only 322 00:21:08 --> 00:21:14 values of lambda for which I will be able to find nonzero 323 00:21:12 --> 00:21:18 values for the a1 and a2. For each of those values of 324 00:21:17 --> 00:21:23 lambda, I now have to find the corresponding a1 and a2. 325 00:21:21 --> 00:21:27 Let's do them one at a time. Let's take first lambda equals 326 00:21:26 --> 00:21:32 negative one. My problem is now to find a1 327 00:21:32 --> 00:21:38 and a2. Where am I going to find them 328 00:21:35 --> 00:21:41 from? Well, from that system of 329 00:21:38 --> 00:21:44 equations over there. I will recopy it over here. 330 00:21:42 --> 00:21:48 What is the system? The hardest part of this is 331 00:21:47 --> 00:21:53 dealing with multiple minus signs, but you had experience 332 00:21:52 --> 00:21:58 with that in determinants so you know all about that. 333 00:21:58 --> 00:22:04 In other words, there is the system of 334 00:22:01 --> 00:22:07 equations over there. Let's recopy them here. 335 00:22:06 --> 00:22:12 Minus 2, minus minus 1 makes minus 1. 336 00:22:09 --> 00:22:15 What's the other coefficient? It is just plain old 2. 337 00:22:15 --> 00:22:21 Good. There is my first equation. 338 00:22:18 --> 00:22:24 And when I substitute lambda equals negative one 339 00:22:24 --> 00:22:30 for the second equation, what do you get? 340 00:22:30 --> 00:22:36 2 a1 plus negative 5 minus negative 1 makes negative 4. 341 00:22:37 --> 00:22:43 There is my system that will find me a1 and a2. 342 00:22:43 --> 00:22:49 What is the first thing you notice about it? 343 00:22:48 --> 00:22:54 You immediately notice that this system is fake because this 344 00:22:56 --> 00:23:02 second equation is twice the first one. 345 00:23:03 --> 00:23:09 Something is wrong. No, something is right. 346 00:23:06 --> 00:23:12 If that did not happen, if the second equation were not 347 00:23:12 --> 00:23:18 a constant multiple of the first one then the only solution of 348 00:23:17 --> 00:23:23 the system would be a1 equals zero, a2 equals zero because the 349 00:23:23 --> 00:23:29 determinant of the coefficients would not be zero. 350 00:23:29 --> 00:23:35 The whole function of this exercise was to find the value 351 00:23:33 --> 00:23:39 of lambda, negative 1, for which the system would be 352 00:23:37 --> 00:23:43 redundant and, therefore, would have a 353 00:23:40 --> 00:23:46 nontrivial solution. Do you get that? 354 00:23:43 --> 00:23:49 In other words, calculate the system out, 355 00:23:47 --> 00:23:53 just as I have done here, you have an automatic check on 356 00:23:51 --> 00:23:57 the method. If one equation is not a 357 00:23:54 --> 00:24:00 constant multiple of the other you made a mistake. 358 00:24:00 --> 00:24:06 You don't have the right value of lambda or you substituted 359 00:24:05 --> 00:24:11 into the system wrong, which is frankly a more common 360 00:24:10 --> 00:24:16 error. Go back, recheck first the 361 00:24:13 --> 00:24:19 substitution, and if convinced that is right 362 00:24:17 --> 00:24:23 then recheck where you got lambda from. 363 00:24:20 --> 00:24:26 But here everything is going fine so we can now find out what 364 00:24:26 --> 00:24:32 the value of a1 and a2 are. You don't have to go through a 365 00:24:32 --> 00:24:38 big song and dance for this since most of the time you will 366 00:24:36 --> 00:24:42 have two-by-two equations and now and then three-by-three. 367 00:24:40 --> 00:24:46 For two-by-two all you do is, since we really have the same 368 00:24:44 --> 00:24:50 equation twice, to get a solution I can assign 369 00:24:47 --> 00:24:53 one of the variables any value and then simply solve for the 370 00:24:51 --> 00:24:57 other. The natural thing to do is to 371 00:24:54 --> 00:25:00 make a2 equal one, then I won't need fractions and 372 00:24:57 --> 00:25:03 then a1 will be a2. So the solution is (2, 373 00:25:01 --> 00:25:07 1). I am only trying to find one 374 00:25:04 --> 00:25:10 solution. Any constant multiple of this 375 00:25:07 --> 00:25:13 would also be a solution, as long as it wasn't zero, 376 00:25:12 --> 00:25:18 zero which is the trivial one. And, therefore, 377 00:25:15 --> 00:25:21 this is a solution to this system of algebraic equations. 378 00:25:20 --> 00:25:26 And the solution to the whole system of differential equations 379 00:25:25 --> 00:25:31 is, this is only the (a1, a2) part. 380 00:25:30 --> 00:25:36 I have to add to it, as a factor, 381 00:25:32 --> 00:25:38 lambda is negative, therefore, e to the minus t. 382 00:25:36 --> 00:25:42 There is our purple thing. 383 00:25:40 --> 00:25:46 384 00:25:50 --> 00:25:56 See how I got it? Starting with the trial 385 00:25:52 --> 00:25:58 solution, I first found out through this procedure what the 386 00:25:57 --> 00:26:03 lambda's have to be. Then I took the lambda and 387 00:26:00 --> 00:26:06 found what the corresponding a1 and a2 that went with it and 388 00:26:04 --> 00:26:10 then made up my solution out of that. 389 00:26:06 --> 00:26:12 Now, quickly I will do the same thing for lambda 390 00:26:11 --> 00:26:17 equals negative 6. Each one of these must be 391 00:26:13 --> 00:26:19 treated separately. They are separate problems and 392 00:26:17 --> 00:26:23 you are looking for separate solutions. 393 00:26:19 --> 00:26:25 Lambda equals negative 6. What do I do? 394 00:26:22 --> 00:26:28 How do my equations look now? Well, the first one is minus 2 395 00:26:25 --> 00:26:31 minus negative 6 makes plus 4. It is 4a1 plus 2a2 equals zero. 396 00:26:32 --> 00:26:38 Then I hold my breath while I 397 00:26:37 --> 00:26:43 calculate the second one to see if it comes out to be a constant 398 00:26:44 --> 00:26:50 multiple. I get 2a1 plus negative 5 minus 399 00:26:49 --> 00:26:55 negative 6, which makes plus 1. And, indeed, 400 00:26:54 --> 00:27:00 one is a constant multiple of the other. 401 00:27:00 --> 00:27:06 I really only have on equation there. 402 00:27:04 --> 00:27:10 I will just write down immediately now what the 403 00:27:09 --> 00:27:15 solution is to the system. Well, the (a1, 404 00:27:13 --> 00:27:19 a2) will be what? Now, it is more natural to make 405 00:27:19 --> 00:27:25 a1 equal 1 and then solve to get an integer for a2. 406 00:27:25 --> 00:27:31 If a1 is 1, then a2 is negative 2. 407 00:27:30 --> 00:27:36 And I should multiply that by e to the negative 6t 408 00:27:34 --> 00:27:40 because negative 6 is the corresponding value. 409 00:27:38 --> 00:27:44 There is my other one. And now there is a 410 00:27:41 --> 00:27:47 superposition principle, which if I get a chance will 411 00:27:44 --> 00:27:50 prove for you at the end of the hour. 412 00:27:47 --> 00:27:53 If not, you will have to do it yourself for homework. 413 00:27:51 --> 00:27:57 Since this is a linear system of equations, 414 00:27:54 --> 00:28:00 once you have two separate solutions, neither a constant 415 00:27:58 --> 00:28:04 multiple of the other, you can multiply each one of 416 00:28:02 --> 00:28:08 these by a constant and it will still be a solution. 417 00:28:08 --> 00:28:14 You can add them together and that will still be a solution, 418 00:28:11 --> 00:28:17 and that gives the general solution. 419 00:28:13 --> 00:28:19 The general solution is the sum of these two, 420 00:28:16 --> 00:28:22 an arbitrary constant. I am going to change the name 421 00:28:19 --> 00:28:25 since I don't want to confuse it with the c1 I used before, 422 00:28:22 --> 00:28:28 times the first solution which is (2, 1) e to the negative t 423 00:28:26 --> 00:28:32 plus c2, another arbitrary constant, times 1 negative 2 e 424 00:28:29 --> 00:28:35 to the minus 6t. 425 00:28:32 --> 00:28:38 Now you notice that is exactly 426 00:28:37 --> 00:28:43 the same solution I got before. The only difference is that I 427 00:28:43 --> 00:28:49 have renamed the arbitrary constants. 428 00:28:47 --> 00:28:53 The relationship between them, c1 over 2, 429 00:28:53 --> 00:28:59 I am now calling c1 tilda, and c2 I am calling c2 tilda. 430 00:29:00 --> 00:29:06 If you have an arbitrary constant, it doesn't matter 431 00:29:04 --> 00:29:10 whether you divide it by two. It is still just an arbitrary a 432 00:29:09 --> 00:29:15 constant. It covers all values, 433 00:29:12 --> 00:29:18 in other words. Well, I think you will agree 434 00:29:16 --> 00:29:22 that is a different procedure, yet it has only one 435 00:29:20 --> 00:29:26 coincidence. It is like elimination goes 436 00:29:24 --> 00:29:30 this way and comes to the answer. 437 00:29:28 --> 00:29:34 And this method goes a completely different route and 438 00:29:31 --> 00:29:37 comes to the answer, except it is not quite like 439 00:29:34 --> 00:29:40 that. They walk like this and then 440 00:29:36 --> 00:29:42 they come within viewing distance of each other to check 441 00:29:40 --> 00:29:46 that both are using the same characteristic equation, 442 00:29:44 --> 00:29:50 and then they again go their separate ways and end up with 443 00:29:48 --> 00:29:54 the same answer. 444 00:29:50 --> 00:29:56 445 00:29:58 --> 00:30:04 There is something special of these values. 446 00:30:00 --> 00:30:06 You cannot get away from those two values of lambda. 447 00:30:03 --> 00:30:09 Somehow they are really intrinsically connected. 448 00:30:06 --> 00:30:12 Occurs the exponential coefficient, and they are 449 00:30:09 --> 00:30:15 intrinsically connected with the problem of the egg that we 450 00:30:12 --> 00:30:18 started with. Now what I would like to do is 451 00:30:15 --> 00:30:21 very quickly sketch how this method looks when I remove all 452 00:30:18 --> 00:30:24 the numbers from it. In some sense, 453 00:30:20 --> 00:30:26 it becomes a little clearer what is going on. 454 00:30:23 --> 00:30:29 And that will give me a chance to introduce the terminology 455 00:30:26 --> 00:30:32 that you need when you talk about it. 456 00:30:30 --> 00:30:36 457 00:30:55 --> 00:31:01 Well, you have notes. Let me try to write it down in 458 00:31:03 --> 00:31:09 general. 459 00:31:05 --> 00:31:11 460 00:31:10 --> 00:31:16 I will first write it out two-by-two. 461 00:31:13 --> 00:31:19 I am just going to sketch. The system looks like (x, 462 00:31:18 --> 00:31:24 y) equals, I will still put it up in colors. 463 00:31:23 --> 00:31:29 464 00:31:30 --> 00:31:36 Except now, instead of using twos and fives, 465 00:31:33 --> 00:31:39 I will use (a, b; c, d). 466 00:31:35 --> 00:31:41 467 00:31:40 --> 00:31:46 The trial solution will look how? 468 00:31:44 --> 00:31:50 The trial is going to be (a1, a2). 469 00:31:48 --> 00:31:54 That I don't have to change the name of. 470 00:31:53 --> 00:31:59 I am going to substitute in, and what the result of 471 00:31:59 --> 00:32:05 substitution is going to be lambda (a1, a2). 472 00:32:06 --> 00:32:12 I am going to skip a step and pretend that the e to the lambda 473 00:32:11 --> 00:32:17 t's have already been canceled out. 474 00:32:16 --> 00:32:22 Is equal to (a, b; c, d) times (a1, 475 00:32:19 --> 00:32:25 a2). What does that correspond to? 476 00:32:22 --> 00:32:28 That corresponds to the system as I wrote it here. 477 00:32:28 --> 00:32:34 And then we wrote it out in terms of two equations. 478 00:32:31 --> 00:32:37 And what was the resulting thing that we ended up with? 479 00:32:35 --> 00:32:41 Well, you write it out, you move the lambda to the 480 00:32:38 --> 00:32:44 other side. And then the homogeneous system 481 00:32:41 --> 00:32:47 is we will look in general how? Well, we could write it out. 482 00:32:45 --> 00:32:51 It is going to look like a minus lambda, 483 00:32:48 --> 00:32:54 b, c, d minus lambda. 484 00:32:50 --> 00:32:56 That is just how it looks there 485 00:32:53 --> 00:32:59 and the general calculation is the same. 486 00:32:56 --> 00:33:02 Times (a1, a2) is equal to zero. 487 00:33:00 --> 00:33:06 488 00:33:05 --> 00:33:11 This is solvable nontrivially. In other words, 489 00:33:13 --> 00:33:19 it has a nontrivial solution if an only if the determinant of 490 00:33:25 --> 00:33:31 coefficients is zero. 491 00:33:30 --> 00:33:36 492 00:33:35 --> 00:33:41 Let's now write that out, calculate out once and for all 493 00:33:39 --> 00:33:45 what that determinant is. I will write it out here. 494 00:33:43 --> 00:33:49 It is a minus lambda times d minus lambda, 495 00:33:46 --> 00:33:52 the product of the diagonal elements, minus the 496 00:33:50 --> 00:33:56 anti-diagonal minus bc is equal to zero. 497 00:33:55 --> 00:34:01 And let's calculate that out. 498 00:34:00 --> 00:34:06 It is lambda squared minus a lambda minus d lambda plus ad, 499 00:34:07 --> 00:34:13 the constant term from here, negative bc from there, 500 00:34:14 --> 00:34:20 plus ad minus bc, 501 00:34:21 --> 00:34:27 where have I seen that before? This equation is the general 502 00:34:29 --> 00:34:35 form using letters of what we calculated using the specific 503 00:34:36 --> 00:34:42 numbers before. Again, I will code it the same 504 00:34:45 --> 00:34:51 way with that color salmon. Now, most of the calculations 505 00:34:54 --> 00:35:00 will be for two-by-two systems. I advise you, 506 00:35:00 --> 00:35:06 in the strongest possible terms, to remember this 507 00:35:03 --> 00:35:09 equation. You could write down this 508 00:35:06 --> 00:35:12 equation immediately for the matrix. 509 00:35:08 --> 00:35:14 You don't have to go through all this stuff. 510 00:35:11 --> 00:35:17 For God's sakes, don't say let the trial 511 00:35:13 --> 00:35:19 solution be blah, blah, blah. 512 00:35:15 --> 00:35:21 You don't want to do that. I don't want you to repeat the 513 00:35:19 --> 00:35:25 derivation of this every time you go through a particular 514 00:35:23 --> 00:35:29 problem. It is just like in solving 515 00:35:25 --> 00:35:31 second order equations. You have a second order 516 00:35:29 --> 00:35:35 equation. You immediately write down its 517 00:35:32 --> 00:35:38 characteristic equation, then you factor it, 518 00:35:35 --> 00:35:41 you find its roots and you construct the solution. 519 00:35:39 --> 00:35:45 It takes a minute. The same thing, 520 00:35:42 --> 00:35:48 this takes a minute, too. 521 00:35:43 --> 00:35:49 What is the constant term? Ad minus bc, 522 00:35:47 --> 00:35:53 what is that? Matrix is (a, 523 00:35:49 --> 00:35:55 b; c, d). Ad minus bc is its determinant. 524 00:35:52 --> 00:35:58 This is the determinant of that matrix. 525 00:35:55 --> 00:36:01 I didn't give the matrix a name, did I? 526 00:36:00 --> 00:36:06 I will now give the matrix a name A. 527 00:36:03 --> 00:36:09 What is this? Well, you are not supposed to 528 00:36:07 --> 00:36:13 know that until now. I will tell you. 529 00:36:10 --> 00:36:16 This is called the trace of A. Put that down in your little 530 00:36:16 --> 00:36:22 books. The abbreviation is trace A, 531 00:36:19 --> 00:36:25 and the word is trace. The trace of a square matrix is 532 00:36:25 --> 00:36:31 the sum of the d elements down its main diagonal. 533 00:36:31 --> 00:36:37 If it were a three-by-three there would be three terms in 534 00:36:35 --> 00:36:41 whatever you are up to. Here it is a plus b, 535 00:36:40 --> 00:36:46 the sum of the diagonal elements. 536 00:36:43 --> 00:36:49 You can immediately write down this characteristic equation. 537 00:36:48 --> 00:36:54 Let's give it a name. This is a characteristic 538 00:36:52 --> 00:36:58 equation of what? Of the matrix, 539 00:36:55 --> 00:37:01 now. Not of the system, 540 00:36:57 --> 00:37:03 of the matrix. You have a two-by-two matrix. 541 00:37:02 --> 00:37:08 You could immediately write down its characteristic 542 00:37:06 --> 00:37:12 equation. Watch out for this sign, 543 00:37:08 --> 00:37:14 minus. That is a very common error to 544 00:37:11 --> 00:37:17 leave out the minus sign because that is the way the formula 545 00:37:15 --> 00:37:21 comes out. Its roots. 546 00:37:17 --> 00:37:23 If it is a quadratic equation it will have roots; 547 00:37:20 --> 00:37:26 lambda1, lambda2 for the moment let's assume are real and 548 00:37:25 --> 00:37:31 distinct. 549 00:37:27 --> 00:37:33 550 00:37:37 --> 00:37:43 For the enrichment of your vocabulary, those are called the 551 00:37:39 --> 00:37:45 eigenvalues. 552 00:37:40 --> 00:37:46 553 00:37:50 --> 00:37:56 They are something which belonged to the matrix A. 554 00:37:53 --> 00:37:59 They are two secret numbers. You can calculate from the 555 00:37:56 --> 00:38:02 coefficients a, b, and c, and d, 556 00:37:58 --> 00:38:04 but they are not in the coefficients. 557 00:38:00 --> 00:38:06 You cannot look at a matrix and see what its eigenvalues are. 558 00:38:05 --> 00:38:11 You have to calculate something. 559 00:38:07 --> 00:38:13 But they are the most important numbers in the matrix. 560 00:38:10 --> 00:38:16 They are hidden, but they are the things that 561 00:38:13 --> 00:38:19 control how this system behaves. Those are called the 562 00:38:17 --> 00:38:23 eigenvalues. Now, there are various purists, 563 00:38:20 --> 00:38:26 there are a fair number of them in the world who do not like 564 00:38:24 --> 00:38:30 this word because it begins German and ends English. 565 00:38:29 --> 00:38:35 Eigenvalues were first introduced by a German 566 00:38:32 --> 00:38:38 mathematician, you know, around the time 567 00:38:35 --> 00:38:41 matrices came into being in or so. 568 00:38:38 --> 00:38:44 A little while after eigenvalues came into being, 569 00:38:41 --> 00:38:47 too. And since all this happened in 570 00:38:44 --> 00:38:50 Germany they were named eigenvalues in German, 571 00:38:47 --> 00:38:53 which begins eigen and ends value. 572 00:38:50 --> 00:38:56 But people who do not like that call them the characteristic 573 00:38:54 --> 00:39:00 values. Unfortunately, 574 00:38:56 --> 00:39:02 it is two words and takes a lot more space to write out. 575 00:39:02 --> 00:39:08 An older generation even calls them something different, 576 00:39:06 --> 00:39:12 which you are not so likely to see nowadays, 577 00:39:10 --> 00:39:16 but you will in slightly older books. 578 00:39:13 --> 00:39:19 You can also call them the proper values. 579 00:39:17 --> 00:39:23 Characteristic is not a translation of eigen, 580 00:39:21 --> 00:39:27 but proper is, but it means it in a funny 581 00:39:25 --> 00:39:31 sense which has almost disappeared nowadays. 582 00:39:30 --> 00:39:36 It means proper in the sense of belong to. 583 00:39:33 --> 00:39:39 The only example I can think of is the word property. 584 00:39:37 --> 00:39:43 Property is something that belongs to you. 585 00:39:40 --> 00:39:46 That is the use of the word proper. 586 00:39:43 --> 00:39:49 It is something that belongs to the matrix. 587 00:39:46 --> 00:39:52 The matrix has its proper values. 588 00:39:49 --> 00:39:55 It does not mean proper in the sense of fitting and proper or I 589 00:39:54 --> 00:40:00 hope you will behave properly when we go to Aunt Agatha's or 590 00:39:59 --> 00:40:05 something like that. But, as I say, 591 00:40:03 --> 00:40:09 by far the most popular thing, slowly the word eigenvalue is 592 00:40:07 --> 00:40:13 pretty much taking over the literature. 593 00:40:11 --> 00:40:17 Just because it's just one word, that is a tremendous 594 00:40:15 --> 00:40:21 advantage. Okay. 595 00:40:16 --> 00:40:22 What now is still to be done? Well, there are those vectors 596 00:40:21 --> 00:40:27 to be found. So the very last step would be 597 00:40:24 --> 00:40:30 to solve the system to find the vectors a1 and a2. 598 00:40:30 --> 00:40:36 599 00:40:35 --> 00:40:41 For each (lambda)i, find the associated vector. 600 00:40:40 --> 00:40:46 The vector, we will call it (alpha)i. 601 00:40:44 --> 00:40:50 That is the a1 and a2. Of course it's going to be 602 00:40:50 --> 00:40:56 indexed. You have to put another 603 00:40:53 --> 00:40:59 subscript on it because there are two of them. 604 00:41:00 --> 00:41:06 And a1 and a2 is stretched a little too far. 605 00:41:04 --> 00:41:10 By solving the system, and the system will be the 606 00:41:09 --> 00:41:15 system which I will write this way, (a minus lambda, 607 00:41:15 --> 00:41:21 b, c, d minus lambda). 608 00:41:19 --> 00:41:25 It is just that system that was 609 00:41:24 --> 00:41:30 over there, but I will recopy it, (a1, a2) equals zero, 610 00:41:30 --> 00:41:36 zero. And these are called the 611 00:41:34 --> 00:41:40 eigenvectors. Each of these is called the 612 00:41:39 --> 00:41:45 eigenvector associated with or belonging to, 613 00:41:44 --> 00:41:50 again, in that sense of property. 614 00:41:48 --> 00:41:54 Eigenvector, let's say belonging to, 615 00:41:52 --> 00:41:58 I see that a little more frequently, belonging to lambda 616 00:41:58 --> 00:42:04 i. So we have the eigenvalues, 617 00:42:01 --> 00:42:07 the eigenvectors and, of course, the people who call 618 00:42:04 --> 00:42:10 them characteristic values also call these guys characteristic 619 00:42:08 --> 00:42:14 vectors. I don't think I have ever seen 620 00:42:11 --> 00:42:17 proper vectors, but that is because I am not 621 00:42:13 --> 00:42:19 old enough. I think that is what they used 622 00:42:16 --> 00:42:22 to be called a long time ago, but not anymore. 623 00:42:20 --> 00:42:26 And then, finally, the general solution will be, 624 00:42:24 --> 00:42:30 by the superposition principle, (x, y) equals the arbitrary 625 00:42:30 --> 00:42:36 constant times the first eigenvector times the eigenvalue 626 00:42:35 --> 00:42:41 times the e to the corresponding eigenvalue. 627 00:42:40 --> 00:42:46 And then the same thing for the second one, (a1, 628 00:42:44 --> 00:42:50 a2), but now the second index will be 2 to indicate that it 629 00:42:50 --> 00:42:56 goes with the eigenvalue e to the lambda 2t. 630 00:42:56 --> 00:43:02 I have done that twice. And now in the remaining five 631 00:43:02 --> 00:43:08 minutes I will do it a third time because it is possible to 632 00:43:06 --> 00:43:12 write this in still a more condensed form. 633 00:43:09 --> 00:43:15 And the advantage of the more condensed form is A, 634 00:43:13 --> 00:43:19 it takes only that much space to write, and B, 635 00:43:16 --> 00:43:22 it applies to systems, not just the two-by-two 636 00:43:20 --> 00:43:26 systems, but to end-by-end systems. 637 00:43:23 --> 00:43:29 The method is exactly the same. Let's write it out as it would 638 00:43:27 --> 00:43:33 apply to end-by-end systems. The vector I started with is 639 00:43:34 --> 00:43:40 (x, y) and so on, but I will simply abbreviate 640 00:43:39 --> 00:43:45 this, as is done in 18.02, by x with an arrow over it. 641 00:43:45 --> 00:43:51 The matrix A I will abbreviate with A, as I did before with 642 00:43:51 --> 00:43:57 capital A. And then the system looks like 643 00:43:56 --> 00:44:02 x prime is equal to -- 644 00:44:00 --> 00:44:06 645 00:44:05 --> 00:44:11 x prime is what? Ax. 646 00:44:06 --> 00:44:12 That is all there is to it. 647 00:44:10 --> 00:44:16 There is our green system. Now notice in this form I did 648 00:44:15 --> 00:44:21 not even tell you whether this a two-by-two matrix or an 649 00:44:20 --> 00:44:26 end-by-end. And in this condensed form it 650 00:44:23 --> 00:44:29 will look the same no matter how many equations you have. 651 00:44:30 --> 00:44:36 Your book deals from the beginning with end-by-end 652 00:44:33 --> 00:44:39 systems. That is, in my view, 653 00:44:36 --> 00:44:42 one of its weaknesses because I don't think most students start 654 00:44:40 --> 00:44:46 with two-by-two. Fortunately, 655 00:44:43 --> 00:44:49 the book double-talks. The theory is end-by-end, 656 00:44:46 --> 00:44:52 but all the examples are two-by-two. 657 00:44:49 --> 00:44:55 So just read the examples. Read the notes instead, 658 00:44:53 --> 00:44:59 which just do two-by-two to start out with. 659 00:44:58 --> 00:45:04 The trial solution is x equals what? 660 00:45:01 --> 00:45:07 An unknown vector alpha times e to the lambda t. 661 00:45:07 --> 00:45:13 Alpha is what we called a1 and 662 00:45:11 --> 00:45:17 a2 before. Plug this into there and cancel 663 00:45:15 --> 00:45:21 the e to the lambda t's. 664 00:45:19 --> 00:45:25 What do you get? Well, this is lambda alpha e to 665 00:45:24 --> 00:45:30 the lambda t equals A alpha e to the lambda t. 666 00:45:29 --> 00:45:35 667 00:45:36 --> 00:45:42 These two cancel. And the system to be solved, 668 00:45:40 --> 00:45:46 A alpha equals lambda alpha. 669 00:45:46 --> 00:45:52 And now the question is how do you solve that system? 670 00:45:51 --> 00:45:57 Well, you can tell if a book is written by a scoundrel or not by 671 00:45:57 --> 00:46:03 how they go -- A book, which is in my opinion 672 00:46:02 --> 00:46:08 completely scoundrel, simply says you subtract one 673 00:46:07 --> 00:46:13 from the other, and without further ado writes 674 00:46:12 --> 00:46:18 A minus lambda, and they tuck a little I in 675 00:46:16 --> 00:46:22 there and write alpha equals zero. 676 00:46:19 --> 00:46:25 Why is the I put in there? Well, this is what you would 677 00:46:25 --> 00:46:31 like to write. What is wrong with this 678 00:46:28 --> 00:46:34 equation? This is not a valid matrix 679 00:46:32 --> 00:46:38 equation because that is a square end-by-end matrix, 680 00:46:36 --> 00:46:42 a square two-by-two matrix if you like. 681 00:46:39 --> 00:46:45 This is a scalar. You cannot subtract the scalar 682 00:46:42 --> 00:46:48 from a matrix. It is not an operation. 683 00:46:45 --> 00:46:51 To subtract matrices they have to be the same size, 684 00:46:49 --> 00:46:55 the same shape. What is done is you make this a 685 00:46:52 --> 00:46:58 two-by-two matrix. This is a two-by-two matrix 686 00:46:56 --> 00:47:02 with lambdas down the main diagonal and I elsewhere. 687 00:47:01 --> 00:47:07 And the justification is that lambda alpha is the same thing 688 00:47:05 --> 00:47:11 as the lambda I times alpha because I is an identity matrix. 689 00:47:10 --> 00:47:16 Now, in fact, jumping from here to here is 690 00:47:13 --> 00:47:19 not something that would occur to anybody. 691 00:47:16 --> 00:47:22 The way it should occur to you to do this is you do this, 692 00:47:21 --> 00:47:27 you write that, you realize it doesn't work, 693 00:47:24 --> 00:47:30 and then you say to yourself I don't understand what these 694 00:47:29 --> 00:47:35 matrices are all about. I think I'd better write it all 695 00:47:33 --> 00:47:39 out. And then you would write it all 696 00:47:36 --> 00:47:42 out and you would write that equation on the left-hand board 697 00:47:39 --> 00:47:45 there. Oh, now I see what it should 698 00:47:41 --> 00:47:47 look like. I should subtract lambda from 699 00:47:44 --> 00:47:50 the main diagonal. That is the way it will come 700 00:47:47 --> 00:47:53 out. And then say, 701 00:47:48 --> 00:47:54 hey, the way to save lambda from the main diagonal is put it 702 00:47:51 --> 00:47:57 in an identity matrix. That will do it for me. 703 00:47:54 --> 00:48:00 In other words, there is a little detour that 704 00:47:57 --> 00:48:03 goes from here to here. And one of the ways I judge 705 00:48:01 --> 00:48:07 books is by how well they explain the passage from this to 706 00:48:05 --> 00:48:11 that. If they don't explain it at all 707 00:48:08 --> 00:48:14 and just write it down, they have never talked to 708 00:48:11 --> 00:48:17 students. They have just written books. 709 00:48:14 --> 00:48:20 Where did we get finally here? The characteristic equation 710 00:48:17 --> 00:48:23 from that, I had forgotten what color. 711 00:48:20 --> 00:48:26 That is in salmon. The characteristic equation, 712 00:48:23 --> 00:48:29 then, is going to be the thing which says that the determinant 713 00:48:27 --> 00:48:33 of that is zero. That is the circumstances under 714 00:48:32 --> 00:48:38 which it is solvable. In general, this is the way the 715 00:48:35 --> 00:48:41 characteristic equation looks. And its roots, 716 00:48:38 --> 00:48:44 once again, are the eigenvalues. 717 00:48:40 --> 00:48:46 And from then you calculate the corresponding eigenvectors. 718 00:48:44 --> 00:48:50 Okay. Go home and practice. 719 00:48:46 --> 00:48:52 In recitation you will practice on both two-by-two and 720 00:48:50 --> 00:48:56 three-by-three cases, and we will talk more next 721 00:48:53 --> 00:48:59 time.