1 00:00:07,030 --> 00:00:07,780 LINAN CHEN: Hello. 2 00:00:07,780 --> 00:00:09,870 Welcome back to recitation. 3 00:00:09,870 --> 00:00:12,050 I'm sure you are becoming more and more 4 00:00:12,050 --> 00:00:15,220 familiar with the determinants of matrices. 5 00:00:15,220 --> 00:00:18,880 In the lecture, we also learned the geometric interpretation 6 00:00:18,880 --> 00:00:20,780 of the determinant. 7 00:00:20,780 --> 00:00:24,660 The absolute value of the determinant of a matrix 8 00:00:24,660 --> 00:00:28,730 is simply equal to the volume of the parallelepiped spanned 9 00:00:28,730 --> 00:00:32,040 by the row vectors of that matrix. 10 00:00:32,040 --> 00:00:35,600 So today, we're going to apply this fact 11 00:00:35,600 --> 00:00:38,420 to solve the following problem. 12 00:00:38,420 --> 00:00:43,990 I have a tetrahedron, T, in this 3D space. 13 00:00:43,990 --> 00:00:50,840 And the vertices of T are given by O, which is the origin, A_1, 14 00:00:50,840 --> 00:00:54,040 A_2, and A_3. 15 00:00:54,040 --> 00:00:56,940 So I have highlighted this tetrahedron 16 00:00:56,940 --> 00:00:59,190 using the blue chalk. 17 00:00:59,190 --> 00:01:03,270 So this is T. 18 00:01:03,270 --> 00:01:07,120 And our first goal is to compute the volume of T 19 00:01:07,120 --> 00:01:10,330 using the determinant. 20 00:01:10,330 --> 00:01:15,260 And the second part is: if I fix A_1 and A_2, 21 00:01:15,260 --> 00:01:20,130 but move A_3 to another point, A_3 prime, which 22 00:01:20,130 --> 00:01:23,380 is given by this coordinate, I ask you 23 00:01:23,380 --> 00:01:26,370 to compute the volume again. 24 00:01:26,370 --> 00:01:27,130 OK. 25 00:01:27,130 --> 00:01:32,120 So since we want to use the fact that the determinant is related 26 00:01:32,120 --> 00:01:36,280 to the volume, we have to figure out which volume 27 00:01:36,280 --> 00:01:37,680 we should be looking at. 28 00:01:37,680 --> 00:01:40,660 We know that the determinant is related 29 00:01:40,660 --> 00:01:43,060 to the volume of a parallelepiped. 30 00:01:43,060 --> 00:01:46,220 But here, we only have a tetrahedron. 31 00:01:46,220 --> 00:01:50,010 So the first goal should be to find out which parallelepiped 32 00:01:50,010 --> 00:01:52,230 you should be working with. 33 00:01:52,230 --> 00:01:56,810 OK, why don't you hit pause and try to work it out yourself. 34 00:01:56,810 --> 00:02:00,510 You can sketch the parallelepiped on this picture. 35 00:02:00,510 --> 00:02:03,780 And I will return in a while and continue working with you. 36 00:02:12,960 --> 00:02:13,640 All right. 37 00:02:13,640 --> 00:02:15,990 How did your computation go? 38 00:02:15,990 --> 00:02:19,520 Now let's complete this picture together. 39 00:02:19,520 --> 00:02:22,930 As we were saying, we need a parallelepiped 40 00:02:22,930 --> 00:02:25,800 so that we can use the fact that the determinant is 41 00:02:25,800 --> 00:02:27,790 related to the volume. 42 00:02:27,790 --> 00:02:30,700 Here, I have a tetrahedron. 43 00:02:30,700 --> 00:02:39,230 And let's look at these three edges, OA_1, OA_2, and OA_3. 44 00:02:39,230 --> 00:02:42,640 All of them meet at the origin. 45 00:02:42,640 --> 00:02:46,720 So why don't we just consider the parallelepiped spanned 46 00:02:46,720 --> 00:02:49,430 by those same three edges? 47 00:02:49,430 --> 00:02:51,870 It's a natural choice, because at least it 48 00:02:51,870 --> 00:02:54,116 shares three edges with T. 49 00:02:54,116 --> 00:02:58,500 OK, now let's move on to this picture here. 50 00:02:58,500 --> 00:03:01,930 As you can see, I have drawn this parallelepiped 51 00:03:01,930 --> 00:03:05,180 in red chalk. 52 00:03:05,180 --> 00:03:11,080 So the blue part is my original T. 53 00:03:11,080 --> 00:03:14,710 And the red part, let me call it P. 54 00:03:14,710 --> 00:03:21,830 It's the parallelepiped spanned by edges OA_1, OA_2, and OA_3. 55 00:03:21,830 --> 00:03:25,870 So that's the parallelepiped that I'm going to work with. 56 00:03:25,870 --> 00:03:30,230 Now the next step is to relate the volume of T 57 00:03:30,230 --> 00:03:34,510 to the volume of P. OK, let's recall together 58 00:03:34,510 --> 00:03:37,820 what is the volume of a tetrahedron. 59 00:03:37,820 --> 00:03:45,100 We note that the volume of a tetrahedron 60 00:03:45,100 --> 00:03:51,470 is going to be equal to 1/3 of the area of the base 61 00:03:51,470 --> 00:03:54,260 times the height, right? 62 00:03:54,260 --> 00:03:59,800 Of course, you can choose any side, any face to be the base. 63 00:03:59,800 --> 00:04:05,770 But for convenience, we're going to choose the triangle OA_1A_2 64 00:04:05,770 --> 00:04:09,540 to be the base of T. So the volume of T 65 00:04:09,540 --> 00:04:15,810 is going to be equal to 1/3 times the area of triangle 66 00:04:15,810 --> 00:04:16,310 OA_1A_2. 67 00:04:18,950 --> 00:04:22,540 So I use this A to indicate the area. 68 00:04:22,540 --> 00:04:24,430 Then times the height. 69 00:04:24,430 --> 00:04:27,760 Well, if I choose this to be my base, 70 00:04:27,760 --> 00:04:31,560 then this A_3 becomes the apex. 71 00:04:31,560 --> 00:04:35,820 Then the height is simply equal to the distance from A_3 72 00:04:35,820 --> 00:04:39,000 to the triangle OA_1A_2. 73 00:04:39,000 --> 00:04:43,220 Let me use letter h to denote this quantity. 74 00:04:43,220 --> 00:04:44,910 So that's the height. 75 00:04:44,910 --> 00:04:47,950 And the volume of T is equal to 1/3 76 00:04:47,950 --> 00:04:52,390 of the product of the area of the base times the height. 77 00:04:52,390 --> 00:04:59,960 OK, this is the volume of T. Now let's see what the volume of P 78 00:04:59,960 --> 00:05:01,250 is. 79 00:05:01,250 --> 00:05:03,790 So P is a parallelepiped. 80 00:05:03,790 --> 00:05:06,760 The volume of a parallelepiped is simply 81 00:05:06,760 --> 00:05:12,800 equal to the area of the base times its height. 82 00:05:12,800 --> 00:05:17,020 This time, which face would you choose to be the base? 83 00:05:17,020 --> 00:05:21,320 Well, of course, you would like to choose this parallelogram 84 00:05:21,320 --> 00:05:26,200 to be the base because it contains the base of T. 85 00:05:26,200 --> 00:05:31,570 If we do so, so we want to choose this parallelogram 86 00:05:31,570 --> 00:05:35,920 to be the base, then what is the area of this? 87 00:05:35,920 --> 00:05:40,590 Well, it clearly contains two copies of the triangle OA_1A_2. 88 00:05:43,460 --> 00:05:46,550 So the area of the parallelogram is simply 89 00:05:46,550 --> 00:05:51,115 equal to twice the area of the triangle OA_1A_2. 90 00:05:56,410 --> 00:05:59,600 Then what is the height of P? 91 00:05:59,600 --> 00:06:03,570 Again, if you choose this face to be the base, 92 00:06:03,570 --> 00:06:07,360 then A_3 becomes the apex again. 93 00:06:07,360 --> 00:06:10,210 Then the height of the parallelepiped 94 00:06:10,210 --> 00:06:13,240 is equal to the distance from A_3 95 00:06:13,240 --> 00:06:17,760 to the base, which is the same as the distance from A_3 96 00:06:17,760 --> 00:06:20,930 to triangle OA_1A_2. 97 00:06:20,930 --> 00:06:25,050 So here, the height is also equal to h. 98 00:06:28,390 --> 00:06:31,860 Now you can compare these two formulae. 99 00:06:31,860 --> 00:06:36,800 You see that volume of T is simply 100 00:06:36,800 --> 00:06:44,640 equal to 1/6 of volume of P. That's 101 00:06:44,640 --> 00:06:47,080 the connection between the volume 102 00:06:47,080 --> 00:06:51,170 of the tetrahedron with the volume of the parallelepiped. 103 00:06:51,170 --> 00:06:53,690 In order to compute the volume of T, 104 00:06:53,690 --> 00:06:57,530 we only need to compute the volume of P. 105 00:06:57,530 --> 00:07:00,380 Now let's compute the volume of this parallelepiped. 106 00:07:03,370 --> 00:07:10,280 We know that it's related to the determinant of a matrix. 107 00:07:10,280 --> 00:07:13,150 And the row vectors of that matrix 108 00:07:13,150 --> 00:07:17,850 are given by these three edges. 109 00:07:17,850 --> 00:07:20,910 So because all of them start from 0, 110 00:07:20,910 --> 00:07:25,090 we only need the coordinate of A_1, A_2, and A_3. 111 00:07:25,090 --> 00:07:30,730 So here, volume of P is equal to the absolute value. 112 00:07:30,730 --> 00:07:33,320 So don't forget the absolute value sign. 113 00:07:33,320 --> 00:07:37,590 Absolute value of the determinant of a three 114 00:07:37,590 --> 00:07:39,180 by three matrix. 115 00:07:39,180 --> 00:07:43,020 So we just need to copy the coordinates of the vertices 116 00:07:43,020 --> 00:07:44,340 down here. 117 00:07:44,340 --> 00:07:49,280 The first one is [2, 2, -1]. 118 00:07:49,280 --> 00:07:50,400 These two are too close. 119 00:07:54,100 --> 00:07:55,240 -1. 120 00:07:55,240 --> 00:08:01,084 And A_2 is [1, 3, 0]. 121 00:08:01,084 --> 00:08:07,070 A3 is [-1, 1, 4]. 122 00:08:07,070 --> 00:08:11,440 The absolute value of this determinant. 123 00:08:11,440 --> 00:08:17,290 And if you compute this, this is a three by three matrix. 124 00:08:17,290 --> 00:08:20,010 The determinant should be easy to compute. 125 00:08:20,010 --> 00:08:25,900 And the result should be 12. 126 00:08:25,900 --> 00:08:29,780 So that's the volume of P, which means 127 00:08:29,780 --> 00:08:35,370 the volume of the tetrahedron T is equal to 12 over 6, 128 00:08:35,370 --> 00:08:38,659 which is 2. 129 00:08:38,659 --> 00:08:41,240 Did you get the correct answer? 130 00:08:41,240 --> 00:08:45,120 OK, so in order to compute the volume of T, 131 00:08:45,120 --> 00:08:48,990 we have related to a parallelepiped, P, 132 00:08:48,990 --> 00:08:52,010 which contains T. All right. 133 00:08:52,010 --> 00:08:54,300 Now let's look at the second part. 134 00:08:54,300 --> 00:08:59,400 The second part says that if I keep A_1 and A_2 unchanged, 135 00:08:59,400 --> 00:09:04,750 but I move A_3 to a new point-- so A3 136 00:09:04,750 --> 00:09:09,630 is going to be moved to a point given by A_3 prime, 137 00:09:09,630 --> 00:09:23,310 and the coordinate is -201, -199, and 104. 138 00:09:23,310 --> 00:09:26,490 And I'm asking you to compute the volume 139 00:09:26,490 --> 00:09:29,180 of the new tetrahedron. 140 00:09:29,180 --> 00:09:32,280 Well as you can see, this point seems 141 00:09:32,280 --> 00:09:35,590 to be far away from the origin. 142 00:09:35,590 --> 00:09:39,610 I'm not even able to draw this point here. 143 00:09:39,610 --> 00:09:42,770 But you can imagine, as this point goes far away 144 00:09:42,770 --> 00:09:45,440 from the origin, this spike is going 145 00:09:45,440 --> 00:09:47,480 to become more and more pointy. 146 00:09:47,480 --> 00:09:50,770 In other words, the entire tetrahedron looks more and more 147 00:09:50,770 --> 00:09:52,010 like a needle. 148 00:09:52,010 --> 00:09:54,610 But nonetheless, we can use the same method 149 00:09:54,610 --> 00:09:56,670 to compute the volume. 150 00:09:56,670 --> 00:10:03,180 So we follow the same idea, the volume of T 151 00:10:03,180 --> 00:10:08,420 is going to be equal to 1/6 of the volume of P. 152 00:10:08,420 --> 00:10:11,350 And in this case, that's going to become 153 00:10:11,350 --> 00:10:15,950 the absolute value of the determinant of a three 154 00:10:15,950 --> 00:10:20,010 by three matrix whose row vectors are 155 00:10:20,010 --> 00:10:23,250 given by these three edges. 156 00:10:23,250 --> 00:10:27,030 So in this case, we again copy down the coordinates 157 00:10:27,030 --> 00:10:29,070 of three vertices. 158 00:10:29,070 --> 00:10:33,250 The first one is [2, 2, -1]. 159 00:10:33,250 --> 00:10:36,300 The second one is [1, 3, 0]. 160 00:10:36,300 --> 00:10:48,850 The third one should become this, so [-201, 199, 104]. 161 00:10:48,850 --> 00:10:49,870 OK. 162 00:10:49,870 --> 00:10:53,840 That should give me the volume of the new tetrahedron. 163 00:10:53,840 --> 00:10:57,480 Let me call it T prime just to differentiate it 164 00:10:57,480 --> 00:10:59,980 from the previous tetrahedron. 165 00:10:59,980 --> 00:11:04,950 And of course, you can compute this determinant explicitly. 166 00:11:04,950 --> 00:11:12,330 If you do so, you will see the answer should be 2 again. 167 00:11:12,330 --> 00:11:16,400 But in fact, there is a way that you can just 168 00:11:16,400 --> 00:11:22,390 read out the answer directly without any real computation. 169 00:11:22,390 --> 00:11:24,620 Let's pay attention to the last row. 170 00:11:24,620 --> 00:11:28,660 In other words, let's pay attention to this new A_3, 171 00:11:28,660 --> 00:11:31,700 well, A_3 prime. 172 00:11:31,700 --> 00:11:34,430 What do you observe here? 173 00:11:34,430 --> 00:11:39,940 A_3 prime, if you consider the difference between A_3 and A_3 174 00:11:39,940 --> 00:11:42,990 prime, in other words, if you consider how much you have 175 00:11:42,990 --> 00:11:48,150 moved your apex, you will see that's equal to -100*A_1. 176 00:11:52,230 --> 00:11:52,870 Right? 177 00:11:52,870 --> 00:11:57,690 So A_3 is [-1, 1, 4]. 178 00:11:57,690 --> 00:11:59,690 A_3 prime is this. 179 00:11:59,690 --> 00:12:04,720 That's exactly equal to -100 times A_1. 180 00:12:04,720 --> 00:12:06,650 What does that mean? 181 00:12:06,650 --> 00:12:10,080 Well from the point of view of the matrix, 182 00:12:10,080 --> 00:12:16,620 you are subtracting from the original third row 100 times 183 00:12:16,620 --> 00:12:18,190 the first row. 184 00:12:18,190 --> 00:12:22,470 But this row operation does not change the determinant. 185 00:12:22,470 --> 00:12:26,090 In other words, you know that this determinant should be 186 00:12:26,090 --> 00:12:27,790 the same as the previous one. 187 00:12:27,790 --> 00:12:31,980 So you can write out 2 directly. 188 00:12:31,980 --> 00:12:36,080 From this picture, we can also see that fact. 189 00:12:36,080 --> 00:12:39,610 So from this picture, we know that this section 190 00:12:39,610 --> 00:12:46,520 is saying that I need to move A_3 in the inverse direction 191 00:12:46,520 --> 00:12:50,730 of A_1 by 100 times A_1. 192 00:12:50,730 --> 00:12:54,500 So you're moving A_3 parallel to A_1. 193 00:12:54,500 --> 00:12:58,410 But it doesn't matter how far you've moved your apex. 194 00:12:58,410 --> 00:13:00,800 You're moving in a way that remains parallel 195 00:13:00,800 --> 00:13:05,160 to the base, which is saying that this movement does not 196 00:13:05,160 --> 00:13:07,210 change the height. 197 00:13:07,210 --> 00:13:12,780 Since A_1 and A_2 are fixed, your base is fixed, 198 00:13:12,780 --> 00:13:14,730 and you're not changing the height. 199 00:13:14,730 --> 00:13:17,660 So of course, the volume is not changing. 200 00:13:17,660 --> 00:13:20,450 That's also a reason, another way 201 00:13:20,450 --> 00:13:22,700 to see that the volume of T prime 202 00:13:22,700 --> 00:13:26,208 is simply equal to the volume of T. 203 00:13:26,208 --> 00:13:29,580 OK, this completes this problem. 204 00:13:29,580 --> 00:13:32,450 I hope this example was helpful to you. 205 00:13:32,450 --> 00:13:36,850 And we should keep in mind that the fact that the determinant 206 00:13:36,850 --> 00:13:39,090 is related to volume sometimes can 207 00:13:39,090 --> 00:13:41,620 become very handy in computing the volume 208 00:13:41,620 --> 00:13:44,210 of certain geometric objects. 209 00:13:44,210 --> 00:13:46,290 Thank you for watching, and I'm looking forward 210 00:13:46,290 --> 00:13:48,390 to seeing you soon.