Augmented Matrices: Reduced Row Echelon Form (update) — Transcript

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1. 0:00welcome to a video that will show how to
2. 0:02transform an augmented Matrix into
3. 0:04reduced R Echelon form in order to solve
4. 0:06a system of
5. 0:08equations let's begin by reviewing reduc
6. 0:11Ro Echelon form a matrix is in reduc Ro
7. 0:14Echelon form if these four conditions
8. 0:16have been met and here are several
9. 0:18examples of matrices in reduced ralon
10. 0:21form step one the first nonzero element
11. 0:25in each row called the leading entry is
12. 0:28one two each leading entry is in a
13. 0:32column to the right of the leading entry
14. 0:35in the previous
15. 0:37Row three rows with all zero elements if
16. 0:41any are below rows having nonzero
17. 0:44elements and then Finly four in each
18. 0:47column that contains a leading entry
19. 0:50which again is a one all other elements
20. 0:53of the column are
21. 0:55zeros so all of these matrices satisfy
22. 0:58these four conditions and are in reduce
23. 1:01Ro Echelon form let's also review how we
24. 1:05transform an augmented Matrix often
25. 1:07referred to as gussian
26. 1:09elimination number one any two rows can
27. 1:12be
28. 1:13interchanged number two the elements of
29. 1:15any row can be multiplied by a nonzero
30. 1:18real number and then number three any
31. 1:20row can be changed by adding or
32. 1:22subtracting the corresponding elements
33. 1:24with another row so let's go ahead and
34. 1:27give it a try on first a system of two
35. 1:30equations with two unknowns we'll start
36. 1:33by writing this as an augmented Matrix
37. 1:35so the first row would be 2 1
38. 1:391 the second row would be 3 -2
39. 1:46-16 so I want to have a zero in this
40. 1:48position and this position let's first
41. 1:51try to get a zero here if this was a
42. 1:54positive two we could add it to row two
43. 1:57to obtain a zero so let's go ahead and
44. 2:00replace Row one with 2 * Row
45. 2:031 plus row two the second row stays the
46. 2:08same so 2 * Row 1 plus row two 2 * 2 + 3
47. 2:13that would be 7 2 * 1 + -2 there's the
48. 2:18zero and 2 * 1 + -16 that's 2 + -16 that
49. 2:25would be
50. 2:27-14 now we want a zero in this position
51. 2:30so since the corresponding element is a
52. 2:32seven the LCM or at least common
53. 2:35multiple of 3 and 7 would be 21 so if we
54. 2:38make this a 21 and this a 21 we can then
55. 2:41subtract these two rows so we'll replace
56. 2:44row two with 7 * Row
57. 2:482 minus 3 * Row 1 Row one stays the same
58. 2:54so we'll have 7 * 3 - 3 * 7 that's 21 -
59. 2:5821 or 0
60. 3:01now we'll have 7 * -2 that's
61. 3:04-14 - 3 * 0 that's
62. 3:08-14 now this is a little more
63. 3:10challenging here we have 7 *
64. 3:13-16 that's
65. 3:17-12 minus 3 * -14 so that'll
66. 3:22become + 42 so we have
67. 3:27-12 + 42
68. 3:30which will give us a -70 and the last
69. 3:33step is to have the main diagonal equal
70. 3:34to ones so we'll multiply Row 1 by 17th
71. 3:38or divide by
72. 3:417 and multiply Row 2 by
73. 3:44-114 or divide by -14 multiplying by
74. 3:4817th we'd have 1 0 and then 17 * -14
75. 3:54that's
76. 3:55-2 this would be a
77. 3:58014 * -14th is
78. 4:011 and then -114 time
79. 4:05-70 or just divide -70 by -14 would give
80. 4:10us pos5 now we convert this back to
81. 4:14equations we have 1X = -2 or x =
82. 4:18-2 and we have 1 y = 5 or Y = 5 and this
83. 4:23system has been solved using reduced row
84. 4:26Echelon
85. 4:28form let's take a look at a system of
86. 4:30three equations in three variables we'll
87. 4:33start with the augmented Matrix first
88. 4:35row would be 4 -1 2 0 second row would
89. 4:39be 2 1
90. 4:43-11 the third row is 2 -2 1 3 let's see
91. 4:48if we can get zeros here and here so if
92. 4:51you take a look at Row one and row two
93. 4:54if this was a -4 we could add it to row
94. 4:56one to get zero so we'll replace row two
95. 4:59with -2 * Row 2 + Row
96. 5:031 and if we took Row three and
97. 5:06subtracted row two this would be a zero
98. 5:08so we'll replace Row three with Row
99. 5:10three minus row two first row stays the
100. 5:14same so multiply this by -2 and then add
101. 5:18it to 4 so -2 * 2 that's -4 + 4 we have
102. 5:230 -2 * 1 plus a - 1 that' be -3
103. 5:30-2 * -1 that's 2 + 2 we have
104. 5:344 -2 * -1 that's POS 22 + 0
105. 5:4022 now for Row three now for Row three
106. 5:44we'll have Row three - Row 2 so 2 - 2 is
107. 5:470 -2 - 1 -3 1 - 1 then be 1 + 1 2 and
108. 5:55then 3 -1 becomes 3 + 11 or 14
109. 6:01now let see if we can get a zero here
110. 6:03and a zero here if we take a look at the
111. 6:06first two rows if this was a positive
112. 6:08three we could add it to row two to get
113. 6:11a zero so we'll multiply this by -3 and
114. 6:14then add it to row two so replace Row
115. 6:18one with -3 * Row 1 + row two and if you
116. 6:23take a look at Row three and row two
117. 6:25these two are the same so if we replace
118. 6:27Row three with Row three minus row two
119. 6:30that would give us a zero here so Row 3
120. 6:34minus row two okay the second row stays
121. 6:37the same so now we're going to multiply
122. 6:40Row one by -3 and then add it to row two
123. 6:43so -3 * 4 that's -12 +
124. 6:480 -3 * -1 that's pos3 + -3 that's 0 -3 *
125. 6:562 that's -6 + 4 it's
126. 7:012 and then we have3 * 0 + 22 that's
127. 7:0622 and now for Row three we'll replace
128. 7:09it with Row three minus row two 0 - 0 is
129. 7:130 -3 -3 becomes -3 + 3 which is
130. 7:18zero here we have 2 - 4 that's
131. 7:22-2 and then we have 14us 22 that's
132. 7:278 let's go ahead and take this Matrix
133. 7:30over to the next screen and
134. 7:33continue now what we want to do is have
135. 7:35a zero here and a zero here and then the
136. 7:38only thing left will be to make the main
137. 7:39diagonal equal to one now we do have to
138. 7:41be a little bit careful here because we
139. 7:43don't want to lose this zero here to get
140. 7:45a zero in this position so we're going
141. 7:47to have to work with Row one and Row
142. 7:48three to make this equal to zero notice
143. 7:51the corresponding elements are the same
144. 7:53number so if we subtract them that would
145. 7:55give us zero so we're going to replace
146. 7:57Row one with Row one minus Row
147. 8:013 and then looking at row two we want a
148. 8:04zero here if this was a ne4 and we added
149. 8:08it to this four we'd have a zero here so
150. 8:09we're going to replace row two with row
151. 8:11two plus 2 * Row 3 the third row stays
152. 8:17the same so now we're going to subtract
153. 8:20Row one and Row three -2 - 0 it's -12 0
154. 8:24- 0 -2 - -2 becom -2 + 2 that's
155. 8:30Z and then 22 -8 becomes 22 + 8 that's
156. 8:3630 and then we're going to replace row
157. 8:38two with row two + 2 * Row 3 so 0 + 2 *
158. 8:430 it's 0 -3 + 2 * 0 that's -3 still and
159. 8:50then 4 + 2 * -2 there's the zero we
160. 8:54wanted and then
161. 8:5622 + 2 * 8 that'll be 22 +
162. 9:02-16 which is equal to 6 okay we're
163. 9:05almost there now we just need to make
164. 9:08the main diagonal consist of ones so
165. 9:10we'll
166. 9:11multiply Row one by -12th multiply Row
167. 9:15Two by - 1/3 and multiply Row 3 by
168. 9:22-2 remember multiplying by -12th is the
169. 9:26same as dividing by -12 so in the first
170. 9:28row would be 1 0 0 30 / -12 is equal to
171. 9:37-2.5 multiplying by - 1/3 is the same as
172. 9:40dividing by -3 so we'd have 0 1 0 -2 and
173. 9:45then multiplying by - one2 is the same
174. 9:48as dividing by -2 so we'd have 0 0 1 and
175. 9:51then
176. 9:54pos4 okay we should be done let's go
177. 9:56ahead and translate this back to
178. 9:57equations the first row would be X =
179. 10:02-2.5 the second row would be y = -2 and
180. 10:07the third row is z = pos4 and now we
181. 10:11have obtained the solution using reduced
182. 10:13row Echelon form the next video we'll
183. 10:16show how you can use your graphing
184. 10:17calculator to check your work thank you
185. 10:19for
186. 10:26watching