Download presentation

Presentation is loading. Please wait.

Published byClinton Freeney Modified over 3 years ago

1
1 Gauss-Jordan Method. How To complete Problem 2.2 # 29 Produced by E. Gretchen Gascon

2
2 The problem

3
3 Plan to solve Step 1 – write a matrix with the coefficients of the terms and as the last column the constant equivalents. Step 2 – use the Gauss-Jordan method to manipulate the matrix so that the solution will become evident

4
4 The matrix: 44-424 21-9 1-231

5
5 2. Keep Row1 fixed 44-424 21-9 1-231 Focus on the first row-first column element 4. We want to make the rest of the column #1 into 0s. To accomplish that we will perform the various row operation. Start with row 2. What would be a LCM of 4 and 2? (This helps to define the formula to be used. Ans: 4. Therefore you multiply 1 times 4 = 4, and 2 times 2 = 4. This is the formula 2*Elements of R2 Plus -1 times the Elements of R1

6
6 3. Replace Each Element of Row 2 44-424 21-9 1-231 Add 2*R2C1 -1*R1C1 R2C1, this would be 2(2) -1(4) 0 Replace all the cells in Row 2 with this formula, just change the column you are working in. 2*R2C2 -1*R1C2 R2C2, 2*R2C3 -1*R1C3 R2C3, 2(-1) - 1(4) -2+-4 -6 2(1) -1(-4) 2+4 6 2*R2C4 -1*R1C4 R2C4, 2(-9) -1(24) -18-24 -42 4-40-2-4-6 2+46-18-24-42

7
7 4. Replace Each Element of Row 3 44-424 0-6+6-42 1-231 Add -4*R3C1 +*R1C1 R3C1, this would be -4(1) +(4) 0 Replace all the cells in Row 2 with this formula, just change the column you are working in. -4*R3C2 +*R1C1 R3C2, -4*R3C3 +*R1C3 R3C3, -4*R3C4 +*R1C4 R3C4, -4+4 0 8+4 -12-4 -4+2412-16 20

8
8 Column 1 Complete 44-424 0-6+6-42 012-1620 Column 1 – row 1 has a value and the rest of the column values are 0.

9
9 Column 2 44-424 0-6+6-42 012-1620 -6 in column 2 is the pivot element. We want to make the rest of the column 0s

10
10 6. Replace Each Element of Row 1 44-424 0-66-42 012-1620 Add 3*R1C1 +2*R2C1 R1C1, this would be 3(4) +0 12 Replace all the cells in Row 1 with this formula, just change the column you are working in. Add 3*R1C2 +2R2C2 R1C2, Add 3*R1C3 +2R2C3 R1C3, Add 3*R1C4 +2R2C4 R1C4, 12+0 12-12-12+12 72-84 1200 -12

11
11 7. Replace Each Element of Row 3 1200-12 0-66-42 012-1620 Add R3C1 +2*R2C1 R3C1, this would be 0 + 3(0) 0 Replace all the cells in Row 1 with this formula, just change the column you are working in. Add R3C2 +2R2C2 R3C2, Add R3C3 +2R2C3 R3C3, Add R3C4 +2R2C4 R3C4, 0+0 12-12 - 16+12 20-84 00-4-64

12
12 Column 2 Complete 1200-12 0-66-42 00-4-64 Column 2 – row 2 has a value and the rest of the column values are 0.

13
13 Work on column 3 1200-12 0-66-42 00-4-64 -4 in column 3 remains, we want all the other elements in column 3 0s. We will accomplish this by using row operations, but the first row already has a 0 in the third column so we can skip that row operation.

14
14 9 Replace Each Element of Row 2 1200-12 0-66-42 00-4-64 Add 2*R2C1 +3*R3C1 R1C1, this would be (0) +3(0) 0 Replace all the cells in Row 1 with this formula, just change the column you are working in. Add 2*R2C2 +3*R3C2 R1C2, Add 2*R2C3 +3*R3C3 R1C3, Add 2*R2C4 +3*R3C4 R1C4, 0+0 0 -12+0 -12 -12+12 0 -84-192 -276

15
15 Column 3 Complete 1200-12 0 0-276 00-4-64 Column 3 – row 3 has a value and the rest of the column values are 0.

16
16 Done with + - rows 1200-12 0 0-276 00-4-64 You have the diagonal elements with values and 0s in the rest of those columns

17
17 Two options: Option # 1 – covert the matrix back to an algebraic equation and solve 1200-12 0 0-276 00-4-64 12x = -12 (from row #1) x = -1 -12y = -276 (from row # 2) y = 23 -4z = -64 (from row # 3) z = 16 We can write these equations because if you remember col#1 was the coefficients of the x, col#2 the coefficients of the y and col#3 the coefficients of the z.

18
18 Option 2: Use the rules for Gauss-Jordan - Divide through by the left most element in each row. 1200-12 0-60-276 00-4-64 R1/r1c1 r1/12 R2/r2c2 r2/(-6) R3/r3c3 r3/(-4) 12/12 1 -12/12 -6/-6 1 -276/(-6) 23 -4/-4 1 -64/(-4) 16

19
19 You are finished 100 01023 00116 At this point you can read the answer from the matrix. x = -1 y = 23 z = 16

20
20 Solutions X = 1 Y = 23 Z = 16

Similar presentations

Presentation is loading. Please wait....

OK

Solving a Sudoku Puzzle

Solving a Sudoku Puzzle

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on content development software Difference between lcd and led display ppt on tv Ppt on 3g wireless networks Ppt on arunachal pradesh culture of india Ppt on hydrogen fuel cell vehicles honda A ppt on nanotechnology Ppt on group development theories Ppt on natural and human nature Ppt on cnc lathe machine Ppt on types of computer