Download presentation

1
**4.4.1 Generalised Row Echelon Form**

Any all-zero rows are at the bottom. Correct ‘step pattern’ of first non-zero row entries.

2
**4.4.1 Generalised Row Echelon Form**

Any all-zero row at the bottom Correct ‘step pattern’ of first non-zero row entries 2 1 3 ROW 1 ROW 2

3
**4.4.1 Generalised Row Echelon Form**

Any all-zero row at the bottom Correct ‘step pattern’ of first non-zero row entries 2 1 3 ROW 1 ROW 2 ROW 3

4
**4.4.1 Generalised Row Echelon Form**

Any all-zero row at the bottom Correct ‘step pattern’ of first non-zero row entries 2 1 3 ROW 1 ROW 2 ROW 3

5
**4.4.1 Generalised Row Echelon Form**

Any all-zero row at the bottom Correct ‘step pattern’ of first non-zero row entries ROW 1 1 ROW 2 2 1 3 2 ROW 3 ROW 4 1

6
**4.4.1 Formal process (Handout 3)**

ann a12 a13 an1 a1n a2n an2 a21 a23 Create zeros a31 a32

7
**4.4.1 Formal process (Handout 3)**

ann a12 a13 an1 a1n a'2n an2 a'23 Create zeros a31 a32

8
**4.4.1 Formal process (Handout 3)**

ann a12 a13 an1 a1n a'2n an2 a'23 Create zeros a'32

9
4.4.1 Formal process a11 a'22 a'33 a'nn a12 a13 a1n a'2n a'n2 a'23 Create zeros a'32

10
**4.4.1 Formal process (Handout 3)**

a'nn a12 a13 a1n a'2n a'n2 a'23 a'32 Create zeros

11
**4.4.1 Formal process (Handout 3)**

a'nn a12 a13 a1n a'2n a'n2 a'23 Create zeros

12
**4.4.1 Formal process (Handout 3)**

a''nn a12 a13 a1n a'2n a'23 Create zeros

13
**4.4.1 Formal process (Handout 3)**

a''nn a12 a13 a1n a'2n a'23 Create zeros

14
**4.4.2 Augmented Matrix notation**

We perform row operations on the matrix and the opposite row: 1 2 -1 x y z 6 5 = Combine both of these into one matrix called the augmented matrix: 1 2 -1 6 5

15
**4.4.3 Row sums A way to check calculations: Do a row operation: e.g. 1**

Add up rows 2. Write totals on right 1 2 -1 6 5 9 3 7 Do a row operation: e.g. 1 -1 2 6 5 9 7 r2 r2 - 2r1 3-2x9

16
**4.4.3 Row sums A way to check calculations: Do a row operation: e.g.**

Add up rows 2. Write totals on right 1 2 -1 6 5 9 3 7 Do a row operation: e.g. 1 -1 2 6 5 9 r2 r2 - 2r1 -1 -3 -11 -15 7 Check row sums: e.g. 0 + (-1) – 3 – 11 = -15

17
**4.5 Examples 1 4 3 2 8 6 -2 x y z 9 18 = Matrix-vector system**

Write in augmented form 1 4 3 2 8 6 -2 x y z 9 18 =

18
4.5 Examples – EXAMPLE 1 Matrix-vector system Write in augmented form Write on row sums -2 1 4 3 2 8 6 9 18 15 36 6 Use the top left entry to create zeros below it First get a zero in the second row: -2 1 3 2 9 4 15 r2 r2 - 4r1 36 – 4x15 6

19
4.5 Examples – EXAMPLE 1 Matrix-vector system Write in augmented form Write on row sums -2 1 4 3 2 8 6 9 18 15 36 6 Use the top left entry to create zeros below it First get a zero in the second row: -2 1 3 2 9 4 15 -24 6 r2 r2 - 4r1

20
4.5 Examples – EXAMPLE 1 Matrix-vector system Write in augmented form Write on row sums -2 1 4 3 2 8 6 9 18 15 36 6 Use the top left entry to create zeros below it First get a zero in the second row: -2 1 3 2 9 4 15 -24 6 r2 r2 - 4r1 -6 -18

21
**4.5 Examples – EXAMPLE 1 Check row sums before continuing...**

-2 1 4 3 2 8 6 9 18 15 36 -2 1 3 2 -6 9 -18 4 15 -24 6 r2 r2 - 4r1 Check row sums before continuing... – 18 = OK! = OK! = OK!

22
**4.5 Examples – EXAMPLE 1 Now get a zero in the third row:**

-2 1 4 3 2 8 6 9 18 15 36 -2 1 3 2 -6 9 -18 4 15 -24 6 r2 r2 - 4r1 Now get a zero in the third row: -11 1 2 3 -6 -5 9 -18 -23 15 -24 -39 r3 r3 - 3r1 Want upper triangular form so swap rows 2 and 3 1 2 3 -6 9 -18 15 -24 -11 -5 -23 -39 r r3

23
**4.5 Examples – EXAMPLE 1 Now solve by backwards substitution:**

2 3 -6 9 -18 15 -24 -5 -11 -23 -39 Now solve by backwards substitution: r3 : -6z = z = 3 r2 : -5y – 11z = -23 -5y -33 =-23 y = -2 r1 : x + 2y +3z = 9 x = 9 x = 4 Hence: x = 4, y = -2, z = 3 is the unique solution.

24
**4.5 Examples 1 -1 x y z 6 = Matrix-vector system**

Write in augmented form 1 -1 x y z 6 =

25
4.5 Examples – EXAMPLE 2 Matrix-vector system Write in augmented form Write on row sums 1 -1 6 1 6 Use the top left entry to create zeros below it First get a zero in the third row: 1 -1 6 -2 r3 r3 - r1

26
4.5 Examples – EXAMPLE 2 r3 r3 - r1 1 -1 6 -2 Use second row to get a zero in the third row: r3 r3 – r2 2 1 -1 6 -8 -6

27
**4.5 Examples – EXAMPLE 2 Solve by backwards substitution: 2 1 -1 6 -8**

-1 6 -8 -6 Solve by backwards substitution: r3 : 2z = -8 z = -4 UNIQUE SOLUTION r2 : y - z = 6 y + 4 = 6 y = 2 r1 : x - y = 1 x - 2 = 1 x = 3

28
4.6 Determinants Question: During the elimination process, what has changed about the determinant of the matrix? Swapping rows multiplies the determinant by (-1) Adding or subtracting multiples of rows does not change the determinant

29
**4.6 Determinants In EXAMPLE 1 we used one swap operation to get from**

2 3 -6 -11 -5 4 8 6 -2 = B A = Calculating the determinant: |A|= (-1) |B| = (-1)x(1)(-5)(-6) = -30 Non-zero, so we got a unique solution

30
**4.6 Determinants In EXAMPLE 2 we used no swaps to get from = B A =**

1 -1 2 = B A = Calculating the determinant: |A|= |B| = (1)(1)(2) = 2 Non-zero, so got a unique solution

31
**4.6 Non-Standard Gaussian Elimination**

In standard Gaussian Elimination the following operation were allowed: Swap two rows; Add or Subtract a multiple of a row from another row. In Non-Standard Gaussian Elimination we are also allowed to do the following: Multiply a row by a constant. E.g. r r3

32
**4.6 Non-Standard Gaussian Elimination**

Quick Example: In Standard G.E. 3 1 5 -1 4 8 3 1 -8/3 4 8 -16/3 r r2 - 5r1/3 This is a bit messy with the fractions. However, in Non-Standard G.E. 3 1 5 -1 4 8 3 1 -8 4 8 -16 r r2 - 5r1 However, in doing this we have multiplied the determinant by 3.

33
**4.7 Backwards substitution: more general case**

Two cases after elimination process: All diagonal entries non-zero, then determinant is non-zero. Hence, get answer by backwards substitution. Is a zero on the diagonal, then determinant is zero. Either get infinite solutions or no solutions.

34
4.7.1 Case of No Solutions Suppose we followed the elimination process and got to: 3 1 2 -1 x y z 11 -3 9 = 3 1 2 -1 11 -3 9 Zeros on the diagonal, so determinant is zero. ROW 3 gives the equation 0x + 0y + 0z = 9 This is impossible. Hence there are no solutions.

35
**4.7.1 Case of Infinite solutions**

Suppose instead that 3 1 2 -1 x y z 11 -3 = 3 1 2 -1 11 -3 ROW 3 now OK: x + 0y + 0z = 0 Have two equations in three unknowns Get infinitely many solutions

36
**4.7.1 Case of Infinite solutions**

Three steps: In the final (echelon-form) of the matrix, circle the first non-zero entry in each row Find the columns that have no circles in. Each column corresponds to a variable. Assign a new name to each of the chosen variables, then use back substitution on the non-zero rows.

37
**4.7.1 Case of Infinite solutions**

3 1 2 -1 x y z 11 -3 = 3 1 2 -1 11 -3 Circle first non-zero row entries Find column with no circles in Column 2 corresponds to the y variable 3. Assign a name to y: let y = α

38
**4.7.1 Case of Infinite solutions**

3 1 2 -1 x y z 11 -3 = 3 1 2 -1 11 -3 Solve by back substitution: r3 : Tells us nothing r2 : -z = z = 3 r1 : 3x + y + 2z = 11 3x + α + 6 = 11 x = (5 – α)/3 So, solution is x = (5 – α)/3, y = α, z = 3 for any α

Similar presentations

OK

Hello Mr. Anderson… We’ve been waiting for you.. Hello Mr. Anderson… We’ve been waiting for you.

Hello Mr. Anderson… We’ve been waiting for you.. Hello Mr. Anderson… We’ve been waiting for you.

© 2018 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google