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Transcript of 1.Any all-zero rows are at the bottom. 2.Correct step pattern of first non-zero row entries. 4.4.1...
![Page 1: 1.Any all-zero rows are at the bottom. 2.Correct step pattern of first non-zero row entries. 4.4.1 Generalised Row Echelon Form.](https://reader034.fdocuments.net/reader034/viewer/2022051515/5515dbff550346cf6f8b4ace/html5/thumbnails/1.jpg)
1. Any all-zero rows are at the bottom.
2. Correct ‘step pattern’ of first non-zero row entries.
4.4.1 Generalised Row Echelon Form
![Page 2: 1.Any all-zero rows are at the bottom. 2.Correct step pattern of first non-zero row entries. 4.4.1 Generalised Row Echelon Form.](https://reader034.fdocuments.net/reader034/viewer/2022051515/5515dbff550346cf6f8b4ace/html5/thumbnails/2.jpg)
2
0
0
1 3
0 1
0 0
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
ROW 2
![Page 3: 1.Any all-zero rows are at the bottom. 2.Correct step pattern of first non-zero row entries. 4.4.1 Generalised Row Echelon Form.](https://reader034.fdocuments.net/reader034/viewer/2022051515/5515dbff550346cf6f8b4ace/html5/thumbnails/3.jpg)
2
0
0
1 3
1 1
0 2
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
ROW 2
ROW 3
![Page 4: 1.Any all-zero rows are at the bottom. 2.Correct step pattern of first non-zero row entries. 4.4.1 Generalised Row Echelon Form.](https://reader034.fdocuments.net/reader034/viewer/2022051515/5515dbff550346cf6f8b4ace/html5/thumbnails/4.jpg)
2
0
0
1 3
1 1
2 2
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
ROW 2
ROW 3
![Page 5: 1.Any all-zero rows are at the bottom. 2.Correct step pattern of first non-zero row entries. 4.4.1 Generalised Row Echelon Form.](https://reader034.fdocuments.net/reader034/viewer/2022051515/5515dbff550346cf6f8b4ace/html5/thumbnails/5.jpg)
2
0
0
1 3
0
10
0
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
ROW 3
2
0
0
0 0 1 0
ROW 2
ROW 4
![Page 6: 1.Any all-zero rows are at the bottom. 2.Correct step pattern of first non-zero row entries. 4.4.1 Generalised Row Echelon Form.](https://reader034.fdocuments.net/reader034/viewer/2022051515/5515dbff550346cf6f8b4ace/html5/thumbnails/6.jpg)
a11
a22
a33
ann
a12 a13
a21
a31
a23
a32
an1
a1n
a2n
an2
4.4.1 Formal process (Handout 3)
Create zeros
![Page 7: 1.Any all-zero rows are at the bottom. 2.Correct step pattern of first non-zero row entries. 4.4.1 Generalised Row Echelon Form.](https://reader034.fdocuments.net/reader034/viewer/2022051515/5515dbff550346cf6f8b4ace/html5/thumbnails/7.jpg)
a11
a'22
a33
ann
a12 a13
0a31
a'23
a32
an1
a1n
a'2n
an2
4.4.1 Formal process (Handout 3)
Create zeros
![Page 8: 1.Any all-zero rows are at the bottom. 2.Correct step pattern of first non-zero row entries. 4.4.1 Generalised Row Echelon Form.](https://reader034.fdocuments.net/reader034/viewer/2022051515/5515dbff550346cf6f8b4ace/html5/thumbnails/8.jpg)
a11
a'22
a'33
ann
a12 a13
0 0
a'23
a'32
an1
a1n
a'2n
an2
4.4.1 Formal process (Handout 3)
Create zeros
![Page 9: 1.Any all-zero rows are at the bottom. 2.Correct step pattern of first non-zero row entries. 4.4.1 Generalised Row Echelon Form.](https://reader034.fdocuments.net/reader034/viewer/2022051515/5515dbff550346cf6f8b4ace/html5/thumbnails/9.jpg)
a11
a'22
a'33
a'nn
a12 a13
0 0
a'23
a'32
0
a1n
a'2n
a'n2
4.4.1 Formal process
Create zeros
![Page 10: 1.Any all-zero rows are at the bottom. 2.Correct step pattern of first non-zero row entries. 4.4.1 Generalised Row Echelon Form.](https://reader034.fdocuments.net/reader034/viewer/2022051515/5515dbff550346cf6f8b4ace/html5/thumbnails/10.jpg)
a11
a'22
a'33
a'nn
a12 a13
0 0
a'23
a'32
0
a1n
a'2n
a'n2
4.4.1 Formal process (Handout 3)
Create zeros
![Page 11: 1.Any all-zero rows are at the bottom. 2.Correct step pattern of first non-zero row entries. 4.4.1 Generalised Row Echelon Form.](https://reader034.fdocuments.net/reader034/viewer/2022051515/5515dbff550346cf6f8b4ace/html5/thumbnails/11.jpg)
a11
a'22
a''33
a'nn
a12 a13
0 0
a'23
0
0
a1n
a'2n
a'n2
4.4.1 Formal process (Handout 3)
Create zeros
![Page 12: 1.Any all-zero rows are at the bottom. 2.Correct step pattern of first non-zero row entries. 4.4.1 Generalised Row Echelon Form.](https://reader034.fdocuments.net/reader034/viewer/2022051515/5515dbff550346cf6f8b4ace/html5/thumbnails/12.jpg)
a11
a'22
a''33
a''nn
a12 a13
0 0
a'23
0
0
a1n
a'2n
0
4.4.1 Formal process (Handout 3)
Create zeros
![Page 13: 1.Any all-zero rows are at the bottom. 2.Correct step pattern of first non-zero row entries. 4.4.1 Generalised Row Echelon Form.](https://reader034.fdocuments.net/reader034/viewer/2022051515/5515dbff550346cf6f8b4ace/html5/thumbnails/13.jpg)
a11
a'22
a''33
a''nn
a12 a13
0 0
a'23
0
0
a1n
a'2n
0
4.4.1 Formal process (Handout 3)
Create zeros
![Page 14: 1.Any all-zero rows are at the bottom. 2.Correct step pattern of first non-zero row entries. 4.4.1 Generalised Row Echelon Form.](https://reader034.fdocuments.net/reader034/viewer/2022051515/5515dbff550346cf6f8b4ace/html5/thumbnails/14.jpg)
4.4.2 Augmented Matrix notation
• We perform row operations on the matrix and the opposite row:
1 2 1
1 1 1 -1
-1 2
x y z
6 1 5
=
• Combine both of these into one matrix called the augmented matrix:
1 2 1
1 1 1 -1
-1 2
6 1 5
![Page 15: 1.Any all-zero rows are at the bottom. 2.Correct step pattern of first non-zero row entries. 4.4.1 Generalised Row Echelon Form.](https://reader034.fdocuments.net/reader034/viewer/2022051515/5515dbff550346cf6f8b4ace/html5/thumbnails/15.jpg)
4.4.3 Row sums
• A way to check calculations:
1 2 1
1 1 1 -1
-1 2
6 1 5
1. Add up rows2. Write totals on right
9 3 7
ROW SUMS
• Do a row operation: e.g.
r2 r2 - 2r1 3-2x9 1
1
1 1
-1 2
6
5
9
7
![Page 16: 1.Any all-zero rows are at the bottom. 2.Correct step pattern of first non-zero row entries. 4.4.1 Generalised Row Echelon Form.](https://reader034.fdocuments.net/reader034/viewer/2022051515/5515dbff550346cf6f8b4ace/html5/thumbnails/16.jpg)
0 -1 -3 -11
4.4.3 Row sums
• A way to check calculations:
1 2 1
1 1 1 -1
-1 2
6 1 5
9 3 7
ROW SUMS
• Do a row operation: e.g.
r2 r2 - 2r1
1
1
1 1
-1 2
6
5
9 -15 7
• Check row sums: e.g. 0 + (-1) – 3 – 11 = -15
1. Add up rows2. Write totals on right
![Page 17: 1.Any all-zero rows are at the bottom. 2.Correct step pattern of first non-zero row entries. 4.4.1 Generalised Row Echelon Form.](https://reader034.fdocuments.net/reader034/viewer/2022051515/5515dbff550346cf6f8b4ace/html5/thumbnails/17.jpg)
4.5 Examples
1 4 3
2 3 8 6
1 -2
x y z
9 18 4
=
1. Matrix-vector system2. Write in augmented
form
![Page 18: 1.Any all-zero rows are at the bottom. 2.Correct step pattern of first non-zero row entries. 4.4.1 Generalised Row Echelon Form.](https://reader034.fdocuments.net/reader034/viewer/2022051515/5515dbff550346cf6f8b4ace/html5/thumbnails/18.jpg)
4.5 Examples – EXAMPLE 1
-2
1 4 3
2 3 8 6
1
9 18 4
1. Matrix-vector system2. Write in augmented
form3. Write on row sums
15 36 6
• Use the top left entry to create zeros below it
r2 r2 - 4r1
-2
1
3
2 3
1
9
4
1536 – 4x15 6
• First get a zero in the second row:
![Page 19: 1.Any all-zero rows are at the bottom. 2.Correct step pattern of first non-zero row entries. 4.4.1 Generalised Row Echelon Form.](https://reader034.fdocuments.net/reader034/viewer/2022051515/5515dbff550346cf6f8b4ace/html5/thumbnails/19.jpg)
4.5 Examples – EXAMPLE 1
-2
1 4 3
2 3 8 6
1
9 18 4
1. Matrix-vector system2. Write in augmented
form3. Write on row sums
15 36 6
• Use the top left entry to create zeros below it
r2 r2 - 4r1
-2
1
3
2 3
1
9
4
15-24 6
• First get a zero in the second row:
![Page 20: 1.Any all-zero rows are at the bottom. 2.Correct step pattern of first non-zero row entries. 4.4.1 Generalised Row Echelon Form.](https://reader034.fdocuments.net/reader034/viewer/2022051515/5515dbff550346cf6f8b4ace/html5/thumbnails/20.jpg)
0 0 -6 -18
4.5 Examples – EXAMPLE 1
-2
1 4 3
2 3 8 6
1
9 18 4
1. Matrix-vector system2. Write in augmented
form3. Write on row sums
15 36 6
• Use the top left entry to create zeros below it
r2 r2 - 4r1
-2
1
3
2 3
1
9
4
15-24 6
• First get a zero in the second row:
![Page 21: 1.Any all-zero rows are at the bottom. 2.Correct step pattern of first non-zero row entries. 4.4.1 Generalised Row Echelon Form.](https://reader034.fdocuments.net/reader034/viewer/2022051515/5515dbff550346cf6f8b4ace/html5/thumbnails/21.jpg)
-2
1 4 3
2 3 8 6
1
9 18 4
15 36 6
r2 r2 - 4r1
-2
1 0 3
2 3 0 -6 1
9-18 4
15-24 6
• Check row sums before continuing...
1 + 2 + 3 + 9 = 15 ... OK!0 + 0 - 6 – 18 = -24 ... OK!3 + 1 - 2 + 4 = 6 ... OK!
4.5 Examples – EXAMPLE 1
![Page 22: 1.Any all-zero rows are at the bottom. 2.Correct step pattern of first non-zero row entries. 4.4.1 Generalised Row Echelon Form.](https://reader034.fdocuments.net/reader034/viewer/2022051515/5515dbff550346cf6f8b4ace/html5/thumbnails/22.jpg)
-2
1 4 3
2 3 8 6
1
9 18 4
15 36 6
r2 r2 - 4r1
-2
1 0 3
2 3 0 -6 1
9-18 4
15-24 6
• Now get a zero in the third row:
r3 r3 - 3r1
-11
1 0 0
2 3 0 -6 -5
9-18
-23
15-24
-39• Want upper triangular form so swap rows 2 and 3
r2 r3
1
0
2 3
0 -6
9
-18
15
-24-11 0 -5 -23 -39
4.5 Examples – EXAMPLE 1
![Page 23: 1.Any all-zero rows are at the bottom. 2.Correct step pattern of first non-zero row entries. 4.4.1 Generalised Row Echelon Form.](https://reader034.fdocuments.net/reader034/viewer/2022051515/5515dbff550346cf6f8b4ace/html5/thumbnails/23.jpg)
1
0
2 3
0 -6
9
-18
15
-24-11 0 -5 -23 -39
• Now solve by backwards substitution:
r3 : -6z = -18 z = 3
r2 : -5y – 11z = -23
• Hence: x = 4, y = -2, z = 3 is the unique solution.
4.5 Examples – EXAMPLE 1
-5y -33 =-23 y = -2
r1 : x + 2y +3z = 9 x - 4 + 9 = 9 x = 4
![Page 24: 1.Any all-zero rows are at the bottom. 2.Correct step pattern of first non-zero row entries. 4.4.1 Generalised Row Echelon Form.](https://reader034.fdocuments.net/reader034/viewer/2022051515/5515dbff550346cf6f8b4ace/html5/thumbnails/24.jpg)
4.5 Examples
1 0 1
-1 0 1 -1
0 1
x y z
16
-1 =
1. Matrix-vector system2. Write in augmented
form
![Page 25: 1.Any all-zero rows are at the bottom. 2.Correct step pattern of first non-zero row entries. 4.4.1 Generalised Row Echelon Form.](https://reader034.fdocuments.net/reader034/viewer/2022051515/5515dbff550346cf6f8b4ace/html5/thumbnails/25.jpg)
4.5 Examples – EXAMPLE 2
1
1 0 1
-1 0 1 -1
0
16
-1
1. Matrix-vector system2. Write in augmented
form3. Write on row sums
1 6 1
• Use the top left entry to create zeros below it
r3 r3 - r1
• First get a zero in the third row:
1
1 0 0
-1 0 1 -11
16
-2
1 6 0
![Page 26: 1.Any all-zero rows are at the bottom. 2.Correct step pattern of first non-zero row entries. 4.4.1 Generalised Row Echelon Form.](https://reader034.fdocuments.net/reader034/viewer/2022051515/5515dbff550346cf6f8b4ace/html5/thumbnails/26.jpg)
• Use second row to get a zero in the third row:
r3 r3 – r2
2
1 0 0
-1 0 1 -1 0
16
-8
16
-6
4.5 Examples – EXAMPLE 2
r3 r3 - r1
1
1 0 1
-1 0 1 -1
0
16
-1
1 6 1 1
1 0 0
-1 0 1 -11
16
-2
1 6 0
![Page 27: 1.Any all-zero rows are at the bottom. 2.Correct step pattern of first non-zero row entries. 4.4.1 Generalised Row Echelon Form.](https://reader034.fdocuments.net/reader034/viewer/2022051515/5515dbff550346cf6f8b4ace/html5/thumbnails/27.jpg)
• Solve by backwards substitution:
r3 : 2z = -8
r2 : y - z = 6
r1 : x - y = 1
UNIQUE SOLUTION
2
1 0 0
-1 0 1 -1 0
16
-8
16
-6
4.5 Examples – EXAMPLE 2
z = -4
y + 4 = 6 y = 2
x - 2 = 1 x = 3
![Page 28: 1.Any all-zero rows are at the bottom. 2.Correct step pattern of first non-zero row entries. 4.4.1 Generalised Row Echelon Form.](https://reader034.fdocuments.net/reader034/viewer/2022051515/5515dbff550346cf6f8b4ace/html5/thumbnails/28.jpg)
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
![Page 29: 1.Any all-zero rows are at the bottom. 2.Correct step pattern of first non-zero row entries. 4.4.1 Generalised Row Echelon Form.](https://reader034.fdocuments.net/reader034/viewer/2022051515/5515dbff550346cf6f8b4ace/html5/thumbnails/29.jpg)
4.6 Determinants
• In EXAMPLE 1 we used one swap operation to get from
1
0
2 3
0 -6-11 0 -5
1 4 3
2 3 8 6
1 -2
|A|= (-1) |B|
• Calculating the determinant:
• Non-zero, so we got a unique solution
= (-1)x(1)(-5)(-6) = -30
A = = B
![Page 30: 1.Any all-zero rows are at the bottom. 2.Correct step pattern of first non-zero row entries. 4.4.1 Generalised Row Echelon Form.](https://reader034.fdocuments.net/reader034/viewer/2022051515/5515dbff550346cf6f8b4ace/html5/thumbnails/30.jpg)
4.6 Determinants
• In EXAMPLE 2 we used no swaps to get from
1
0
-1 0
0 2 -1 0 1
1 0 1
-1 0 1 -10 1
• Calculating the determinant:
• Non-zero, so got a unique solution
= (1)(1)(2) = 2
A = = B
|A|= |B|
![Page 31: 1.Any all-zero rows are at the bottom. 2.Correct step pattern of first non-zero row entries. 4.4.1 Generalised Row Echelon Form.](https://reader034.fdocuments.net/reader034/viewer/2022051515/5515dbff550346cf6f8b4ace/html5/thumbnails/31.jpg)
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.
r3 2r3
![Page 32: 1.Any all-zero rows are at the bottom. 2.Correct step pattern of first non-zero row entries. 4.4.1 Generalised Row Echelon Form.](https://reader034.fdocuments.net/reader034/viewer/2022051515/5515dbff550346cf6f8b4ace/html5/thumbnails/32.jpg)
4.6 Non-Standard Gaussian Elimination
• Quick Example: In Standard G.E.
3 15 -1
44
88
r2 r2 - 5r1/3 3 10 -8/3
4 8-16/3-8/3
• This is a bit messy with the fractions. However, in Non-Standard G.E.
3 15 -1
44
88
r2 3r2 - 5r1 3 10 -8
4 8-16-8
• However, in doing this we have multiplied the determinant by 3.
![Page 33: 1.Any all-zero rows are at the bottom. 2.Correct step pattern of first non-zero row entries. 4.4.1 Generalised Row Echelon Form.](https://reader034.fdocuments.net/reader034/viewer/2022051515/5515dbff550346cf6f8b4ace/html5/thumbnails/33.jpg)
4.7 Backwards substitution: more general case
• Two cases after elimination process:
1. All diagonal entries non-zero, then determinant is non-zero. Hence, get answer by backwards substitution.
2. Is a zero on the diagonal, then determinant is zero. Either get infinite solutions or no solutions.
![Page 34: 1.Any all-zero rows are at the bottom. 2.Correct step pattern of first non-zero row entries. 4.4.1 Generalised Row Echelon Form.](https://reader034.fdocuments.net/reader034/viewer/2022051515/5515dbff550346cf6f8b4ace/html5/thumbnails/34.jpg)
4.7.1 Case of No Solutions
3 0 0
1 2 0 -1
0 0
x y z
11 -3 9
= 3 0 0
1 2 0 -1
0 0
11 -3 9
• Suppose we followed the elimination process and got to:
• Zeros on the diagonal, so determinant is zero.
• ROW 3 gives the equation
• This is impossible. Hence there are no solutions.
0x + 0y + 0z = 9
![Page 35: 1.Any all-zero rows are at the bottom. 2.Correct step pattern of first non-zero row entries. 4.4.1 Generalised Row Echelon Form.](https://reader034.fdocuments.net/reader034/viewer/2022051515/5515dbff550346cf6f8b4ace/html5/thumbnails/35.jpg)
4.7.1 Case of Infinite solutions
• Suppose instead that
3 0 0
1 2 0 -1
0 0
x y z
11 -3 0
= 3 0 0
1 2 0 -1
0 0
11 -3 0
• ROW 3 now OK: 0x + 0y + 0z = 0
• Have two equations in three unknowns
• Get infinitely many solutions
![Page 36: 1.Any all-zero rows are at the bottom. 2.Correct step pattern of first non-zero row entries. 4.4.1 Generalised Row Echelon Form.](https://reader034.fdocuments.net/reader034/viewer/2022051515/5515dbff550346cf6f8b4ace/html5/thumbnails/36.jpg)
4.7.1 Case of Infinite solutions
• Three steps:
1. In the final (echelon-form) of the matrix, circle the first non-zero entry in each row
2. Find the columns that have no circles in. Each column corresponds to a variable.
3. Assign a new name to each of the chosen variables, then use back substitution on the non-zero rows.
![Page 37: 1.Any all-zero rows are at the bottom. 2.Correct step pattern of first non-zero row entries. 4.4.1 Generalised Row Echelon Form.](https://reader034.fdocuments.net/reader034/viewer/2022051515/5515dbff550346cf6f8b4ace/html5/thumbnails/37.jpg)
4.7.1 Case of Infinite solutions
3 0 0
1 2 0 -1
0 0
11 -3 0
1. Circle first non-zero row entries2. Find column with no circles in
3 0 0
1 2 0 -1
0 0
x y z
11 -3 0
=
Column 2 corresponds to the y variable
3. Assign a name to y: let y = α
![Page 38: 1.Any all-zero rows are at the bottom. 2.Correct step pattern of first non-zero row entries. 4.4.1 Generalised Row Echelon Form.](https://reader034.fdocuments.net/reader034/viewer/2022051515/5515dbff550346cf6f8b4ace/html5/thumbnails/38.jpg)
4.7.1 Case of Infinite solutions
3 0 0
1 2 0 -1
0 0
11 -3 0
3 0 0
1 2 0 -1
0 0
x y z
11 -3 0
=
r3 : Tells us nothing
r2 : -z = -3 z = 3 r1 : 3x + y + 2z = 11
• Solve by back substitution:
• So, solution is x = (5 – α)/3, y = α, z = 3 for any α
3x + α + 6 = 11 x = (5 – α)/3