1.any all-zero rows are at the bottom. 2.correct step pattern of first non-zero row entries. 4.4.1...

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

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

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

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

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

a11

a22

a33

ann

a12 a13

a21

a31

a23

a32

an1

a1n

a2n

an2

4.4.1 Formal process (Handout 3)

Create zeros

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

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

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

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

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

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

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

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

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

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

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

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:

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:

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:

-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

-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

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

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

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

• 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

• 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

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

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

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|

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

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.

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.

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

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

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.

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

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