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 from2 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 solutions3 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 solutions3 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