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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.

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Presentation on theme: "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."— Presentation transcript:

1 1.Any all-zero rows are at the bottom. 2.Correct step pattern of first non-zero row entries Generalised Row Echelon Form

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

3 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

4 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

5 Generalised Row Echelon Form Any all-zero row at the bottom Correct step pattern of first non-zero row entries ROW 1 ROW ROW 2 ROW 4

6 a 11 a 22 a 33 a nn a 12 a 13 a 21 a 31 a 23 a 32 a n1 a 1n a 2n a n Formal process (Handout 3) Create zeros

7 a 11 a' 22 a 33 a nn a 12 a 13 0 a 31 a' 23 a 32 a n1 a 1n a' 2n a n Formal process (Handout 3) Create zeros

8 a 11 a' 22 a' 33 a nn a 12 a a' 23 a' 32 a n1 a 1n a' 2n a n Formal process (Handout 3) Create zeros

9 a 11 a' 22 a' 33 a' nn a 12 a a' 23 a' 32 0 a 1n a' 2n a' n Formal process Create zeros

10 a 11 a' 22 a' 33 a' nn a 12 a a' 23 a' 32 0 a 1n a' 2n a' n Formal process (Handout 3) Create zeros

11 a 11 a' 22 a'' 33 a' nn a 12 a a' a 1n a' 2n a' n Formal process (Handout 3) Create zeros

12 a 11 a' 22 a'' 33 a'' nn a 12 a a' a 1n a' 2n Formal process (Handout 3) Create zeros

13 a 11 a' 22 a'' 33 a'' nn a 12 a a' a 1n a' 2n Formal process (Handout 3) Create zeros

14 4.4.2 Augmented Matrix notation We perform row operations on the matrix and the opposite row: x y z = Combine both of these into one matrix called the augmented matrix:

15 4.4.3 Row sums A way to check calculations: Add up rows 2. Write totals on right ROW SUMS Do a row operation: e.g. r 2 r 2 - 2r 1 3-2x

16 Row sums A way to check calculations: ROW SUMS Do a row operation: e.g. r 2 r 2 - 2r Check row sums: e.g. 0 + (-1) – 3 – 11 = Add up rows 2. Write totals on right

17 4.5 Examples x y z = 1.Matrix-vector system 2.Write in augmented form

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

19 4.5 Examples – EXAMPLE Matrix-vector system 2.Write in augmented form 3.Write on row sums Use the top left entry to create zeros below it r 2 r 2 - 4r First get a zero in the second row:

20 Examples – EXAMPLE Matrix-vector system 2.Write in augmented form 3.Write on row sums Use the top left entry to create zeros below it r 2 r 2 - 4r First get a zero in the second row:

21 r 2 r 2 - 4r Check row sums before continuing = OK! – 18 = OK! = 6... OK! 4.5 Examples – EXAMPLE 1

22 r 2 r 2 - 4r Now get a zero in the third row: r 3 r 3 - 3r Want upper triangular form so swap rows 2 and 3 r 2 r Examples – EXAMPLE 1

23 Now solve by backwards substitution: r 3 : -6z = -18 z = 3 r 2 : -5y – 11z = -23 Hence: x = 4, y = -2, z = 3 is the unique solution. 4.5 Examples – EXAMPLE 1 -5y -33 =-23y = -2 r 1 : x + 2y +3z = 9x = 9x = 4

24 4.5 Examples x y z 1 6 = 1.Matrix-vector system 2.Write in augmented form

25 4.5 Examples – EXAMPLE Matrix-vector system 2.Write in augmented form 3.Write on row sums Use the top left entry to create zeros below it r 3 r 3 - r 1 First get a zero in the third row:

26 Use second row to get a zero in the third row: r 3 r 3 – r Examples – EXAMPLE 2 r 3 r 3 - r

27 Solve by backwards substitution: r 3 : 2z = -8 r 2 : y - z = 6 r 1 : x - y = 1 UNIQUE SOLUTION Examples – EXAMPLE 2 z = -4 y + 4 = 6y = 2 x - 2 = 1x = 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 |A|= (-1) |B| Calculating the determinant: Non-zero, so we got a unique solution = (-1)x(1)(-5)(-6) = -30 A = = B

30 4.6 Determinants In EXAMPLE 2 we used no swaps to get from Calculating the determinant: Non-zero, so got a unique solution = (1)(1)(2) = 2 A = = B |A|= |B|

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 3 2r 3

32 4.6 Non-Standard Gaussian Elimination Quick Example: In Standard G.E r 2 r 2 - 5r 1 / / /3-8/3 This is a bit messy with the fractions. However, in Non-Standard G.E r 2 3r 2 - 5r However, in doing this we have multiplied the determinant by 3.

33 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.

34 4.7.1 Case of No Solutions x y z = 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

35 4.7.1 Case of Infinite solutions Suppose instead that x y z = ROW 3 now OK: 0x + 0y + 0z = 0 Have two equations in three unknowns Get infinitely many solutions

36 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.

37 4.7.1 Case of Infinite solutions Circle first non-zero row entries 2.Find column with no circles in x y z = Column 2 corresponds to the y variable 3.Assign a name to y: let y = α

38 4.7.1 Case of Infinite solutions x y z = r 3 : Tells us nothing r 2 : -z = -3 z = 3 r 1 : 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


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