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**4.4.1 Generalised Row Echelon Form**

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

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**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

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**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

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**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

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**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

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**4.4.1 Formal process (Handout 3)**

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

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**4.4.1 Formal process (Handout 3)**

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

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**4.4.1 Formal process (Handout 3)**

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

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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

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**4.4.1 Formal process (Handout 3)**

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

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**4.4.1 Formal process (Handout 3)**

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

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**4.4.1 Formal process (Handout 3)**

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

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**4.4.1 Formal process (Handout 3)**

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

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**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

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**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

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**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

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**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 =

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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

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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

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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

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**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!

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**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

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**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.

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

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

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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

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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

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**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

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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

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**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

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**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

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**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

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**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.

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**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.

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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.

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**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

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**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.

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**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

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**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

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