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Row Echelon Form and Augmented Matrix Notation

Learn about the generalised row echelon form, augmented matrix notation, and how to perform row operations in order to create zeros and solve matrix-vector systems. Understand the impact of row operations on determinants.

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Row Echelon Form and Augmented Matrix Notation

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  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 ROW 1 ROW 2 1 3 2 0 0 1 0 0 0

  3. 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 1 3 2 0 1 1 0 2 0

  4. 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 1 3 2 0 1 1 0 2 2

  5. 4.4.1 Generalised Row Echelon Form • Any all-zero row at the bottom • Correct ‘step pattern’ of first non-zero row entries 0 0 0 1 ROW 3 ROW 1 ROW 2 ROW 4 1 3 2 2 0 0 0 0 0 0 0 1

  6. 4.4.1 Formal process (Handout 3) a12 a13 a1n a11 a21 a23 a2n a22 a31 a32 a33 Create zeros an1 an2 ann

  7. 4.4.1 Formal process (Handout 3) a12 a13 a1n a11 a'23 0 a'2n a'22 a31 a32 a33 Create zeros an1 an2 ann

  8. 4.4.1 Formal process (Handout 3) a12 a13 a1n a11 a'23 0 a'2n a'22 a'32 0 a'33 Create zeros an1 an2 ann

  9. 4.4.1 Formal process a12 a13 a1n a11 a'23 0 a'2n a'22 a'32 0 a'33 Create zeros 0 a'n2 a'nn

  10. 4.4.1 Formal process (Handout 3) a12 a13 a1n a11 a'23 0 a'2n a'22 a'32 0 a'33 Create zeros 0 a'n2 a'nn

  11. 4.4.1 Formal process (Handout 3) a12 a13 a1n a11 a'23 0 a'2n a'22 0 0 a''33 Create zeros 0 a'n2 a'nn

  12. 4.4.1 Formal process (Handout 3) a12 a13 a1n a11 a'23 0 a'2n a'22 0 0 a''33 Create zeros 0 0 a''nn

  13. 4.4.1 Formal process (Handout 3) a12 a13 a1n a11 a'23 0 a'2n a'22 0 0 a''33 Create zeros 0 0 a''nn

  14. 4.4.2 Augmented Matrix notation • We perform row operations on the matrix and the opposite row: • Combine both of these into one matrix called the augmented matrix: 1 1 1 1 6 x 6 1 1 2 2 1 1 -1 -1 1 y = 1 1 1 -1 -1 2 2 z 5 5

  15. 4.4.3 Row sums • A way to check calculations: ROW SUMS Add up rows 2. Write totals on right 9 3 7 • Do a row operation: e.g. 9 3-2x9 1 1 1 1 6 6 1 1 7 2 1 -1 1 r2 r2 - 2r1 1 1 -1 -1 2 2 5 5

  16. 4.4.3 Row sums • A way to check calculations: ROW SUMS Add up rows 2. Write totals on right 9 3 7 • Do a row operation: e.g. 9 -15 0 -1 -3 -11 1 1 1 1 6 6 1 1 7 2 1 -1 1 r2 r2 - 2r1 1 1 -1 -1 2 2 5 5 • Check row sums: e.g. 0 + (-1) – 3 – 11 = -15

  17. 4.5 Examples Matrix-vector system Write in augmented form 2 3 x 9 1 4 8 6 y = 18 z 3 1 -2 4

  18. 4.5 Examples – EXAMPLE 1 Matrix-vector system Write in augmented form Write on row sums 15 36 6 • Use the top left entry to create zeros below it • First get a zero in the second row: 2 2 3 3 9 9 1 1 r2 r2 - 4r1 15 4 8 6 18 36 – 4x15 3 3 1 1 -2 -2 4 4 6

  19. 4.5 Examples – EXAMPLE 1 Matrix-vector system Write in augmented form Write on row sums 15 15 -24 36 6 6 • Use the top left entry to create zeros below it • First get a zero in the second row: 2 2 3 3 9 9 1 1 r2 r2 - 4r1 4 8 6 18 3 3 1 1 -2 -2 4 4

  20. 4.5 Examples – EXAMPLE 1 Matrix-vector system Write in augmented form Write on row sums 15 15 -24 36 6 6 • Use the top left entry to create zeros below it • First get a zero in the second row: 2 2 3 3 9 9 1 1 r2 r2 - 4r1 4 8 6 18 0 0 -6 -18 3 3 1 1 -2 -2 4 4

  21. 4.5 Examples – EXAMPLE 1 15 15 36 -24 6 6 • Check row sums before continuing... 1 + 2 + 3 + 9 = 15 ... OK! 0 + 0 - 6 – 18 = -24 ... OK! 3 + 1 - 2 + 4 = 6 ... OK! 2 2 3 3 9 9 1 1 r2 r2 - 4r1 0 4 0 8 6 -6 18 -18 3 3 1 1 -2 -2 4 4

  22. 4.5 Examples – EXAMPLE 1 15 15 15 15 36 -24 -24 -24 -39 6 6 • Now get a zero in the third row: • Want upper triangular form so swap rows 2 and 3 2 2 2 2 3 3 3 3 9 9 9 9 1 1 1 1 r3 r3 - 3r1 r2 r2 - 4r1 r2 r3 4 0 0 8 0 0 6 -6 -6 18 -18 -18 0 0 -6 -18 -39 0 -5 -11 -23 3 0 3 -5 1 1 -11 -2 -2 4 -23 4

  23. 4.5 Examples – EXAMPLE 1 15 -39 0 -5 -11 -23 -24 • Now solve by backwards substitution: r3 : -6z = -18 z = 3 r2 : -5y – 11z = -23 -5y -33 =-23 y = -2 2 3 9 1 r1 : x + 2y +3z = 9 x - 4 + 9 = 9 x = 4 • Hence: x = 4, y = -2, z = 3 is the unique solution. 0 0 -6 -18

  24. 4.5 Examples Matrix-vector system Write in augmented form -1 0 x 1 1 0 1 -1 y = 6 z 1 0 1 -1

  25. 4.5 Examples – EXAMPLE 2 Matrix-vector system Write in augmented form Write on row sums 1 1 6 6 0 1 • Use the top left entry to create zeros below it • First get a zero in the third row: -1 -1 0 0 1 1 1 1 r3 r3 - r1 0 0 1 1 -1 -1 6 6 0 1 1 1 1 -2 -1 0

  26. 4.5 Examples – EXAMPLE 2 1 1 1 6 6 6 0 -6 1 • Use second row to get a zero in the third row: -1 -1 -1 0 0 0 1 1 1 1 1 1 r3 r3 - r1 r3 r3 – r2 0 0 0 1 1 1 -1 -1 -1 6 6 6 2 0 0 1 0 1 1 1 -2 -1 0 -8

  27. 4.5 Examples – EXAMPLE 2 1 6 -6 • Solve by backwards substitution: r3 : 2z = -8 z = -4 -1 0 1 1 UNIQUE SOLUTION y + 4 = 6 y = 2 r2 : y - z = 6 0 1 -1 6 2 0 0 -8 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 from 0 -5 -11 = B A = • Calculating the determinant: |A|= (-1) |B| = (-1)x(1)(-5)(-6) = -30 2 2 3 3 1 1 4 8 6 0 0 -6 3 1 -2 • Non-zero, so we got a unique solution

  30. 4.6 Determinants • In EXAMPLE 2 we used no swaps to get from 0 1 -1 = B A = • Calculating the determinant: |A|= |B| = (1)(1)(2) = 2 -1 -1 0 0 1 1 0 1 -1 2 0 0 1 0 1 • 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. r3 2r3

  32. 4.6 Non-Standard Gaussian Elimination • Quick Example: In Standard G.E. 1 4 8 1 4 8 3 3 -8/3 -16/3 5 -1 4 8 0 -8/3 • This is a bit messy with the fractions. However, in Non-Standard G.E. 1 4 8 1 4 8 3 3 -8 -16 5 -1 4 8 0 -8 • However, in doing this we have multiplied the determinant by 3. r2 3r2 - 5r1 r2 r2 - 5r1/3

  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: 11 -3 9 1 2 3 • Zeros on the diagonal, so determinant is zero. 0 0 -1 0 0 0 • ROW 3 gives the equation 1 x 2 11 3 0 0 -1 y = -3 0x + 0y + 0z = 9 0 z 0 0 9 • This is impossible. Hence there are no solutions.

  35. 4.7.1 Case of Infinite solutions • Suppose instead that 11 -3 0 • ROW 3 now OK: 0x + 0y + 0z = 0 1 2 3 0 0 -1 0 0 0 • Have two equations in three unknowns 1 x 2 11 3 0 0 -1 y = -3 • Get infinitely many solutions 0 z 0 0 0

  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 solutions 11 -3 0 Circle first non-zero row entries Find column with no circles in 1 2 3 0 0 -1 0 0 0 Column 2 corresponds to the y variable 1 x 2 11 3 0 0 -1 y = -3 3. Assign a name to y: let y = α 0 z 0 0 0

  38. 4.7.1 Case of Infinite solutions 11 -3 0 • Solve by back substitution: 1 2 3 r3 : Tells us nothing 0 0 -1 r2 : -z = -3 z = 3 0 0 0 r1 : 3x + y + 2z = 11 3x + α + 6 = 11 1 x 2 11 3 x = (5 – α)/3 0 0 -1 y = -3 0 z 0 0 0 • So, solution is x = (5 – α)/3, y = α, z = 3 for any α

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