5.3 system of linear equations and matrices

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5.3 system of linear equations and matrices

  1. 1. Systems of Linear Equations http://www.lahc.edu/math/precalculus/math_260a.html
  2. 2. Systems of Linear Equations We need one piece of information to solve for one unknown quantity.
  3. 3. Systems of Linear Equations We need one piece of information to solve for one unknown quantity. If 2 hamburgers cost 4 dollars and x is the cost of a hamburger, then 2x = 4 or x = $2.
  4. 4. Systems of Linear Equations We need one piece of information to solve for one unknown quantity. If 2 hamburgers cost 4 dollars and x is the cost of a hamburger, then 2x = 4 or x = $2. If there are two unknowns x and y, we need two pieces of information to solve them.
  5. 5. Systems of Linear Equations We need one piece of information to solve for one unknown quantity. If 2 hamburgers cost 4 dollars and x is the cost of a hamburger, then 2x = 4 or x = $2. If there are two unknowns x and y, we need two pieces of information to solve them. Example A: Let x = cost of a hamburger, y = cost of a fry. If two hamburgers and a fry cost $7, and one hamburger and one fry cost $5. what is the price of each?
  6. 6. Systems of Linear Equations We need one piece of information to solve for one unknown quantity. If 2 hamburgers cost 4 dollars and x is the cost of a hamburger, then 2x = 4 or x = $2. If there are two unknowns x and y, we need two pieces of information to solve them. Example A: Let x = cost of a hamburger, y = cost of a fry. If two hamburgers and a fry cost $7, and one hamburger and one fry cost $5. what is the price of each? We translate these information in to two equations:
  7. 7. Systems of Linear Equations We need one piece of information to solve for one unknown quantity. If 2 hamburgers cost 4 dollars and x is the cost of a hamburger, then 2x = 4 or x = $2. If there are two unknowns x and y, we need two pieces of information to solve them. Example A: Let x = cost of a hamburger, y = cost of a fry. If two hamburgers and a fry cost $7, and one hamburger and one fry cost $5. what is the price of each? 2x + y = 7 We translate these information in to two equations: Eq. 1
  8. 8. Systems of Linear Equations We need one piece of information to solve for one unknown quantity. If 2 hamburgers cost 4 dollars and x is the cost of a hamburger, then 2x = 4 or x = $2. If there are two unknowns x and y, we need two pieces of information to solve them. Example A: Let x = cost of a hamburger, y = cost of a fry. If two hamburgers and a fry cost $7, and one hamburger and one fry cost $5. what is the price of each? 2x + y = 7 x + y = 5{ Eq. 1 Eq. 2 We translate these information in to two equations:
  9. 9. Systems of Linear Equations We need one piece of information to solve for one unknown quantity. If 2 hamburgers cost 4 dollars and x is the cost of a hamburger, then 2x = 4 or x = $2. If there are two unknowns x and y, we need two pieces of information to solve them. Example A: Let x = cost of a hamburger, y = cost of a fry. If two hamburgers and a fry cost $7, and one hamburger and one fry cost $5. what is the price of each? 2x + y = 7 x + y = 5{ A group of equations such as this is called a system of equations. Eq. 1 Eq. 2 We translate these information in to two equations:
  10. 10. Systems of Linear Equations We are to find values for x and for y that will satisfy all equations in the system.
  11. 11. Systems of Linear Equations We are to find values for x and for y that will satisfy all equations in the system. Such a set of values is called a solution for the system.
  12. 12. Systems of Linear Equations We are to find values for x and for y that will satisfy all equations in the system. Such a set of values is called a solution for the system. To solve the system, we note that the difference in the two orders is one hamburger and the difference in price is $2.
  13. 13. Systems of Linear Equations We are to find values for x and for y that will satisfy all equations in the system. Such a set of values is called a solution for the system. To solve the system, we note that the difference in the two orders is one hamburger and the difference in price is $2. So the price of the hamburger is $2.
  14. 14. Systems of Linear Equations We are to find values for x and for y that will satisfy all equations in the system. Such a set of values is called a solution for the system. To solve the system, we note that the difference in the two orders is one hamburger and the difference in price is $2. So the price of the hamburger is $2. In algebra, we subtract the equations in the system: 2x + y = 7 x + y = 5) Eq. 1 Eq. 2
  15. 15. Systems of Linear Equations We are to find values for x and for y that will satisfy all equations in the system. Such a set of values is called a solution for the system. To solve the system, we note that the difference in the two orders is one hamburger and the difference in price is $2. So the price of the hamburger is $2. In algebra, we subtract the equations in the system: 2x + y = 7 x + y = 5) x = 2 Eq. 1 Eq. 2
  16. 16. Systems of Linear Equations We are to find values for x and for y that will satisfy all equations in the system. Such a set of values is called a solution for the system. To solve the system, we note that the difference in the two orders is one hamburger and the difference in price is $2. So the price of the hamburger is $2. In algebra, we subtract the equations in the system: 2x + y = 7 x + y = 5) x = 2 Eq. 1 Eq. 2 Put x = 2 back into Eq. 2, we get y = 3.
  17. 17. Systems of Linear Equations We are to find values for x and for y that will satisfy all equations in the system. Such a set of values is called a solution for the system. To solve the system, we note that the difference in the two orders is one hamburger and the difference in price is $2. So the price of the hamburger is $2. In algebra, we subtract the equations in the system: 2x + y = 7 x + y = 5) x = 2 Eq. 1 Eq. 2 Put x = 2 back into Eq. 2, we get y = 3. Hence a hamburger is $2 and a fry is $3.
  18. 18. Systems of Linear Equations We are to find values for x and for y that will satisfy all equations in the system. Such a set of values is called a solution for the system. To solve the system, we note that the difference in the two orders is one hamburger and the difference in price is $2. So the price of the hamburger is $2. In algebra, we subtract the equations in the system: 2x + y = 7 x + y = 5) x = 2 Eq. 1 Eq. 2 Put x = 2 back into Eq. 2, we get y = 3. Hence a hamburger is $2 and a fry is $3. This is called the elimination method - we eliminate variables by adding or subtracting two equations.
  19. 19. Elimination method reduces the number of variables in the system one at a time. Systems of Linear Equations
  20. 20. Elimination method reduces the number of variables in the system one at a time. The method changes a given system into a smaller system. Systems of Linear Equations
  21. 21. Elimination method reduces the number of variables in the system one at a time. The method changes a given system into a smaller system. We do this successively until the solution is clear. Systems of Linear Equations
  22. 22. Elimination method reduces the number of variables in the system one at a time. The method changes a given system into a smaller system. We do this successively until the solution is clear. Systems of Linear Equations Example B: Let x = cost of a hamburger, y = cost of a fry, z = cost of a soda. 2 hamburgers, 3 fries and 3 soda cost $13. 1 hamburger, 2 fries and 2 soda cost $8. 3 hamburgers, 2 fries, 3 soda cost $13. Find x, y, and z.
  23. 23. Elimination method reduces the number of variables in the system one at a time. The method changes a given system into a smaller system. We do this successively until the solution is clear. Systems of Linear Equations We translate the information into a system of three equations with three unknowns Example B: Let x = cost of a hamburger, y = cost of a fry, z = cost of a soda. 2 hamburgers, 3 fries and 3 soda cost $13. 1 hamburger, 2 fries and 2 soda cost $8. 3 hamburgers, 2 fries, 3 soda cost $13. Find x, y, and z.
  24. 24. 2x + 3y + 3z = 13 x + 2y + 2z = 8 3x + 2y + 3z = 13 { E 1 E 2 E 3 Systems of Linear Equations
  25. 25. 2x + 3y + 3z = 13 x + 2y + 2z = 8 3x + 2y + 3z = 13 { E 1 E 2 E 3 Select x to eliminate since there is 1x in E2. Systems of Linear Equations
  26. 26. 2x + 3y + 3z = 13 x + 2y + 2z = 8 3x + 2y + 3z = 13 { E 1 E 2 E 3 Select x to eliminate since there is 1x in E2. -2*E 2 + E1: Systems of Linear Equations
  27. 27. 2x + 3y + 3z = 13 x + 2y + 2z = 8 3x + 2y + 3z = 13 { E 1 E 2 E 3 Select x to eliminate since there is 1x in E2. -2*E 2 + E1: -2x – 4y – 4z = -16 +) 2x + 3y + 3z = 13 Systems of Linear Equations
  28. 28. 2x + 3y + 3z = 13 x + 2y + 2z = 8 3x + 2y + 3z = 13 { E 1 E 2 E 3 Select x to eliminate since there is 1x in E2. -2*E 2 + E1: -2x – 4y – 4z = -16 +) 2x + 3y + 3z = 13 0 – y – z = - 3 Systems of Linear Equations
  29. 29. 2x + 3y + 3z = 13 x + 2y + 2z = 8 3x + 2y + 3z = 13 { E 1 E 2 E 3 Select x to eliminate since there is 1x in E2. -2*E 2 + E1: -2x – 4y – 4z = -16 +) 2x + 3y + 3z = 13 0 – y – z = - 3 -3*E 2 + E3: Systems of Linear Equations
  30. 30. 2x + 3y + 3z = 13 x + 2y + 2z = 8 3x + 2y + 3z = 13 { E 1 E 2 E 3 Select x to eliminate since there is 1x in E2. -2*E 2 + E1: -2x – 4y – 4z = -16 +) 2x + 3y + 3z = 13 0 – y – z = - 3 -3*E 2 + E3: -3x – 6y – 6z = -24 +) 3x + 2y + 3z = 13 Systems of Linear Equations
  31. 31. 2x + 3y + 3z = 13 x + 2y + 2z = 8 3x + 2y + 3z = 13 { E 1 E 2 E 3 Select x to eliminate since there is 1x in E2. -2*E 2 + E1: -2x – 4y – 4z = -16 +) 2x + 3y + 3z = 13 0 – y – z = - 3 -3*E 2 + E3: -3x – 6y – 6z = -24 +) 3x + 2y + 3z = 13 0 – 4y – 3z = -11 Systems of Linear Equations
  32. 32. 2x + 3y + 3z = 13 x + 2y + 2z = 8 3x + 2y + 3z = 13 { E 1 E 2 E 3 Select x to eliminate since there is 1x in E2. -2*E 2 + E1: -2x – 4y – 4z = -16 +) 2x + 3y + 3z = 13 0 – y – z = - 3 -3*E 2 + E3: -3x – 6y – 6z = -24 +) 3x + 2y + 3z = 13 0 – 4y – 3z = -11 Group these into a system of two equations with two unknowns. Systems of Linear Equations
  33. 33. 2x + 3y + 3z = 13 x + 2y + 2z = 8 3x + 2y + 3z = 13 { E 1 E 2 E 3 Select x to eliminate since there is 1x in E2. -2*E 2 + E1: -2x – 4y – 4z = -16 +) 2x + 3y + 3z = 13 0 – y – z = - 3 -3*E 2 + E3: -3x – 6y – 6z = -24 +) 3x + 2y + 3z = 13 0 – 4y – 3z = -11 Group these into a system of two equations with two unknowns. Systems of Linear Equations Hence we reduced the system to a simpler one.
  34. 34. – y – z = - 3 – 4y – 3z = -11{ Systems of Linear Equations
  35. 35. – y – z = - 3 – 4y – 3z = -11{ 4y + 3z = 11 { y + z = 3 Eq.4 Eq. 5 Systems of Linear Equations *(-1)
  36. 36. – y – z = - 3 – 4y – 3z = -11{ 4y + 3z = 11 { y + z = 3 Eq.4 Eq. 5 Systems of Linear Equations *(-1) Let's eliminate y.
  37. 37. – y – z = - 3 – 4y – 3z = -11{ 4y + 3z = 11 { y + z = 3 Eq.4 Eq. 5 -3*E 4 + E5: Systems of Linear Equations *(-1) Let's eliminate y.
  38. 38. – y – z = - 3 – 4y – 3z = -11{ 4y + 3z = 11 { y + z = 3 Eq.4 Eq. 5 -3*E 4 + E5: +) 4y + 3z = 11 -3y – 3z = -9 Systems of Linear Equations *(-1) Let's eliminate y.
  39. 39. – y – z = - 3 – 4y – 3z = -11{ 4y + 3z = 11 { y + z = 3 Eq.4 Eq. 5 -3*E 4 + E5: +) 4y + 3z = 11 -3y – 3z = -9 y + 0 = 2 y = 2 Systems of Linear Equations *(-1) Let's eliminate y.
  40. 40. – y – z = - 3 – 4y – 3z = -11{ 4y + 3z = 11 { y + z = 3 Eq.4 Eq. 5 -3*E 4 + E5: +) 4y + 3z = 11 -3y – 3z = -9 y + 0 = 2 y = 2 To get z, set 2 for y in E4: Systems of Linear Equations *(-1) Let's eliminate y.
  41. 41. – y – z = - 3 – 4y – 3z = -11{ 4y + 3z = 11 { y + z = 3 Eq.4 Eq. 5 -3*E 4 + E5: +) 4y + 3z = 11 -3y – 3z = -9 y + 0 = 2 y = 2 To get z, set 2 for y in E4: 2 + z = 3  z = 1 Systems of Linear Equations *(-1) Let's eliminate y.
  42. 42. – y – z = - 3 – 4y – 3z = -11{ 4y + 3z = 11 { y + z = 3 Eq.4 Eq. 5 -3*E 4 + E5: +) 4y + 3z = 11 -3y – 3z = -9 y + 0 = 2 y = 2 To get z, set 2 for y in E4: 2 + z = 3  z = 1 For x, set 2 for y , set 1 for z in E2: Systems of Linear Equations *(-1) Let's eliminate y.
  43. 43. – y – z = - 3 – 4y – 3z = -11{ 4y + 3z = 11 { y + z = 3 Eq.4 Eq. 5 -3*E 4 + E5: +) 4y + 3z = 11 -3y – 3z = -9 y + 0 = 2 y = 2 To get z, set 2 for y in E4: 2 + z = 3  z = 1 For x, set 2 for y , set 1 for z in E2: x + 2*2 + 2*1 = 8  Systems of Linear Equations *(-1) Let's eliminate y.
  44. 44. – y – z = - 3 – 4y – 3z = -11{ 4y + 3z = 11 { y + z = 3 Eq.4 Eq. 5 -3*E 4 + E5: +) 4y + 3z = 11 -3y – 3z = -9 y + 0 = 2 y = 2 To get z, set 2 for y in E4: 2 + z = 3  z = 1 For x, set 2 for y , set 1 for z in E2: x + 2*2 + 2*1 = 8  x + 6 = 8  x = 2 Systems of Linear Equations *(-1) Let's eliminate y.
  45. 45. – y – z = - 3 – 4y – 3z = -11{ 4y + 3z = 11 { y + z = 3 Eq.4 Eq. 5 -3*E 4 + E5: +) 4y + 3z = 11 -3y – 3z = -9 y + 0 = 2 y = 2 To get z, set 2 for y in E4: 2 + z = 3  z = 1 For x, set 2 for y , set 1 for z in E2: x + 2*2 + 2*1 = 8  x + 6 = 8  x = 2 Systems of Linear Equations *(-1) Hence the hamburger cost $2, the fry cost $2 and the drink cost $1. Let's eliminate y.
  46. 46. A matrix is a rectangular table of numbers. Matrix Notation
  47. 47. A matrix is a rectangular table of numbers. For example 5 2 -3 4 -1 0 -1 2 -1 6 -2 3 11 9 -4 Matrix Notation are matrices.
  48. 48. A matrix is a rectangular table of numbers. For example 5 2 -3 4 -1 0 -1 2 -1 6 -2 3 11 9 -4 Matrix Notation are matrices. Systems of linear equations can be put into matrices and solved using matrix notation.
  49. 49. A matrix is a rectangular table of numbers. For example 5 2 -3 4 -1 0 -1 2 -1 6 -2 3 11 9 -4 Matrix Notation are matrices. Systems of linear equations can be put into matrices and solved using matrix notation. x + 4y = -7 2x – 3y = 8{ For example, the system 1 4 -7 2 -3 8 may be written as the matrix: x y #
  50. 50. A matrix is a rectangular table of numbers. For example 5 2 -3 4 -1 0 -1 2 -1 6 -2 3 11 9 -4 Matrix Notation are matrices. Systems of linear equations can be put into matrices and solved using matrix notation. x + 4y = -7 2x – 3y = 8{ For example, the system 1 4 -7 2 -3 8 may be written as the matrix: x y # This is called the augmented matrix for the system.
  51. 51. A matrix is a rectangular table of numbers. For example 5 2 -3 4 -1 0 -1 2 -1 6 -2 3 11 9 -4 Matrix Notation are matrices. Systems of linear equations can be put into matrices and solved using matrix notation. x + 4y = -7 2x – 3y = 8{ For example, the system 1 4 -7 2 -3 8 may be written as the matrix: x y # This is called the augmented matrix for the system. Each row of the matrix corresponds to an equation, each column corresponds to a variable and the last column corresponds to numbers.
  52. 52. Matrix Notation An augmented matrix can be easily converted back to a system .
  53. 53. Matrix Notation An augmented matrix can be easily converted back to a system . For example, the matrix 2 -3 8 1 4 -7 is the system x + 4y = -7{2x – 3y = 8
  54. 54. Operations of the equations correspond to operations of rows in the matrices. Matrix Notation An augmented matrix can be easily converted back to a system . For example, the matrix 2 -3 8 1 4 -7 is the system x + 4y = -7{2x – 3y = 8
  55. 55. Operations of the equations correspond to operations of rows in the matrices. Three such operations are important. These are the elementary row operations: Matrix Notation An augmented matrix can be easily converted back to a system . For example, the matrix 2 -3 8 1 4 -7 is the system x + 4y = -7{2x – 3y = 8
  56. 56. Matrix Notation I. Switch two rows. An augmented matrix can be easily converted back to a system . For example, the matrix 2 -3 8 1 4 -7 is the system x + 4y = -7{2x – 3y = 8 Operations of the equations correspond to operations of rows in the matrices. Three such operations are important. These are the elementary row operations:
  57. 57. Matrix Notation I. Switch two rows. Switching row i with row j is notated as Ri Rj. An augmented matrix can be easily converted back to a system . For example, the matrix 2 -3 8 1 4 -7 is the system x + 4y = -7{2x – 3y = 8 Operations of the equations correspond to operations of rows in the matrices. Three such operations are important. These are the elementary row operations:
  58. 58. Matrix Notation 1 4 -7 2 -3 8 For example, R1 R2 I. Switch two rows. Switching row i with row j is notated as Ri Rj. An augmented matrix can be easily converted back to a system . For example, the matrix 2 -3 8 1 4 -7 is the system x + 4y = -7{2x – 3y = 8 Operations of the equations correspond to operations of rows in the matrices. Three such operations are important. These are the elementary row operations:
  59. 59. Matrix Notation 1 4 -7 2 -3 8 For example, R1 R2 2 -3 8 1 4 -7 I. Switch two rows. Switching row i with row j is notated as Ri Rj. An augmented matrix can be easily converted back to a system . For example, the matrix 2 -3 8 1 4 -7 is the system x + 4y = -7{2x – 3y = 8 Operations of the equations correspond to operations of rows in the matrices. Three such operations are important. These are the elementary row operations:
  60. 60. Matrix Notation II. Multiply a row by a constant k = 0.
  61. 61. Matrix Notation II. Multiply a row by a constant k = 0. Multiply row i by k is notated as k*Ri.
  62. 62. Matrix Notation II. Multiply a row by a constant k = 0. Multiply row i by k is notated as k*Ri. 1 4 -7 2 -3 8 -3*R2 For example,
  63. 63. Matrix Notation II. Multiply a row by a constant k = 0. Multiply row i by k is notated as k*Ri. 1 4 -7 2 -3 8 -3*R2 1 4 -7 -6 9 -24 For example,
  64. 64. III. Add the multiple of a row to another row. Matrix Notation II. Multiply a row by a constant k = 0. Multiply row i by k is notated as k*Ri. 1 4 -7 2 -3 8 -3*R2 1 4 -7 -6 9 -24 For example,
  65. 65. III. Add the multiple of a row to another row. k times row i added to row j is notated as “k*Ri add  Rj”. Matrix Notation II. Multiply a row by a constant k = 0. Multiply row i by k is notated as k*Ri. 1 4 -7 2 -3 8 -3*R2 1 4 -7 -6 9 -24 For example,
  66. 66. III. Add the multiple of a row to another row. k times row i added to row j is notated as “k*Ri add  Rj”. 1 4 -7 2 -3 8 For example, -2*R1 add  R2 Matrix Notation II. Multiply a row by a constant k = 0. Multiply row i by k is notated as k*Ri. 1 4 -7 2 -3 8 -3*R2 1 4 -7 -6 9 -24 For example,
  67. 67. III. Add the multiple of a row to another row. k times row i added to row j is notated as “k*Ri add  Rj”. 1 4 -7 2 -3 8 For example, -2*R1 add  R2 -2 -8 14 write a copy of -2*R1 Matrix Notation II. Multiply a row by a constant k = 0. Multiply row i by k is notated as k*Ri. 1 4 -7 2 -3 8 -3*R2 1 4 -7 -6 9 -24 For example,
  68. 68. III. Add the multiple of a row to another row. k times row i added to row j is notated as “k*Ri add  Rj”. 1 4 -7 2 -3 8 For example, -2*R1 add  R2 1 4 -7 0 -11 22 -2 -8 14 write a copy of -2*R1 Matrix Notation II. Multiply a row by a constant k = 0. Multiply row i by k is notated as k*Ri. 1 4 -7 2 -3 8 -3*R2 1 4 -7 -6 9 -24 For example,
  69. 69. III. Add the multiple of a row to another row. k times row i added to row j is notated as “k*Ri add  Rj”. 1 4 -7 2 -3 8 For example, -2*R1 add  R2 1 4 -7 0 -11 22 -2 -8 14 write a copy of -2*R1 Fact: Performing elementary row operations on a matrix does not change the solution of the system. Matrix Notation II. Multiply a row by a constant k = 0. Multiply row i by k is notated as k*Ri. 1 4 -7 2 -3 8 -3*R2 1 4 -7 -6 9 -24 For example,
  70. 70. Matrix Notation The elimination method may be carried out with matrix notation.
  71. 71. Elimination Method in Matrix Notation: Matrix Notation The elimination method may be carried out with matrix notation.
  72. 72. Matrix Notation The elimination method may be carried out with matrix notation. Elimination Method in Matrix Notation: 1. Apply row operations to transform the matrix to the upper diagonal form where all the entries below the main diagonal (the lower left triangular region) are 0 .
  73. 73. * * * * * * * * * * * * Row operations Matrix Notation The elimination method may be carried out with matrix notation. Elimination Method in Matrix Notation: 1. Apply row operations to transform the matrix to the upper diagonal form where all the entries below the main diagonal (the lower left triangular region) are 0 .
  74. 74. * * * * * * * * * * * * Row operations * * * * * * * * * 0 0 0 Matrix Notation The elimination method may be carried out with matrix notation. Elimination Method in Matrix Notation: 1. Apply row operations to transform the matrix to the upper diagonal form where all the entries below the main diagonal (the lower left triangular region) are 0 .
  75. 75. * * * * * * * * * * * * Row operations * * * * * * * * * 0 0 0 2. Starting from the bottom row, get the answer for one of the variable. Matrix Notation The elimination method may be carried out with matrix notation. Elimination Method in Matrix Notation: 1. Apply row operations to transform the matrix to the upper diagonal form where all the entries below the main diagonal (the lower left triangular region) are 0 .
  76. 76. * * * * * * * * * * * * Row operations * * * * * * * * * 0 0 0 2. Starting from the bottom row, get the answer for one of the variable. Then go up one row, use the solution already obtained to get another answer of another variable. Matrix Notation The elimination method may be carried out with matrix notation. Elimination Method in Matrix Notation: 1. Apply row operations to transform the matrix to the upper diagonal form where all the entries below the main diagonal (the lower left triangular region) are 0 .
  77. 77. Elimination Method in Matrix Notation: 1. Apply row operations to transform the matrix to the upper diagonal form where all the entries below the main diagonal (the lower left triangular region) are 0 . * * * * * * * * * * * * Row operations * * * * * * * * * 0 0 0 2. Starting from the bottom row, get the answer for one of the variable. Then go up one row, use the solution already obtained to get another answer of another variable. Repeat the process, working from the bottom row to the top row to extract all solutions. Matrix Notation The elimination method may be carried out with matrix notation.
  78. 78. x + 4y = -7 2x – 3y = 8{ Eq. 1 Eq. 2 Example C: Solve using matrix notation: Matrix Notation
  79. 79. 1 4 -7 2 -3 8 x + 4y = -7 2x – 3y = 8{ Eq. 1 Eq. 2 Example C: Solve using matrix notation: Put the system into a matrix: Matrix Notation
  80. 80. 1 4 -7 2 -3 8 -2*R1 add  R2 x + 4y = -7 2x – 3y = 8{ Eq. 1 Eq. 2 Example C: Solve using matrix notation: Put the system into a matrix: Matrix Notation
  81. 81. 1 4 -7 2 -3 8 -2*R1 add  R2 -2 -8 14 x + 4y = -7 2x – 3y = 8{ Eq. 1 Eq. 2 Example C: Solve using matrix notation: Put the system into a matrix: Matrix Notation
  82. 82. 1 4 -7 2 -3 8 -2*R1 add  R2 1 4 -7 0 -11 22 -2 -8 14 x + 4y = -7 2x – 3y = 8{ Eq. 1 Eq. 2 Example C: Solve using matrix notation: Put the system into a matrix: Matrix Notation
  83. 83. 1 4 -7 2 -3 8 -2*R1 add  R2 1 4 -7 0 -11 22 -2 -8 14 x + 4y = -7 2x – 3y = 8{ Eq. 1 Eq. 2 Example C: Solve using matrix notation: Put the system into a matrix: Start from the bottom row: -11y = 22  y = -2 Matrix Notation
  84. 84. 1 4 -7 2 -3 8 -2*R1 add  R2 1 4 -7 0 -11 22 -2 -8 14 x + 4y = -7 2x – 3y = 8{ Eq. 1 Eq. 2 Example C: Solve using matrix notation: Put the system into a matrix: Start from the bottom row: -11y = 22  y = -2 Go up one row to R1 and set y = -2: Matrix Notation
  85. 85. 1 4 -7 2 -3 8 -2*R1 add  R2 1 4 -7 0 -11 22 -2 -8 14 x + 4y = -7 2x – 3y = 8{ Eq. 1 Eq. 2 Example C: Solve using matrix notation: Put the system into a matrix: Start from the bottom row: -11y = 22  y = -2 Go up one row to R1 and set y = -2: x + 4y = -7 x + 4(-2) = -7 Matrix Notation
  86. 86. 1 4 -7 2 -3 8 -2*R1 add  R2 1 4 -7 0 -11 22 -2 -8 14 x + 4y = -7 2x – 3y = 8{ Eq. 1 Eq. 2 Example C: Solve using matrix notation: Put the system into a matrix: Start from the bottom row: -11y = 22  y = -2 Go up one row to R1 and set y = -2: x + 4y = -7 x + 4(-2) = -7  x = 1 Matrix Notation Hence the solution is {x = 1, y = -2}.
  87. 87. 1 4 -7 2 -3 8 -2*R1 add  R2 1 4 -7 0 -11 22 -2 -8 14 x + 4y = -7 2x – 3y = 8{ Eq. 1 Eq. 2 Example C: Solve using matrix notation: Put the system into a matrix: Start from the bottom row: -11y = 22  y = -2 Go up one row to R1 and set y = -2: x + 4y = -7 x + 4(-2) = -7  x = 1 Matrix Notation Hence the solution is {x = 1, y = -2}. It may also be written as the ordered pair (1, -2).
  88. 88. Example D: Solve using matrix notation: 2x + 3y + 3z = 13 x + 2y + 2z = 8 3x + 2y + 3z = 13 { E 1 E 2 E 3 Matrix Notation
  89. 89. Example D: Solve using matrix notation: 2x + 3y + 3z = 13 x + 2y + 2z = 8 3x + 2y + 3z = 13 { E 1 E 2 E 3 2 3 3 13 Put the system into a matrix: 1 2 2 8 3 2 3 13 Matrix Notation
  90. 90. Example D: Solve using matrix notation: 2x + 3y + 3z = 13 x + 2y + 2z = 8 3x + 2y + 3z = 13 { E 1 E 2 E 3 2 3 3 13 R1 R2 Put the system into a matrix: 1 2 2 8 3 2 3 13 2 3 3 13 1 2 2 8 3 2 3 13 Matrix Notation
  91. 91. Example D: Solve using matrix notation: 2x + 3y + 3z = 13 x + 2y + 2z = 8 3x + 2y + 3z = 13 { E 1 E 2 E 3 2 3 3 13 R1 R2 -2 -4 -4 -16Put the system into a matrix: 1 2 2 8 3 2 3 13 2 3 3 13 1 2 2 8 3 2 3 13 -2*R1 add R2 Matrix Notation
  92. 92. Example D: Solve using matrix notation: 2x + 3y + 3z = 13 x + 2y + 2z = 8 3x + 2y + 3z = 13 { E 1 E 2 E 3 2 3 3 13 R1 R2 -2 -4 -4 -16Put the system into a matrix: 1 2 2 8 3 2 3 13 2 3 3 13 1 2 2 8 3 2 3 13 -2*R1 add R2 0 -1 -1 -3 1 2 2 8 3 2 3 13 Matrix Notation
  93. 93. Example D: Solve using matrix notation: 2x + 3y + 3z = 13 x + 2y + 2z = 8 3x + 2y + 3z = 13 { E 1 E 2 E 3 2 3 3 13 R1 R2 -2 -4 -4 -16Put the system into a matrix: 1 2 2 8 3 2 3 13 2 3 3 13 1 2 2 8 3 2 3 13 -2*R1 add R2 0 -1 -1 -3 1 2 2 8 3 2 3 13 -3* R1 add R3 -3 -6 -6 -24 Matrix Notation
  94. 94. Example D: Solve using matrix notation: 2x + 3y + 3z = 13 x + 2y + 2z = 8 3x + 2y + 3z = 13 { E 1 E 2 E 3 2 3 3 13 R1 R2 -2 -4 -4 -16Put the system into a matrix: 1 2 2 8 3 2 3 13 2 3 3 13 1 2 2 8 3 2 3 13 -2*R1 add R2 0 -1 -1 -3 1 2 2 8 3 2 3 13 -3* R1 add R3 -3 -6 -6 -24 0 -1 -1 -3 1 2 2 8 0 -4 -3 -11 Matrix Notation
  95. 95. Example D: Solve using matrix notation: 2x + 3y + 3z = 13 x + 2y + 2z = 8 3x + 2y + 3z = 13 { E 1 E 2 E 3 2 3 3 13 R1 R2 -2 -4 -4 -16Put the system into a matrix: 1 2 2 8 3 2 3 13 2 3 3 13 1 2 2 8 3 2 3 13 -2*R1 add R2 0 -1 -1 -3 1 2 2 8 3 2 3 13 -3* R1 add R3 -3 -6 -6 -24 0 -1 -1 -3 1 2 2 8 0 -4 -3 -11 -4*R2 add R3 0 4 4 12 Matrix Notation
  96. 96. Example D: Solve using matrix notation: 2x + 3y + 3z = 13 x + 2y + 2z = 8 3x + 2y + 3z = 13 { E 1 E 2 E 3 2 3 3 13 R1 R2 -2 -4 -4 -16Put the system into a matrix: 1 2 2 8 3 2 3 13 2 3 3 13 1 2 2 8 3 2 3 13 -2*R1 add R2 0 -1 -1 -3 1 2 2 8 3 2 3 13 -3* R1 add R3 -3 -6 -6 -24 0 -1 -1 -3 1 2 2 8 0 -4 -3 -11 -4*R2 add R3 0 4 4 12 0 -1 -1 -3 1 2 2 8 0 0 1 1 Matrix Notation
  97. 97. Example D: Solve using matrix notation: 2x + 3y + 3z = 13 x + 2y + 2z = 8 3x + 2y + 3z = 13 { E 1 E 2 E 3 2 3 3 13 R1 R2 -2 -4 -4 -16Put the system into a matrix: 1 2 2 8 3 2 3 13 2 3 3 13 1 2 2 8 3 2 3 13 -2*R1 add R2 0 -1 -1 -3 1 2 2 8 3 2 3 13 -3* R1 add R3 -3 -6 -6 -24 0 -1 -1 -3 1 2 2 8 0 -4 -3 -11 -4*R2 add R3 0 4 4 12 0 -1 -1 -3 1 2 2 8 0 0 1 1 Matrix Notation
  98. 98. 0 -1 -1 -3 1 2 2 8 0 0 1 1 Matrix Notation Extract the solution starting from the bottom row.
  99. 99. 0 -1 -1 -3 1 2 2 8 0 0 1 1 From R3, we get z = 1. Matrix Notation Extract the solution starting from the bottom row.
  100. 100. 0 -1 -1 -3 1 2 2 8 0 0 1 1 From R3, we get z = 1. Go up one row to R2, we get -y – z = -3 -y – (1) = -3 Matrix Notation Extract the solution starting from the bottom row.
  101. 101. 0 -1 -1 -3 1 2 2 8 0 0 1 1 From R3, we get z = 1. Go up one row to R2, we get -y – z = -3 -y – (1) = -3 3 – 1 = y 2 = y Matrix Notation Extract the solution starting from the bottom row.
  102. 102. 0 -1 -1 -3 1 2 2 8 0 0 1 1 From R3, we get z = 1. Go up one row to R2, we get -y – z = -3 -y – (1) = -3 3 – 1 = y 2 = y Up to R1, we get x + 2y + 2z = 8 x + 2(2) + 2(1) = 8 Matrix Notation Extract the solution starting from the bottom row.
  103. 103. 0 -1 -1 -3 1 2 2 8 0 0 1 1 From R3, we get z = 1. Go up one row to R2, we get -y – z = -3 -y – (1) = -3 3 – 1 = y 2 = y Up to R1, we get x + 2y + 2z = 8 x + 2(2) + 2(1) = 8 x + 6 = 8 x = 2 Matrix Notation Extract the solution starting from the bottom row.
  104. 104. 0 -1 -1 -3 1 2 2 8 0 0 1 1 From R3, we get z = 1. Go up one row to R2, we get -y – z = -3 -y – (1) = -3 3 – 1 = y 2 = y Up to R1, we get x + 2y + 2z = 8 x + 2(2) + 2(1) = 8 x + 6 = 8 x = 2 So the solution is (2, 2, 1). Matrix Notation Extract the solution starting from the bottom row.

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