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Systems of Linear Equations: An Introduction Unique Solutions Underdetermined and PowerPoint Presentation
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Systems of Linear Equations: An Introduction Unique Solutions Underdetermined and

Systems of Linear Equations: An Introduction Unique Solutions Underdetermined and

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Systems of Linear Equations: An Introduction Unique Solutions Underdetermined and

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  1. 2 • Systems of Linear Equations: • An Introduction • Unique Solutions • Underdetermined and Overdetermined Systems • Multiplication of Matrices • The Inverse of a Square Matrix • Leontief Input-Output Model Systems of Linear Equations and Matrices

  2. 2.1 Systems of Linear Equations: An Introduction

  3. Systems of Equations • Recall that a system of two linear equations in two variables may be written in the general form where a, b, c, d, h, and k are real numbers and neither a and b nor c and d are both zero. • Recall that the graph of each equation in the system is a straight line in the plane, so that geometrically, the solution to the system is the point(s) of intersection of the two straight lines L1 and L2, represented by the first and second equations of the system.

  4. Systems of Equations • Given the two straight lines L1 and L2, one and only one of the following may occur: 1.L1 and L2 intersect at exactly one point. y L1 Unique solution (x1, y1) y1 (x1, y1) x x1 L2

  5. Systems of Equations • Given the two straight lines L1 and L2, one and only one of the following may occur: 2.L1 and L2 are coincident. y L1, L2 Infinitely many solutions x

  6. Systems of Equations • Given the two straight lines L1 and L2, one and only one of the following may occur: 3.L1 and L2 are parallel. y L1 L2 No solution x

  7. Example: A System of Equations With Exactly One Solution • Consider the system • Solving the first equation for y in terms of x, we obtain • Substituting this expression for y into the second equation yields

  8. Example: A System of Equations With Exactly One Solution • Finally, substituting this value of x into the expression fory obtained earlier gives • Therefore, the unique solution of the system is given by x = 2 and y = 3.

  9. Example: A System of Equations With Exactly One Solution • Geometrically, the two lines represented by the two equations that make up the system intersect at the point(2, 3): y 6 5 4 3 2 1 –1 (2, 3) x 1 2 3 4 5 6

  10. Example: A System of Equations With Infinitely Many Solutions • Consider the system • Solving the first equation for y in terms of x, we obtain • Substituting this expression for y into the second equation yields which is a true statement. • This result follows from the fact that the second equation is equivalent to the first.

  11. Example: A System of Equations With Infinitely Many Solutions • Thus, any order pair of numbers(x, y) satisfying the equation y =2x – 1 constitutes a solution to the system. • By assigning the valuet to x, where t is any real number, we find that y =2t – 1 and so the ordered pair (t, 2t – 1) is a solution to the system. • The variable t is called a parameter. • For example: • Setting t = 0, gives the point (0, –1) as asolution of the system. • Setting t = 1, gives the point (1, 1) as another solution of the system.

  12. Example: A System of Equations With Infinitely Many Solutions • Since t represents any real number, there are infinitely many solutions of the system. • Geometrically, the two equations in the system represent the same line, and all solutions of the system are points lying on the line: y 6 5 4 3 2 1 –1 x 1 2 3 4 5 6

  13. Example: A System of Equations That Has No Solution • Consider the system • Solving the first equation for y in terms of x, we obtain • Substituting this expression for y into the second equation yields which is clearlyimpossible. • Thus, there is no solution to the system of equations.

  14. Example: A System of Equations That Has No Solution • To interpret the situation geometrically, cast both equations in the slope-intercept form, obtaining y = 2x – 1 and y = 2x – 4 which shows that the lines are parallel. • Graphically: y 6 5 4 3 2 1 –1 x 1 2 3 4 5 6

  15. 2.2 Systems of Linear Equations: Unique Solutions

  16. The Gauss-Jordan Method • The Gauss-Jordan elimination method is a technique for solving systems of linear equations of any size. • The operations of the Gauss-Jordan method are • Interchange any two equations. • Replace an equation by a nonzero constant multiple of itself. • Replace an equation by the sum of that equation and a constant multiple of any other equation.

  17. Example • Solve the following system of equations: Solution • First, we transform this system into an equivalent system in which the coefficient of x in the first equation is 1: Multiply the equation by 1/2 Toggle slides back and forth to compare before and changes Example 1, page 77

  18. Example • Solve the following system of equations: Solution • First, we transform this system into an equivalent system in which the coefficient of x in the first equation is 1: Multiply the first equation by 1/2 Toggle slides back and forth to compare before and changes

  19. Example • Solve the following system of equations: Solution • Next, we eliminate the variable x from all equations except the first: Toggle slides back and forth to compare before and changes Replace by the sum of – 3 X the first equation + the second equation

  20. Example • Solve the following system of equations: Solution • Next, we eliminate the variable x from all equations except the first: Toggle slides back and forth to compare before and changes Replace by the sum of – 3 ☓the first equation + the second equation

  21. Example • Solve the following system of equations: Solution • Next, we eliminate the variable x from all equations except the first: Toggle slides back and forth to compare before and changes Replace by the sum of the first equation + the third equation

  22. Example • Solve the following system of equations: Solution • Next, we eliminate the variable x from all equations except the first: Toggle slides back and forth to compare before and changes Replace by the sum of the first equation + the third equation

  23. Example • Solve the following system of equations: Solution • Then we transform so that the coefficient of y in the second equation is 1: Toggle slides back and forth to compare before and changes Multiply the second equation by 1/2

  24. Example • Solve the following system of equations: Solution • Then we transform so that the coefficient of y in the second equation is 1: Toggle slides back and forth to compare before and changes Multiply the second equation by 1/2

  25. Example • Solve the following system of equations: Solution • We now eliminatey from all equations except the second: Replace by the sum of the first equation +(–2) ☓ the second equation Toggle slides back and forth to compare before and changes

  26. Example • Solve the following system of equations: Solution • We now eliminatey from all equations except the second: Replace by the sum of the first equation + (–2) ☓ the second equation Toggle slides back and forth to compare before and changes

  27. Example • Solve the following system of equations: Solution • We now eliminatey from all equations except the second: Toggle slides back and forth to compare before and changes Replace by the sum of the third equation +(–3) ☓ the second equation

  28. Example • Solve the following system of equations: Solution • We now eliminatey from all equations except the second: Toggle slides back and forth to compare before and changes Replace by the sum of the third equation +(–3) ☓ the second equation

  29. Example • Solve the following system of equations: Solution • Now we transform so that the coefficient of z in the third equation is 1: Toggle slides back and forth to compare before and changes Multiply the third equation by 1/11

  30. Example • Solve the following system of equations: Solution • Now we transform so that the coefficient of z in the third equation is 1: Toggle slides back and forth to compare before and changes Multiply the third equation by 1/11

  31. Example • Solve the following system of equations: Solution • We now eliminatez from all equations except the third: Replace by the sum of the first equation + (–7) ☓ the third equation Toggle slides back and forth to compare before and changes

  32. Example • Solve the following system of equations: Solution • We now eliminatez from all equations except the third: Replace by the sum of the first equation + (–7) ☓ the third equation Toggle slides back and forth to compare before and changes

  33. Example • Solve the following system of equations: Solution • We now eliminatez from all equations except the third: Toggle slides back and forth to compare before and changes Replace by the sum of the second equation + 2☓the third equation

  34. Example • Solve the following system of equations: Solution • We now eliminatez from all equations except the third: Toggle slides back and forth to compare before and changes Replace by the sum of the second equation + 2☓the third equation

  35. Example • Solve the following system of equations: Solution • Thus, the solution to the system is x = 3, y = 1, and z = 2.

  36. Augmented Matrices • Matrices are rectangular arrays of numbers that can aid us by eliminating the need to write the variables at each step of the reduction. • For example, the system may be represented by the augmentedmatrix Coefficient Matrix

  37. Matrices and Gauss-Jordan • Every step in the Gauss-Jordan elimination method can be expressed with matrices, rather than systems of equations, thus simplifying the whole process: • Steps expressed as systems of equations: • Steps expressed as augmented matrices: Toggle slides back and forth to compare before and changes Example 2, page 78

  38. Matrices and Gauss-Jordan • Every step in the Gauss-Jordan elimination method can be expressed with matrices, rather than systems of equations, thus simplifying the whole process: • Steps expressed as systems of equations: • Steps expressed as augmented matrices: Toggle slides back and forth to compare before and changes Example 2, page 78

  39. Matrices and Gauss-Jordan • Every step in the Gauss-Jordan elimination method can be expressed with matrices, rather than systems of equations, thus simplifying the whole process: • Steps expressed as systems of equations: • Steps expressed as augmented matrices: Toggle slides back and forth to compare before and changes Example 2, page 78

  40. Matrices and Gauss-Jordan • Every step in the Gauss-Jordan elimination method can be expressed with matrices, rather than systems of equations, thus simplifying the whole process: • Steps expressed as systems of equations: • Steps expressed as augmented matrices: Toggle slides back and forth to compare before and changes Example 2, page 78

  41. Matrices and Gauss-Jordan • Every step in the Gauss-Jordan elimination method can be expressed with matrices, rather than systems of equations, thus simplifying the whole process: • Steps expressed as systems of equations: • Steps expressed as augmented matrices: Toggle slides back and forth to compare before and changes Example 2, page 78

  42. Matrices and Gauss-Jordan • Every step in the Gauss-Jordan elimination method can be expressed with matrices, rather than systems of equations, thus simplifying the whole process: • Steps expressed as systems of equations: • Steps expressed as augmented matrices: Toggle slides back and forth to compare before and changes Example 2, page 78

  43. Matrices and Gauss-Jordan • Every step in the Gauss-Jordan elimination method can be expressed with matrices, rather than systems of equations, thus simplifying the whole process: • Steps expressed as systems of equations: • Steps expressed as augmented matrices: Toggle slides back and forth to compare before and changes Example 2, page 78

  44. Matrices and Gauss-Jordan • Every step in the Gauss-Jordan elimination method can be expressed with matrices, rather than systems of equations, thus simplifying the whole process: • Steps expressed as systems of equations: • Steps expressed as augmented matrices: Toggle slides back and forth to compare before and changes Example 2, page 78

  45. Matrices and Gauss-Jordan • Every step in the Gauss-Jordan elimination method can be expressed with matrices, rather than systems of equations, thus simplifying the whole process: • Steps expressed as systems of equations: • Steps expressed as augmented matrices: Toggle slides back and forth to compare before and changes Example 2, page 78

  46. Matrices and Gauss-Jordan • Every step in the Gauss-Jordan elimination method can be expressed with matrices, rather than systems of equations, thus simplifying the whole process: • Steps expressed as systems of equations: • Steps expressed as augmented matrices: Toggle slides back and forth to compare before and changes Row Reduced Form of the Matrix Example 2, page 78

  47. Row-Reduced Form of a Matrix • Each row consisting entirely of zeros lies below all rows having nonzero entries. • The firstnonzero entry in each nonzero row is 1 (called a leading1). • In any two successive (nonzero) rows, the leading1 in the lower row lies to the right of the leading1 in the upper row. • If a column contains a leading1, then the other entries in that column are zeros.

  48. Row Operations • Interchange any two rows. • Replace any row by a nonzero constant multiple of itself. • Replace any row by the sum of that row and a constant multiple of any other row.

  49. Terminology for theGauss-Jordan Elimination Method Unit Column • A column in a coefficient matrix is in unit form if one of the entries in the column is a 1 and the other entries are zeros. Pivoting • The sequence of row operations that transforms a given column in an augmented matrix into a unit column.

  50. Notation for Row Operations • Letting Ri denote the ithrow of a matrix, we write Operation 1: Ri↔ Rjto mean: Interchange row i with row j. Operation 2:cRito mean: replace row i with ctimes row i. Operation 3: Ri + aRj to mean: Replace row i with the sum of row i and atimes row j.