Gauss and Gauss Jordan Elimination

Transcription

1 Gauss and Gauss Jordan Elimination

2 Row-echelon form: (,, ) A matri is said to be in row echelon form if it has the following three properties. () All row consisting entirel of zeros occur at the bottom of the matri. () All entries in a column below a leading entr i.e. are zero.. () For two successive nonzero rows, the leading in the higher row is farther to the left than the leading in the lower row. Reduced row-echelon form: (,,, ) If a matri is in echelon form and satisf following additional condition, then it is in reduced echelon from. () Ever column that contains a leading has zeros everwhere else.

3 Echelon & Reduced Echelon Forms: Notes A matri ma be row reduced (i.e. transformed b elementar row operations) into more than one matri in echelon form, using different sequence of row operations. In summar, echelon form obtained from a given matri ma not necessaril be unique. However, the reduced echelon from one obtains from a matri is unique. A matri in reduced row-echelon form is of necessit in row-echelon form, but not conversel.

4 Eample Row-Echelon & Reduced Row-Echelon form Reduced row-echelon form:,, Row-echelon form:,,

5 Eample More on Row-Echelon and Reduced Row- Echelon form All matrices of the following tpes are in row-echelon form ( an real numbers substituted for the s. ) :,,,,,, All matrices of the following tpes are in reduced row-echelon form ( an real numbers substituted for the s. ) :

6 form) echelon row - (reduced form) (row - echelon Eercise(Row-echelon form or reduced row-echelon form) form) (row - echelon form) echelon row - (reduced

7 Eample (a) Solutions of Linear Sstems Suppose that the augmented matri for a sstem of linear equations have been reduced b row operations to the given reduced row-echelon form. Solve the sstem. (a) Solution (a) the corresponding sstem of equations is : - z

8 Eample (b) Solutions of Linear Sstems (b) Solution (b). The corresponding sstem of equations is : - leading variables free variables

9 Eample (b) Solutions of Linear Sstems We see that the free variable can be assigned an arbitrar value, sa t, which then determines values of the leading variables.. There are infinitel man solutions, and the general solution is given b the formulas t, t, t, t

10 Eample (c) Solutions of Linear Sstems (c) - Solution (c). The th row of zeros leads to the equation places no restrictions on the solutions (wh?). Thus, we can omit this equation.

11 Solution (c) Eample (c) Solutions of Linear Sstems. Solving for the leading variables (,, ) in terms of the free variables (, ) :. The free variable can be assigned an arbitrar value, there are infinitel man solutions, and the general solution is given b the formulas s - t s -t -t, t,

12 Eample (d) Solutions of Linear Sstems (d) Solution (d): the last equation in the corresponding sstem of equation is Since this equation cannot be satisfied, there is no solution to the sstem.

13 Elimination Methods (/) There should be a step-b-step elimination procedure that can be used to reduce an matri to reduced rowechelon form. 8

14 Elimination Methods (/) Step. Locate the leftmost pivot column i.e. a coulmn that does not consist entirel of zeros. 8 Leftmost pivot or nonzero column Step. Interchange the top row with another row, to bring a nonzero entr (Pivot) to top of the Pivot column found in Step. 8 The st and nd rows in the preceding matri were interchanged.

15 Elimination Methods (/) Step. If the entr that is now at the top of the column found in Step is a, multipl the first row b /a in order to introduce a leading. The st row of the preceding matri was multiplied b /. Step. Add suitable multiples of the top row to the rows below so that all entries below the leading become zeros. 9 - times the st row of the preceding matri was added to the rd row.

16 Elimination Methods (/) Step. Now cover the top row in the matri and begin again with Step applied to the sub-matri that remains. Continue in this wa until the entire matri is in row-echelon form. Pivot 9 Leftmost nonzero or pivot column in the sub-matri 9 The st row in the submatri was multiplied b -/ to introduce a leading.

17 Elimination Methods (/) Step (cont.) - times the st row of the submatri was added to the nd row of the sub-matri to introduce a zero below the leading. The top row in the sub-matri was covered, and we returned again Step. The first (and onl) row in the new submetri was multiplied b to introduce a leading. Leftmost nonzero or pivot column in the new sub-matri The entire matri is now in row-echelon form. Pivot

18 Elimination Methods (/) Step. Beginning with last nonzero row and working upward, add suitable multiples of each row to the rows above to introduce zeros above the leading s. / times the rd row of the preceding matri was added to the nd row. - times the rd row was added to the st row. The last matri is in reduced row-echelon form. times the nd row was added to the st row.

19 The first nonzero column Produce leading Zeros elements below leading leading Produce leading The first nonzero column Summar: Elimination Methods Eample 8 8 r ) ( r 9 ) ( r Submatri

20 Zeros elements below leading Zeros elsewhere leading Produce leading leading Eample (Contd.) 9 ) ( r ) ( r () r ) ( r ) ( r () r Submatri form) (row - echelon form) echelon row - (reduced

21 Summar: Elimination Methods Step~Step: the above procedure produces a row-echelon form and is called Gaussian elimination. Step~Step: the above procedure produces a reduced row-echelon form and is called Gaussian-Jordan elimination. Two matrices are said to be row equivalent if one matri can be obtained from the other using elementar row operations Ever matri has a unique reduced row-echelon form but a row-echelon form of a given matri is not unique.

22 Gaussian elimination and back substitution Method To solve a sstem of linear equations, we first simplif the sstem through succession of elementar row operations, until we arrive at an augmented matri in row-echelon form. This process is called the gaussian elimination. and we ma solve the simplest equation. Then we successivel substitute the results into more and more complicated equations to get the whole solution. This process is called the backward substitution.

23 Solve the sstem Eample + - = - + = + + = Note that the variable appears in all equations. We tr to eliminate so that it appears in onl equation. Specificall, b multipling - to the nd equation and then adding to the st equation, is eliminated from the st equation. - = = + + =

24 Eample (contd.) Net the operation (-)[equation ] + [equation ] (i.e. multipl - to the nd equation and then add to the rd equation) further eliminates from the rd equation. - = = - = The sstem is as simple as we can get as far as is concerned. Net we tr to simplif the sstem regarding. The operation (-)[equation ] + [equation ] eliminates from the rd equation. - = = =

25 Eample (contd.) B echanging the st and the nd equations (denoted [equation ] [equation ]), we rearrange the equations from the most complicated down to the simplest, i.e. - + = - = - = Finall, the coefficients can be further simplified b multipling / and / to the nd and the rd equations (denoted (/)[equation ] and (/)[equation ]). - + = - = - = OR This completes the simplification process Augmented Matri in Row-Echelon form

26 Eample (contd.) Now we start solving the sstem b backward substitution. From the rd equation (the simplest equation), we have = Substituting this into the nd equation (the net simplest), we get = - + = - + => =. Further substituting into the st equation (the most complicated), we get = + - = + - => =. We conclude that the sstem has a unique solution =, =, =.

27 Eample (contd.) In the eample, we have used the elementar row operations in following order to simplif given sstem. A constant multiple of one row is added to another row Two rows are interchanged A row is multiplied b a nonzero constant In general, an sstem of linear equations can be simplified, b these three operations, to a simple sstem in which equations are arranged from the most complicated to the simplest. This process is called the Gaussian elimination method. After completing the gaussian elimination, we ma solve the simplest equation. Then we successivel substitute the results into more and more complicated equations to get the whole solution. This process is called the backward substitution.

28 Eample Solve b Gaussian elimination and back-substitution. Solution We convert the augmented matri to the row-echelon form The sstem corresponding to this matri is 9 z z z 9 9, 9, z z z

29 Eample Solution Solving for the leading variables 9 z z, z, Substituting the bottom equation into those above, z, Substituting the nd equation into the top,, z

30 Eercise: 8 z z z Let us consider the set of linearl independent equations. Augmented matri for the set is: 8

31 Eercise: (Contd.) 8 8 z z z Step : Eliminate from the nd and rd equation..... z z z

32 Eercise: (Contd.) Step : Eliminate from the rd equation. R +R R. 8 8z z.z Step :.R R -R R.. 8 (/8)R R.z 8.z z

33 Eercise: (Contd.).. 8.z 8.z z From Row, z = From Row, -.z = - or, -. () = or, = - From Row, +.z = 8 or, (- ) +. () = 8 or, =

34 Eercise Solve the linear sstem using Gaussian elimination and back substitution method Answer: = - / = = - / = + + = =

35 Eercise Solve the linear sstem using Gaussian elimination and back substitution method + - = - + = - + = This is the simplest shape (row echelon form) we can get b row operations. The corresponding sstem is - + = - = - = Since the rd equation = is a contradiction impling the original sstem has no solution.

36 Eercise A garden suppl centre bus flower seed in bulk then mies and packages the seeds for home garden use. The suppl center provides different mies of flower seeds: Wild Thing, Momm Dearest and Medicine Chest. ) Wild Thing seed mi contains % of wild flower seed, % of Echinacea seed and % of Chrsanthemum seed. ) Momm Dearest mi consists % Chrsanthemum seed and % wild flower seed. ) The Medicine Chest mi contains 9% Echinacea seed, % other seeds. In a single order, the store received grams of wild flower seed, grams of Echinacea seed and grams of Chrsanthemum seed. Use matrices and complete Gauss-Jordan Elimination to determine how much of each miture the store can prepare. Assume that the garden center has an ample suppl of other seeds used in Medicine chest on hand.

37 Solution: Eercise: (contd.) Assign variables to the amount of each mi that will be produced. Let X = Amount of Wild Thing Let Y = Amount of Momm Dearest Let Z = Amount of Medicine Chest Perform a balance on each of the components that are available. Wild flower Echinacea Chrsanthemum.X +.Y + Z = g.x + Y +.9Z = g.x +.Y + Z = g In matri form, this can be written as A b

38 Eercise: (contd.) Before the matrices are populated, it is (sometimes) helpful to re-arrange the equations to reduce the number of steps in the Gauss Elimination. To do this (if there seems like an eas solution), attempt to move zeros to the bottom left, and tr to maintain the first row with non-zeros ecept for the last entr, since row is used to reduce other rows. B moving the last column (Z) to the front, and switching the first and second row, the new set of equations becomes: Echinacea Wild flower Chrsanthemum.9Z +.X + Y = g Z +.X +.Y = g Z +.X +.Y = g

39 Eercise: (contd.) Appl the Gauss Elimination: 9 Z =, X =, and Y =

40 Summar: Gauss Elimination for Solving A Sstem of Equations. Write the augmented matri of the sstem.. Use elementar row operations to construct a row equivalent matri in row-echelon form.. Write the sstem of equations corresponding to the matri in row-echelon form.. Use back-substitution to find the solutions to this sstem.

41 Gauss-Jordan Elimination In Gauss-Jordan elimination, we continue the reduction of the augmented matri until we get a row equivalent matri in reduced rowechelon form. (r-e form where ever column with a leading has rest zeros) a b c

42 Eample: 8 z z z Let us consider the set of linearl independent equations. Augmented matri for the set is: 8

43 Eample: (Contd.) 8 8 z z z Step : Eliminate from the nd and rd equation..... z z z

44 Eample: (Contd.) Step : Eliminate from the rd equation. R +R R. 8 8z z.z Step :.R R -R R.. 8 (/8)R R.z 8.z z

45 Eample: (Contd.) Step : Eliminate z from the nd equation.. 8 ( Row ) (.) ( Row ).z.z z 8. 8.z z 8

46 Eample: (Contd.) ) ( () ) ( Row New Row Row z z Step -: Eliminate from the st equation.. z z

47 Eample: (Contd.) ) (.) ( ) ( Row New Row Row.. z z Step -: Eliminate z from the st equation z

48 Eample : Gauss-Jordan Elimination Solve b Gauss-Jordan Elimination Solution: The augmented matri for the sstem is

49 Eample (Contd.) Adding - times the st row to the nd and th rows gives Multipling the nd row b - and then adding - times the new nd row to the rd row and - times the new nd row to the th row gives

50 Eample (contd.) Interchanging the rd and th rows and then multipling the rd row of the resulting matri b / gives the row-echelon form. Adding - times the rd row to the nd row and then adding times the nd row of the resulting matri to the st row ields the reduced row-echelon form. - -

51 Eample (contd.) The corresponding sstem of equations is Solution The augmented matri for the sstem is We assign the free variables, and the general solution is given b the formulas:,,,,, t s s r t s r

52 Eercise: Solve the linear sstem using (a) Gaussian elimination and (b) Gauss-Jordanelimination 9 Answer and

53 Eercise : Solve a sstem b Gauss-Jordan elimination method (onl one solution) 9 z z Sol: augmented matri 9 9 ) (, r r 9 r 9) ( ) (,, r r r z form) (row - echelon form) row -echelon (reduced

54 Eercise :Solve a sstem b Gauss-Jordan elimination method (infinitel man solutions) Sol: augmented matri ( ) ( ) ( ), r, r, r r ( ) (reduced row - echelon form) the corresponding sstemof equations is leading variable: free variable:,,since onl is common to both equations, we can assign itsvalues arbitraril.

55 Eercise: (Contd.) Let t,,, t R t t t So this sstem has infinitel man solutions.

56 Homogeneous sstems of linear equations A sstem of linear equations is said to be homogeneous if all the constant terms are zero. n mn m m m n n n n n n a a a a a a a a a a a a a a a a A homogenous sstem of linear equations is alwas consistent, since =, =,. n = will satisf each equation in the sstem.

57 Homogeneous sstems of linear equations (Contd.) Trivial solution n Nontrivial solution: other solutions Notes: () Ever homogeneous sstem of linear equations is consistent. () If the homogenous sstem has fewer equations than variables, then it must have an infinite number of solutions. () For a homogeneous sstem, eactl one of the following is true. (a) The sstem has onl the trivial solution. OR (b) The sstem has infinitel man nontrivial solutions in addition to the trivial solution.

58 Solve the following homogeneous sstem Sol: augmented matri r leading variable : free variable : Let t t, t, () ( ) r,, r t, t R When t,, () (trivial solution) (reduced row - echelon form)

59 Eample Solve the following homogeneous sstem of linear equations b using Gauss-Jordan elimination. Solution The augmented matri Reducing this matri to reduced rowechelon form

60 Eample (contd.) Solution (cont) The corresponding sstem of equation Solving for the leading variables is Thus the general solution is Note: the trivial solution is obtained when s= t =. t t s t s,,,,

61 Eample (Contd.) None of the three row operations alters the final column of zeros, so the sstem of equations corresponding to the reduced row-echelon form of the augmented matri must also be a homogeneous sstem.

62 Application of Linear Sstems A mathematical model is an equation or sstem of equations that represent a real life situation. One main reason to solve sstems of linear equations is to solve real world problems.

63 Applications of Sstems To solve problems using a sstem. Determine the unknown quantities. Let different variables represent those quantities. Write a sstem of equations one for each variable Eample : In a recent ear, the national average spent on two athletes, one female and one male, was $ for Division I-A schools. However, average ependitures for a male athlete eceeded those for a female athlete b $9. Determine how much was spent per athlete for each gender.

64 Applications of Sstems Solution Let = average ependitures per male = average ependitures per female Average spent on one male and one female 9 8 () (), Average Ependiture per male: $8, and per female: from () = 8 9 = $.

65 Eercise: A man walks at a rate of miles per hour and jogs at a rate of miles per hour. He walks and jogs a total distance of. miles in.9 hours. How long does the man jog? Solution: Let represent the amount of time spent walking and represent the amount of time spent jogging. Since the total time spent walking and jogging is.9 hours, we have the equation + =.9. We are given the total distance traveled as. miles. Since Distance = Rate Time, distance walking = and distance jogging =. Then total distance is represented b + =..

66 We can solve the sstem using substitution.. Solve the first equation for. Substitute this epression into the second equation.. Solve second equation for Eercise: (contd.). Find the value b substituting this value back into the first equation.. Answer the question: Time spent jogging is. hours. Solution:.9. (.9 )

67 Bob and Mar went to the movies. Bob purchased medium bags of popcorns and large drinks and paid $.. Mar purchased medium bags of popcorns and large drinks and ended up paing $.. What was the price of a medium bag of popcorn and what was the price of a large drink? Denote p the price of a medium bag of popcorn and d the price of a large drink. Then we have p + d = p + d =

68 p + d = p + d = -p - d = - p + d = (multipl the first b -) (add the first to the second) -p = - p =. Substituting back and solving for d we get d =.. Medium bag of popcorn costs $. and a large drink costs $..

69 Eercise: A restaurant serves two tpes of fish dinnerssmall for $.99 each and large for $8.99. One da, there were total orders of fish, and the total receipts for these orders was $.. How man small dinners and how man large dinners were ordered? Answer: small orders and large orders

70 Application of Homogenous equation in Chemistr Balance the chemical equation Solution: We would like to determine,, and so that Now for a balance equation, the number of atoms of each element must be same on each side of above equation, so we have Carbon (C): = Hdrogen (H): = OR Ogen (O): = + O H CO O H C O H CO O H C

71 Eercise: (Contd.). Augmented matri can be written as. Now performing row operations to obtain reduced row-echelon form

72 . The corresponding sstem of equations is : leading variables free variable Eercise: (Contd.). Reduced row-echelon matri :

73 Eercise: (Contd.). We see that the free variable ( ). There are infinitel man solutions, and the general solution is given b the formulas. can be assigned an arbitrar value, sa t, which then determines values of the leading variables. t t t Note that in this case t must be a positive integer chosen in such a manner that, and are positive integers, sa t = gives =, =, =, =. The balanced chemical equation is: CH O CO HO

74 Eercises. Balance the chemical equation Fe O C Fe CO Answer: Fe O C Fe CO. Balance the chemical equation Cu HNO Cu( NO ) H O NO Answer: Cu 8HNO Cu( NO ) H O NO

75 Eercises. Balance the chemical equation Fe O C Fe CO Answer: Fe O C Fe CO. Balance the chemical equation Cu HNO Cu( NO ) H O NO Answer: Cu 8HNO Cu( NO ) H O NO

76 Gaussian elimination: The procedure for reducing a matri to a rowechelon form. Gauss-Jordan elimination: Notes The procedure for reducing a matri to a reduced rowechelon form. Summar () Ever matri has an unique reduced row echelon form. () A row-echelon form of a given matri is not unique. (Different sequences of row operations can produce different row-echelon forms.)

77 Summar (Contd.) A sstem of linear equations is said to be homogeneous if all the constant terms are zero. Ever homogeneous sstem of linear equations is consistent A homogeneous sstem of linear equations with more unknowns than equations has infinitel man solutions