Systems of Equations 1. Systems of Linear Equations

Transcription

1 Lecture 1

2 Systems of Equations 1. Systems of Linear Equations [We will see examples of how linear equations arise here, and how they are solved:] Example 1: In a lab experiment, a researcher wants to provide a rabbit 7 units of vitamin A, and 10 units of vitamin C. She has two foods, Food 1 and Food 2 to give them. Each gram of food 1 contains 3 units of vitamin A, and 4 units of vitamin C. Each gram of food 2 contains 1 unit of vitamin A and 2 units of vitamin C. How many grams of each food should the rabbit be fed?

3 Solution: Let B œ " B œ # Systems of Equations the number of grams of food 1 the number of grams of food 2 Then total number of units of vitamin Eœ$B "B " # total number of units of vitamin Gœ%B #B " #

4 Systems of Equations Thus have: $B" "B# œ ( (1) %B" #B# œ "! (2) To solve for B" and B# ß follow Gaussian elimination process: 1. Get 1 in first position of first equation (multiply both sides by 1/3) " ( B" $ B# œ $ (3) œ "Î$ (1). %B" #B# œ "! (2)

5 Systems of Equations 2. Eliminate all terms underneath the first one in the first equation (multiply first equation by % and add to second equation) Ð %B %Î$ B œ #)Î$Ñ " # " 7 B" 3 B# œ 3 (3) %B" #B# œ "! (2) " 7 B" 3 B# œ 3 (3) #! #Î$ B # œ $ (4) œ % (3) (2)

6 Systems of Equations 3. Now get 1 in the first position of the second equation (multiply $ second row by ): " 7 B" 3 B# œ $ (3) $! B# œ " (5) œ (3) # #

7 Systems of Equations 4. Finally guarantee that all entries above every " leading position are 0 - we want $ B # to be 0. Multiply second equation by " $ and add to first equation: B"! œ #! B# œ" " (6) œ $ (5) (3) $ (5) œ (3) # Ê B# œ "à B" œ # [Thus feed 2 ounces of food 1 and 1 ounce of food 2.]

8 Systems of Equations Definition 1: An equation is linear in the variables BßBßáßB " # 8 if it is of the form +B " " +B # # á +B 8 8 œ,ß where +ß+ßáß+ and, are given constants. " # 8 [as in the above example] A collection of such equations is called a system of linear equations. A set of fixed numbers BßB " #ßáßB8 which makes all of the equations true is called a solution of the system.

9 Input-output analysis Example 2 (Input-Output Analysis in economics): An economy is divided into three sectors: Coal, Steel, Electricity. Each industry depends on the others for raw materials. To make $1 of coal, takes no coal, $.10 of steel, $.10 of electricity. To make $1 of steel, it takes $.20 of coal, $.10 of steel, and $.20 of electricity. To make $1 of electricity, it takes $.40 of coal; $.20 of steel, and $.10 of electricity.

10 Input-output analysis If we want the economy to output $1 billion of coal, $.7 billion of steel, and $2.9 billion of electricity, how much coal, steel, and electricity will we need to use up? Let B " œ total amount of coal produced B # œ total amount of steel produced B $ œ total amount of electricity produced

11 Input-output analysis Then: Since we produce B " dollars of coal, we use:! ÐB" Ñ of coal ÐÞ"!Ñ ÐB" Ñ of steel ÐÞ"!Ñ ÐB" Ñ of electricity Since we produce B # dollars of steel, we use: Þ#! ÐB# Ñ of coal Þ"! ÐB# Ñ of steel Þ#! ÐB# Ñ of electricity Since we produce B 3 dollars of electricity, we use: Þ%! ÐB$ Ñ of coal Þ#! ÐB$ Ñ of steel Þ"! ÐB$ Ñ of electricity

12 Input-output analysis Thus, the amount of coal output = (amount of coal produced) - (amount of coal used) œb! B Þ# B Þ% B " " # $ œb Þ#B Þ%B " $ $ Similarly, amount of steel output = (amount of steel produced) - (amount of steel used) œb Þ"B Þ"B Þ#B # " # $ œ Þ"B Þ*B Þ#B " # $

13 Input-output analysis And amount of electric output is, similarly: B Þ"B Þ#B Þ"B $ " # $ œ Þ"B Þ#B Þ*B " # $ Thus we have: B Þ#B Þ%B œ " " # $ Þ"B Þ*B Þ#B œþ( " # $ Þ"B Þ#B Þ*B œ#þ* " # $

14 Solving the system To solve for B"ß B#ß B$ [generalization of above method] Step 1: Step 2: Make sure that coefficient of B " in first row is 1 Clear out B [it is] coefficients below the first B À " " [Below we can keep tabs on the equations using an augmented matrix; then just work with the matrix]

15 Solving the system Multiply first row by.1 and add to second row; ÐÞ"B Þ!#B!Þ%B œ Þ"Ñ " # $ B" Þ#B# Þ%B$ œ " (1) Þ"B" Þ*B# Þ#B$ œþ( (2) Þ"B" Þ#B# Þ*B$ œ #Þ* (3) Corresponding augmented matrix: Ô " Þ# Þ% l " Þ" Þ* Þ# l Þ( Õ Þ" Þ# Þ* l #Þ* Ø Now add above red equation to second equation:

16 Solving the system B Þ#B Þ%B œ "!B" Þ))B# Þ#%B$ œ.) (4) œ (2) Þ" (1) Þ"B Þ#B Þ*B œ#þ* " # $ (1) " # $ (3) or Ô " Þ# Þ% l "! Þ)) Þ#% l Þ) Õ Þ" Þ# Þ* l #Þ* Ø

17 Solving the system Now multiply first row by.1 and add to third row: ÐÞ"B Þ!#B!Þ%B œ Þ"Ñ " # $ B" Þ#B# Þ%B$ œ " (1) Ê!B" Þ))B# #%B$ œ ) (4)!B" Þ##B# Þ)'B$ œ $ (5) œ (3) Þ" (1) Ô " Þ# Þ% l "! Þ)) Þ#% l Þ) Õ! Þ## Þ)' l $ Ø or

18 Solving the system Step 3: Make sure coefficient of B # in second row is 1: multiply 1 row 2 by :.88 B" Þ#B# Þ%B$ œ " (1)!B" B# Þ#(#(B$ œ Þ*!*" (6) œ (4)!B" Þ##B# Þ)'B$ œ $ (5) or Ô " Þ# Þ% l "! " Þ#(#( l Þ*" Õ! Þ## Þ)' l $ Ø " Þ))

19 Step 4: Solving the system Clear coefficients below x in second row: 2 Multiply second row by.22 and add to third row: B" Þ#B# Þ%B$ œ " (1) Ð!B" Þ##B# Þ!'!!B$ œ Þ#!!!Ñ!B" B# Þ#(#(B$ œ Þ*!*" (6)!B"!B# Þ)!!!B$ œ $Þ#!! (7) = (5) + (6) ## or Ô " Þ# Þ% l "! " Þ#(#( l Þ*" Õ!! Þ) l $Þ#!! Ø

20 Step 5: Solving the system Make sure coefficient of B $ in third row is 1: 1 Multiply third row by :.8 B Þ#B Þ%B œ "!B B Þ#(#(B œ Þ*!*"!B"!B# B$ œ % " # $ (1) " # $ (6) (7) œ (5) " Þ)!!! or Ô " Þ# Þ% l "! " Þ#(#( l Þ*" Õ!! " l % Ø (") (6) (7)

21 Solving the system Continue: take all leading 1's and turn all numbers above them to 0. Ô " Þ#! l #Þ' (") Þ% (7) œ (8)! "! l # (6) Þ#(#( (7) œ (9) Õ!! " l % Ø (7)

22 Solving the system Ô "!! l $ ( 8) Þ 2 ( 9) œ ( 10)! "! l # ( 9) Õ!! " l % Ø (7) Now the system is: B œ! " B œ # # B œ % $

23 Solving the system Conclusion: if we want to output 1,.8 and 2.9 billion dollars worth of coal, steel, and electricity, we have to produce 3, 2, and 4 billion dollars of these respectively. This procedure is called Gaussian Elimination. [note that larger numbers of components in economies will require larger numbers of variables in the system]

24 Solving the system General system of equations: + B + B á + B œ, "" " "# # "8 8 " + B + B á + B œ, #" " ## # #8 8 # + B + B á + B œ, $" " $# # $8 8 $ ã + B + B á + B œ, 7" " 7# # [note the + 34 are just names for constants; the theory of what can happen to such systems of equations motivates linear algebra and lots of mathematics.]

25 Solving the system 2. An example where the number of solutions is infinite: B" #B# %B$ œ % Ê " # % l % " % # l ' B " + %B# #B$ = ' (1) (2) Ê " # % l %! # # l # (1) (3) œ (2) (1)

26 Infinite number of solutions Ê " # % l %! " " l " (1) " # (3) œ (4) Ê "! ' l #! " " l " [ write out again:] B"!B# 'B$ œ #!B" B# B$ œ "

27 Infinite number of solutions Equivalently: B" œ # 'B$ B# œ " B$ [ note all steps are reversible] : Notice that here B can be any number, and then B and are determined in terms of it. Solution: B # $ " [for example B" œ # 'B$ B# œ " B$ B$ œ any number will work] B œ "ß B œ #ß B œ % $ # "

28 Infinite number of solutions Definition 2: A variable like B $ in terms of which other variables are defined is known as a free variable.

29 No solutions Example with no solutions: B" #B# œ % (1) #B" %B# œ ' (2) Ê " # l % # % l ' (1) (2) Ê " # l % (1)!! l # (2) # (1) œ (3) [ write out again:] B #B œ % " #

30 No solutions!b"!b# œ # system is inconsistent, i.e., it has no solutions. Note: last row has the form: c!!! l # d [can see this by looking at graphs of the two original equations: The lines B" #B# œ % and 2 B" %B# = ' are parallel and have no intersection; hence no common solution of the equations:]

31 No solutions

32 No solutions Theorem 1: A system of linear equations is consistent (i.e., it has a solution) if and only if no row of the reduced matrix has the form c!! á! l, d where, is a non-zero number. If the system is consistent it has either a unique solution if there are no free variables, or an infinite number of solutions if there is at least one free variable. Sketch of Proof: in book

33 3. Terminology: Def 3: A system of equations is consistent if it has some solution; it is inconsistent if there is no solution. A linear system in which the right hand sides are all zero is a homogeneous system. [ Ex: B $B œ! ; B $B œ! ] " # " # Two systems are equivalent if they have exactly the same sets of solutions BßáßB " 8, i.e. if every set of BßáßB " 8 which solves the first system also solves the second.

34 4. Matrices: [ Have seen example above. ] Definition 4: A numbers. matrix is a rectangular array of Example 3: Ô $ "Î# $Î# & % ( ) * Õ "" "# & $ Ø [Have seen these things; all will be done by example:]

35 Addition: " " $ " " # œ!! & % $Î# # # & % # "$Î# ' [i.e., add corresponding entries] Scalar multiplication (multiplication by a constant): # " & ' $ "& $ œ $ % "Î# * "# $Î#

36 Multiplication: [note that the number of columns in first matrix must be number of rows in second] Ô # "Î# # # " * $ # " œ " " $ Þ Õ Ø " #&Î# " %

37 System solutions - review 1. Review of solving systems by elimination: Example 1: #B $C %D œ ) B #C Dœ! $B C #Dœ"" Recall: standard procedure is Gaussian elimination: Ô # $ % l ) " # " l! Õ$ " # l "" Ø

38 System solutions - review Ô " $Î# # l % " # " l! Õ$ " # l "" Ø Ä Ô " $Î# # l %! "Î# " l % Õ! ""Î# % l " Ø

39 Ä System solutions - review Ô " $Î# # l %! " # l ) Õ! ""Î# % l " Ø Ä Ô " $Î# # l %! " # l ) Õ! " )Î"" l #Î"" Ø

40 Ä System solutions - review Ô " $Î# # l %! " # l ) Õ!! $!Î"" l *!Î"" Ø Ä Ô " $Î# # l %! " # l ) Õ!! " l $ Ø [called row echelon form; can now solve for B 3 ] [ can go further: reduced row echelon form] :

41 System solutions - review Ä Ô " $Î#! l #! "! l # Õ!! " l $ Ø Ä Ô "!! l "! "! l # Õ!! " l $ Ø [note difference between echelon and reduced echelon forms]

42 System solutions - review Definition 1: The process of reducing a matrix to echelon form is Gaussian elimination; the process of reducing it to reduced echelon form is Gauss- Jordan elimination. [then can read off B 3 more easily]

43 Elementary row operations Elementary Row Operations: 1. Interchange rows: Type I 2. Multiply a row by non-zero constant: Type II 3. Add - times row 3 to row 4: Type III Definition 2: A matrix is in row echelon form if: (1) All rows of 0s are at the bottom (2) The first nonzero entry in the first row is a 1 (3) For two successive rows the first entry on the second row is to the right of the first entry on the first row.

44 Elementary row operations A matrix is in addition in reduced row echelon form if also: (4) For any column which contains the leading entry of some row, all of the other entries in that column are 0. Definition 3: A matrix is row equivalent to F if it can be obtained from F through a series of ERO's. FOR NOW: Proof of the following theorem is deferred till later.

45 Elementary row operations Theorem 2: Every matrix is row equivalent to one and only one matrix in reduced echelon form. Proof: This proof is given in Appendix A; it will follow after we have covered the theory of Chapter 4. Theorem 3: If two systems have row equivalent augmented matrices, they are equivalent. Proof: Since we can get from one system to the second by ERO, every solution of the first system must solve the second. Since we can reverse the ERO to get from the second to the first, it follows

46 Elementary row operations that every solution of the second must solve the first. Thus the systems are equivalent. Definition 4: Row reduction is the process of applying elementary row operations to obtain the row echelon or reduced row echelon form of a matrix. Typical form of augmented matrix after row reduction:

47 Elementary row operations Ô " # $ " " " $ " l $!! " $ # " # " l " Ö Ù!!! " & % $ # l " Õ!!!!!! " # l # Ø General form of corresponding system: B" #B# $B$ B% B& B' $B( B) œ $ B$ $B% #B& B' #B( B) œ " B% &B& %B' $B ( + #B) œ " B #B œ # ( )

48 Elementary row operations Definition 5: The leading entry in each row is called a pivot entry. The columns containing pivot entries are called pivot columns. Exercise: Now solve for the pivot variables in terms of the non-pivot variables; note this can be done easily at this point [nevertheless - note this can be hard for large systems] [So: get into reduced row echelon form]

49 Elementary row operations Ô " # $ " " "! ( l $!! " $ # "! $ l $!!! " & %! ) l & Ö Ù!!!!!! " # l # Õ Ø Ô " # $! % $! " l #!! "! "$ ""! #" l "# Ö Ù!!! " & %! ) l & Õ!!!!!! " # l # Ø

50 Elementary row operations Ô " #!! $& $!! '# l $%!! "! "$ ""! #" l "# Ö Ù!!! " & %! ) l & Õ!!!!!! " # l # Ø Pivot entries in 1st, 3d, 4th, 7th columns Ê pivot variables: BßBßBßB " $ % ( Ê remaining (free) variables BßBßBßBÞ # & ' ) Now system is: B" #B#!B$!B% $&B& $!B'!B( '#B) œ $% B$!B% "$B& ""B'!B( #"B) œ "# B% &B& %B'!B( )B) œ & B #B œ # ( )

51 Elementary row operations Express pivot variables in terms of free ones: B" œ #B# $&B& $!B' '#B) $% B$ œ "$B& ""B' #"B) "# B% œ &B& %B' )B) & B œ #B # ( ) General solution: pick any values for free variables; pivot (non-free) values are determined]

52 Elementary row operations Note we have replaced z œd" ßD# áby trajectory z œd" ßD# ßá defined by Recall: D œ D 5 if W 5! or D 5! 5 œ.! otherwise Excursion is a segment of original random walk from a negative value to a max value. or: a segment of modified walk from 0 to max value.