Finite Math Lecture Slides

Set Theory

A set is an unordered collection of objects without repetitions.

Example{a, b, c} = {b, c , a}

We usually use capital letters to denote sets.

A = {a, b, c}B = {b, c, d , e}

Sets are the nouns of Set Theory, they’re what it’s about.

Set Theory




A = {a, b, c}B = {b, c, d , e}


Set Theory




A = {a, b, c}B = {b, c, d , e}


Set Theory




A = {a, b, c}B = {b, c, d , e}


Relations

In mathematics a relation is like a verb:

it “relates” two things in a way that can be either true or false.

In high school algebra the common relations are =, <, >.

In Set Theory the most common relations are =, ⊂, ⊃, ∈.

Relations





Relations





⊂

The expressionA ⊂ B

means that every object (or element) of A is also in B.

Example:A = {a, b, c}B = {a, b, c , d , e}A ⊂ BB ⊃ AB 6⊂ A

Also,A ⊂ AB ⊂ B,

so ⊂ is analagous to ≤.

⊂

The expressionA ⊂ B

means that every object (or element) of A is also in B.

Example:A = {a, b, c}B = {a, b, c , d , e}A ⊂ BB ⊃ AB 6⊂ A

Also,A ⊂ AB ⊂ B,

so ⊂ is analagous to ≤.

∈

The expressiona ∈ A

means that the object a is contained in A.

Example:A = {a, b, c}a ∈ Ab ∈ Ad 6∈ A

∈

The expressiona ∈ A

means that the object a is contained in A.

Example:A = {a, b, c}a ∈ Ab ∈ Ad 6∈ A

Operations

An operation is something you do to objects to get new ones.

In algebra the common operations are +, −, ·, ÷, etc.

In Set Theory the common operations are ∪, ∩, ′, and ×.

Operations

An operation is something you do to objects to get new ones.

In algebra the common operations are +, −, ·, ÷, etc.

In Set Theory the common operations are ∪, ∩, ′, and ×.

∪

The expressionA ∪ B

means “combine all elements of A with those of B”.

Example:A = {a, b, c}B = {b, c, d , e}

A ∪ B = {a, b, c , d , e}

Loosely, ∪ means “or”.

∪




A ∪ B = {a, b, c , d , e}


∪




A ∪ B = {a, b, c , d , e}


∩

The expressionA ∩ B

means “take only those elements of A which are also in B”.

Example:A = {a, b, c}B = {b, c , d , e}

A ∩ B = {b, c}

Loosely, ∩ means “and”.

But the English word “and” sometimes means “or”!

∩




A ∩ B = {b, c}



∩




A ∩ B = {b, c}



U

The letter U is used to denote the universe,

the set of every object that matters in a problem.

Example: If a problem is about English letters, the universe mightbe

U = {a, b, c , d , e, f , g , h, i , j , k , l ,m, n, o, p, q, r , s, t, u, v ,w , x , y , z}.

But if it’s about Greek letters the universe would be

U = {α, β, γ, δ, ε, ζ, η, θ, ι, κ, λ, µ, ν, ξ, o, π, ρ, σ, τ, υ, φ, χ, ψ, ω}.

If it’s about rolling dice the universe might be

U = {1, 2, 3, 4, 5, 6}.

U

The letter U is used to denote the universe,

the set of every object that matters in a problem.

Example: If a problem is about English letters, the universe mightbe

U = {a, b, c , d , e, f , g , h, i , j , k , l ,m, n, o, p, q, r , s, t, u, v ,w , x , y , z}.

But if it’s about Greek letters the universe would be

U = {α, β, γ, δ, ε, ζ, η, θ, ι, κ, λ, µ, ν, ξ, o, π, ρ, σ, τ, υ, φ, χ, ψ, ω}.

If it’s about rolling dice the universe might be

U = {1, 2, 3, 4, 5, 6}.

′

The expressionA′

is read “A complement” and denotes the objects of U not in A.

Example:U = {1, 2, 3, 4, 5, 6}E = {2, 4, 6}

E ′ = {1, 3, 5}

Loosely, ′ means “not”.

′

The expressionA′


Example:U = {1, 2, 3, 4, 5, 6}E = {2, 4, 6}

E ′ = {1, 3, 5}


′

The expressionA′


Example:U = {1, 2, 3, 4, 5, 6}E = {2, 4, 6}

E ′ = {1, 3, 5}


Example: Suppose that

U = all animals W = whalesF = fish M = mammalsA = aquatic animals S = seals.

Translate the following English phrases into Set Theory.

1. Whales are aquatic mammals.

2. Whales are not fish.

1.W = A ∩MW ⊂ A ∩MW = A ∪MW ⊂ A ∪M






1.W = A ∩MW ⊂ A ∩MW = A ∪MW ⊂ A ∪M






1.Solution:

W ⊂ A ∩M






2.W 6= FW 6⊂ FW ⊂ F ′

W ∩ F = ∅ = {}






2.W 6= FW 6⊂ FW ⊂ F ′

W ∩ F = ∅ = {}






2.Solution:

W ⊂ F ′

W ∩ F = ∅

Venn Diagrams

A Venn diagram is a visual representation of sets.

How can we describe the shaded region?

(A′ ∩ B ′) ∪ (A ∩ B ′) ∪ (A′ ∩ B)

(A ∩ B)′


(A′ ∩ B ′) ∪ (A ∩ B ′) ∪ (A′ ∩ B)

(A ∩ B)′


(A ∩ B ′ ∩ C ) ∪ (A ∩ B ∩ C ) ∪ (A′ ∩ B ∩ C )

(A ∪ B) ∩ C


(A ∩ B ′ ∩ C ) ∪ (A ∩ B ∩ C ) ∪ (A′ ∩ B ∩ C )

(A ∪ B) ∩ C

Using Venn Diagrams to Count

Venn diagrams can be helpful in some counting problems.

We’ll use the notation n[A] to denote the number of elements in A.

Example:A = {a, r , y , p, j}

n[A] = 5

There are two important principles:

AP The addition principle, also called the partitionprinciple;

CP the complement principle.

Disjoint

Sets are disjoint if they don’t overlap:

A ∩ B = {} = ∅.

Example:

disjoint: {a, d , t, p}, {b, f , o, v , z}not disjoint: {a, d , t, p}, {b, f , p, v , z}

APThe addition principle, or partition principle, says that the size ofthe union of two disjoint sets is the sum of the sizes of each one.

If A ∩ B = ∅ then n[A ∪ B] = n[A] + n[B].





Example: What is n[A]?

n[A] = 4 + 3 = 7


n[A] = 4 + 3 = 7

Example: What is n[A ∪ B]?

n[A ∪ B] = 3 + 5 + 12 = 20

Example: What is n[A ∪ B]?

n[A ∪ B] = 3 + 5 + 12 = 20

CP

The complement principle is based on the observation thatU = A ∪ A′.

n[A] + n[A′] = n[U]

n[A] = n[U]− n[A′]n[A′] = n[U]− n[A].

So if you know n[U] and either n[A] or n[A′] you can figure out theother.

CP


n[A] + n[A′] = n[U]n[A] = n[U]− n[A′]n[A′] = n[U]− n[A].


CP


n[A] + n[A′] = n[U]n[A] = n[U]− n[A′]n[A′] = n[U]− n[A].



n[A] = 15− (5 + 7) = 3


n[A] = 15− (5 + 7) = 3

Example: What is n[A′ ∩ B ′ ∩ C ′]?

n[A] = 40− (3 + 5 + 6 + 12) = 14

Example: What is n[A′ ∩ B ′ ∩ C ′]?

n[A] = 40− (3 + 5 + 6 + 12) = 14

Play VennDoku athttp://mypage.iu.edu/˜dabrowsa/FiniteJS.html!

Example: There are 100 members of Beta Epsilon Rho fraternity,of whom 60 are business majors and 50 are psych majors; 20 aremajoring in both business and psych. How many members of thefraternity are majoring in neither business nor psych?

Venn Diagram

Example: There are 100 members of Beta Epsilon Rho fraternity,of whom 60 are business majors and 50 are psych majors; 20 aremajoring in both business and psych. How many members of thefraternity are majoring in neither business nor psych?

Venn Diagram

Example: The 100 members of Beta Epsilon Rho fraternity areordering pizza. Pepperoni is requested by 80 members, andBrussels sprouts by 15; 12 members want neither pepperoni norBrussels sprouts. How many want both pepperoni and Brusselssprouts?

Venn Diagram

Example: The 100 members of Beta Epsilon Rho fraternity areordering pizza. Pepperoni is requested by 80 members, andBrussels sprouts by 15; 12 members want neither pepperoni norBrussels sprouts. How many want both pepperoni and Brusselssprouts?

Venn Diagram

Note that 12+? = 100, so ? = 88.

Now 80+? = 88, so ? = 8.

Now work across.

Answer: 7 people like pizza with pepperoni and Brussels sprouts.

Alternate Method

n[B] + n[P] = n[B ∪ P] + n[B ∩ P]15 + 80 = 88 +

= 7

Alternate Method

n[B] + n[P] = n[B ∪ P] + n[B ∩ P]

15 + 80 = 88 += 7

Alternate Method

n[B] + n[P] = n[B ∪ P] + n[B ∩ P]15 + 80 = 88 +

= 7

Alternate Method

n[B] + n[P] = n[B ∪ P] + n[B ∩ P]15 + 80 = 88 +

= 7

Inclusion-Exclusion Formula

In general,

n[A] + n[B] = n[A ∪ B] + n[A ∩ B]n[A ∪ B] = n[A] + n[B]− n[A ∩ B]n[A ∩ B] = n[A] + n[B]− n[A ∪ B].

Example: According to a survey of the members of an investmentclub, 150 own stock in Alcoa, 180 own stock in Boeing, 100 ownstock in Coke, 70 own stock in Alcoa and Boeing, 30 own stock inAlcoa and Coke, 50 own stock in Boeing and Coke, 20 own stockin all three companies, and 60 own stock in none of these ofcompanies.

1. How many members were surveyed?

2. How many of those surveyed do not own stock in Coke?

3. How many own stock in Boeing but not Alcoa?

4. How many own stock in Alcoa or Boeing and in Coke?

5. How many own stock in Alcoa or in Boeing and Coke?

6. How many own stock in Alcoa or Boeing but not in Coke?

7. How many own stock in exactly 1 company?

8. How many own stock in exactly 2 companies?

Example: According to a survey of the members of an investmentclub, 150 own stock in Alcoa, 180 own stock in Boeing, 100 ownstock in Coke, 70 own stock in Alcoa and Boeing, 30 own stock inAlcoa and Coke, 50 own stock in Boeing and Coke, 20 own stockin all three companies, and 60 own stock in none of these ofcompanies.

Venn Diagram

Work from the middle out.

Keep going.

Finish up.

Now we can start answering the questions.

How many members were surveyed?360

How many of those surveyed do not own stock in Coke?CP: 360− 100 = 260

How many own stock in Boeing but not Alcoa?n[B ∩ A′] = 80 + 30 = 110

How many own stock in Alcoa or Boeing and in Coke?n[(A ∪ B) ∩ C ] = 10 + 20 + 30 = 60

How many own stock in Alcoa or in Boeing and Coke?n[A ∪ (B ∩ C )] = 150 + 30 = 180

How many own stock in Alcoa or Boeing but not in Coke?n[(A ∪ B) ∩ C ′] = 70 + 50 + 80 = 200

How many own stock in exactly 1 company?70 + 80 + 40 = 190

How many own stock in exactly 2 companies?

10 + 30 + 50 = 90

How many own stock in exactly 2 companies?10 + 30 + 50 = 90

Example: Of the 360 members of the Investment Club, 180 ownstock in Alcoa (A), 180 own stock in Boeing (B), 120 own stock inCoke (C), 100 own stock in Alcoa and Boeing, 90 own stock inAlcoa and Coke, 70 own stock in Boeing and Coke, and 80 ownstock in none of these companies.

1. How many members own stock in all three companies?

2. How many members own stock in exactly one of Boeing andCoke?

3. How many own stock in Coke or in Boeing and Alcoa?

4. How many do not own stock in Boeing and Coke?

5. How many do not own stock in Alcoa or Coke?

Example: Of the 360 members of the Investment Club, 180 ownstock in Alcoa (A), 180 own stock in Boeing (B), 120 own stock inCoke (C), 100 own stock in Alcoa and Boeing, 90 own stock inAlcoa and Coke, 70 own stock in Boeing and Coke, and 80 ownstock in none of these companies.

Use the Inclusion-Exclusion formula on A ∪ B.

n[A ∪ B] = n[A] + n[B]− n[A ∩ B]= 180 + 180− 100 = 260

Now360 = 260 + 80 + ?

? = 20.

Now focus on C :

120 = 20 + 70 + ?? = 30.

Now focus on A ∩ C :

90 = 30 + ?? = 60.

Now that we have the center it’s easy to work from the middle out.

How many members own stock in all three companies? 60

How many members own stock in all three companies?

60

How many members own stock in all three companies? 60

How many members own stock in exactly one of Boeing and Coke?

40 + 70 + 30 + 20 = 160

How many members own stock in exactly one of Boeing and Coke?40 + 70 + 30 + 20 = 160

How many own stock in Coke or in Boeing and Alcoa?

120 + 40 = 160

How many own stock in Coke or in Boeing and Alcoa?120 + 40 = 160

How many do not own stock in Boeing and Coke?

CP: 360− 70 = 290

How many do not own stock in Boeing and Coke?CP: 360− 70 = 290

How many do not own stock in Alcoa or Coke?

CP: 360− 210 = 150

How many do not own stock in Alcoa or Coke?CP: 360− 210 = 150

Partitions

A partition of a set A isa division of A up into nonoverlapping pieces.

Example: IfA = {1, 2, 3, 4, 5, 6}

thenA1 = {1, 6}A2 = {2, 3, 5}A3 = {4}

is a partition of A.

Of course the AP (PP) automatically applies:

n[A] = n[A1] + n[A2] + n[A3].

Example: Suppose that n[A] = 120 and A is partitioned into 4subsets W , X , Y , and Z . Suppose further that W has 12elements, X and Y are the same size, and Z is twice as large as X .Find the sizes of X , Y , and Z .

A partition can always be diagrammed as a pizza.

Fill in the sizes using a variable if necessary.








X is the smallest piece, start with that and call its size x .

Y has the same size, and Zhas size 2x .

Now use the AP:

120 = 12 + x + x + 2x

108 = 4x

27 = x .




Now use the AP:

120 = 12 + x + x + 2x

108 = 4x

27 = x .




Now use the AP:

120 = 12 + x + x + 2x

108 = 4x

27 = x .


Now we can determine the other sizes.

x = 27

n[X ] = 27

n[Y ] = 27

n[Z ] = 54

Outcomes: What Matters?

Example 1: A sale table at a jewelry store has 5 bracelets, 4necklaces, and 8 pairs of earrings, all different and all at the sameprice. Shelly selects 3 items to purchase. How many outcomes arepossible?

Example 2: A sale table has 6 red scarves, 4 blue scarves, 6 greenscarves, and 6 purple scarves. Emily buys one of the red scarves,Kristen buys one of the blue scarves, Sara buys one of the greenscarves, and Laura buys one of the purple scarves. How manyoutcomes are possible?

Example 3: 6 IU students and 5 Purdue students go to the SummerIntern Job Fair in Indianapolis. Bank One, Neimann Marcus,Conseco, and the Lilly company each get one summer intern fromthis group. How many outcomes are possible?

Example 4: A sale catalogue lists 6 rock CD’s and 4 country CD’s.Each of Kim, Stacey, and Lynn picks out her favorite CD from thislist. How many outcomes are possible?

All the examples examples involve selecting some objects.

The details involve:

1. What objects are you choosing?(Domain)

2. Do different objects have different qualities that matter?(Qualities)

3. Do you do different things with different selected objects?(Roles)

4. Can the same object be selected multiple times?(Replacement)

















































Example: A sale table at a jewelry store has 5 bracelets, 4necklaces, and 8 pairs of earrings, all different and all at the sameprice. Gladys selects 3 items to purchase. How many outcomes arepossible?

Analysis:

1. Domain: 17 items of jewelry.

2. Qualities: none that matter.

3. Roles: none.

4. Replacement: no.


Analysis:



3. Roles: none.

4. Replacement: no.


Analysis:



3. Roles: none.

4. Replacement: no.


Analysis:



3. Roles: none.

4. Replacement: no.


Analysis:



3. Roles: none.

4. Replacement: no.


Analysis:



3. Roles: none.

4. Replacement: no.


Analysis:



3. Roles: none.

4. Replacement: no.


Analysis:



3. Roles: none.

4. Replacement: no.


Analysis:



3. Roles: none.

4. Replacement: no.

Example: A sale table has 6 red scarves, 4 blue scarves, 6 greenscarves, and 6 purple scarves. Emily buys one of the red scarves,Kristen buys one of the blue scarves, Sara buys one of the greenscarves, and Laura buys one of the purple scarves. How manyoutcomes are possible?

Analysis:

1. Domain: 22 scarves.

2. Qualities: red, blue, green, and purple.

3. Roles: Emily, Kristen, Sara, and Laura.

4. Replacement: no.


Analysis:




4. Replacement: no.


Analysis:




4. Replacement: no.


Analysis:




4. Replacement: no.


Analysis:




4. Replacement: no.


Analysis:




4. Replacement: no.


Analysis:




4. Replacement: no.


Analysis:




4. Replacement: no.


Analysis:




4. Replacement: no.

Example: 6 IU students and 5 Purdue students go to the SummerIntern Job Fair in Indianapolis. Bank One, Neimann Marcus,Conseco, and the Lilly company each get one summer intern fromthis group. How many outcomes are possible?

Analysis:

1. Domain: 11 students.


3. Roles: Bank One, Neimann Marcus, Conseco, and Lilly.

4. Replacement: no.


Analysis:




4. Replacement: no.


Analysis:




4. Replacement: no.


Analysis:




4. Replacement: no.


Analysis:




4. Replacement: no.


Analysis:




4. Replacement: no.


Analysis:




4. Replacement: no.


Analysis:




4. Replacement: no.


Analysis:




4. Replacement: no.

Example: A sale catalogue lists 6 rock CD’s and 4 country CD’s.Each of Kim, Stacey, and Lynn picks out her favorite CD from thislist. How many outcomes are possible?

Analysis:

1. Domain: 10 CDs.


3. Roles: Kim, Stacey, and Lynn.

4. Replacement: yes.


Analysis:

1. Domain: 10 CDs.





Analysis:

1. Domain: 10 CDs.





Analysis:

1. Domain: 10 CDs.





Analysis:

1. Domain: 10 CDs.





Analysis:

1. Domain: 10 CDs.





Analysis:

1. Domain: 10 CDs.





Analysis:

1. Domain: 10 CDs.





Analysis:

1. Domain: 10 CDs.




Shorthand

We’ll need a shorthand to use for analyzing problems.

We’ll use D to stand for all possible choices of object.

We’ll use S to stand for all possible outcomes.

So if you have to pick one tomato out of 40, let

D = 40 tomatoes

and then the set S of all possible outcomes can be represented by

role: Tomatoquality: D

how many: 1.

That was the simplest possible example.

Shorthand





D = 40 tomatoes



how many: 1.


Shorthand





D = 40 tomatoes



how many: 1.


Shorthand





D = 40 tomatoes



how many: 1.


Now suppose you must choose two tomatoes.

Roles don’t matter (both tomatoes suffer the same fate).

No Replacements (you can’t eat the same tomato twice).

We’ll use to indicate that there are no replacements.

ThenT = 40 tomatoes

and S is described by

T

2.

Note the role was omitted,because there is only one.





ThenT = 40 tomatoes


T

2.






ThenT = 40 tomatoes


T

2.






ThenT = 40 tomatoes


T

2.






ThenT = 40 tomatoes


T

2.


Now suppose you also need an onion,

and that there are 50 onions to choose from.

That’s a different quality .

I’ll denote the tomatoes by T and the onions by O.

Then S is described by

T O

2 1.






T O

2 1.






T O

2 1.

Now suppose that you need one tomato for a marinara sauce, andthe other tomato and onion for a salad.

Now there are two roles, sauce and salad.


Sauce SaladT T O

1 1 1

.

Now suppose that you need one tomato for a marinara sauce, andthe other tomato and onion for a salad.

Now there are two roles, sauce and salad.


Sauce SaladT T O

1 1 1

.


Analysis:



3. Roles: none.

4. Replacement: no.

Therefore,D = 17 articles of jewelry

and S = D

3.


Analysis:



3. Roles: none.

4. Replacement: no.

Therefore,D = 17 articles of jewelry

and S = D

3.


Analysis:

1. Domain: 22 scarves.2. Qualities: red, blue, green, and purple.3. Roles: Emily, Kristen, Sara, and Laura.4. Replacement: no.

Therefore,

R = 6 red scarves B = 4 blue scarvesG = 6 green scarves P = 6 purple scarves

and S =

Emily Kristen Sara LauraR B G P

1 1 1 1

.


Analysis:

1. Domain: 22 scarves.2. Qualities: red, blue, green, and purple.3. Roles: Emily, Kristen, Sara, and Laura.4. Replacement: no.

Therefore,

R = 6 red scarves B = 4 blue scarvesG = 6 green scarves P = 6 purple scarves

and S =

Emily Kristen Sara LauraR B G P

1 1 1 1

.


Analysis:




4. Replacement: no.

Therefore,D = 11 students

and S =

Bank1 Neimann Conseco LillyD D D D

1 1 1 1

.


Analysis:




4. Replacement: no.

Therefore,D = 11 students

and S =

Bank1 Neimann Conseco LillyD D D D

1 1 1 1

.


Analysis:

1. Domain: 10 CDs.




Therefore,D = 10 CDs

and S =

Kim Stacey LynnD D D

1 1 1

.


Analysis:

1. Domain: 10 CDs.




Therefore,D = 10 CDs

and S =

Kim Stacey LynnD D D

1 1 1

.

ParadigmsParadigm: City Council Committee: A city council has 5Democratic members (Anne, Bill, Cathy, Dan, and Ellen) and 3Republican members (Frank, Gina, and Hank). The mayor is toappoint a committee consisting of 3 council members.

Analysis:

1. Domain: 8 council members

2. Qualities: none that matter

3. Roles: none that matter

4. Replacement: no

Therefore,D = 8 council members

and S =

D

3.


Analysis:




4. Replacement: no


and S =

D

3.


Analysis:




4. Replacement: no


and S =

D

3.

Paradigm: Soup, Salad, Entree, Veg: For dinner, a customer at arestaurant must select one of 3 soups, one of 5 salads, one of 4entrees, and one of 6 vegetables (see the menu below).

Analysis:

1. Domain: 18 menu items2. Qualities: 3 soups, 5 salads, 4 entrees, 6 vegetables3. Roles: soup, salad, entree, vegetable4. Replacement: no (n/a)

Therefore,

P = 3 soups L = 5 saladsE = 4 entrees V = 6 vegetables

and S =

P L E V

1 1 1 1.

Another notation for this is the Cartesian Product:

S = P × L× E × V


Analysis:


Therefore,


and S =

P L E V

1 1 1 1.


S = P × L× E × V


Analysis:


Therefore,


and S =

P L E V

1 1 1 1.


S = P × L× E × V


Analysis:


Therefore,


and S =

P L E V

1 1 1 1.


S = P × L× E × V

Paradigm: Coffee, Tea, Coke, Sprite: John, Mary, and Bill musteach select one of four beverages for dinner (for instance, coffee,tea, coke, or sprite).

Analysis:

1. Domain: 4 beverages


3. Roles: John, Mary, and Bill

4. Replacement: yes

Therefore,D = 4 beverages

and S =

John Mary BillD D D

1 1 1

.


Analysis:




4. Replacement: yes


and S =

John Mary BillD D D

1 1 1

.


Analysis:




4. Replacement: yes


and S =

John Mary BillD D D

1 1 1

.

Paradigm: Casting a Play: A play has three female roles: thegrandmother, the mother, and the daughter. Six (6) women —Anne, Barb, Carol, Debra, Ellen, and Fran — audition for theseroles. (One should presume that no one can fill two roles, becauseall three characters may have to be on stage simultaneously.)

Analysis:

1. Domain: 6 Actresses


3. Roles: Grandmother, Mother, Daughter

4. Replacement: no

Therefore,D = 6 Actresses

and S =

Grandmother Mother DaughterD D D

1 1 1

.


Analysis:




4. Replacement: no


and S =


1 1 1

.


Analysis:




4. Replacement: no


and S =


1 1 1

.

Paradigm

City Council Soup, Salad, Coffee, Tea, CastingCommittee Entree, Veg Coke, Sprite a Play

Qualities no yes no no

Roles no yes yes yes

Replacement no no (n/a) yes no

Example: At dinner one night at summer camp, 8 campers and acounselor are sitting at a long table. The counselor must select 3of the campers to carry the dishes to the cleanup window. Howmany overall outcomes are possible?

Analysis:

1. Domain: 8 campers(Counseler? Apply Creative Tunnel Vision.)


3. Roles: none

4. Replacement: no

City Council Committee Paradigm

Therefore,D = 8 campers

and S = D

3.


Analysis:



3. Roles: none

4. Replacement: no



and S = D

3.


Analysis:



3. Roles: none

4. Replacement: no



and S = D

3.


Analysis:



3. Roles: none

4. Replacement: no



and S = D

3.


Alternate Analysis:

1. Domain: 9 people at dinner

2. Qualities: 8 campers, 1 counseler

3. Roles: none

4. Replacement: no

Therefore,D1 = 8 campersD2 = 1 counseler

and S = D1

3.

Example: A voter must select one city council candidate, oneschool board candidate, and one county commissioner candidate.There are 4 candidates for the city council, 9 candidates for theschool board, and 5 candidates for county commissioner.

Analysis:

1. Domain: 18 candidates

2. Qualities: city council, school board, county commission

3. Roles: city council, school board, county commission

4. Replacement: no

Soup, Salad, Entree, Veg Paradigm

Therefore,

D1 = 4 city council candidatesD2 = 9 school board candidatesD3 = 5 county commission candidates

D1 D2 D3

1 1 1.


Analysis:




4. Replacement: no


Therefore,


D1 D2 D3

1 1 1.


Analysis:




4. Replacement: no


Therefore,


D1 D2 D3

1 1 1.


Analysis:




4. Replacement: no


Therefore,


D1 D2 D3

1 1 1.

Example: Mary, Paul and Ryan are late for breakfast in the dorm.When they arrive, the pastry tray has only 8 different donuts (e.g.,one glazed donut, one bear claw, one apple fritter, etc.), 4 differentDanishes (e.g., one cheese, one cheery, etc.), and 4 differentmuffins (e.g., one blueberry, one apple, etc.). Each of Mary, Paul,and Ryan takes one of these pastries.

Analysis:

1. Domain: 16 pastries


3. Roles: Mary, Paul, Ryan

4. Replacement: no

Casting a Play Paradigm

Therefore, D = 16 pastries, and S =

Mary Paul RyanD D D

1 1 1

.


Analysis:




4. Replacement: no



Mary Paul RyanD D D

1 1 1

.


Analysis:




4. Replacement: no



Mary Paul RyanD D D

1 1 1

.


Analysis:




4. Replacement: no



Mary Paul RyanD D D

1 1 1

.

Example: Claire, Emily, and Laura are at a store that sells works bylocal artists. The store has 6 oil paintings, 5 water color sketches,and 9 textile pieces for sale. Clair purchases an oil painting, Emilypurchases a water color sketch, and Laura purchases a textile piece.

Analysis:

1. Domain: 20 artworks

2. Qualities: oil, water color, textile

3. Roles: Clair, Emily, Laura

4. Replacement: no


Therefore,

O = 6 oil paintings W = 5 water colors T = 9 textiles

and S =

Clair Emily LauraO W T

1 1 1

.


Analysis:




4. Replacement: no


Therefore,


and S =


1 1 1

.


Analysis:




4. Replacement: no


Therefore,


and S =


1 1 1

.


Analysis:




4. Replacement: no


Therefore,


and S =


1 1 1

.

Example: The Dean of Students has 4 Associate Deans, 5Assistant Deans, and 9 program specialists on his staff. To checkfor alcohol violations this weekend, the Dean is going to select oneof these staff members to visit the Sigma Lambda house at 11:00pm on Saturday, one of these staff members to visit the Delta Phihouse at 11:00 pm on Saturday, and one of these staff members tovisit the Phi Sigma house at 11:00 pm on Saturday.

Analysis:

1. Domain: 19 staff members


3. Roles: ΣΛ, ∆Φ, ΦΣ

4. Replacement: no


Therefore, D = 18 staff members, and S =

ΣΛ ∆Φ ΦΣD D D

1 1 1

.


Analysis:




4. Replacement: no




1 1 1

.


Analysis:




4. Replacement: no




1 1 1

.


Analysis:




4. Replacement: no




1 1 1

.

Example: Laura, Susan, and Chris have just enrolled in a universityin Chicago. Each one must open a checking/savings account at alocal institution. They may choose from 8 (regular) banks, 4savings banks, and 4 credit unions.

Analysis:

1. Domain: 16 banks


3. Roles: Laura, Susan, Chris

4. Replacement: yes

Coffee, Tea, Coke, Sprite Paradigm

Therefore,B = 16 banks

and S =

Laura Susan ChrisB B B

1 1 1

.


Analysis:

1. Domain: 16 banks



4. Replacement: yes



and S =


1 1 1

.


Analysis:

1. Domain: 16 banks



4. Replacement: yes



and S =


1 1 1

.


Analysis:

1. Domain: 16 banks



4. Replacement: yes



and S =


1 1 1

.

Example: Alyssa, Brandy,and Carl are having dinner at a Chineserestaurant. The menu lists 9 beef dishes, 6 pork dishes, and 6chicken dishes. Alyssa, Brandy, and Carl have decided that theywill share 3 dishes. They jointly select one beef dish, one porkdish, and one chicken dish.

Analysis:

1. Domain: 21 dishes

2. Qualities: beef, pork, chicken

3. Roles: beef, pork, chicken

4. Replacement: yes(?), n/a


Therefore,

B = 9 beef dishes P = 6 pork dishes C = 6 chicken dishes

and S =

B P C

1 1 1.


Analysis:






Therefore,


and S =

B P C

1 1 1.


Analysis:






Therefore,


and S =

B P C

1 1 1.


Analysis:






Therefore,


and S =

B P C

1 1 1.

Example: Allyson, Beth, and Chelsea, who are sharing anapartment, have gone to a bookstore to purchase 3 cookbooks.The store had 5 different general cookbooks, 3 different Italiancookbooks, and 6 different French cookbooks. Together theyselect 3 of these cookbooks.

Analysis:

1. Domain: 14 cookbooks


3. Roles: none

4. Replacement: no


Therefore,C = 14 cookbooks

and S =

C

3.


Analysis:



3. Roles: none

4. Replacement: no



and S =

C

3.


Analysis:



3. Roles: none

4. Replacement: no



and S =

C

3.


Analysis:



3. Roles: none

4. Replacement: no



and S =

C

3.

Calculating Sizes

Now that you know how to analyze problems,the next thing is to learn:

How do you use your analysis to count the number of outcomes?


S = 3 soups L = 5 saladsE = 4 entrees V = 6 vegetables

and S =

P L E V

1 1 1 1# choices: 3 5 4 6

.

Answer: n[S ] = size of S is

n[S ] = 3 · 5 · 4 · 6 = 360.


S = 3 soups L = 5 saladsE = 4 entrees V = 6 vegetables

and S =

P L E V

1 1 1 1# choices: 3 5 4 6

.

Answer: n[S ] = size of S is

n[S ] = 3 · 5 · 4 · 6 = 360.


D = 4 beverages

and S =


1 1 1# choices: 4 4 4

.

Answer:n[S ] = 4 · 4 · 4 = 43 = 64.

In general, if d = size of domain and r = # roles,

n[S ] = d r .


D = 4 beverages

and S =


1 1 1# choices: 4 4 4

.

Answer:n[S ] = 4 · 4 · 4 = 43 = 64.


n[S ] = d r .


D = 4 beverages

and S =


1 1 1# choices: 4 4 4

.

Answer:n[S ] = 4 · 4 · 4 = 43 = 64.


n[S ] = d r .


D = 6 Actresses

and S =


1 1 1# choices: 6 5 4

.

Answer:n[S ] = 6 · 5 · 4 = P(6, 3) = 120.


n[S ] = P(d , r).


D = 6 Actresses

and S =


1 1 1# choices: 6 5 4

.

Answer:n[S ] = 6 · 5 · 4 = P(6, 3) = 120.


n[S ] = P(d , r).


D = 6 Actresses

and S =


1 1 1# choices: 6 5 4

.

Answer:n[S ] = 6 · 5 · 4 = P(6, 3) = 120.


n[S ] = P(d , r).


D = 8 council members

and S =

D

3# choices: C(8, 3) = 56

.

Answer:n[S ] = C(8, 3) = 56.

In general, if d = size of domain and n = # chosen,

n[S ] = C(d , n).



and S =

D

3# choices: C(8, 3) = 56

.

Answer:n[S ] = C(8, 3) = 56.


n[S ] = C(d , n).



and S =

D

3# choices: C(8, 3) = 56

.

Answer:n[S ] = C(8, 3) = 56.


n[S ] = C(d , n).

CombinationsThe C function calculates the number of combinations.

A combination is an unordered list without repetitions.

(A permutation is an ordered list.)

An outcome in the City Council Committee Paradigm is a combo.

Mathematically,C(d , r) = P(d ,r)

P(r ,r) .

For example,

C(8, 3) = 8·7·63·2·1 = 8·7·66

63·62·1 = 56

C(8, 5) = 8·7·6·5·45·4·3·2·1 = 8·7·66·65·64

65·64·63·62·1 = 56.

In general,

C(d , r) = C(d , d − r).






P(r ,r) .

For example,

C(8, 3) = 8·7·63·2·1 = 8·7·66

63·62·1 = 56

C(8, 5) = 8·7·6·5·45·4·3·2·1 = 8·7·66·65·64

65·64·63·62·1 = 56.

In general,

C(d , r) = C(d , d − r).






P(r ,r) .

For example,

C(8, 3) = 8·7·63·2·1 = 8·7·66

63·62·1 = 56

C(8, 5) = 8·7·6·5·45·4·3·2·1 = 8·7·66·65·64

65·64·63·62·1 = 56.

In general,

C(d , r) = C(d , d − r).



D = 8 campers

and S = D

3C(8, 3) = 56

.

Answer:n[S ] = C(8, 3) = 56



D = 8 campers

and S = D

3C(8, 3) = 56

.

Answer:n[S ] = C(8, 3) = 56



D = 8 campers

and S = D

3C(8, 3) = 56

.

Answer:n[S ] = C(8, 3) = 56



and S =

City Council School Board County CommissionD1 D2 D3

1 1 14 9 5


Answer:n[S ] = 4 · 9 · 5 = 180



and S =


1 1 14 9 5


Answer:n[S ] = 4 · 9 · 5 = 180



and S =


1 1 14 9 5


Answer:n[S ] = 4 · 9 · 5 = 180


D = 16 pastries

and S =

Mary Paul RyanD D D

1 1 116 15 14.


Answer:n[S ] = P(16, 3) = 16 · 15 · 14 = 3360


D = 16 pastries

and S =

Mary Paul RyanD D D

1 1 116 15 14.


Answer:n[S ] = P(16, 3) = 16 · 15 · 14 = 3360


D = 16 pastries

and S =

Mary Paul RyanD D D

1 1 116 15 14.


Answer:n[S ] = P(16, 3) = 16 · 15 · 14 = 3360

Example: Claire, Emily, and Laura are at a store that sells worksby local artists. The store has 6 oil paintings, 5 water colorsketches, and 9 textile pieces for sale. All of the items are differentfrom one another. Clair purchases an oil painting, Emily purchasesa water color sketch, and Laura purchases a textile piece.

O = 6 oil paintingsW = 5 water colorsT = 9 textiles

and S =

Claire Emily LauraO W T

1 1 16 5 9


Answer:n[S ] = 6 · 5 · 9 = 270



and S =


1 1 16 5 9


Answer:n[S ] = 6 · 5 · 9 = 270



and S =


1 1 16 5 9


Answer:n[S ] = 6 · 5 · 9 = 270


D = 18 staff members

and S =


1 1 118 17 16


Answer:n[S ] = P(18, 3) = 18 · 17 · 16 = 4896



and S =


1 1 118 17 16


Answer:n[S ] = P(18, 3) = 18 · 17 · 16 = 4896



and S =


1 1 118 17 16


Answer:n[S ] = P(18, 3) = 18 · 17 · 16 = 4896


D = 16 banks

and S =


1 1 116 16 16


Answer:n[S ] = 163 = 4096


D = 16 banks

and S =


1 1 116 16 16


Answer:n[S ] = 163 = 4096


D = 16 banks

and S =


1 1 116 16 16


Answer:n[S ] = 163 = 4096


B = 9 beef dishesP = 6 pork dishesC = 6 chicken dishes

and S =

B P C

1 1 19 6 6


Answer:n[S ] = 9 · 6 · 6 = 324



and S =

B P C

1 1 19 6 6


Answer:n[S ] = 9 · 6 · 6 = 324



and S =

B P C

1 1 19 6 6


Answer:n[S ] = 9 · 6 · 6 = 324


D = 14 cookbooks

and S =

D

3C(14, 3).


Answer:n[S ] = C(14, 3) = 14·13·12

3·2·1 = 364


D = 14 cookbooks

and S =

D

3C(14, 3).


Answer:n[S ] = C(14, 3) = 14·13·12

3·2·1 = 364


D = 14 cookbooks

and S =

D

3C(14, 3).


Answer:n[S ] = C(14, 3) = 14·13·12

3·2·1 = 364

Example: A city council has 5 Democratic members and 3Republican members. The mayor is to assign 1 Democratic councilmember to the Police Com. and 1 Republican council member tothe Zoning Com. How many outcomes are possible?

Analysis:

1. Domain: 8 council members2. Qualities: Dem, Rep3. Roles: Police, Zoning4. Replacement: no


D = 5 Dems R = 3 Reps

and S =

Police ZoningD R

1 15 3.

Answer: n[S ] = 5 · 3 = 15


Analysis:




and S =

Police ZoningD R

1 15 3.

Answer: n[S ] = 5 · 3 = 15


Analysis:




and S =

Police ZoningD R

1 15 3.

Answer: n[S ] = 5 · 3 = 15

Example: A city council has 5 Democratic and 3 Republicanmembers. The mayor is to assign 1 council member to the PoliceCommission and 1 council member to the Zoning Commission.(The same person could be assigned to both commissions.) Howmany outcomes are possible?

Analysis:



3. Roles: Police, Zoning

4. Replacement: yes

Coffee, Tea, Coke, Sprite ParadigmLet C = “8 council members”, then S is

Police ZoningC C

1 18 8 = 64.


Analysis:




4. Replacement: yes


Let C = “8 council members”, then S is

Police ZoningC C

1 18 8 = 64.


Analysis:




4. Replacement: yes

Coffee, Tea, Coke, Sprite ParadigmLet C = “8 council members”, then S is

Police ZoningC C

1 18 8 = 64.

Example: A city council has 5 Democratic members and 3Republican members. The mayor is to assign 1 council member tothe Police Commission and 1 council member to the ZoningCommission. No one can be assigned to two commissions. Howmany outcomes are possible?

Analysis:




4. Replacement: no

Casting a Play ParadigmLet C = “8 council members”, then S is

Police ZoningC C

1 18 7 = 56.


Analysis:




4. Replacement: no



Police ZoningC C

1 18 7 = 56.


Analysis:




4. Replacement: no

Casting a Play ParadigmLet C = “8 council members”, then S is

Police ZoningC C

1 18 7 = 56.

Example: A city council has 5 Democratic members and 3Republican members. The mayor is to assign 2 council members tothe Police Commission. How many outcomes are possible?

Analysis:




4. Replacement: no

City Council Committee ParadigmLet C = “8 council members”, then S is

PoliceC

2C(8, 2) = 28.


Analysis:




4. Replacement: no



PoliceC

2C(8, 2) = 28.


Analysis:




4. Replacement: no

City Council Committee ParadigmLet C = “8 council members”, then S is

PoliceC

2C(8, 2) = 28.

Example: A city council has 5 Democratic members and 3Republican members. In how many different ways can the mayorappoint a committee consisting of 3 Democratic council membersand 2 Republican council members?

Analysis:

1. Domain: 8 council members2. Qualities: Dem, Rep3. Roles: Dem, Rep4. Replacement: no

Doesn’t exactly fit any paradigm


and S is

CommitteeD R

3 2C(5, 3) C(3, 2) = 30.


Analysis:




and S is

CommitteeD R

3 2C(5, 3) C(3, 2) = 30.


Analysis:




and S is

CommitteeD R

3 2C(5, 3) C(3, 2) = 30.


Analysis:




and S is

CommitteeD R

3 2C(5, 3) C(3, 2) = 30.

Example: A small fraternity consists of 7 seniors, 11 juniors, and 9sophomores. In how many different ways could a steeringcommittee consisting of 3 seniors, 2 juniors, and 1 sophomore beselected.

Analysis:

1. Domain: 27 frat bros

2. Qualities: Senior, Junior, Sophomore

3. Roles: Senior, Junior, Sophomore

4. Replacement: no

N = 7 Seniors J = 11 Juniors M = 9 Sophomores

and S is

CommitteeN J M

3 2 1C(7, 3) C(11, 2) C(9, 1) = 17325.


Analysis:




4. Replacement: no


and S is

CommitteeN J M

3 2 1C(7, 3) C(11, 2) C(9, 1) = 17325.


Analysis:




4. Replacement: no


and S is

CommitteeN J M

3 2 1C(7, 3) C(11, 2) C(9, 1) = 17325.

Example: The Board of Directors of a company has 12 members. If3 are to be assigned to the budget committee, 2 are to be assignedto the personnel committee, and 4 are to be assigned to thegovernment relations committee, and no one can serve on morethan one committee, how many different outcomes are possible?

Analysis:

1. Domain: 12 Directors


3. Roles: Budget, Personnel, GovRel

4. Replacement: no

Let D = “12 Directors”, then S is

Budget Personnel GovRelD D D

3 2 4C(12, 3) C(9, 2) C(7, 4) = 277200.


Analysis:




4. Replacement: no



3 2 4C(12, 3) C(9, 2) C(7, 4) = 277200.


Analysis:




4. Replacement: no



3 2 4C(12, 3) C(9, 2) C(7, 4) = 277200.

Example: The Board of Directors of a company has 12 members. If3 are to be assigned to the budget committee, 2 are to be assignedto the personnel committee, and 4 are to be assigned to thegovernment relations committee, and one can serve on more thanone committee, how many different outcomes are possible?

Analysis:




4. Replacement: within committee: no; between committees: yes



3 2 4C(12, 3) C(12, 2) C(12, 4) = 7187400.


Analysis:







3 2 4C(12, 3) C(12, 2) C(12, 4) = 7187400.


Analysis:







3 2 4C(12, 3) C(12, 2) C(12, 4) = 7187400.

Example: (Anagrams) How many “words” can be formed using theletters in the word “card”?

Analysis:

1. Domain: 4 letters (c, a, r, d)


3. Roles: Position (1–4)

4. Replacement: no

D = 4 Letters

and S is

1st 2nd 3rd 4thD D D D

1 1 1 14 3 2 1 = 24


Analysis:




4. Replacement: no

D = 4 Letters

and S is


1 1 1 14 3 2 1 = 24


Analysis:




4. Replacement: no

D = 4 Letters

and S is


1 1 1 14 3 2 1 = 24

Alternate SolutionExample: (Anagrams) How many “words” can be formed using theletters in the word “card”?

Analysis:

1. Domain: 4 Positions


3. Roles: c, a, r, d

4. Replacement: no

D = 4 Positions

and S is

C A R DD D D D

1 1 1 14 3 2 1 = 24


Analysis:




4. Replacement: no

D = 4 Positions

and S is

C A R DD D D D

1 1 1 14 3 2 1 = 24


Analysis:




4. Replacement: no

D = 4 Positions

and S is

C A R DD D D D

1 1 1 14 3 2 1 = 24

Example: How many “words” can be formed using the letters in theword “cool”?

Analysis:



3. Roles: c, o, l

4. Replacement: no

D = 4 Positions

and S is

C O LD D D

1 2 1C(4, 1) C(3, 2) C(1, 1) = 12


Analysis:



3. Roles: c, o, l

4. Replacement: no

D = 4 Positions

and S is

C O LD D D

1 2 1C(4, 1) C(3, 2) C(1, 1) = 12


Analysis:



3. Roles: c, o, l

4. Replacement: no

D = 4 Positions

and S is

C O LD D D

1 2 1C(4, 1) C(3, 2) C(1, 1) = 12

Example: How many “words” can be formed using the letters in theword “bananas”?

Analysis:



3. Roles: b, a, n, s

4. Replacement: no

D = 7 Positions

and S is

B A N SD D D D

1 3 2 1C(7, 1) C(6, 3) C(3, 2) C(1, 1) = 420


Analysis:




4. Replacement: no

D = 7 Positions

and S is

B A N SD D D D

1 3 2 1C(7, 1) C(6, 3) C(3, 2) C(1, 1) = 420


Analysis:




4. Replacement: no

D = 7 Positions

and S is

B A N SD D D D

1 3 2 1C(7, 1) C(6, 3) C(3, 2) C(1, 1) = 420

Compound Problems

Some problems are like Russian dolls:

problems inside problems.

You’ll need to use multiple strategies on a single problem.

Example: The city council has 5 Democratic and 3 Republicanmembers. The mayor is to appoint 3 or 4 of the council membersto a committee. How many different outcomes are possible?

Analysis:



3. Roles: none

4. Replacement: no

Let D = “8 council members”, then S isD

3C(8, 3) 56

4C(8, 4) 70

126

.


Analysis:



3. Roles: none

4. Replacement: no


3C(8, 3) 56

4C(8, 4) 70

126

.


Analysis:



3. Roles: none

4. Replacement: no


3C(8, 3) 56

4C(8, 4) 70

126

.

Example: The city council has 5 Democratic and 3 Republicanmembers. The mayor is to appoint 3 of the council members to abipartisan committee. How many different outcomes are possible?

Analysis:

1. Domain: 8 council members2. Qualities: Dem, Rep3. Roles: none that matter4. Replacement: no

D = 5 Democrats R = 3 Republicans

and S is D R

1 2C(5, 1) C(3, 2) 15

2 1C(5, 2) C(3, 1) 30

45


Analysis:



and S is D R

1 2C(5, 1) C(3, 2) 15

2 1C(5, 2) C(3, 1) 30

45


Analysis:



and S is D R

1 2C(5, 1) C(3, 2) 15

2 1C(5, 2) C(3, 1) 30

45

Example: The city council has 5 Democratic and 3 Republicanmembers. The mayor is to appoint 3 of the council members to acommittee with at most 1 Democrat. How many differentoutcomes are possible?

Analysis:



and then S is D R

0 3C(5, 0) C(3, 3) 1

1 2C(5, 1) C(3, 2) 15

16


Analysis:



and then S is D R

0 3C(5, 0) C(3, 3) 1

1 2C(5, 1) C(3, 2) 15

16


Analysis:



and then S is D R

0 3C(5, 0) C(3, 3) 1

1 2C(5, 1) C(3, 2) 15

16

Example: The city council has 5 Democratic and 3 Republicanmembers. The mayor is to appoint 3 of the council members to acommittee with at least 1 Democrat. How many differentoutcomes are possible?

Analysis: House Subcommittee Paradigm

5 Dems 3 Reps

1 2C(5, 1) C(3, 2) 15

2 1C(5, 2) C(3, 1) 30

3 0C(5, 3) C(3, 0) 10

55



5 Dems 3 Reps

1 2C(5, 1) C(3, 2) 15

2 1C(5, 2) C(3, 1) 30

3 0C(5, 3) C(3, 0) 10

55



5 Dems 3 Reps

1 2C(5, 1) C(3, 2) 15

2 1C(5, 2) C(3, 1) 30

3 0C(5, 3) C(3, 0) 10

55

Alternate Method: by Complement


Analysis:

ST = all outcomes with no restrictionsSR = outcomes with at least 1 DemS ′R = outcomes with no Dems

n[ST ] = n[SR ] + n[S ′R ]n[ST ]− n[S ′R ] = n[SR ]

C(8, 3)− C(3, 3) =56− 1 = 55



Analysis:



C(8, 3)− C(3, 3) =56− 1 = 55



Analysis:


n[ST ] = n[SR ] + n[S ′R ]

n[ST ]− n[S ′R ] = n[SR ]C(8, 3)− C(3, 3) =

56− 1 = 55



Analysis:



C(8, 3)− C(3, 3) =56− 1 = 55



Analysis:



C(8, 3)−

C(3, 3) =56− 1 = 55



Analysis:



C(8, 3)− C(3, 3) =56− 1 = 55

The Complement Principle

The symbol ′ denotes the complement of a set:

A′ = S A = { x | x ∈ S & x 6∈ A }.

Then:

In general,

n[A] = n[S ]− n[A′].

The Complement Principle (CP) is worth considering if theproblem contains phrases like “at least 1” or “at most 5”.

The Complement Principle

The symbol ′ denotes the complement of a set:

A′ = S A = { x | x ∈ S & x 6∈ A }.

Then:

In general,

n[A] = n[S ]− n[A′].

The Complement Principle (CP) is worth considering if theproblem contains phrases like “at least 1” or “at most 5”.

Example: The House committee on Education has 8 Democraticand 9 Republican members. They want to form a subcommittee of7 to investigate corruption in Finite Math, and the subcommitteemust have at least 1 Democrat. How many different outcomes arepossible?


S = all outcomes with no restrictionsE = outcomes with at least 1 Dem

E ′ = outcomes with no Dems= outcomes with all Reps

Then by the CP,

n[E ] = n[S ]− n[E ′]= C(17, 7)− C(9, 7)= 19412.





Then by the CP,

n[E ] = n[S ]− n[E ′]= C(17, 7)− C(9, 7)= 19412.





Then by the CP,

n[E ] = n[S ]− n[E ′]

= C(17, 7)− C(9, 7)= 19412.





Then by the CP,

n[E ] = n[S ]− n[E ′]= C(17, 7)− C(9, 7)= 19412.

Example: A class has 5 sophomores, 7 juniors, and 6 seniors. If 3students are selected to give a report, how many different outcomesare possible in which the students are not all in the same year?

Analysis:

S = all outcomes with no restrictionsE = outcomes with students not all in same year

E ′ = outcomes with students all in same yearn[E ] = n[S ]− n[E ′] (CP)

And we easily get

n[S ] = C(18, 3) = 816.

But what is n[E ′]?

Example: A class has 5 sophomores, 7 juniors, and 6 seniors. If 3students are selected to give a report, how many different outcomesare possible in which the students are not all in the same year?

Analysis:



And we easily get

n[S ] = C(18, 3) = 816.

But what is n[E ′]?

Subproblem: A class has 5 sophomores, 7 juniors, and 6 seniors. If3 students are selected to give a report, how many differentoutcomes are possible in which all students are in the same year?

Analysis:

Year5 Soph 7 Jun 6 Sen size

Alt

ern

ativ

es

3 0 0C(5, 3) = 10 1 1 10

0 3 01 C(7, 3) = 35 1 350 0 31 1 C(6, 3) = 20 20

65


Analysis:


Alt

ern

ativ

es

3 0 0C(5, 3) = 10 1 1 10

0 3 01 C(7, 3) = 35 1 350 0 31 1 C(6, 3) = 20 20

65


Analysis:


Alt

ern

ativ

es

3 0 0C(5, 3) = 10 1 1 10

0 3 01 C(7, 3) = 35 1 350 0 31 1 C(6, 3) = 20 20

65

Example: (cont.) A class has 5 sophomores, 7 juniors, and 6seniors. If 3 students are selected to give a report, how manydifferent outcomes are possible in which the students are not all inthe same year?

Analysis:



n[S ] = 816n[E ′] = 65n[E ] = 816− 65 = 751

Example: (cont.) A class has 5 sophomores, 7 juniors, and 6seniors. If 3 students are selected to give a report, how manydifferent outcomes are possible in which the students are not all inthe same year?

Analysis:



n[S ] = 816n[E ′] = 65n[E ] = 816− 65 = 751

Example: The math chairperson has a meeting at 11 to decide onthe recipient of this year’s Outstanding Finite Student award. Healso has time for meetings at 10, 12, and 2. He has three othertasks to do today: meet with the winner of the award, draft abudget proposal, and finalize the summer teaching schedule. Howmany ways are there to schedule his day?

Analysis:

1. Objects: 3 tasks

2. Qualities: award and nonaward

3. Roles: 10, 12, 2

4. Replacement: no


Analysis:

1. Objects: 3 tasks


3. Roles: 10, 12, 2

4. Replacement: no


Analysis:

1. Objects: 3 tasks


3. Roles: 10, 12, 2

4. Replacement: no

Analysis:

1. Objects: 3 tasks


3. Roles: 10, 12, 2

4. Replacement: no

D = 3 tasksA = meeting with award winnerN = 2 other meetings

S :

10 12 2D D D

1 1 13 2 1 6

Analysis:

1. Objects: 3 tasks


3. Roles: 10, 12, 2

4. Replacement: no


S :

10 12 2D D D

1 1 13 2 1 6

Analysis:

1. Objects: 3 tasks


3. Roles: 10, 12, 2

4. Replacement: no


E = “feasible schedules”:

10 12 2N D D

1 1 12 2 1 4

Example: A department chairperson has 4 tasks to do today: selectan award recipient; have a meeting with this recipient; write aletter to the dean; and prepare the department’s budget request fornext year. If he schedules these tasks one after another, how manydifferent schedules are possible. (Note that he cannot meet withthe committee until after he has selected its members.)

Analysis:

1. Objects: 4 tasks


3. Roles: 4 time slots

4. Replacement: no

We can break this problem down into alternatives.


Analysis:

1. Objects: 4 tasks



4. Replacement: no



Analysis:

1. Objects: 4 tasks



4. Replacement: no


C = meeting to choose award winnerN = 3 other meetings, including one with awardee

and E = “feasible schedules” is

1st 2nd 3rd 4th

C N N N1 3 2 1 6

N C N N2 1 2 1 4N N C N2 1 1 1 2N N N C2 1 0 1 0

12

Since we are always picking just one task for each slotI have omitted all the 1’s.

C = meeting to choose award winnerN = 3 other meetings, including one with awardee

and E = “feasible schedules” is

1st 2nd 3rd 4th

C N N N1 3 2 1 6N C N N2 1 2 1 4N N C N2 1 1 1 2N N N C2 1 0 1 0

12

Since we are always picking just one task for each slotI have omitted all the 1’s.

Example: The Personnel Manager at a company must schedule 4job interviews and 3 different committee meetings for tomorrow. Ifshe decides to schedule the job interviews one after another and toschedule the committee meetings one after another, how manydifferent schedules are possible?

Analysis:

1. Objects: 7 tasks

2. Qualities: interviews, meetings

3. Roles: 2 time blocks

4. Replacement: no

Example: The Personnel Manager at a company must schedule 4job interviews and 3 different committee meetings for tomorrow. Ifshe decides to schedule the job interviews one after another and toschedule the committee meetings one after another, how manydifferent schedules are possible?

Analysis:

1. Objects: 7 tasks

2. Qualities: interviews, meetings

3. Roles: 2 time blocks

4. Replacement: no

LetI = 4 job interviews

M = 3 committee meetings

then E = “desired schedules” is

Block 1 Block 2I M4 3

P(4, 4) P(3, 3) 144M I3 4

P(3, 3) P(4, 4) 144

288

Example: A student has 4 novels by Thomas Hardy, 2 novels byGeorge Eliot, and 5 novels by Charles Dickens. In how manydifferent ways could she arrange these books on a book shelf (fromleft to right) if each author’s books are kept together?

Analysis:

1. Objects: 11 books

2. Qualities: Hardy, Eliot, Dickens

3. Roles: 3 Blocks

4. Replacement: no

H = 4 Hardy novelsE = 2 Eliot novelsD = 5 Dickens novels


Analysis:



3. Roles: 3 Blocks

4. Replacement: no



Analysis:



3. Roles: 3 Blocks

4. Replacement: no


Block 1 Block 2 Block 3H E DP(4, 4) P(2, 2) P(5, 5) 5760H D EP(4, 4) P(5, 5) P(2, 2) 5760E H DP(2, 2) P(4, 4) P(5, 5) 5760E D HP(2, 2) P(5, 5) P(4, 4) 5760D H EP(5, 5) P(4, 4) P(2, 2) 5760D E HP(5, 5) P(2, 2) P(4, 4) 5760

34560

Alternate Solution

Note that in each row the qualities are anagrams of HED.

So the number of rows must be C(3, 1)C(2, 1)C(1, 1) = 6.

Since each row has the same size,

n[E ] = 6 · 5760 = 34560.

This is an example of the Same Size Principle (SSP).

Alternate Solution




n[E ] = 6 · 5760 = 34560.


Alternate Solution




n[E ] = 6 · 5760 = 34560.


Alternate Solution




n[E ] = 6 · 5760 = 34560.


Block 1 Block 2 Block 3

6

H E D

P(4, 4) P(2, 2) P(5, 5) 5760H D EP(4, 4) P(5, 5) P(2, 2) 5760E H DP(2, 2) P(4, 4) P(5, 5) 5760E D HP(2, 2) P(5, 5) P(4, 4) 5760D H EP(5, 5) P(4, 4) P(2, 2) 5760D E HP(5, 5) P(2, 2) P(4, 4) 5760

6 · 5760 = 34560

Example: In how many ways could Anne, Bill, Carol, Dan, and Ellenbe arranged in a line for a picture if Anne and Dan insist onstanding next to one another?

Analysis:

1. Objects: 5 people

2. Qualities: Ann, Dan, Others

3. Roles: 5 positions

4. Replacement: no

A = AnnD = DanP = 3 people besides Ann and Dan

Example: In how many ways could Anne, Bill, Carol, Dan, and Ellenbe arranged in a line for a picture if Anne and Dan insist onstanding next to one another?

Analysis:

1. Objects: 5 people

2. Qualities: Ann, Dan, Others

3. Roles: 5 positions

4. Replacement: no

A = AnnD = DanP = 3 people besides Ann and Dan

Then E = “desired orders” is

1st 2nd 3rd 4th 5th

8

A D P P P1 1 3 2 1 6

D A P P P1 1 3 2 1 6P A D P P3 1 1 2 1 6P D A P PP P A D PP P D A PP P P A DP P P D A

8 · 6 = 48

Alternative Method

There are 4 blocks of people:AD and B, C , and E .

1st 2nd 3rd 4th

P(4, 4) = 24

AD B C E2 1 1 1 2

E AD C B1 2 1 1 2

......

......

...

24 · 2 = 48

Alternative Method


1st 2nd 3rd 4th

P(4, 4) = 24

AD B C E

2 1 1 1 2

E AD C B1 2 1 1 2

......

......

...

24 · 2 = 48

Alternative Method


1st 2nd 3rd 4th

P(4, 4) = 24

AD B C E2 1 1 1 2

E AD C B1 2 1 1 2

......

......

...

24 · 2 = 48

Alternative Method


1st 2nd 3rd 4th

P(4, 4) = 24

AD B C E2 1 1 1 2

E AD C B1 2 1 1 2

......

......

...

24 · 2 = 48

Alternative Method


1st 2nd 3rd 4th

P(4, 4) = 24

AD B C E2 1 1 1 2

E AD C B1 2 1 1 2

......

......

...

24 · 2 = 48

Example: A committee has 5 freshmen, 5 sophomores, and 5juniors. In how many different ways could a panel consisting of 2people from each of 2 classes be selected?

1. Objects: 15 students

2. Qualities: Freshmen, Sophomores, Juniors


4. Replacement: no

Freshmen Sophomores Juniors

2 2 0C(5, 2) C(5, 2) C(5, 0) 100

2 0 2C(5, 2) C(5, 0) C(5, 2) 100

0 2 2C(5, 0) C(5, 2) C(5, 2) 100

300

Note the three cases are all “anagrams” of 2, 2, 0.





4. Replacement: no


2 2 0C(5, 2) C(5, 2) C(5, 0) 100

2 0 2C(5, 2) C(5, 0) C(5, 2) 100

0 2 2C(5, 0) C(5, 2) C(5, 2) 100

300






4. Replacement: no


2 2 0C(5, 2) C(5, 2) C(5, 0) 100

2 0 2C(5, 2) C(5, 0) C(5, 2) 100

0 2 2C(5, 0) C(5, 2) C(5, 2) 100

300


Example: A committee has 5 freshmen, 5 sophomores, 5 juniors,and 5 seniors. In how many different ways could a panel be chosenif it must consist of 2 people from one class and 1 person eachfrom two other classes?

Analysis:


2. Qualities: Freshmen, Sophomores, Juniors, Seniors


4. Replacement: no

Example: A committee has 5 freshmen, 5 sophomores, 5 juniors,and 5 seniors. In how many different ways could a panel be chosenif it must consist of 2 people from one class and 1 person eachfrom two other classes?

Analysis:


2. Qualities: Freshmen, Sophomores, Juniors, Seniors


4. Replacement: no

The “desired panels” can be described as follows.

Freshmen Sophomores Juniors Seniors

12

2 1 1 0C(5, 2) C(5, 1) C(5, 1) C(5, 0) 250

1 1 0 2C(5, 1) C(5, 1) C(5, 0) C(5, 2) 250

......

......

...

3000

Why 12?

# anagrams of 2110 = C(4, 1)C(3, 2)C(1, 1) = 12

The “desired panels” can be described as follows.

Freshmen Sophomores Juniors Seniors

12

2 1 1 0C(5, 2) C(5, 1) C(5, 1) C(5, 0) 250

1 1 0 2C(5, 1) C(5, 1) C(5, 0) C(5, 2) 250

......

......

...

3000

Why 12?

# anagrams of 2110 = C(4, 1)C(3, 2)C(1, 1) = 12

Example: How many different hands consisting of 5 cards from aregular deck of 52 cards (i.e., four suits — spades, hearts,diamonds, and clubs — with thirteen cards each and no jokers)have 3 cards from one suit and 2 cards from one other suit?

Analysis:

1. Objects: 52 cards

2. Qualities: 13 ♠, 13 ♥, 13 ♦, 13 ♣3. Roles: none that matter

4. Replacement: no

Example: How many different hands consisting of 5 cards from aregular deck of 52 cards (i.e., four suits — spades, hearts,diamonds, and clubs — with thirteen cards each and no jokers)have 3 cards from one suit and 2 cards from one other suit?

Analysis:

1. Objects: 52 cards

2. Qualities: 13 ♠, 13 ♥, 13 ♦, 13 ♣3. Roles: none that matter

4. Replacement: no

The desired hands can be described as follows.

♠ ♥ ♦ ♣

12

3 2 0 0C(13, 3) C(13, 2) C(13, 0) C(13, 0) 22308

2 0 3 0C(13, 2) C(13, 0) C(13, 3) C(13, 0) 22308

......

......

...

267696

12 = # anagrams of 3200 = C(4, 1)C(3, 1)C(2, 2)

Example: A small retail chain has vacancies for 4 store managers allin different cities. Because the positions are in different cities, noone could hold more than one of these positions. The company has6 female assistant managers and 5 male assistant managers whoare eligible for promotion to these store manager positions. Howmany outcomes are there in which 2 women and 2 men arepromoted?

Analysis:

1. Objects: 11 managers

2. Qualities: 6 female, 5 male

3. Roles: 4 stores

4. Replacement: no

Example: A small retail chain has vacancies for 4 store managers allin different cities. Because the positions are in different cities, noone could hold more than one of these positions. The company has6 female assistant managers and 5 male assistant managers whoare eligible for promotion to these store manager positions. Howmany outcomes are there in which 2 women and 2 men arepromoted?

Analysis:

1. Objects: 11 managers

2. Qualities: 6 female, 5 male

3. Roles: 4 stores

4. Replacement: no

F = 6 femalesM = 5 males

Store 1 Store 2 Store 3 Store 4

C(4, 2)C(2, 2) = 6

F F M M

6 5 5 4 600F M M F6 5 4 5 600

......

......

...

3600

Example: 9 seniors and 4 juniors in English each submitted onepoem for the English Department’s annual poetry competition.The department will award 1st, 2nd, 3rd, 4th, and 5th prizes. Inhow many overall outcomes would seniors be selected for exactly 3prizes?

Analysis:


2. Qualities: 9 Seniors, 4 Juniors

3. Roles: 5 prizes

4. Replacement: no

Example: 9 seniors and 4 juniors in English each submitted onepoem for the English Department’s annual poetry competition.The department will award 1st, 2nd, 3rd, 4th, and 5th prizes. Inhow many overall outcomes would seniors be selected for exactly 3prizes?

Analysis:


2. Qualities: 9 Seniors, 4 Juniors

3. Roles: 5 prizes

4. Replacement: no

N = 9 SeniorsJ = 4 Juniors

1st 2nd 3rd 4th 5th

C(5, 3) = 10

N N N J J

9 8 7 4 3 6048N J N N J9 4 8 7 3 6048

......

......

......

60480

Equally Likely OutcomesWhen all outcomes are equally likely, probabilities are easy tocompute.

Example: There are 15 people in a club, including Adrienne andBen, who run the door prize committee. Two people are randomlychosen to receive door prizes, and it turns out to be Adrienne andBen! What is the probability of that happening?

Solution: For the sample space,

n[S ] = C(15, 2) = 105.

The event we’re interested in is

E = A and B both get prizes

and n[E ] = 1. Then

Pr[E ] =1

105< 1%.




n[S ] = C(15, 2) = 105.



and n[E ] = 1.

Then

Pr[E ] =1

105< 1%.




n[S ] = C(15, 2) = 105.



and n[E ] = 1. Then

Pr[E ] =1

105< 1%.

Formula for ELO Probability

In general,

Pr[E ] =n[E ]

n[S ].

Example: You are dealt 5 cards at random from a standard deck ofcards. What is the probability the hand is a full house (3 cards ofone rank and 2 of another)?

Analysis:S = all 5 card handsE = full houses

n[S ] = C(52, 5) = 2598960n[E ] = ?


Analysis:S = all 5 card handsE = full houses

n[S ] = C(52, 5) = 2598960n[E ] = ?

n[E ] = ?:

2 3 4 5 6 7 8 9 10 J Q K A

156

3 2 0 0 0 0 0 0 0 0 0 0 04 6 24

0 0 0 0 2 0 0 0 0 0 0 0 36 4 24

......

......

......

......

......

......

...

3744

Note

4 = C(4, 3) 6 = C(4, 2)156 = C(13, 1)C(12, 1)C(11, 11) 3744 = 156 · 24.

156 is the number of anagrams of 3200000000000.


Answer:Since n[E ] = 3744 and n[S ] = 2598960, therefore

Pr[E ] =n[E ]

n[S ]=

3744

2598960= .001441.

Conditional Probability

Partial information about an outcome can change the probability.

Example: If you know that someone won $60 million in thepowerball last night, the probability the winner was an IU studentis about

107, 160

6, 516, 922= .01644 = 1.644%.

But if you know the winner lived in Bloomington the probability

goes up to about1

3= 33%.

Conditional Probability

Partial information about an outcome can change the probability.

Example: If you know that someone won $60 million in thepowerball last night, the probability the winner was an IU studentis about

107, 160

6, 516, 922= .01644 = 1.644%.

But if you know the winner lived in Bloomington the probability

goes up to about1

3= 33%.

Example: A special die has the numbers 1–3 in yellow and 4–6 inred, otherwise the die is normal and fair. You roll the die, but youdon’t have your glasses on, so all you tell is that the number is red.What is the probability it is even?

Solution: Normally the probability of an even number is1

2, but

here we have the extra information that the number is red.

Of the 3 red numbers, 2 are even so the answer is2

3.

Pr[ E |R ] =2

3



2, but



3.

Pr[ E |R ] =2

3



2, but



3.

Pr[ E |R ] =2

3

Let E be the even numbers and R the red ones.

We know the number is red.

We’re wondering whether it is even.

So the answer is

Pr[ E |R ] =©

=2

3.

In general,

Assuming ELO,

Pr[ B |A ] =n[B ∩ A]

n[A].

Example: A city council has 5 Democratic members and 3Republican members. A committee consisting of 3 members isselected at random. If the selected committee has at least 1Democrat, then what is the probability that it has at least 1Republican also?

Solution: Let E be the event that the committee has at least 1Democrat, and let F be the event that it has at least 1Republican. Then we want

Pr[ F |E ] =n[F ∩ E ]

n[E ].

So there are two subproblems: n[E ] and n[F ∩ E ].



Pr[ F |E ] =n[F ∩ E ]

n[E ].




Pr[ F |E ] =n[F ∩ E ]

n[E ].



n[E ]:

CP:

E = the committee has at least 1 DemocratE ′ = the committee has at no Democrats

= the committee has 3 Republicans

n[E ′] = C(3, 3) = 1n[S ] = C(8, 3) = 56n[E ] = 56− 1 = 55


n[E ]:

CP:

E = the committee has at least 1 DemocratE ′ = the committee has at no Democrats

= the committee has 3 Republicans

n[E ′] = C(3, 3) = 1n[S ] = C(8, 3) = 56n[E ] = 56− 1 = 55


n[F ∩ E ]:

5 Dem 3 Rep size

1 2C(5, 1) = 5 C(3, 2) = 3 15

2 1C(5, 2) = 10 C(3, 1) = 3 30

45


Therefore

Pr[ F |E ] =n[F ∩ E ]

n[E ]=

45

55.

Example: A city council has 5 Democratic members and 3Republican members. The mayor appoints one randomly selectedmember to the Police Commission and another randomly selectedmember to the Zoning Commission.

1. What is the probability that the member appointed to theZoning Commission is a Democrat?

2. What is the probability that the member appointed to theZoning Commission is a Democrat given that the memberappointed to the Police Commission is a Democrat?

3. What is the probability that the member appointed to theZoning Commission is a Democrat given that the memberappointed to the Police Commission is a Republican?

Answers:

(1) 58 ; (2) 4

7 ; (3) 57 .

Example: A city council has 5 Democratic members and 3Republican members. The mayor appoints one randomly selectedmember to the Police Commission and another randomly selectedmember to the Zoning Commission.

1. What is the probability that the member appointed to theZoning Commission is a Democrat?

2. What is the probability that the member appointed to theZoning Commission is a Democrat given that the memberappointed to the Police Commission is a Democrat?

3. What is the probability that the member appointed to theZoning Commission is a Democrat given that the memberappointed to the Police Commission is a Republican?

Answers: (1) 58 ; (2) 4

7 ; (3) 57 .

Example: A city council has 5 Democratic members and 3Republican members. The mayor appoints one randomly selectedmember to the Police Commission and another randomly selectedmember to the Zoning Commission.What is the probability that the member appointed to the ZoningCommission is a Democrat?

Solution:

S = all possible appointees to two commissionsE = appointee to Zoning Com is a Dem

n[S ] = P(8, 2) = 56n[E ] = 5 · 7 = 35

Pr[E ] =35

56=

5

8


Solution:


n[S ] = P(8, 2) = 56n[E ] = 5 · 7 = 35

Pr[E ] =35

56=

5

8


Solution:


n[S ] = P(8, 2) = 56n[E ] = 5 · 7 = 35

Pr[E ] =35

56=

5

8

Example: A city council has 5 Democratic members and 3Republican members. The mayor appoints one randomly selectedmember to the Police Commission and another randomly selectedmember to the Zoning Commission.What is the probability that the member appointed to the ZoningCommission is a Democrat given that the member appointed tothe Police Commission is a Democrat?

Solution:

E = appointee to Zoning Com is a DemF = appointee to Police Com is a Dem

We want Pr[ E |F ] =n[E ∩ F ]

n[F ].

n[E ∩ F ] = 5 · 4 = 20n[F ] = 5 · 7 = 35

Pr[ E |F ] =20

35=

4

7

Example: A city council has 5 Democratic members and 3Republican members. The mayor appoints one randomly selectedmember to the Police Commission and another randomly selectedmember to the Zoning Commission.What is the probability that the member appointed to the ZoningCommission is a Democrat given that the member appointed tothe Police Commission is a Democrat?

Solution:


We want Pr[ E |F ] =n[E ∩ F ]

n[F ].

n[E ∩ F ] = 5 · 4 = 20n[F ] = 5 · 7 = 35

Pr[ E |F ] =20

35=

4

7

Example: A city council has 5 Democratic members and 3Republican members. The mayor appoints one randomly selectedmember to the Police Commission and another randomly selectedmember to the Zoning Commission.What is the probability that the member appointed to the ZoningCommission is a Democrat given that the member appointed tothe Police Commission is a Republican?

Solution:


We want Pr[ E |F ′ ] =n[E ∩ F ′]

n[F ′].

n[E ∩ F ′] = 5 · 3 = 15n[F ′] = 3 · 7 = 21

Pr[ E |F ′ ] =15

21=

5

7

Example: A city council has 5 Democratic members and 3Republican members. The mayor appoints one randomly selectedmember to the Police Commission and another randomly selectedmember to the Zoning Commission.What is the probability that the member appointed to the ZoningCommission is a Democrat given that the member appointed tothe Police Commission is a Republican?

Solution:


We want Pr[ E |F ′ ] =n[E ∩ F ′]

n[F ′].

n[E ∩ F ′] = 5 · 3 = 15n[F ′] = 3 · 7 = 21

Pr[ E |F ′ ] =15

21=

5

7

Independence

If the truth of an event has no influence on the probability ofanother event being true, we say those two events are independent.

Example:Suppose E is the event that you win the lottery today;

and F is the event that it rains today.

Pr[ E |F ] = Pr[E ]Pr[ F |E ] = Pr[F ]

The two equations are equivalent, and in fact they’re bothequivalent to

Pr[E ∩ F ] = Pr[E ]Pr[F ].

Independence

If the truth of an event has no influence on the probability ofanother event being true, we say those two events are independent.

Example:Suppose E is the event that you win the lottery today;

and F is the event that it rains today.

Pr[ E |F ] = Pr[E ]Pr[ F |E ] = Pr[F ]

The two equations are equivalent, and in fact they’re bothequivalent to

Pr[E ∩ F ] = Pr[E ]Pr[F ].

Pr[E ∩ F ] = Pr[E ] · Pr[F ]

n[E ∩ F ]

n[S ]=

n[E ]

n[S ]· Pr[F ]

n[E ∩ F ] = n[E ] · Pr[F ]

n[E ∩ F ]

n[E ]= Pr[F ]

Pr[ F |E ] = Pr[F ]

Example: A special die has the numbers 1–3 in yellow and 4–6 inred, otherwise the die is normal and fair. Let E be the event thenumber is even and R the event the number is red. Are E and Rindependent?

Solution: Since the die is fair we know

Pr[E ] = 1/2Pr[R] = 1/2.

In an earlier problem we calculated

Pr[ E |R ] = 2/3.

Since Pr[E ] 6= Pr[ E |R ], then E and R are not independent.



Pr[E ] = 1/2Pr[R] = 1/2.


Pr[ E |R ] = 2/3.




Pr[E ] = 1/2Pr[R] = 1/2.


Pr[ E |R ] = 2/3.




Pr[E ] = 1/2Pr[R] = 1/2.


Pr[ E |R ] = 2/3.



Alternate Solution: Since the die is fair we know

Pr[E ] = 1/2Pr[R] = 1/2.

Also

Pr[E ∩ R] =n[E ∩ R]

n[S ]=

2

6= 1/3.

But Pr[E ∩ R] 6= Pr[E ]Pr[R], so E and R are not independent.



Pr[E ] = 1/2Pr[R] = 1/2.

Also

Pr[E ∩ R] =n[E ∩ R]

n[S ]=

2

6= 1/3.




Pr[E ] = 1/2Pr[R] = 1/2.

Also

Pr[E ∩ R] =n[E ∩ R]

n[S ]=

2

6= 1/3.


Example: 120 tennis balls have been numbered 1–120. In addition,1/3 of the balls are green and 2/3 are white. How many greenballs with even numbers are there?

Solution: There’s not enough information given to answer thequestion.

Example: 120 tennis balls have been numbered 1–120. In addition,1/3 of the balls are green and 2/3 are white. How many greenballs with even numbers are there?

Solution: There’s not enough information given to answer thequestion.

Example: 120 tennis balls have been numbered 1–120. In addition,1/3 of the balls are green and 2/3 are white. Suppose the events

G = ball is greenE = ball is even

are known to be independent. How many green balls with evennumbers are there?

Solution: We know that

Pr[G ] = 1/3Pr[E ] = 1/2.

Since G and E are independent,

Pr[G ∩ E ] = Pr[G ] Pr[E ] = 1/3 · 1/2 = 1/6.

Therefore

n[G ∩ E ] = Pr[G ∩ E ] n[S ] = 1/6 · 120 = 20.

(Because Pr[A] =n[A]

n[S ].)





Pr[G ] = 1/3Pr[E ] = 1/2.


Pr[G ∩ E ] = Pr[G ] Pr[E ] = 1/3 · 1/2 = 1/6.

Therefore

n[G ∩ E ] = Pr[G ∩ E ] n[S ] = 1/6 · 120 = 20.


n[S ].)





Pr[G ] = 1/3Pr[E ] = 1/2.


Pr[G ∩ E ] = Pr[G ] Pr[E ] = 1/3 · 1/2 = 1/6.

Therefore

n[G ∩ E ] = Pr[G ∩ E ] n[S ] = 1/6 · 120 = 20.


n[S ].)





Pr[G ] = 1/3Pr[E ] = 1/2.


Pr[G ∩ E ] = Pr[G ] Pr[E ] = 1/3 · 1/2 = 1/6.

Therefore

n[G ∩ E ] = Pr[G ∩ E ] n[S ] = 1/6 · 120 = 20.


n[S ].)





Pr[G ] = 1/3Pr[E ] = 1/2.


Pr[G ∩ E ] = Pr[G ] Pr[E ] = 1/3 · 1/2 = 1/6.

Therefore

n[G ∩ E ] = Pr[G ∩ E ] n[S ] = 1/6 · 120 = 20.


n[S ].)

Example: Suppose you shuffle a deck of cards, then cut the deck toreveal one card. You do this 10 times. What is the probability youget exactly 3 face cards?

Solution:In a standard deck there are 12 face cards (F)

and 40 nonface cards (N).

There are 10 slots to fill, each with one card.

There are many cases to consider, for example:

F F F N N N N N N NF F N F N N N N N N...

Example: Suppose you shuffle a deck of cards, then cut the deck toreveal one card. You do this 10 times. What is the probability youget exactly 3 face cards?

Solution:In a standard deck there are 12 face cards (F)

and 40 nonface cards (N).

There are 10 slots to fill, each with one card.

There are many cases to consider, for example:

F F F N N N N N N NF F N F N N N N N N...

The number of cases is the number of anagrams of“FFFNNNNNNN”:

C(10, 3) C(7, 7) = C(10, 3).

The number of outcomes in each case is 123407,so the total number of outcomes is C(10, 3)123407.

The size of the sample space is

n[S ] = 5210.

Therefore the probability is

C(10, 3)123407

5210.


C(10, 3) C(7, 7) = C(10, 3).

The number of outcomes in each case is 123407,

so the total number of outcomes is C(10, 3)123407.


n[S ] = 5210.


C(10, 3)123407

5210.


C(10, 3) C(7, 7) = C(10, 3).



n[S ] = 5210.


C(10, 3)123407

5210.


C(10, 3) C(7, 7) = C(10, 3).



n[S ] = 5210.


C(10, 3)123407

5210.


C(10, 3) C(7, 7) = C(10, 3).



n[S ] = 5210.


C(10, 3)123407

5210.

C(10, 3)123 407

5210

This can be rewritten:

C(10, 3)123 407

523 527= C(10, 3)

123

523407

527

= C(10, 3)

(12

52

)3(40

52

)7

.

Note that

Pr[F ] =12

52

Pr[N] =40

52.

C(10, 3)123 407

5210

This can be rewritten:

C(10, 3)123 407

523 527= C(10, 3)

123

523407

527

= C(10, 3)

(12

52

)3(40

52

)7

.

Note that

Pr[F ] =12

52

Pr[N] =40

52.

Bernoulli Processes

Situations like that occur often.

A Bernoulli Process is a repeated process of selecting an objectwith replacement, in which we keep track of the number of “good”objects selected.

In the previous problem the good objects were face cards.

Bernoulli FormulaSuppose there are N objects to choose from, of which G areconsidered “good”. We select n objects with replacement.

N = # objectsG = # good objects

N − G = # bad objectsn = # selectionsg = # good selections

n − g = # bad selections

p =G

N= probability of getting a good object

1− p =N − G

N= probability of getting a bad object

Then the probability of exactly g good objects se-lected is

C(n, g)G g (N − G )(n−g)

Nn= C(n, g)pg (1− p)(n−g).

The probability of g good objects in n picks with replacement is

C(n, g)(prob of good) (# good picks)(prob of bad) (# bad picks).

So in the last problem, the probability of 3 face cards in 10 picks is

C(10, 3)

(12

52

)3(40

52

)7

.

The probability of g good objects in n picks with replacement is

C(n, g)(prob of good) (# good picks)(prob of bad) (# bad picks).

So in the last problem, the probability of 3 face cards in 10 picks is

C(10, 3)

(12

52

)3(40

52

)7

.

Example: A fair standard die is rolled 12 times. What’s theprobability of getting exactly 6 even numbers?

Solution:

n = # selections = 12g = # good selections = 6

n − g = # bad selections = 6p = probability of getting a good object = 1/2

1− p = probability of getting a bad object = 1/2

The probability of g = 6 even numbers and 6 odd numbers inn = 12 rolls is

C(n, g)(prob of good) (# good picks)(prob of bad) (# bad picks)

= C(12, 6)(1/2)6(1/2)

6

= 0.2256.


Solution:






= C(12, 6)(1/2)6(1/2)

6

= 0.2256.


Solution:






= C(12, 6)(1/2)6(1/2)

6

= 0.2256.


Solution:






= C(12, 6)(1/2)6(1/2)

6

= 0.2256.

Example: A roulette wheel has 38 slots numbered 0, 00, and 1–36.Suppose a fair roulette wheel is spun 100 times. What’s theprobability of getting 17 exactly three times?

Solution:




The probability of three 17’s is


= C(100, 3)(1/38)3(37/38)

97

= .2218.


Solution:






= C(100, 3)(1/38)3(37/38)

97

= .2218.


Solution:






= C(100, 3)(1/38)3(37/38)

97

= .2218.


Solution:






= C(100, 3)(1/38)3(37/38)

97

= .2218.

Example: Suppose a fair roulette wheel is spun 100 times. What’sthe probability of getting 17 at most three times?

Solution: “At most 3” means 0, 1, 2, or 3.

So repeat the previous method for the cases g = 0, 1, 2, 3 and add.

C(100, 0)(1/38)0(37/38)

100

+ C(100, 1)(1/38)1(37/38)

99

+ C(100, 2)(1/38)2(37/38)

98

+ C(100, 3)(1/38)3(37/38)

97

= .7302




C(100, 0)(1/38)0(37/38)

100

+ C(100, 1)(1/38)1(37/38)

99

+ C(100, 2)(1/38)2(37/38)

98

+ C(100, 3)(1/38)3(37/38)

97

= .7302




C(100, 0)(1/38)0(37/38)

100

+ C(100, 1)(1/38)1(37/38)

99

+ C(100, 2)(1/38)2(37/38)

98

+ C(100, 3)(1/38)3(37/38)

97

= .7302

NonELO Probability

If a sample space does not have ELO, then we have to make a fewadjustments.

The Bernoulli formula can still be used.

But any formula that uses n[ ] to measure the size of an outcomemust use Pr[ ] instead.

ELO:

Pr[ E |F ] =n[E ∩ F ]

n[F ]

NonELO:

Pr[ E |F ] =Pr[E ∩ F ]

Pr[F ]

The nonELO version can be used even in ELO problems,but not vice-versa.

NonELO Probability




ELO:

Pr[ E |F ] =n[E ∩ F ]

n[F ]

NonELO:

Pr[ E |F ] =Pr[E ∩ F ]

Pr[F ]


NonELO Probability




ELO:

Pr[ E |F ] =n[E ∩ F ]

n[F ]

NonELO:

Pr[ E |F ] =Pr[E ∩ F ]

Pr[F ]


NonELO Probability




ELO:

Pr[ E |F ] =n[E ∩ F ]

n[F ]

NonELO:

Pr[ E |F ] =Pr[E ∩ F ]

Pr[F ]


The MP for counting outcomes can be translatedusing the conditional probability formula.

Pr[E ∩ F ]

Pr[F ]= Pr[ E |F ]

Pr[E ∩ F ] = Pr[F ] Pr[ E |F ]

Pr[E and F both happen] = Pr[F happens first]·Pr[then E happens]


Pr[E ∩ F ]

Pr[F ]= Pr[ E |F ]

Pr[E ∩ F ] = Pr[F ] Pr[ E |F ]



Pr[E ∩ F ]

Pr[F ]= Pr[ E |F ]

Pr[E ∩ F ] = Pr[F ] Pr[ E |F ]


Example: A standard deck of cards is shuffled and then 2 cards aredrawn without replacement. What’s the probability that both cardsare Aces?

Solution: Let

E = first card is an AceF = second card is an Ace.

Then

Pr[E ] =4

52

Pr[ F |E ] =3

51

Pr[F ] =4

52

Therefore

Pr[E ∩ F ] = Pr[E ] Pr[ F |E ] =4

52

3

51= .004525.


Solution: Let


Then

Pr[E ] =4

52

Pr[ F |E ] =3

51

Pr[F ] =4

52

Therefore

Pr[E ∩ F ] = Pr[E ] Pr[ F |E ] =4

52

3

51= .004525.


Solution: Let


Then

Pr[E ] =4

52

Pr[ F |E ] =3

51

Pr[F ] =4

52

Therefore

Pr[E ∩ F ] = Pr[E ] Pr[ F |E ] =4

52

3

51= .004525.


Solution: Let


Then

Pr[E ] =4

52

Pr[ F |E ] =3

51

Pr[F ] =4

52

Therefore

Pr[E ∩ F ] = Pr[E ] Pr[ F |E ] =4

52

3

51= .004525.

Venn diagrams can still be used,but sizes must be indicated by probability rather than # elements.

Example: Suppose that Pr[A] = .6, Pr[B] = .3, andPr[A ∪ B] = .8. Find Pr[ B |A ].

Venn Diagram:



Venn Diagram:



Venn Diagram:

Then

Pr[ B |A ] =Pr[B ∩ A]

Pr[A]

=.1

.6= 1/6.

Then

Pr[ B |A ] =Pr[B ∩ A]

Pr[A]

=.1

.6= 1/6.

Then

Pr[ B |A ] =Pr[B ∩ A]

Pr[A]

=.1

.6= 1/6.

Tree Diagrams: The Last ResortIf nothing else works you can try a tree diagram.

Tree diagrams are often useful for multistage processes in whichlater stages depend on earlier ones.

Example: Suppose you have 5 candies in your pocket, 2 of whichare butterscotch and the other 3 cherry. You grab one candy atrandom and eat it. Of course you are only aware of the flavor ofthe candy.

Tree Diagram: There are two possible outcomes.









“Begin”, “B”, and “C” are called nodes,the connecting lines are called edges.

We label the edges with probabilities.

For clarity, I’ll indicate under each nodethe number of candies of each flavor available.

Now suppose you grab and eat another candy.Now the possibilities look like this.

Notice that the probabilities at the first stageare ordinary probabilities.

If B1 is the event the first candy was butterscotch, thenPr[B1] = 2/5.

But the probabilities at the second stageare conditional probabilities.

If C2 is the event the second candy was cherry, thenPr[ C2 |B1 ] = 3/4.

Everywhere, the probabilities from a single node sum to 1.

If we keep track of the flavor of each candy as we eat it,there are 4 possible outcomes.

The probability of each outcome can be computed using the MP:Pr[E ]Pr[ F |E ] = Pr[E ∩ F ].

The probability of each outcome can be computed using the MP:Pr[E ]Pr[ F |E ] = Pr[E ∩ F ].

The sum of those four probabilities should be 1.

Now we can answer questions.Example: What’s the probability of eating two cherry candies?

Solution:

Answer: 6/20

Example: What’s the probability of eating one of each flavor?

Solution:

Answer: 6/20 + 6/20 = 12/20

Example: What’s the conditional probability that the second ischerry given that the first was butterscotch?

Solution:

Answer:3

4

Example: What’s the conditional probability that the second ischerry given that the first was butterscotch?

Solution:

Answer:3

4

Example: What’s the conditional probability that the first wascherry given that the second was butterscotch?

Solution:

Pr[ 1C | 2B ] =Pr[1C ∩ 2B]

Pr[2B]=

6/202/20 + 6/20

=6

8=

3

4


Solution:

Pr[ 1C | 2B ] =Pr[1C ∩ 2B]

Pr[2B]

=6/20

2/20 + 6/20=

6

8=

3

4


Solution:

Pr[ 1C | 2B ] =Pr[1C ∩ 2B]

Pr[2B]=

6/202/20 + 6/20

=6

8=

3

4


Solution:

Pr[ 1C | 2B ] =Pr[1C ∩ 2B]

Pr[2B]=

6/202/20 + 6/20

=6

8=

3

4

Random VariablesA random variable is a variable whose value depends onthe outcome of a random process.

Example: Let X be the value that comes up when a die is rolled.

Whether a die is physically rolled ora computer generates a random number,the only important thing is the probability distribution (p.d.f.),the table of values of X and their probabilities.

X Pr[X ]

1 1/62 1/63 1/64 1/65 1/66 1/6

Random VariablesA random variable is a variable whose value depends onthe outcome of a random process.

Example: Let X be the value that comes up when a die is rolled.

Whether a die is physically rolled ora computer generates a random number,the only important thing is the probability distribution (p.d.f.),the table of values of X and their probabilities.

X Pr[X ]

1 1/62 1/63 1/64 1/65 1/66 1/6

Example: Suppose there are 3 red marbles and 2 green marbles ina bowl. Two marbles are taken simultaneously at random from thebowl. Let X be the number of green marbles taken. Write out thep.d.f. of X .

X Pr[X ]

0 ?1 ?2 ?

Each individual probability like Pr[X = 1]can be computed using the C function.

Pr[X = 1] =C(3, 1) C(2, 1)

C(5, 2)=

6

10=

3

5


X Pr[X ]

0 ?1 ?2 ?


Pr[X = 1] =C(3, 1) C(2, 1)

C(5, 2)=

6

10=

3

5


X Pr[X ]

0 ?1 ?2 ?


Pr[X = 1] =C(3, 1) C(2, 1)

C(5, 2)=

6

10=

3

5


X Pr[X ]

0 ?1 ?2 ?


Pr[X = 1] =C(3, 1) C(2, 1)

C(5, 2)=

6

10=

3

5


X Pr[X ]

0 C(3,2)C(5,2)

3/10

1 C(3,1)C(2,1)C(5,2)

6/10

2 C(2,2)C(5,2)

1/10

1

Note the sum of the probabilities must be 1.


X Pr[X ]

0 C(3,2)C(5,2)

3/10

1 C(3,1)C(2,1)C(5,2)

6/10

2 C(2,2)C(5,2)

1/10

1

Note the sum of the probabilities must be 1.

Let’s play a game.

We both put $1 in the pot.Then we roll a fair die.If the die comes up 6, you win the pot.If anything else comes up, I win.

Is this game fair? No.Exactly how unfair is it?



Is this game fair?

No.Exactly how unfair is it?



Is this game fair? No.Exactly how unfair is it?


Let X be my net winnings.

Then X can take only the values ±1.

Here’s the p.d.f.

X Pr[X ]

-1 1/61 5/6


Let X be my net winnings.

Then X can take only the values ±1.

Here’s the p.d.f.

X Pr[X ]

-1 1/61 5/6

X Pr[X ]

-1 1/61 5/6

The expected value of X , denoted E(X ), is the weighted averageof the X values:

E(X ) = −1 · (1/6) + 1 · (5/6) = 4/6 = 2/3.

This means that on average I win 66.6¢ every time we play.

If we play 100 times, I should win about $67.

X Pr[X ]

-1 1/61 5/6


E(X ) = −1 · (1/6) + 1 · (5/6) = 4/6 = 2/3.



X Pr[X ]

-1 1/61 5/6


E(X ) = −1 · (1/6) + 1 · (5/6) = 4/6 = 2/3.



The easiest way to compute E(X ) is to add an extra column to thep.d.f.

X Pr[X ] X · Pr[X ]

-1 1/6 −1/61 5/6

5/6

4/6 = E(X )

Then E(X ) is the sum of the entries in the X · Pr[X ] column.

Example: Suppose there are 3 red marbles and 2 green marbles ina bowl. Two marbles are taken simultaneously at random from thebowl. Let X be the number of green marbles taken. Find theexpected value of X .

Solution: Recall the p.d.f.


0 3/10 0

1 6/106/10

2 1/102/10

8/10

Add the extra column.

So E(X ) = 8/10 = 4/5.



X Pr[X ]

0 3/10

1 6/10

2 1/10


So E(X ) = 8/10 = 4/5.




0 3/10 0

1 6/106/10

2 1/102/10

8/10


So E(X ) = 8/10 = 4/5.




0 3/10 0

1 6/106/10

2 1/102/10

8/10


So E(X ) = 8/10 = 4/5.


E(X ) = 4/5

Shouldn’t that have been “obvious”?

Of all the marbles, 2/5 are green.So 2/5 of the 2 marbles we take should be green.

E(X ) = 2/5 · 2


E(X ) = 4/5

Shouldn’t that have been “obvious”?

Of all the marbles, 2/5 are green.So 2/5 of the 2 marbles we take should be green.

E(X ) = 2/5 · 2

This shortcut works for sampling problems,i.e. when objects are chosen from the universe with ELO,with or without replacement.

In general,

If n samples are taken from a universewhere the fraction of “good” objects isp, and X is the number of good objectsin the sample, then

E(X ) = p · n.

This shortcut works for sampling problems,i.e. when objects are chosen from the universe with ELO,with or without replacement.

In general,

If n samples are taken from a universewhere the fraction of “good” objects isp, and X is the number of good objectsin the sample, then

E(X ) = p · n.

Example: You sit at the roulette wheel for 36 hours, betting 1000times on your lucky number, 17. What is the expected number oftimes that you win?

Solution:

E(X ) =1

38· 1000 = 26.32

Example: You sit at the roulette wheel for 36 hours, betting 1000times on your lucky number, 17. What is the expected number oftimes that you win?

Solution:

E(X ) =1

38· 1000 = 26.32

Linear Algebra

A linear equation is one in whichvariables are used in the simplest possible way:multiplied by a constant and added to the other terms.

ExamplesLinear:

2x + 3y = 4

.1a−√

5b = 137c

Nonlinear:x

y= 1

xy = 52√

x − 3y2 = z

Linear Algebra


ExamplesLinear:

2x + 3y = 4

.1a−√

5b = 137c

Nonlinear:x

y= 1

xy = 52√

x − 3y2 = z

Linear Algebra


ExamplesLinear:

2x + 3y = 4

.1a−√

5b = 137c

Nonlinear:x

y= 1

xy = 52√

x − 3y2 = z

Linear equations get their name from their graphs.

Here is the graph of 3x − 2y = 6.

-6

-4

-2

0

2

4

6

-6 -4 -2 0 2 4 6

x

y

Each point on the blue line has coordinates that satisfy 3x −2y = 6.

Linear equations get their name from their graphs.

Here is the graph of 3x − 2y = 6.

-6

-4

-2

0

2

4

6

-6 -4 -2 0 2 4 6

x

y

Each point on the blue line has coordinates that satisfy 3x −2y = 6.

The points where the graph crosses the axes are called intercepts.

-6

-4

-2

0

2

4

6

-6 -4 -2 0 2 4 6

x

y

For the line 3x − 2y = 6the x-intercept is (2, 0) andthe y -intercept is (0,−3).

The points where the graph crosses the axes are called intercepts.

-6

-4

-2

0

2

4

6

-6 -4 -2 0 2 4 6

x

y

For the line 3x − 2y = 6the x-intercept is (2, 0) andthe y -intercept is (0,−3).

To find the intercepts for a linear equation:

x-intercept This is where the line crosses the x-axis, so y = 0.Plug in y = 0 and solve for x .

y -intercept This is where the line crosses the y -axis, so x = 0.Plug in x = 0 and solve for y .

Example: For 3x − 2y = 6,

x : 3x − 2(0) = 6 =⇒ x = 2y : 3(0)− 2y = 6 =⇒ y = −3.

To find the intercepts for a linear equation:

x-intercept This is where the line crosses the x-axis, so y = 0.Plug in y = 0 and solve for x .

y -intercept This is where the line crosses the y -axis, so x = 0.Plug in x = 0 and solve for y .

Example: For 3x − 2y = 6,

x : 3x − 2(0) = 6 =⇒ x = 2y : 3(0)− 2y = 6 =⇒ y = −3.

If you’re asked to graph a linear equation,all you need are two distinct points satisfying the equation(for example, the intercepts).Just plot them and draw a line through them.

Example: Graph the linear equation x + 3y = 4.

Solution: Here it’s easy to find one solution: (1, 1).You can find others without much trouble, e.g. (−2, 2).Or you can find the intercepts:

x : x + 3(0) = 4 =⇒ x = 4y : 0 + 3y = 4 =⇒ y = 4/3.

That gives us (4, 0) and (0, 4/3).

If you’re asked to graph a linear equation,all you need are two distinct points satisfying the equation(for example, the intercepts).Just plot them and draw a line through them.

Example: Graph the linear equation x + 3y = 4.

Solution: Here it’s easy to find one solution: (1, 1).You can find others without much trouble, e.g. (−2, 2).Or you can find the intercepts:

x : x + 3(0) = 4 =⇒ x = 4y : 0 + 3y = 4 =⇒ y = 4/3.

That gives us (4, 0) and (0, 4/3).

It doesn’t matter which two points you use since they all lie on thesame line.

-6

-4

-2

0

2

4

6

-6 -4 -2 0 2 4 6

But you might want to plot extra points as a check.

It doesn’t matter which two points you use since they all lie on thesame line.

-6

-4

-2

0

2

4

6

-6 -4 -2 0 2 4 6

But you might want to plot extra points as a check.

Slope

Slope is a measure of the steepness of a line.

If the line rises from left to right, slope is +.

If the line falls from left to right, slope is −.

The formula for slope is

slope =rise

run=

vertical change

horizontal change.

Slope





slope =rise

run=

vertical change

horizontal change.

Slope





slope =rise

run=

vertical change

horizontal change.

Consider the line we just graphed.

-6

-4

-2

0

2

4

6

-6 -4 -2 0 2 4 6

As the line travels from (−2, 2)to (4, 0):the vertical change is

0− 2 = −2

and the horizontal change is

4− (−2) = 6.

So the slope of the line is

rise

run=−2

6= −1

3.

We could have used any two pointsbecause they all give the same result.


-6

-4

-2

0

2

4

6

-6 -4 -2 0 2 4 6


0− 2 = −2


4− (−2) = 6.


rise

run=−2

6= −1

3.



-6

-4

-2

0

2

4

6

-6 -4 -2 0 2 4 6


0− 2 = −2


4− (−2) = 6.


rise

run=−2

6= −1

3.


Jack and Jill have a date on a hill.

Jack is climbing up the line from (1,5) to Jill’s picnic site at (4,14).

Jack’s path;

1. Straight line; use Pythagorean Theorem to find distance;2. Horizontal and vertical components.

2.1 Horizontal: 4− 1 = 32.2 Vertical: 14− 5 = 9



Jack’s path;

1. Straight line; use Pythagorean Theorem to find distance;

2. Horizontal and vertical components.




Jack’s path;





Jack’s path;


2.1 Horizontal: 4− 1 = 3

2.2 Vertical: 14− 5 = 9



Jack’s path;



In general, if a line passes through the points

(x1, y1) and (x2, y2)

then its slope is

Slope = m =y2 − y1x2 − x1

=y1 − y2x1 − x2

.


(x1, y1) and (x2, y2)

then its slope is


=y1 − y2x1 − x2

.


(x1, y1) and (x2, y2)

then its slope is


=y1 − y2x1 − x2

.

You may also remember this rule from high school:

The slope of the line Ax + By = C is −B/A.

Example: Find the slope of the line 2x − 3y = 5.

Solution: We have A = 2, B = −3, and C = 5, so

m = −(−3/2) = 3/2.





m = −(−3/2) = 3/2.





m = −(−3/2) = 3/2.

What’s the formula for the line with the following properties?

1. The vertical intercept is 2, i.e. it passes through (0, 2);

2. The slope is 3.

Fact 2 means the height grows like 3x :one unit of x produces three units of y .So

y = 3x + ?.

To find the value of ?, use Fact 1:

2 = 3 · 0 + ?.

So ? has a value of 2.

y = 3x + 2

In general, the equation of a line can be written

y = mx + b (Slope-Intercept Form)

where m is the slope and b is from the vertical intercept.



2. The slope is 3.


y = 3x + ?.


2 = 3 · 0 + ?.


y = 3x + 2






2. The slope is 3.

Fact 2 means the height grows like 3x :one unit of x produces three units of y .

Soy = 3x + ?.


2 = 3 · 0 + ?.


y = 3x + 2






2. The slope is 3.


y = 3x + ?.


2 = 3 · 0 + ?.


y = 3x + 2






2. The slope is 3.


y = 3x + ?.


2 = 3 · 0 + ?.


y = 3x + 2






2. The slope is 3.


y = 3x + ?.


2 = 3 · 0 + ?.


y = 3x + 2






2. The slope is 3.


y = 3x + ?.


2 = 3 · 0 + ?.


y = 3x + 2






2. The slope is 3.


y = 3x + ?.


2 = 3 · 0 + ?.


y = 3x + 2




Here’s another form of the linear equation:

Start with the definition of slope:


.

Now rearrange terms to get rid of the fraction:

m(x2 − x1) = y2 − y1.

Since this holds for any (x2, y2) on the line, we can think of it asthe defining equation of the line:

m(x − x1) = y − y1.

This is called point-slope form.

Both forms have their strengths and weaknesses for problemsolving.




.


m(x2 − x1) = y2 − y1.


m(x − x1) = y − y1.






.


m(x2 − x1) = y2 − y1.


m(x − x1) = y − y1.






.


m(x2 − x1) = y2 − y1.


m(x − x1) = y − y1.






.


m(x2 − x1) = y2 − y1.


m(x − x1) = y − y1.






.


m(x2 − x1) = y2 − y1.


m(x − x1) = y − y1.






.


m(x2 − x1) = y2 − y1.


m(x − x1) = y − y1.



Types of problems:

1. Given information about a line, figure out its equation.

Point-Slope form is usually best.

2. Given the equation of a line, determine facts about it.

Slope-Intercept form is usually best.

Types of problems:





Types of problems:





Types of problems:





Types of problems:





Types of problems:





Types of problems:





Example:

Given the equation −4y + 2x + 8 = 0, find the slope andy -intercept.


Solution Rewrite in Slope-Intercept form:

−4y = −2x − 8

y =1

2x + 2.

Now we can read off the info we need:

m =1

2b = 2

Example:




−4y = −2x − 8

y =1

2x + 2.


m =1

2b = 2

Example:




−4y = −2x − 8

y =1

2x + 2.


m =1

2b = 2

Example:




−4y = −2x − 8

y =1

2x + 2.


m =1

2b = 2

Example:

A line passes through the points (4, 5) and (2,−1). What is theequation of the line?


Solution:First find the slope.

m =y2 − y1x2 − x1

=−1− 5

2− 4=−6

−2= 3

Now apply the point-slope form.

y − 5 = 3(x − 4)

Example:




m =y2 − y1x2 − x1

=−1− 5

2− 4=−6

−2= 3


y − 5 = 3(x − 4)

Example:




m =y2 − y1x2 − x1

=−1− 5

2− 4=−6

−2= 3


y − 5 = 3(x − 4)

Example:




m =y2 − y1x2 − x1

=−1− 5

2− 4=−6

−2= 3


y − 5 = 3(x − 4)

Example:




m =y2 − y1x2 − x1

=−1− 5

2− 4=−6

−2= 3


y − 5 = 3(x − 4)

There are exceptions.

Example:

The population was 30,700 and grew at a rate of 850 per year.

1. Give the linear equation for p and t (years since 2000).

2. What is the population in the year 2010?

3. When will the population be 45,000?

Solution

1. p = 850t + 30, 700.

2. Plug in t = 10.

p = 850 · 10 + 30,700 = 39,200

3. Plug in p = 45,000.

45,000 = 850t + 30,70014,300 = 850t

t ≈ 16.8


Example:





Solution

1. p = 850t + 30, 700.

2. Plug in t = 10.

p = 850 · 10 + 30,700 = 39,200

3. Plug in p = 45,000.

45,000 = 850t + 30,70014,300 = 850t

t ≈ 16.8


Example:





Solution

1. p = 850t + 30, 700.

2. Plug in t = 10.

p = 850 · 10 + 30,700 = 39,200

3. Plug in p = 45,000.

45,000 = 850t + 30,70014,300 = 850t

t ≈ 16.8


Example:





Solution

1. p = 850t + 30, 700.

2. Plug in t = 10.

p = 850 · 10 + 30,700 = 39,200

3. Plug in p = 45,000.

45,000 = 850t + 30,70014,300 = 850t

t ≈ 16.8


Example:





Solution

1. p = 850t + 30, 700.

2. Plug in t = 10.

p = 850 · 10 + 30,700 = 39,200

3. Plug in p = 45,000.

45,000 = 850t + 30,70014,300 = 850t

t ≈ 16.8


Example:





Solution

1. p = 850t + 30, 700.

2. Plug in t = 10.

p = 850 · 10 + 30,700 = 39,200

3. Plug in p = 45,000.

45,000 = 850t + 30,70014,300 = 850t

t ≈ 16.8


Example:





Solution

1. p = 850t + 30, 700.

2. Plug in t = 10.

p = 850 · 10 + 30,700 = 39,200

3. Plug in p = 45,000.

45,000 = 850t + 30,70014,300 = 850t

t ≈ 16.8


Example:





Solution

1. p = 850t + 30, 700.

2. Plug in t = 10.

p = 850 · 10 + 30,700 = 39,200

3. Plug in p = 45,000.

45,000 = 850t + 30,70014,300 = 850t

t ≈ 16.8


Example:





Solution

1. p = 850t + 30, 700.

2. Plug in t = 10.

p = 850 · 10 + 30,700 = 39,200

3. Plug in p = 45,000.

45,000 = 850t + 30,70014,300 = 850t

t ≈ 16.8


Example:





Solution

1. p = 850t + 30, 700.

2. Plug in t = 10.

p = 850 · 10 + 30,700 = 39,200

3. Plug in p = 45,000.

45,000 = 850t + 30,70014,300 = 850t

t ≈ 16.8


Example:





Solution

1. p = 850t + 30, 700.

2. Plug in t = 10.

p = 850 · 10 + 30,700 = 39,200

3. Plug in p = 45,000.

45,000 = 850t + 30,700

14,300 = 850tt ≈ 16.8


Example:





Solution

1. p = 850t + 30, 700.

2. Plug in t = 10.

p = 850 · 10 + 30,700 = 39,200

3. Plug in p = 45,000.

45,000 = 850t + 30,70014,300 = 850t

t ≈ 16.8


Example:





Solution

1. p = 850t + 30, 700.

2. Plug in t = 10.

p = 850 · 10 + 30,700 = 39,200

3. Plug in p = 45,000.

45,000 = 850t + 30,70014,300 = 850t

t ≈ 16.8

Example:

Latitude # Species

11◦ 3444◦ 26

(a) First find the slope:

m =N2 − N1

l2 − l1=

34− 26

11− 44= − 8

33.

Now apply point-slope form:

N − 26 = − 8

33(l − 44).

Example:

Latitude # Species

11◦ 3444◦ 26


m =N2 − N1

l2 − l1=

34− 26

11− 44= − 8

33.


N − 26 = − 8

33(l − 44).

Example:

Latitude # Species

11◦ 3444◦ 26


m =N2 − N1

l2 − l1=

34− 26

11− 44= − 8

33.


N − 26 = − 8

33(l − 44).

Example:

Latitude # Species

11◦ 3444◦ 26


m =N2 − N1

l2 − l1=

34− 26

11− 44= − 8

33.


N − 26 = − 8

33(l − 44).

N − 26 = − 8

33(l − 44).

(b) Find the slope and y -intercept.

Let’s first rewrite the equation in slope-intercept form:

N = − 8

33l +

8

3344 + 26 = − 8

33l +

110

3.

Now we can say:

m = − 833 species per degree;

b = 1103 species at the equator.

(c) Graph the line.

N − 26 = − 8

33(l − 44).

(b) Find the slope and y -intercept.Let’s first rewrite the equation in slope-intercept form:

N = − 8

33l +

8

3344 + 26 = − 8

33l +

110

3.

Now we can say:



(c) Graph the line.

N − 26 = − 8

33(l − 44).


N = − 8

33l +

8

3344 + 26 = − 8

33l +

110

3.

Now we can say:



(c) Graph the line.

N − 26 = − 8

33(l − 44).


N = − 8

33l +

8

3344 + 26 = − 8

33l +

110

3.

Now we can say:



(c) Graph the line.

N − 26 = − 8

33(l − 44).


N = − 8

33l +

8

3344 + 26 = − 8

33l +

110

3.

Now we can say:



(c) Graph the line.

Problem: A coffee shop sells coffee to go in three sizes.

Assuming the relationship between price and size is linear, whatdoes the 12 oz. coffee cost?





Size in oz. Price in $x y

8 1.0012 ?20 2.08

(It’s OK to reverse x and y , just be consistent.)

Solution: First find the slope:

m =2.08− 1.00

20− 8=

1.08

12= 0.09.

Thereforey − 1.00 = 0.09(x − 8)

andy = 0.09x + 0.28.

Finally, when x = 12,

y = 0.09 · 12 + 0.28 = $1.36.


8 1.0012 ?20 2.08



m =2.08− 1.00

20− 8=

1.08

12= 0.09.

Thereforey − 1.00 = 0.09(x − 8)

andy = 0.09x + 0.28.


y = 0.09 · 12 + 0.28 = $1.36.


8 1.0012 ?20 2.08



m =2.08− 1.00

20− 8=

1.08

12= 0.09.

Thereforey − 1.00 = 0.09(x − 8)

andy = 0.09x + 0.28.


y = 0.09 · 12 + 0.28 = $1.36.


8 1.0012 ?20 2.08



m =2.08− 1.00

20− 8=

1.08

12= 0.09.

Thereforey − 1.00 = 0.09(x − 8)

andy = 0.09x + 0.28.


y = 0.09 · 12 + 0.28 = $1.36.


8 1.0012 ?20 2.08



m =2.08− 1.00

20− 8=

1.08

12= 0.09.

Thereforey − 1.00 = 0.09(x − 8)

andy = 0.09x + 0.28.


y = 0.09 · 12 + 0.28 = $1.36.


8 1.0012 ?20 2.08



m =2.08− 1.00

20− 8=

1.08

12= 0.09.

Thereforey − 1.00 = 0.09(x − 8)

andy = 0.09x + 0.28.


y = 0.09 · 12 + 0.28 = $1.36.

y = 0.09x + 0.28.

Problem: What do m = 0.09 and b = 0.28 (the y -intercept)mean in this problem?

Solution: The slope is rise/run, in this case $/oz.

So evidently coffee costs 9¢ per ounce.

The value of b is the value that y takes when x = 0.

So if you buy 0 ounces of coffee it still costs you 28¢.

Evidently that is the price of renting a mug.

b = fixed costsmx = variable costs

y = 0.09x + 0.28.








y = 0.09x + 0.28.








y = 0.09x + 0.28.








y = 0.09x + 0.28.








y = 0.09x + 0.28.








y = 0.09x + 0.28.








y = 0.09x + 0.28.








Systems of Equations

Sometimes it’s easier to set up story problems using more than onevariable.

Example: Suppose that a mother is 25 years older than her son,and that in 15 years she will be twice as old as her son. How oldare they now?

Solution: Letx = Mother’s agey = Son’s age.

The first phrase says that

x = y + 25,

and the second phrase says that

x + 15 = 2(y + 15).






x = y + 25,


x + 15 = 2(y + 15).






x = y + 25,


x + 15 = 2(y + 15).






x = y + 25,


x + 15 = 2(y + 15).






x = y + 25,


x + 15 = 2(y + 15).






x = y + 25,


x + 15 = 2(y + 15).

This gives us the system of Equations

x = y + 25x + 15 = 2(y + 15).

This particular system can be solved easilyusing the substitution method.

(y + 25) + 15 = 2(y + 15)

y + 40 = 2y + 30

−y − 30 = −y − 30

10 = y

Then x = 35.


x = y + 25x + 15 = 2(y + 15).


(y + 25) + 15 = 2(y + 15)

y + 40 = 2y + 30

−y − 30 = −y − 30

10 = y

Then x = 35.


x = y + 25x + 15 = 2(y + 15).


(y + 25) + 15 = 2(y + 15)

y + 40 = 2y + 30

−y − 30 = −y − 30

10 = y

Then x = 35.


x = y + 25x + 15 = 2(y + 15).


(y + 25) + 15 = 2(y + 15)

y + 40 = 2y + 30

−y − 30 = −y − 30

10 = y

Then x = 35.


x = y + 25x + 15 = 2(y + 15).


(y + 25) + 15 = 2(y + 15)

y + 40 = 2y + 30

−y − 30 = −y − 30

10 = y

Then x = 35.


x = y + 25x + 15 = 2(y + 15).


(y + 25) + 15 = 2(y + 15)

y + 40 = 2y + 30

−y − 30 = −y − 30

10 = y

Then x = 35.


x = y + 25x + 15 = 2(y + 15).


(y + 25) + 15 = 2(y + 15)

y + 40 = 2y + 30

−y − 30 = −y − 30

10 = y

Then x = 35.

Substitution Method

1. Solve for one variable in one equation.

2. Plug that result into the other equation.

3. Solve for the remaining variable.

4. Go back to (1) to evaluate the first variable.

Example:x + y = 2x − y = 0

1. x = 2− y

2. (2− y)− y = 0

3.2− 2y = 0−2y = −2

y = 1

4. x = 2− y = 2− 1 = 1

Substitution Method






1. x = 2− y

2. (2− y)− y = 0

3.2− 2y = 0−2y = −2

y = 1

4. x = 2− y = 2− 1 = 1

Substitution Method






1. x = 2− y

2. (2− y)− y = 0

3.2− 2y = 0−2y = −2

y = 1

4. x = 2− y = 2− 1 = 1

Substitution Method






1. x = 2− y

2. (2− y)− y = 0

3.2− 2y = 0−2y = −2

y = 1

4. x = 2− y = 2− 1 = 1

Substitution Method






1. x = 2− y

2. (2− y)− y = 0

3.2− 2y = 0−2y = −2

y = 1

4. x = 2− y = 2− 1 = 1

Substitution Method






1. x = 2− y

2. (2− y)− y = 0

3.2− 2y = 0−2y = −2

y = 1

4. x = 2− y = 2− 1 = 1

Elimination Method

1. Put both equations in standard form (variables on left).

2. Multiply either or both equations by constant.

3. Add the equations together, eliminating one variable.


5. Plug that into the first equation and solve for the first variable.


1. Already done.

2. Already done.

3. 2x = 2.

4. x = 1.

5.1 + y = 2

y = 1

Elimination Method







1. Already done.

2. Already done.

3. 2x = 2.

4. x = 1.

5.1 + y = 2

y = 1

Elimination Method







1. Already done.

2. Already done.

3. 2x = 2.

4. x = 1.

5.1 + y = 2

y = 1

Elimination Method







1. Already done.

2. Already done.

3. 2x = 2.

4. x = 1.

5.1 + y = 2

y = 1

Elimination Method







1. Already done.

2. Already done.

3. 2x = 2.

4. x = 1.

5.1 + y = 2

y = 1

Elimination Method







1. Already done.

2. Already done.

3. 2x = 2.

4. x = 1.

5.1 + y = 2

y = 1

Example: Solve the system

2x = 4− 3y4y = 5− 3x .

Solution:

1.2x + 3y = 43x + 4y = 5.

2. Multiply the first equation by 3, the second by −2.

6x + 9y = 12−6x − 8y = −10.

3. y = 24. y = 25.

2x = 4− 3(2)2x = −2

x = −1


2x = 4− 3y4y = 5− 3x .

Solution:

1.2x + 3y = 43x + 4y = 5.


6x + 9y = 12−6x − 8y = −10.

3. y = 24. y = 25.

2x = 4− 3(2)2x = −2

x = −1


2x = 4− 3y4y = 5− 3x .

Solution:

1.2x + 3y = 43x + 4y = 5.


6x + 9y = 12−6x − 8y = −10.

3. y = 24. y = 25.

2x = 4− 3(2)2x = −2

x = −1


2x = 4− 3y4y = 5− 3x .

Solution:

1.2x + 3y = 43x + 4y = 5.


6x + 9y = 12−6x − 8y = −10.

3. y = 24. y = 2

5.2x = 4− 3(2)2x = −2

x = −1


2x = 4− 3y4y = 5− 3x .

Solution:

1.2x + 3y = 43x + 4y = 5.


6x + 9y = 12−6x − 8y = −10.

3. y = 24. y = 25.

2x = 4− 3(2)2x = −2

x = −1

Graphic Systems

2x = 4− 3y4y = 5− 3x .

0.5

1

1.5

2

2.5

3

-2 -1.5 -1 -0.5 0 0.5

Solution: (−1, 2)The solution is the point of intersection.

Graphic Systems

2x = 4− 3y4y = 5− 3x .

0.5

1

1.5

2

2.5

3

-2 -1.5 -1 -0.5 0 0.5

Solution: (−1, 2)The solution is the point of intersection.

System WeirdnessThings can go wrong.

Example:x + y = 2

y = 4− x

Substitution produces

x + (4− x) = 24 = 2

Contradiction.

0

0.5

1

1.5

2

2.5

3

0 0.5 1 1.5 2 2.5 3

So this system of equationshas no solution,because the lines are parallel.


Example:x + y = 2

y = 4− x


x + (4− x) = 24 = 2

Contradiction.

0

0.5

1

1.5

2

2.5

3

0 0.5 1 1.5 2 2.5 3



Example:x + y = 2

y = 4− x


x + (4− x) = 2

4 = 2

Contradiction.

0

0.5

1

1.5

2

2.5

3

0 0.5 1 1.5 2 2.5 3



Example:x + y = 2

y = 4− x


x + (4− x) = 24 = 2

Contradiction.

0

0.5

1

1.5

2

2.5

3

0 0.5 1 1.5 2 2.5 3



Example:x + y = 2

y = 4− x


x + (4− x) = 24 = 2

Contradiction.

0

0.5

1

1.5

2

2.5

3

0 0.5 1 1.5 2 2.5 3


Here’s another flavor of weirdness.

x + y = 2y = 2− x

This time substitution produces

x + (2− x) = 22 = 2.

This is an identity.

The interpretation is thatthe two equations must have been equivalent.

So the lines are actually the same.

So in this case there are infinitely many solutions:every point on the line satisfies both equations.


x + y = 2y = 2− x


x + (2− x) = 2

2 = 2.






x + y = 2y = 2− x


x + (2− x) = 22 = 2.






x + y = 2y = 2− x


x + (2− x) = 22 = 2.






x + y = 2y = 2− x


x + (2− x) = 22 = 2.






x + y = 2y = 2− x


x + (2− x) = 22 = 2.






x + y = 2y = 2− x


x + (2− x) = 22 = 2.





x + y = 2y = 2− x

Infinitely many solutions.

Even in this case we can do a little more:

provide a recipe for producing as many solutions as desired.

Note that the second equation gives y as a function of x .

y = 2− x

This is a recipe for y in terms of x .

If we think of x as the input, then y is the output.

x is called the free variable or arbitrary parameter.

x + y = 2y = 2− x





y = 2− x




x + y = 2y = 2− x





y = 2− x




x + y = 2y = 2− x





y = 2− x




x + y = 2y = 2− x





y = 2− x




x + y = 2y = 2− x





y = 2− x




x + y = 2y = 2− x





y = 2− x




Reduction Method

Reduction Method = Elimination Method - Cleverness

Or alternately,

Reduction is Elimination by brute force.

This is not pretty.

Reduction Method


Or alternately,


This is not pretty.

Reduction Method


Or alternately,


This is not pretty.

Example2x + 3y = 43x + 4y = 5

Let’s change that 2x into a 1x .

12(2x + 3y = 4) x + 3

2y = 2=⇒

3x + 4y = 5 3x + 4y = 5

Note that we always write both equations together,but we modify only one at a time.

Example2x + 3y = 43x + 4y = 5


12(2x + 3y = 4) x + 3

2y = 2=⇒

3x + 4y = 5 3x + 4y = 5


Example2x + 3y = 43x + 4y = 5


12(2x + 3y = 4) x + 3

2y = 2=⇒

3x + 4y = 5 3x + 4y = 5


Example2x + 3y = 43x + 4y = 5


12(2x + 3y = 4) x + 3

2y = 2=⇒

3x + 4y = 5 3x + 4y = 5


x + 32y = 2

3x + 4y = 5

Now let’s eliminate the x .

−3(x + 32y = 2) −3x − 9

2y = −6=⇒

3x + 4y = 5 3x + 82y = 5

−12y = −1

Let’s regard the result as a new improved equation 2,and let’s keep the simpler old equation 1.

x + 32y = 2

−12y = −1

x + 32y = 2

3x + 4y = 5


−3(x + 32y = 2) −3x − 9

2y = −6=⇒

3x + 4y = 5 3x + 82y = 5

−12y = −1


x + 32y = 2

−12y = −1

x + 32y = 2

3x + 4y = 5


−3(x + 32y = 2) −3x − 9

2y = −6=⇒

3x + 4y = 5 3x + 82y = 5

−12y = −1


x + 32y = 2

−12y = −1

x + 32y = 2

3x + 4y = 5


−3(x + 32y = 2) −3x − 9

2y = −6=⇒

3x + 4y = 5 3x + 82y = 5

−12y = −1


x + 32y = 2

−12y = −1

x + 32y = 2

−12y = −1

Now let’s finish solving for y .

x + 32y = 2 x + 3

2y = 2=⇒

−2(−1

2y = −1) y = 2

x + 32y = 2

−12y = −1

Now let’s finish solving for y .

x + 32y = 2 x + 3

2y = 2=⇒

−2(−1

2y = −1) y = 2

x + 32y = 2

y = 2

Finally, let’s eliminate y from equation 1.

x + 32y = 2 x + 3

2y = 2=⇒

−32(y = 2) −3

2y = −3

x = −1

So the complete simplification (and solution) is

x = −1y = 2.

x + 32y = 2

y = 2


x + 32y = 2 x + 3

2y = 2=⇒

−32(y = 2) −3

2y = −3

x = −1


x = −1y = 2.

x + 32y = 2

y = 2


x + 32y = 2 x + 3

2y = 2=⇒

−32(y = 2) −3

2y = −3

x = −1


x = −1y = 2.

x + 32y = 2

y = 2


x + 32y = 2 x + 3

2y = 2=⇒

−32(y = 2) −3

2y = −3

x = −1


x = −1y = 2.

Note there were only two things we had to know to do:·→ Multiply an equation by a constant.+→ Add a multiple of one equation to another.

You can also do:

↔ Reverse the equations.

Note there were only two things we had to know to do:·→ Multiply an equation by a constant.+→ Add a multiple of one equation to another.

You can also do:

↔ Reverse the equations.

Now let’s work on speed.

Shorthand notation: Instead of

2x + 3y = 43x + 4y = 5

we’ll write [2 3 43 4 5

].

[2 3 43 4 5

].

The first step: get a 1 in upper left.

1/2·→ R1 :

[1 3/2 23 4 5

]

Now use the 1 to cancel the 3.

−3R1+→ R2 :

[1 3/2 20 −1/2 −1

]Now get a 1 in the lower right.

−2·→ R2 :

[1 3/2 20 1 2

]Finally use the 1 in the lower right to cancel the 3/2.

−3/2R2+→ R1 :

[1 0 −10 1 2

]

[2 3 43 4 5

].


1/2·→ R1 :

[1 3/2 23 4 5

]Now use the 1 to cancel the 3.

−3R1+→ R2 :

[1 3/2 20 −1/2 −1

]

Now get a 1 in the lower right.

−2·→ R2 :

[1 3/2 20 1 2


−3/2R2+→ R1 :

[1 0 −10 1 2

]

[2 3 43 4 5

].


1/2·→ R1 :

[1 3/2 23 4 5


−3R1+→ R2 :

[1 3/2 20 −1/2 −1


−2·→ R2 :

[1 3/2 20 1 2

]

Finally use the 1 in the lower right to cancel the 3/2.

−3/2R2+→ R1 :

[1 0 −10 1 2

]

[2 3 43 4 5

].


1/2·→ R1 :

[1 3/2 23 4 5


−3R1+→ R2 :

[1 3/2 20 −1/2 −1


−2·→ R2 :

[1 3/2 20 1 2


−3/2R2+→ R1 :

[1 0 −10 1 2

]

Note that the goal is to get it looking like[1 0 ∗0 1 ∗

].

We do it one column at a time, left to right.

In each column,

first get the 1,

then use the 1 to get the 0s.


].


In each column,

first get the 1,



].


In each column,

first get the 1,


Operations

You use the three operations·→,

+→, ↔ to get what you want.

1. To get a 1,·→ usually;↔ sometimes.

2. to get a 0, use+→, using 1s to make 0s.

Operations



1. To get a 1,

·→ usually;↔ sometimes.


Operations



1. To get a 1,·→ usually;

↔ sometimes.


Operations





Operations






2x + 2y + 6z = −1x + y + 2z = 1

2x + 3y + z = 2

using matrix methods.

Solution: First translate the problem into matrix form. 2 2 6 −11 1 2 12 3 1 2

.


2x + 2y + 6z = −1x + y + 2z = 1

2x + 3y + z = 2

using matrix methods.

Solution: First translate the problem into matrix form. 2 2 6 −11 1 2 12 3 1 2

.

2 2 6 −11 1 2 12 3 1 2

.Now we need a 1 in the upper left.

We could use 1/2·→ R1,

but that creates a fraction, and there is an alternative.

R1 ↔ R2 :

1 1 2 12 2 6 −12 3 1 2

.

2 2 6 −11 1 2 12 3 1 2




R1 ↔ R2 :

1 1 2 12 2 6 −12 3 1 2

.

2 2 6 −11 1 2 12 3 1 2




R1 ↔ R2 :

1 1 2 12 2 6 −12 3 1 2

.

1 1 2 12 2 6 −12 3 1 2

Now get 0s down the rest of the first column.

−2R1+→ R2 :

1 1 2 10 0 2 −32 3 1 2

−2R1+→ R3 :

1 1 2 10 0 2 −30 1 −3 0

1 1 2 12 2 6 −12 3 1 2


−2R1+→ R2 :

1 1 2 10 0 2 −32 3 1 2

−2R1+→ R3 :

1 1 2 10 0 2 −30 1 −3 0

1 1 2 12 2 6 −12 3 1 2


−2R1+→ R2 :

1 1 2 10 0 2 −32 3 1 2

−2R1+→ R3 :

1 1 2 10 0 2 −30 1 −3 0

1 1 2 10 0 2 −30 1 −3 0

Now we need a 1 in the 2,2 position.

Normally we’d use·→, but that won’t work here.

So instead we use

R2 ↔ R3 :

1 1 2 10 1 −3 00 0 2 −3

.Now finish the second column:

−1R2+→ R1 :

1 0 5 10 1 −3 00 0 2 −3

.

1 1 2 10 0 2 −30 1 −3 0



So instead we use

R2 ↔ R3 :

1 1 2 10 1 −3 00 0 2 −3

.

Now finish the second column:

−1R2+→ R1 :

1 0 5 10 1 −3 00 0 2 −3

.

1 1 2 10 0 2 −30 1 −3 0



So instead we use

R2 ↔ R3 :

1 1 2 10 1 −3 00 0 2 −3

.Now finish the second column:

−1R2+→ R1 :

1 0 5 10 1 −3 00 0 2 −3

.

1 0 5 10 1 −3 00 0 2 −3

.Finally the third column,

where we need a 1 in the 3,3 position.

1/2·→ R3 :

1 0 5 10 1 −3 00 0 1 −3/2

.

1 0 5 10 1 −3 00 0 2 −3

.Finally the third column,

where we need a 1 in the 3,3 position.

1/2·→ R3 :

1 0 5 10 1 −3 00 0 1 −3/2

.

1 0 5 10 1 −3 00 0 1 −3/2

Then clear out the upper entries.

3R3+→ R2 :

1 0 5 10 1 0 −9/20 0 1 −3/2

−5R3+→ R1 :

1 0 0 17/20 1 0 −9/20 0 1 −3/2

So we get the solution

x = 17/2y = −9/2

z = −3/2.

1 0 5 10 1 −3 00 0 1 −3/2


3R3+→ R2 :

1 0 5 10 1 0 −9/20 0 1 −3/2

−5R3+→ R1 :

1 0 0 17/20 1 0 −9/20 0 1 −3/2


x = 17/2y = −9/2

z = −3/2.

1 0 5 10 1 −3 00 0 1 −3/2


3R3+→ R2 :

1 0 5 10 1 0 −9/20 0 1 −3/2

−5R3+→ R1 :

1 0 0 17/20 1 0 −9/20 0 1 −3/2


x = 17/2y = −9/2

z = −3/2.

1 0 5 10 1 −3 00 0 1 −3/2


3R3+→ R2 :

1 0 5 10 1 0 −9/20 0 1 −3/2

−5R3+→ R1 :

1 0 0 17/20 1 0 −9/20 0 1 −3/2


x = 17/2y = −9/2

z = −3/2.

Example: Here is a variation on the last problem.

2x + 2y + 6z = −1x + y + 2z = 1

2x + 2y + z = 2

Start off the same way.

R1 ↔ R2 :

1 1 2 12 2 6 −12 2 1 2

− 2R1+→ R2 :

1 1 2 10 0 2 −32 2 1 2

−2R1+→ R3 :

1 1 2 10 0 2 −30 0 −3 0


2x + 2y + 6z = −1x + y + 2z = 1

2x + 2y + z = 2


R1 ↔ R2 :

1 1 2 12 2 6 −12 2 1 2

− 2R1+→ R2 :

1 1 2 10 0 2 −32 2 1 2

−2R1+→ R3 :

1 1 2 10 0 2 −30 0 −3 0


2x + 2y + 6z = −1x + y + 2z = 1

2x + 2y + z = 2


R1 ↔ R2 :

1 1 2 12 2 6 −12 2 1 2

− 2R1+→ R2 :

1 1 2 10 0 2 −32 2 1 2

−2R1+→ R3 :

1 1 2 10 0 2 −30 0 −3 0


2x + 2y + 6z = −1x + y + 2z = 1

2x + 2y + z = 2


R1 ↔ R2 :

1 1 2 12 2 6 −12 2 1 2

− 2R1+→ R2 :

1 1 2 10 0 2 −32 2 1 2

−2R1+→ R3 :

1 1 2 10 0 2 −30 0 −3 0

1 1 2 10 0 2 −30 0 −3 0

Now we have a another problem.

There’s no way to make progress on column 2.

So we have no choice but to skip it and move on to column 3.

But where should the 1 go?

Just below the previous 1 in column 1.

1/2·→ R2 :

1 1 2 10 0 1 −3/20 0 −3 0

Those are called leading ones.

1 1 2 10 0 2 −30 0 −3 0






1/2·→ R2 :

1 1 2 10 0 1 −3/20 0 −3 0


1 1 2 10 0 2 −30 0 −3 0






1/2·→ R2 :

1 1 2 10 0 1 −3/20 0 −3 0


1 1 2 10 0 2 −30 0 −3 0






1/2·→ R2 :

1 1 2 10 0 1 −3/20 0 −3 0


1 1 2 10 0 2 −30 0 −3 0






1/2·→ R2 :

1 1 2 10 0 1 −3/20 0 −3 0


1 1 2 10 0 2 −30 0 −3 0






1/2·→ R2 :

1 1 2 10 0 1 −3/20 0 −3 0


1 1 2 10 0 1 −3/20 0 −3 0

Now the 0s.

−2R2+→ R1 :

1 1 0 40 0 1 −3/20 0 −3 0

3R2+→ R3 :

1 1 0 40 0 1 −3/20 0 0 −9/2

1 1 2 10 0 1 −3/20 0 −3 0

Now the 0s.

−2R2+→ R1 :

1 1 0 40 0 1 −3/20 0 −3 0

3R2+→ R3 :

1 1 0 40 0 1 −3/20 0 0 −9/2

1 1 0 40 0 1 −3/20 0 0 −9/2

Now we’re done with the row reduction,

so we can translate back to equation form.

x + y = 4z = −3/20 = −9/2

That last equation is trouble.

Conclusion: The system has no solution.

1 1 0 40 0 1 −3/20 0 0 −9/2



x + y = 4z = −3/20 = −9/2



1 1 0 40 0 1 −3/20 0 0 −9/2



x + y = 4z = −3/20 = −9/2



1 1 0 40 0 1 −3/20 0 0 −9/2



x + y = 4z = −3/20 = −9/2



Suppose, hypothetically, that −9/2 had been 0 instead. 1 1 0 40 0 1 −3/20 0 0 0

Then we getx + y = 4

z = −3/20 = 0

and the third equation is OK.

On the other hand,it doesn’t tell us anything useful,so let’s ignore it.


Then we get

x + y = 4z = −3/20 = 0




Then we get

x + y = 4z = −3/20 = 0



x + y = 4z = −3/2

Note that we now have more variables than equations.

That usually means infinitely many solutions.

Solve for the leading variables.

x = 4− yz = −3/2

We say that y is arbitrary or free.

It can take any value you want.

But once you decide on a value for y ,then x and z are determined by the equations.

x + y = 4z = −3/2




x = 4− yz = −3/2




x + y = 4z = −3/2




x = 4− yz = −3/2




x + y = 4z = −3/2




x = 4− yz = −3/2




x = 4− yz = −3/2

For example, suppose y = 17.

Thenx = 4− 17 = −13

giving the complete solution

x = −13y = 17

z = −3/2.

x = 4− yz = −3/2

For example, suppose y = 17.Then

x = 4− 17 = −13


x = −13y = 17

z = −3/2.

x = 4− yz = −3/2

For example, suppose y = 17.Then

x = 4− 17 = −13


x = −13y = 17

z = −3/2.

Row Reduced Form

To recap the last example:

Sometimes you cannot achieve your goal of 1 0 0 ∗0 1 0 ∗0 0 1 ∗

.

In that case you have to settle for row-reduced form.

1. Each row has a leading 1;

2. every column with a leading 1 has 0s elsewhere;

3. leading 1s do not zigzag.

Row Reduced Form



.In that case you have to settle for row-reduced form.




Row Reduced Form



.In that case you have to settle for row-reduced form.




Row-reduced: [1 0 2 30 1 4 5

][

1 2 0 30 0 1 4

][

1 2 3 4]

Not row-reduced: [0 1 4 51 0 2 3

][

1 1 0 00 1 0 1

][−1 0 2 3

0 2 4 5

]

Row-reduced: [1 0 2 30 1 4 5

][

1 2 0 30 0 1 4

][

1 2 3 4]


]

[1 1 0 00 1 0 1

][−1 0 2 3

0 2 4 5

]

Row-reduced: [1 0 2 30 1 4 5

][

1 2 0 30 0 1 4

][

1 2 3 4]


][

1 1 0 00 1 0 1

]

[−1 0 2 3

0 2 4 5

]

Row-reduced: [1 0 2 30 1 4 5

][

1 2 0 30 0 1 4

][

1 2 3 4]


][

1 1 0 00 1 0 1

][−1 0 2 3

0 2 4 5

]

Having gotten your matrx in row-reduced form:

1. check for nonsense like 0 = 5, which means “no solution”;

2. if the equations all look good, then solve for the leadingvariables;

3. the remaining variables (now on the right) are free.

Chapter 1: Matrices

A matrix is a rectangular array of numbers.[1 2 34 5 6

]

The size of a matrix is described by the number of rows andcolumns.

These numbers are called the dimensions of the matrix.

(Rows are horizontal; columns are vertical.)

This matrix is 2× 3. (# rows × # columns)

Chapter 1: Matrices

A matrix is a rectangular array of numbers.[1 2 34 5 6

]The size of a matrix is described by the number of rows andcolumns.

These numbers are called the dimensions of the matrix.

(Rows are horizontal; columns are vertical.)

This matrix is 2× 3. (# rows × # columns)

Matrices are usually denoted by capital letters.

A =

[1 2 34 5 6

]

Individual entries in a matrix are denoted by the correspondinglower-case letter, with subscripts to indicate the position.

a1,1 = 1 a1,2 = 2 a1,3 = 3a2,1 = 4 a2,2 = 5 a2,3 = 6

The commas can be omitted from the subscripts if it doesn’tcause confusion.

a11 = 1 a12 = 2 a13 = 3a21 = 4 a22 = 5 a23 = 6


A =

[1 2 34 5 6

]Individual entries in a matrix are denoted by the corresponding

lower-case letter, with subscripts to indicate the position.

a1,1 = 1 a1,2 = 2 a1,3 = 3a2,1 = 4 a2,2 = 5 a2,3 = 6


a11 = 1 a12 = 2 a13 = 3a21 = 4 a22 = 5 a23 = 6


A =

[1 2 34 5 6

]Individual entries in a matrix are denoted by the corresponding

lower-case letter, with subscripts to indicate the position.

a1,1 = 1 a1,2 = 2 a1,3 = 3a2,1 = 4 a2,2 = 5 a2,3 = 6


a11 = 1 a12 = 2 a13 = 3a21 = 4 a22 = 5 a23 = 6

A vector is a skinny matrix:

a matrix with only 1 row or only 1 column.

row vector: [1 2 3]

column vector:

123

Matrix Operations: +,−

Total no-brainers.

Example:[1 2 34 5 6

]−[−1 1 7

1 3 −4

]=

[2 1 −43 2 10

]

If the dimensions don’t match, the result is undefined.

Example: [1 24 5

]+

[−1 1 −7

1 3 −4

]= undefined


Total no-brainers.

Example:[1 2 34 5 6

]−[−1 1 7

1 3 −4

]=

[2 1 −43 2 10

]


Example: [1 24 5

]+

[−1 1 −7

1 3 −4

]= undefined


Total no-brainers.

Example:[1 2 34 5 6

]−[−1 1 7

1 3 −4

]=

[2 1 −43 2 10

]


Example: [1 24 5

]+

[−1 1 −7

1 3 −4

]= undefined

Matrix Operations: Scalar ·

There are 2 kinds of matrix multiplication, this is theeasy one.

Example:

3

[1 −2−3 4

]=

[3 −6−9 12

]An ordinary number can be called a scalar.

Scalar multiplication is always defined.



Example:

3

[1 −2−3 4

]=

[3 −6−9 12

]

An ordinary number can be called a scalar.




Example:

3

[1 −2−3 4

]=

[3 −6−9 12

]An ordinary number can be called a scalar.


Linear CombinationsIt’s common to combine addition and scalar multiplication.


A =

[2 0−1 3

]B =

[−3 5

2 −1

].

What is 2A− 3B?

Solution:

2A− 3B = 2

[2 0−1 3

]− 3

[−3 5

2 −1

]

=

[4 0−2 6

]−

[−9 15

6 −3

]

=

[13 −15−8 9

].



A =

[2 0−1 3

]B =

[−3 5

2 −1

].

What is 2A− 3B?

Solution:

2A− 3B = 2

[2 0−1 3

]− 3

[−3 5

2 −1

]

=

[4 0−2 6

]−

[−9 15

6 −3

]

=

[13 −15−8 9

].



A =

[2 0−1 3

]B =

[−3 5

2 −1

].

What is 2A− 3B?

Solution:

2A− 3B = 2

[2 0−1 3

]− 3

[−3 5

2 −1

]

=

[4 0−2 6

]−

[−9 15

6 −3

]

=

[13 −15−8 9

].



A =

[2 0−1 3

]B =

[−3 5

2 −1

].

What is 2A− 3B?

Solution:

2A− 3B = 2

[2 0−1 3

]− 3

[−3 5

2 −1

]

=

[4 0−2 6

]−

[−9 15

6 −3

]

=

[13 −15−8 9

].


A =

[2 0−1 3

]B =

[−3 5

2 −1

].

What is 2A− 3B?

Alternate Solution:

2A− 3B = 2

[2 0−1 3

]+ −3

[−3 5

2 −1

]

=

[4 0−2 6

]+

[9 −15−6 3

]

=

[13 −15−8 9

].


A =

[2 0−1 3

]B =

[−3 5

2 −1

].

What is 2A− 3B?

Alternate Solution:

2A− 3B = 2

[2 0−1 3

]+ −3

[−3 5

2 −1

]

=

[4 0−2 6

]+

[9 −15−6 3

]

=

[13 −15−8 9

].


A =

[2 0−1 3

]B =

[−3 5

2 −1

].

What is 2A− 3B?

Alternate Solution:

2A− 3B = 2

[2 0−1 3

]+ −3

[−3 5

2 −1

]

=

[4 0−2 6

]+

[9 −15−6 3

]

=

[13 −15−8 9

].

Matrix Multiplication

This one is confusing.

Imagine that vectors are cannons,and numbers are cannon balls.

like this: 2−1

2

↓

[4 2 3

]→ .




like this: 2−1

2

↓

[4 2 3

]→ .




Think of this problem

[4 2 3

]·

−2−1

2




like this: 2−1

2

↓

[4 2 3

]→ .




like this: 2−1

2

↓

[4 2 3

]→ .

−2

−1

2

↓

[4 2 3

]→

−2

−1

[4 2

]

−2

−1

↓

[4 2

]→

−2

−1

↓

[4 2

]→ 6

−2

[4

]6

−2

↓

[4

]→

−2

↓

[4

]→ −2 + 6

[ ]−2 + 6

↓

[ ]→

↓

[ ]→ −8− 2 + 6

Thus, [4 2 3

]·

−2−1

2

=[−4

]

Example:

[−3 −1 0

]·

1−5

8

= ?

Solution: 1−5

8

↓[

−3 −1 0]→

[−3 · 1− 1(−5) + 0 · 8

]=[

2]

Thus, [4 2 3

]·

−2−1

2

=[−4

]Example:

[−3 −1 0

]·

1−5

8

= ?

Solution: 1−5

8

↓[

−3 −1 0]→

[−3 · 1− 1(−5) + 0 · 8

]=[

2]

Thus, [4 2 3

]·

−2−1

2

=[−4

]Example:

[−3 −1 0

]·

1−5

8

= ?

Solution: 1−5

8

↓[

−3 −1 0]→

[−3 · 1− 1(−5) + 0 · 8

]=[

2]

Thus, [4 2 3

]·

−2−1

2

=[−4

]Example:

[−3 −1 0

]·

1−5

8

= ?

Solution: 1−5

8

↓[

−3 −1 0]→

[−3 · 1− 1(−5) + 0 · 8

]

=[

2]

Thus, [4 2 3

]·

−2−1

2

=[−4

]Example:

[−3 −1 0

]·

1−5

8

= ?

Solution: 1−5

8

↓[

−3 −1 0]→

[−3 · 1− 1(−5) + 0 · 8

]=[

2]

Note that both vectors must be of the same length!

Example: [1 2

] 456

= Undefined

The same idea works for double barreled cannons.[−1 2

3 −4

]↓ ↓[

2 −3]→

[ ]

Fire! [−1 2

−4

]↓[

2]→

[−9

]

Fire! [2−4

]↓[ ]

→[−11

]

Reload. [2−4

]↓[

2 −3]→

[−11

]

Fire, and fire again. [ ]↓[ ]

→[−11 16

]

So we get

[2 −3

] [ −1 23 −4

]=[−11 16

].

In fact this works for almost any sizes of matrix:

The only restriction is that

the length of the rows (2nd dimension) in the left matrix=

the length of the columns (1st dimension) in the right matrix.

Example: [1 2 3

] 4 56 78 9

Can we multiply a 1× 3 matrix by a 3× 2 matrix?Yes, because

the 2nd dimension of the first=

the 1st dimension of the second.

Moreover the result is 1× 2.

(1× 3)(3× 2) = (1×63)( 63× 2) = 1× 2





Example: [1 2 3

] 4 56 78 9

Can we multiply a 1× 3 matrix by a 3× 2 matrix?

Yes, because




(1× 3)(3× 2) = (1×63)( 63× 2) = 1× 2





Example: [1 2 3

] 4 56 78 9





(1× 3)(3× 2) = (1×63)( 63× 2) = 1× 2





Example: [1 2 3

] 4 56 78 9





(1× 3)(3× 2) = (1×63)( 63× 2) = 1× 2

If both dimensions are the same we say the matrix is square.[−2 1

4 −3

]↓ ↓[

2 −30 1

]→→

[ ]

First row on left against first column on right.[−2 1

4 −3

]↓[

2 −30 1

]→

[ ]

Fire both rounds. [1−3

]↓[

0 1

]→

[−16

]

Second row on left against second column on right.[1−3

]↓[

0 1

]→

[−16

]

Fire both rounds. [ ]↓[ ]

→

[−16

−3

]

Reload. [−2 1

4 −3

][

2 −30 1

] [−16

−3

]

First row on left against second column on right.[−2 1

4 −3

]↓[

2 −30 1

]→

[−16

−3

]

Fire. [−2

4

]↓[

0 1

]→

[−16 11

−3

]

Second row on left against first column on right.[−2

4

]↓[

0 1

]→

[−16 11

−3

]

Fire. [ ]↓[ ]

→

[−16 11

4 −3

]

So [2 −30 1

] [−2 1

4 −3

]=

[−16 11

4 −3

].

Fire. [ ]↓[ ]

→

[−16 11

4 −3

]So [

2 −30 1

] [−2 1

4 −3

]=

[−16 11

4 −3

].

Example: [3 −11 2

] [−1 −5

3 2

]= ?

Solution: [−1 −5

3 2

][

3 −11 2

] [−6 −17

5 −1

]Note: [

31

](−1) +

[−1

2

](3) =

[−6

5

][

31

](−5) +

[−1

2

](2) =

[−17−1

]So the each column in the answer is a linear combination of thecolumns in the left factor.

Example: [3 −11 2

] [−1 −5

3 2

]= ?


3 2

][

3 −11 2

] [−6 −17

5 −1

]

Note: [31

](−1) +

[−1

2

](3) =

[−6

5

][

31

](−5) +

[−1

2

](2) =

[−17−1


Example: [3 −11 2

] [−1 −5

3 2

]= ?


3 2

][

3 −11 2

] [−6 −17

5 −1

]Note: [

31

](−1) +

[−1

2

](3) =

[−6

5

][

31

](−5) +

[−1

2

](2) =

[−17−1


Matrix multiplication is not commutative!

This means that usually

AB 6= BA.

This is very unlike ordinary arithmetic.

You must take care with the order in which matrices are multiplied!



AB 6= BA.





AB 6= BA.



Special Matrices

In ordinary arithmetic there are two special numbers:

0, the additive identity, for which

0 + x = x ;

and 1, the multiplicative identity, for which

1 · x = x .

There are special matrices with the same properties:

the zero matrix 0 with the property that

0 + X = X ;

and the identity matrix I with the property that

I · X = X and X · I = X .

All the entries of the zero matrix 0 are 0.


X =

[1 2 34 5 6

].

Since X is 2× 3, the corresponding zero matrix should be

0 =

[0 0 00 0 0

].

Then0 + X = X + 0 = X .



X =

[1 2 34 5 6

].


0 =

[0 0 00 0 0

].

Then0 + X = X + 0 = X .



X =

[1 2 34 5 6

].


0 =

[0 0 00 0 0

].

Then0 + X = X + 0 = X .

The identity matrix is trickier.

It also comes in different sizes, but is always square.

I1 =[

1],

I2 =

[1 00 1

],

I3 =

1 0 00 1 00 0 1

,etc.


X =

[1 2 34 5 6

].

Then

I2 X = X and X I3 = X .(2× 2) (2× 3) (2× 3) (3× 3)

Notice that we need a different sized I on each side of X .

But since the size of I is determined by its neighbor,we can omit the subscript.

IX = X and X I = X .


X =

[1 2 34 5 6

].

Then

I2 X = X and X I3 = X .(2× 2) (2× 3) (2× 3) (3× 3)

Notice that we need a different sized I on each side of X .

But since the size of I is determined by its neighbor,we can omit the subscript.

IX = X and X I = X .

Matrix Inverses

In ordinary algebra the multiplicative inverse is used constantly.For example,

2x = 612 · 2x = 1

2 · 61x = 3

x = 3.

Wouldn’t it be nice if we had the same for matrices?

[1 23 4

] [xy

]=

[56

]

[−2 1

32 −1

2

] [1 23 4

] [xy

]=

[−2 1

32 −1

2

] [56

][

1 00 1

] [xy

]=

[−4

92

][

xy

]=

[−4

92

]

[1 23 4

] [xy

]=

[56

][−2 1

32 −1

2

] [1 23 4

] [xy

]=

[−2 1

32 −1

2

] [56

]

[1 00 1

] [xy

]=

[−4

92

][

xy

]=

[−4

92

]

[1 23 4

] [xy

]=

[56

][−2 1

32 −1

2

] [1 23 4

] [xy

]=

[−2 1

32 −1

2

] [56

][

1 00 1

] [xy

]=

[−4

92

]

[xy

]=

[−4

92

]

[1 23 4

] [xy

]=

[56

][−2 1

32 −1

2

] [1 23 4

] [xy

]=

[−2 1

32 −1

2

] [56

][

1 00 1

] [xy

]=

[−4

92

][

xy

]=

[−4

92

]

[−2 1

32 −1

2

]is called the inverse matrix of

[1 23 4

]because

[−2 1

32 −1

2

] [1 23 4

]=

[1 00 1

]and [

1 23 4

] [−2 1

32 −1

2

]=

[1 00 1

].

As shorthand we can write[1 23 4

]−1=

[−2 1

32 −1

2

].

Note that also [−2 1

32 −1

2

]−1=

[1 23 4

].

[−2 1

32 −1

2


[1 23 4

]because

[−2 1

32 −1

2

] [1 23 4

]=

[1 00 1

]and [

1 23 4

] [−2 1

32 −1

2

]=

[1 00 1

].


]−1=

[−2 1

32 −1

2

].


32 −1

2

]−1=

[1 23 4

].

[−2 1

32 −1

2


[1 23 4

]because

[−2 1

32 −1

2

] [1 23 4

]=

[1 00 1

]and [

1 23 4

] [−2 1

32 −1

2

]=

[1 00 1

].


]−1=

[−2 1

32 −1

2

].


32 −1

2

]−1=

[1 23 4

].

In general, ifAB = I = BA

then we say B is the inverse of A.

Almost every square matrix has an inverse.

Only square matrices can have inverses,

but not every one does.

If a matrix does not have an inverse,we say it is singular or noninvertible.

Examples:

[1 2−2 −4

],

[1 2 34 5 6

],

[1 00 0

],

1 2 34 5 67 8 9

In general, ifAB = I = BA

then we say B is the inverse of A.

Almost every square matrix has an inverse.

Only square matrices can have inverses,

but not every one does.

If a matrix does not have an inverse,we say it is singular or noninvertible.

Examples:

[1 2−2 −4

],

[1 2 34 5 6

],

[1 00 0

],

1 2 34 5 67 8 9

How do you calculate the inverse of a matrix?

Row reduction.

Each of those row operations are equivalent to a multiplication.

3·→ R1 :

3 0 00 1 00 0 1

·

R2 ↔ R3 :

1 0 00 0 10 1 0

·

5R2+→ R3 :

1 0 00 1 00 5 1

·


Row reduction.


3·→ R1 :

3 0 00 1 00 0 1

·

R2 ↔ R3 :

1 0 00 0 10 1 0

·

5R2+→ R3 :

1 0 00 1 00 5 1

·


Row reduction.


3·→ R1 :

3 0 00 1 00 0 1

·

R2 ↔ R3 :

1 0 00 0 10 1 0

·

5R2+→ R3 :

1 0 00 1 00 5 1

·


Row reduction.


3·→ R1 :

3 0 00 1 00 0 1

·

R2 ↔ R3 :

1 0 00 0 10 1 0

·

5R2+→ R3 :

1 0 00 1 00 5 1

·

So row reduction is equivalent to multiplication.

That means if −1 2 34 5 67 8 9

RR−→

1 0 00 1 00 0 1

then then same result can be gotten by matrix multiplication:

M7M6M5M4M3M2M1

−1 0 12 −1 21 0 1

=

1 0 00 1 00 0 1

with the Ms equivalent to the row reduction.

Then −1 0 12 −1 21 0 1

−1 = M7M6M5M4M3M2M1.


That means if −1 2 34 5 67 8 9

RR−→

1 0 00 1 00 0 1


M7M6M5M4M3M2M1

−1 0 12 −1 21 0 1

=

1 0 00 1 00 0 1


Then −1 0 12 −1 21 0 1

−1 = M7M6M5M4M3M2M1.


That means if −1 2 34 5 67 8 9

RR−→

1 0 00 1 00 0 1


M7M6M5M4M3M2M1

−1 0 12 −1 21 0 1

=

1 0 00 1 00 0 1


Then −1 0 12 −1 21 0 1

−1 = M7M6M5M4M3M2M1.

The only problem is:

how do we keep track of the Ms while doing the row reduction?

Simple trick:Row reduce −1 0 1 1 0 0

2 −1 2 0 1 01 0 1 0 0 1

instead.

Then the right side keeps track of the row operations performed onthe left side.

Calculating the Matrix Inverse

If A is a square matrix, then to calculate its inverse:

1. Set up the augmented matrix[A I

].

2. Row reduce.

3. If you succeed in getting [I B

]then B = A−1.

4. If you fail to get the identity on the left, then A is notinvertible.

Example: Find the matrix inverse of

A =

[1 23 4

].

Solution: Start off with the augmented matrix[1 2 1 03 4 0 1

]then row reduce.

Example: Find the matrix inverse of

A =

[1 23 4

].

Solution: Start off with the augmented matrix[1 2 1 03 4 0 1

]then row reduce.

[1 2 1 03 4 0 1

]

−3R1+→ R2 :

[1 2 1 00 −2 −3 1

]

−1/2·→ R2 :

[1 2 1 00 1 3/2 −1/2

]

−2R2+→ R1 :

[1 0 −2 10 1 3/2 −1/2

]So

A−1 =

[−2 13/2 −1/2

]You can check your answer: is AA−1 = I?

[1 2 1 03 4 0 1

]

−3R1+→ R2 :

[1 2 1 00 −2 −3 1

]

−1/2·→ R2 :

[1 2 1 00 1 3/2 −1/2

]

−2R2+→ R1 :

[1 0 −2 10 1 3/2 −1/2

]So

A−1 =

[−2 13/2 −1/2


[1 2 1 03 4 0 1

]

−3R1+→ R2 :

[1 2 1 00 −2 −3 1

]

−1/2·→ R2 :

[1 2 1 00 1 3/2 −1/2

]

−2R2+→ R1 :

[1 0 −2 10 1 3/2 −1/2

]

So

A−1 =

[−2 13/2 −1/2


[1 2 1 03 4 0 1

]

−3R1+→ R2 :

[1 2 1 00 −2 −3 1

]

−1/2·→ R2 :

[1 2 1 00 1 3/2 −1/2

]

−2R2+→ R1 :

[1 0 −2 10 1 3/2 −1/2

]So

A−1 =

[−2 13/2 −1/2

]

You can check your answer: is AA−1 = I?

[1 2 1 03 4 0 1

]

−3R1+→ R2 :

[1 2 1 00 −2 −3 1

]

−1/2·→ R2 :

[1 2 1 00 1 3/2 −1/2

]

−2R2+→ R1 :

[1 0 −2 10 1 3/2 −1/2

]So

A−1 =

[−2 13/2 −1/2


Solving Systems with Inverses

Note that the matrix equation −1 2 34 5 67 8 9

xyz

=

236

is equivalent to the system of equations

−x + 2y + 3z = 34x + 5y + 6z = 67x + 8y + 9z = 9.

So matrix inverses can also be used to solve systems.

Example: Solve the following system.

2x + z = 2−y + 2z = −1

x + z = 3.

Solution: First translate the system into matrix equation form. 2 0 10 −1 21 0 1

xyz

=

2−1

3


2x + z = 2−y + 2z = −1

x + z = 3.


xyz

=

2−1

3

2 0 10 −1 21 0 1

xyz

=

2−1

3

Now compute the inverse matrix. 2 0 1 1 0 0

0 −1 2 0 1 01 0 1 0 0 1

R1 ↔ R3

1 0 1 0 0 10 −1 2 0 1 02 0 1 1 0 0

−2R1+→ R3

1 0 1 0 0 10 −1 2 0 1 00 0 −1 1 0 −2

2 0 10 −1 21 0 1

xyz

=

2−1

3


0 −1 2 0 1 01 0 1 0 0 1

R1 ↔ R3

1 0 1 0 0 10 −1 2 0 1 02 0 1 1 0 0

−2R1+→ R3

1 0 1 0 0 10 −1 2 0 1 00 0 −1 1 0 −2

2 0 10 −1 21 0 1

xyz

=

2−1

3


0 −1 2 0 1 01 0 1 0 0 1

R1 ↔ R3

1 0 1 0 0 10 −1 2 0 1 02 0 1 1 0 0

−2R1+→ R3

1 0 1 0 0 10 −1 2 0 1 00 0 −1 1 0 −2

1 0 1 0 0 10 −1 2 0 1 00 0 −1 1 0 −2

−1·→ R2

1 0 1 0 0 10 1 −2 0 −1 00 0 −1 1 0 −2

−1·→ R3

1 0 1 0 0 10 1 −2 0 −1 00 0 1 −1 0 2

−1R3+→ R1

1 0 0 1 0 −10 1 −2 0 −1 00 0 1 −1 0 2

2R3+→ R2

1 0 0 1 0 −10 1 0 −2 −1 40 0 1 −1 0 2

1 0 1 0 0 10 −1 2 0 1 00 0 −1 1 0 −2

−1·→ R2

1 0 1 0 0 10 1 −2 0 −1 00 0 −1 1 0 −2

−1·→ R3

1 0 1 0 0 10 1 −2 0 −1 00 0 1 −1 0 2

−1R3+→ R1

1 0 0 1 0 −10 1 −2 0 −1 00 0 1 −1 0 2

2R3+→ R2

1 0 0 1 0 −10 1 0 −2 −1 40 0 1 −1 0 2

1 0 1 0 0 10 −1 2 0 1 00 0 −1 1 0 −2

−1·→ R2

1 0 1 0 0 10 1 −2 0 −1 00 0 −1 1 0 −2

−1·→ R3

1 0 1 0 0 10 1 −2 0 −1 00 0 1 −1 0 2

−1R3+→ R1

1 0 0 1 0 −10 1 −2 0 −1 00 0 1 −1 0 2

2R3+→ R2

1 0 0 1 0 −10 1 0 −2 −1 40 0 1 −1 0 2

1 0 1 0 0 10 −1 2 0 1 00 0 −1 1 0 −2

−1·→ R2

1 0 1 0 0 10 1 −2 0 −1 00 0 −1 1 0 −2

−1·→ R3

1 0 1 0 0 10 1 −2 0 −1 00 0 1 −1 0 2

−1R3+→ R1

1 0 0 1 0 −10 1 −2 0 −1 00 0 1 −1 0 2

2R3+→ R2

1 0 0 1 0 −10 1 0 −2 −1 40 0 1 −1 0 2

1 0 1 0 0 10 −1 2 0 1 00 0 −1 1 0 −2

−1·→ R2

1 0 1 0 0 10 1 −2 0 −1 00 0 −1 1 0 −2

−1·→ R3

1 0 1 0 0 10 1 −2 0 −1 00 0 1 −1 0 2

−1R3+→ R1

1 0 0 1 0 −10 1 −2 0 −1 00 0 1 −1 0 2

2R3+→ R2

1 0 0 1 0 −10 1 0 −2 −1 40 0 1 −1 0 2

So the inverse is 1 0 −1−2 −1 4−1 0 2

.Now we can solve the matrix equation. 1 0 −1−2 −1 4−1 0 2

2 0 10 −1 21 0 1

xyz

=

1 0 −1−2 −1 4−1 0 2

2−1

3

xyz

=

−194

So the solution to the original system is

x = −1y = 9

z = 4.

So the inverse is 1 0 −1−2 −1 4−1 0 2


2 0 10 −1 21 0 1

xyz

=

1 0 −1−2 −1 4−1 0 2

2−1

3

x

yz

=

−194


x = −1y = 9

z = 4.

So the inverse is 1 0 −1−2 −1 4−1 0 2


2 0 10 −1 21 0 1

xyz

=

1 0 −1−2 −1 4−1 0 2

2−1

3

x

yz

=

−194


x = −1y = 9

z = 4.


2x + z = 3−y + 2z = 2

x + z = 1.


xyz

=

321

Now reuse the inverse from the last example.


2x + z = 3−y + 2z = 2

x + z = 1.


xyz

=

321



2x + z = 3−y + 2z = 2

x + z = 1.


xyz

=

321


1 0 −1−2 −1 4−1 0 2

2 0 10 −1 21 0 1

xyz

=

1 0 −1−2 −1 4−1 0 2

321

xyz

=

2−4−1

So the solution to the system is

x = 2y = −4

z = −1.

1 0 −1−2 −1 4−1 0 2

2 0 10 −1 21 0 1

xyz

=

1 0 −1−2 −1 4−1 0 2

321

x

yz

=

2−4−1


x = 2y = −4

z = −1.

1 0 −1−2 −1 4−1 0 2

2 0 10 −1 21 0 1

xyz

=

1 0 −1−2 −1 4−1 0 2

321

x

yz

=

2−4−1


x = 2y = −4

z = −1.

Augmented matrix versus Matrix Equation

A system can be solved either by

1. converting it to an augmented matrix and row reducing; or

2. converting it to a matrix equation and using a matrix inverse.

Method 1 is usually easier, and it’s more general:

it even handles cases where there are infinitely many solutions.

Method 2 only makes sense if you’ll reuse the inverse.















Recipes

Example: Suppose that a recipe for a dozen biscuits calls for 4cups of flour and 2 eggs; a recipe for a dozen pancakes calls for 5cups of flour and 1 egg. Now suppose you need to make 2 dozenbiscuits and 3 dozen pancakes. How much flour and how manyeggs do you need?

Solution: Define

R1 =

[42

]R2 =

[51

].

Then the flour and eggs required are given by

2R1 + 3R2 = 2

[42

]+ 3

[51

].

This can also be expressed as[4 52 1

] [23

].

RecipesExample: Suppose that a recipe for a dozen biscuits calls for 4cups of flour and 2 eggs; a recipe for a dozen pancakes calls for 5cups of flour and 1 egg. Now suppose you need to make 2 dozenbiscuits and 3 dozen pancakes. How much flour and how manyeggs do you need?

Solution: Define

R1 =

[42

]R2 =

[51

].


2R1 + 3R2 = 2

[42

]+ 3

[51

].


] [23

].


Solution: Define

R1 =

[42

]R2 =

[51

].


2R1 + 3R2 = 2

[42

]+ 3

[51

].


] [23

].


Solution: Define

R1 =

[42

]R2 =

[51

].


2R1 + 3R2 = 2

[42

]+ 3

[51

].


] [23

].


Solution: Define

R1 =

[42

]R2 =

[51

].


2R1 + 3R2 = 2

[42

]+ 3

[51

].


] [23

].

A =

[4 52 1

]is called the production matrix.

X =

[23

]is called the production schedule.

R denotes the matrix of required flour and eggs, the resource vector.

The fundamental equation relating these is

R = AX .

In our case,

R = AX =

[4 52 1

] [23

]=

[23

7

].

This is called a linear production model.

A =

[4 52 1


X =

[23




R = AX .

In our case,

R = AX =

[4 52 1

] [23

]=

[23

7

].


A =

[4 52 1


X =

[23




R = AX .

In our case,

R = AX =

[4 52 1

] [23

]=

[23

7

].


Note in our problem

R =

[r1r2

]where

r1 = flour in cupsr2 = eggs

X =

[x1

x2

]where

x1 = biscuits in dozensx2 = pancakes in dozens.

What would be different if biscuits and pancakes were countedindividually?

Note in our problem

R =

[r1r2

]where


X =

[x1

x2

]where

x1 = biscuits in dozensx2 = pancakes in dozens.

What would be different if biscuits and pancakes were countedindividually?

R =

[r1r2

]where


X =

[x1

x2

]where

x1 = # biscuitsx2 = # pancakes

A =

[4/12

5/122/12

1/12

]=

[1/3

5/121/6

1/12

]

What if eggs were in dozens and flour in kilograms?(Assume 1 cup flour is 125 grams.)

R =

[r1r2

]where


X =

[x1

x2

]where


A =

[4/12

5/122/12

1/12

]=

[1/3

5/121/6

1/12

]

What if eggs were in dozens and flour in kilograms?(Assume 1 cup flour is 125 grams.)

R =

[r1r2

]where

r1 = flour in kg.r2 = eggs in dozens

X =

[x1

x2

]where


A =

[.125 · 1/3 .125 · 5/121/12 · 1/6 1/12 · 1/12

]=

[.042 .0521/72

1/144

]

Example: Solve for X in 1 23 45 6

[ x1

x2

]=

102437

.

Solution: Row reducing 1 2 103 4 245 6 37

yields 1 0 0

0 1 00 0 1

which translates into

x1 = 0x2 = 00 = 1

which has no solution.


[ x1

x2

]=

102437

.Solution: Row reducing 1 2 10

3 4 245 6 37

yields 1 0 0

0 1 00 0 1


x1 = 0x2 = 00 = 1



[ x1

x2

]=

102437

.Solution: Row reducing 1 2 10

3 4 245 6 37

yields 1 0 0

0 1 00 0 1


x1 = 0x2 = 00 = 1



[ x1

x2

]=

102438

.

Solution: Row reduction yields

1 0 40 1 30 0 0

,which translates into

x1 = 4x2 = 30 = 0.


[ x1

x2

]=

102438

.Solution: Row reduction yields

1 0 40 1 30 0 0

,


x1 = 4x2 = 30 = 0.


[ x1

x2

]=

102438

.Solution: Row reduction yields

1 0 40 1 30 0 0

,which translates into

x1 = 4x2 = 30 = 0.

Example: Solve for X in[1 −2−2 4

] [x1

x2

]=

[3−6

].

Solution: Row reducing [1 −2 3−2 4 −6

]yields [

1 −2 30 0 0

],


1x1 − 2x2 = 30 + 0 = 0.

Solving the first equation for x1 gives

x1 = 2x2 + 3

and we interpret x2 as a free parameter.


] [x1

x2

]=

[3−6

].


]yields [

1 −2 30 0 0

],


1x1 − 2x2 = 30 + 0 = 0.


x1 = 2x2 + 3



] [x1

x2

]=

[3−6

].


]yields [

1 −2 30 0 0

],


1x1 − 2x2 = 30 + 0 = 0.


x1 = 2x2 + 3



] [x1

x2

]=

[3−6

].


]yields [

1 −2 30 0 0

],


1x1 − 2x2 = 30 + 0 = 0.


x1 = 2x2 + 3


x1 = 2x2 + 3x2 = free

A free parameter is a variable that can take any value.

For example, suppose x2 = −17.

Then x1 = 2(−17) + 3 = −31.

And that is a valid solution to the initial problem:

1x1 − 2x2 = 3−2x1 + 4x2 = −6

1 · (−31) − 2 · (−17) = 3−2 · (−31) + 4 · (−17) = −6.

x1 = 2x2 + 3x2 = free



Then x1 = 2(−17) + 3 = −31.


1x1 − 2x2 = 3−2x1 + 4x2 = −6

1 · (−31) − 2 · (−17) = 3−2 · (−31) + 4 · (−17) = −6.

x1 = 2x2 + 3x2 = free



Then x1 = 2(−17) + 3 = −31.


1x1 − 2x2 = 3−2x1 + 4x2 = −6

1 · (−31) − 2 · (−17) = 3−2 · (−31) + 4 · (−17) = −6.


] [x1

x2

]=

[36

].

Solution: Row reducing [1 −2 3−2 4 6

]yields [

1 −2 00 0 1

],


1x1 − 2x2 = 00 + 0 = 1,



] [x1

x2

]=

[36

].


]yields [

1 −2 00 0 1

],


1x1 − 2x2 = 00 + 0 = 1,



] [x1

x2

]=

[36

].


]yields [

1 −2 00 0 1

],


1x1 − 2x2 = 00 + 0 = 1,



] [x1

x2

]=

[36

].


]yields [

1 −2 00 0 1

],


1x1 − 2x2 = 00 + 0 = 1,


Example: Solve for X in

[1 2 34 5 6

] x1

x2

x3

=

[2124

].

Solution: Row reducing [1 2 3 214 5 6 24

]yields [

1 0 −1 −190 1 2 20

],


x1 − x3 = −19x2 + 2x3 = 20.


[1 2 34 5 6

] x1

x2

x3

=

[2124

].


]yields [

1 0 −1 −190 1 2 20

],


x1 − x3 = −19x2 + 2x3 = 20.


[1 2 34 5 6

] x1

x2

x3

=

[2124

].


]yields [

1 0 −1 −190 1 2 20

],


x1 − x3 = −19x2 + 2x3 = 20.

x1 − x3 = −19x2 + 2x3 = 20.

Here we will solve for x1 and x2

x1 = x3 − 19x2 = −2x3 + 20

and interpret x3 as a free variable.

For example, if x3 = 5 then

x1 = 5− 19 = −14x2 = −2(5) + 20 = 10x3 = 5

is a solution.

x1 − x3 = −19x2 + 2x3 = 20.

Here we will solve for x1 and x2

x1 = x3 − 19x2 = −2x3 + 20

and interpret x3 as a free variable.

For example, if x3 = 5 then

x1 = 5− 19 = −14x2 = −2(5) + 20 = 10x3 = 5

is a solution.

Number of Solutions

General Observations:

1. zeros on left & nonzero on right =⇒ no solution; otherwise

2. #rows = #variables =⇒ unique solution;

3. #rows < #variables =⇒ infinitely many solutions.

Number of Solutions





Number of Solutions





Number of Solutions





Row reduction produces reduced row echelon form. 1 0 40 1 30 0 0

.RREF is the closest to I.

RREF satisfies the following.


2. each column with a leading 1 has 0’s elsewhere;

3. the leading 1’s form a staircase (no zigzags).



















Leontiev ModelsNow we apply linear production models to self-sustainingeconomies.

Think of an Amish oat farmer who plows with horses.

He uses horse manure to fertilize his oat fields,and oats to feed his horses.

Suppose that it takes 1/10 ton of oats and half a ton of manure togrow one ton of oats.And it takes one ton of oats to produce a ton of manure.

Then ifx1 = # tons of oatsx2 = # tons of manure

we get

A =

[0.1 10.5 0

].






we get

A =

[0.1 10.5 0

].




Suppose that it takes 1/10 ton of oats and half a ton of manure togrow one ton of oats.

And it takes one ton of oats to produce a ton of manure.


we get

A =

[0.1 10.5 0

].






we get

A =

[0.1 10.5 0

].

x1 = # tons of oatsx2 = # tons of manure

OutputsOats Manure

Inp

uts Oats 0.1 1Manure 0.5 0

A =

[0.1 10.5 0

]

So if

X =

[x1x2

]then the resources consumed are

R = AX =

[0.1 10.5 0

] [x1x2

].

Now the farmer would like to sell what he can:

excess oats to Quaker,excess manure to other farmers.

So if

X =

[x1x2

]then the resources consumed are

R = AX =

[0.1 10.5 0

] [x1x2

].

Now the farmer would like to sell what he can:

excess oats to Quaker,excess manure to other farmers.

How much does he have at the end of the harvestthat doesn’t need himself for next year?

D = X − AX

D is called the demand vectorand denotes the excess of each product that can be soldto meet external Demand.

The equation above is the fundamental one for Leontiev theory.


D = X − AX




D = X − AX



The Fundamental Equation of Leontiev Theory

This equation can be written in three different ways,and each has its own uses.

D = X − AX (1)

D = (I− A)X (2)(I− A)−1 = X (3)

1. Form (1) is easy to understand;

2. form (2) can be solved for X using row reduction;

3. form (3) has been solved using a matrix inverse.



D = X − AX (1)D = (I− A)X (2)

(I− A)−1 = X (3)






D = X − AX (1)D = (I− A)X (2)

(I− A)−1 = X (3)






D = X − AX (1)D = (I− A)X (2)

(I− A)−1 = X (3)




Back to the Amish oat farmer:

I− A =

[1 00 1

]−[

0.1 10.5 0

]=

[0.9 −1−0.5 1

].

Suppose the farmer is currently growing 100 tons of oats and“producing” 60 tons of horse manure. How much will he be ableto sell?

X =

[100

60

],

D = (I− A)X

=

[0.9 −1−0.5 1

] [100

60

]

=

[3010

]


I− A =

[1 00 1

]−[

0.1 10.5 0

]=

[0.9 −1−0.5 1

].


X =

[100

60

],

D = (I− A)X

=

[0.9 −1−0.5 1

] [100

60

]

=

[3010

]


I− A =

[1 00 1

]−[

0.1 10.5 0

]=

[0.9 −1−0.5 1

].


X =

[100

60

],

D = (I− A)X

=

[0.9 −1−0.5 1

] [100

60

]

=

[3010

]


I− A =

[1 00 1

]−[

0.1 10.5 0

]=

[0.9 −1−0.5 1

].


X =

[100

60

],

D = (I− A)X

=

[0.9 −1−0.5 1

] [100

60

]

=

[3010

]


I− A =

[1 00 1

]−[

0.1 10.5 0

]=

[0.9 −1−0.5 1

].


X =

[100

60

],

D = (I− A)X

=

[0.9 −1−0.5 1

] [100

60

]

=

[3010

]

Harder Question: Suppose the farmer wants to be able to sell 50tons of oats and 15 tons of manure. How much of each should heproduce?

Solution: Here D is given but X is the unknown. We can use eitherform (2) or form (3). Here I use form (2).

D =

[5015

],

D = (I− A)X[5015

]=

[0.9 −1−0.5 1

]X

This translates into the augmented matrix[0.9 −1 50−0.5 1 15

].



D =

[5015

],

D = (I− A)X[5015

]=

[0.9 −1−0.5 1

]X


].



D =

[5015

],

D = (I− A)X[5015

]=

[0.9 −1−0.5 1

]X


].



D =

[5015

],

D = (I− A)X[5015

]=

[0.9 −1−0.5 1

]X


].

[0.9 −1 50−0.5 1 15

].

Now row reduce.

R1 ↔ R2 :

[−0.5 1 15

0.9 −1 50

]

−2·→ R1 :

[1 −2 −30

0.9 −1 50

]

−0.9R1+→ R2 :

[1 −2 −300 0.8 77

]

5/4·→ R2 :

[1 −2 −300 1 385/4

]

2R2+→ R1 :

[1 0 445/20 1 385/4

]

[0.9 −1 50−0.5 1 15

].

Now row reduce.

R1 ↔ R2 :

[−0.5 1 15

0.9 −1 50

]

−2·→ R1 :

[1 −2 −30

0.9 −1 50

]

−0.9R1+→ R2 :

[1 −2 −300 0.8 77

]

5/4·→ R2 :

[1 −2 −300 1 385/4

]

2R2+→ R1 :

[1 0 445/20 1 385/4

]

[0.9 −1 50−0.5 1 15

].

Now row reduce.

R1 ↔ R2 :

[−0.5 1 15

0.9 −1 50

]

−2·→ R1 :

[1 −2 −30

0.9 −1 50

]

−0.9R1+→ R2 :

[1 −2 −300 0.8 77

]

5/4·→ R2 :

[1 −2 −300 1 385/4

]

2R2+→ R1 :

[1 0 445/20 1 385/4

]

[0.9 −1 50−0.5 1 15

].

Now row reduce.

R1 ↔ R2 :

[−0.5 1 15

0.9 −1 50

]

−2·→ R1 :

[1 −2 −30

0.9 −1 50

]

−0.9R1+→ R2 :

[1 −2 −300 0.8 77

]

5/4·→ R2 :

[1 −2 −300 1 385/4

]

2R2+→ R1 :

[1 0 445/20 1 385/4

]

[0.9 −1 50−0.5 1 15

].

Now row reduce.

R1 ↔ R2 :

[−0.5 1 15

0.9 −1 50

]

−2·→ R1 :

[1 −2 −30

0.9 −1 50

]

−0.9R1+→ R2 :

[1 −2 −300 0.8 77

]

5/4·→ R2 :

[1 −2 −300 1 385/4

]

2R2+→ R1 :

[1 0 445/20 1 385/4

]

[1 0 445/20 1 385/4

]So the solution is

# tons of oats = x1 = 445/2 = 222.5

# tons of manure = x2 = 385/4 = 96.25.

Linear Programming

Bad news: this chapter is all about story problems.

Good news: the only math you need is high school algebra.

Bad news: solving a single problem is a 3 step process that cantake 20 minutes.

Good news: all 3 steps are easy and most do well on theseproblems.

Linear Programming





Linear Programming





Linear Programming





Example: Suppose you’re taking a dozen 10 year olds out to seeIce Age IV. You figure you’ll need 44 ounces of popcorn and 56ounces of soda to see them through the movie. The snack baroffers two snack combos that look interesting: the Gulpalicious,which offers a 12 oz. popcorn with an 8 oz. soda for $8, and theSlurpalicious, which offers an 8 oz. popcorn and a 12 oz. soda for$7. How many of each combo should you buy in order to get allthe popcorn and soda you need at the lowest price?

Strategy: There’s a lot of info here.

First identify exactly what the question is asking for.


Strategy: There’s a lot of info here.

First identify exactly what the question is asking for.


Strategy: So the answer should be two numbers:

x = # Gulpaliciousesy = # Slurpaliciouses.


Strategy: So the answer should be two numbers:

x = # Gulpaliciousesy = # Slurpaliciouses.


Strategy: In all these problems there will be some quantity you’retrying to optimize.

You’re not given a definite target for this quantity,you just want to do as well as you can.

What is that quantity here?










Strategy: Here we want the lowest price possible.

This is called the objective function,because it’s actual value depends on x and y .

Here the objective is

minimize 8x + 7y .





minimize 8x + 7y .





minimize 8x + 7y .


Strategy: There are other quantities for which we do have targets.

The precise amounts of popcorn and soda depends on x and y ,but each must satisfy at least the given requirements.

# oz.s popcorn ≥ 44# oz.s soda ≥ 56










Strategy: We don’t know exactly how much of each we’ll get,but we can express the quantities in terms of x and y .

12x + 8y = # oz.s popcorn ≥ 448x + 12y = # oz.s soda ≥ 56


Strategy: Some people find it helpful in these “Materials &Products” problems to make a table.

Gulp. Slurp. Total

popcorn 12 oz. 8 oz. 44 oz.soda 8 oz. 12 oz. 56 oz.

Gulp. Slurp. Total


From this table we get the inequalities

12x + 8y ≥ 448x + 12y ≥ 56.

The direction of the inequalities is becausethe numbers on the right are minimum requirements.

In other problems they may go the other way.

Gulp. Slurp. Total



12x + 8y ≥ 448x + 12y ≥ 56.



Gulp. Slurp. Total



12x + 8y ≥ 448x + 12y ≥ 56.




Strategy:

You can’t buy a negative number of snack combos.

Therefore x and y must satisfy the natural constraints:

x ≥ 0y ≥ 0.

The 4 inequalities together are called the constraints.


Strategy:

You can’t buy a negative number of snack combos.

Therefore x and y must satisfy the natural constraints:

x ≥ 0y ≥ 0.

The 4 inequalities together are called the constraints.

Putting together everything we’ve done so far, we get the following.

x = # Gulpaliciousesy = # Slurpaliciouses

We want tominimize 8x + 7y

subject to the constraints

x ≥ 0y ≥ 0

12x + 8y ≥ 448x + 12y ≥ 56.

This is called the mathematical formulation of the problem.

Putting together everything we’ve done so far, we get the following.

x = # Gulpaliciousesy = # Slurpaliciouses

We want tominimize 8x + 7y

subject to the constraints

x ≥ 0y ≥ 0

12x + 8y ≥ 448x + 12y ≥ 56.

This is called the mathematical formulation of the problem.

Formulation of a Linear Programming Problem

1. Define the variables, including the units of measure;

2. write out the objective function in terms of the variables, andspecify whether you want to maximize or minimize it;

3. write out the system of constraints, including the naturalconstraints if applicable.

Example: (7.1 #12) A purchasing agent for a college finds itnecessary to decide how much hard and soft chalk should bepurchased each month. He knows that a typical instructor will usetwo pieces of soft chalk or one piece of hard chalk for each class.In addition, he has observed that hard chalk always amounts to atleast one-quarter of the total used. Finally, his supplier has limitedhim to a purchase of at most 60 gross (8640 pieces) of soft chalk.There are 3600 classes to be taught each month. If soft chalk is$1.50 per gross and hard chalk is $3.50 per gross, how much ofeach should be purchased to meet the needs and to keep costs to aminimum?


Variables:x = # pieces of soft chalky = # pieces of hard chalk

Other units could be used, e.g. gross.


Objective Function: Since gross = dozen dozen = 144,

minimize Cost = 1.5( x

144

)+ 3.5

( y

144

)=

(1.5

144

)x +

(3.5

144

)y .


Constraints: The natural constraints apply here:

x ≥ 0y ≥ 0.


Constraints:

classes per piece soft chalk = 1/2classes per piece hard chalk = 1

# classes with all soft chalk = 1/2x# classes with all hard chalk = 1y

total # classes with all chalk = 1/2x + y ≥ 3600


Constraints:





Constraints:





Constraints:

total chalk used =

x + yone-quarter the total chalk used = 1

4(x + y)

y ≥ 14(x + y)


Constraints:

total chalk used = x + y

one-quarter the total chalk used = 14(x + y)

y ≥ 14(x + y)


Constraints:

total chalk used = x + yone-quarter the total chalk used =

14(x + y)

y ≥ 14(x + y)


Constraints:

total chalk used = x + yone-quarter the total chalk used = 1

4(x + y)

y ≥ 14(x + y)


Constraints:


4(x + y)

y ≥

14(x + y)


Constraints:


4(x + y)

y ≥ 14(x + y)


Constraints:x ≤ 8640

Variables:x = # pieces of soft chalky = # pieces of hard chalk

Objective Function:

minimize

(1.5

144

)x +

(3.5

144

)y

Constraints:x ≥ 0y ≥ 0

1/2x + y ≥ 3600y ≥ 1

4(x + y)x ≤ 8640

Graphing InequalitiesTo “graph the inequality” x + 2y ≤ 4 meansto shade the region of the plane containing the points that satisfy it.

Here is the graph of the equation x + 2y = 4.

-4

-3

-2

-1

0

1

2

3

4

5

-4 -3 -2 -1 0 1 2 3 4 5

All the points onthe line satisfythe inequality.

Graphing InequalitiesTo “graph the inequality” x + 2y ≤ 4 meansto shade the region of the plane containing the points that satisfy it.

Here is the graph of the equation x + 2y = 4.

-4

-3

-2

-1

0

1

2

3

4

5

-4 -3 -2 -1 0 1 2 3 4 5

All the points onthe line satisfythe inequality.

What other points satisfy x + 2y ≤ 4?

Let’s try out a few.

(2, 3) : 2 + 2(3) ≤ 4 is false, so no.

(2,−3) : 2 + 2(−3) ≤ 4 is true, so yes.(−3, 2) : −3 + 2(2) ≤ 4 is true, so yes.(−1, 4) : −1 + 2(4) ≤ 4 is false, so no.

(−3,−2) : −3 + 2(−2) ≤ 4 is true, so yes.(1, 1) : 1 + 2(1) ≤ 4 is true, so yes.

-4

-3

-2

-1

0

1

2

3

4

5

-4 -3 -2 -1 0 1 2 3 4 5

There’s a pattern here.



(2, 3) : 2 + 2(3) ≤ 4 is false, so no.(2,−3) : 2 + 2(−3) ≤ 4 is true, so yes.

(−3, 2) : −3 + 2(2) ≤ 4 is true, so yes.(−1, 4) : −1 + 2(4) ≤ 4 is false, so no.


-4

-3

-2

-1

0

1

2

3

4

5

-4 -3 -2 -1 0 1 2 3 4 5




(2, 3) : 2 + 2(3) ≤ 4 is false, so no.(2,−3) : 2 + 2(−3) ≤ 4 is true, so yes.(−3, 2) : −3 + 2(2) ≤ 4 is true, so yes.

(−1, 4) : −1 + 2(4) ≤ 4 is false, so no.(−3,−2) : −3 + 2(−2) ≤ 4 is true, so yes.

(1, 1) : 1 + 2(1) ≤ 4 is true, so yes.

-4

-3

-2

-1

0

1

2

3

4

5

-4 -3 -2 -1 0 1 2 3 4 5




(2, 3) : 2 + 2(3) ≤ 4 is false, so no.(2,−3) : 2 + 2(−3) ≤ 4 is true, so yes.(−3, 2) : −3 + 2(2) ≤ 4 is true, so yes.(−1, 4) : −1 + 2(4) ≤ 4 is false, so no.


-4

-3

-2

-1

0

1

2

3

4

5

-4 -3 -2 -1 0 1 2 3 4 5





(−3,−2) : −3 + 2(−2) ≤ 4 is true, so yes.

(1, 1) : 1 + 2(1) ≤ 4 is true, so yes.

-4

-3

-2

-1

0

1

2

3

4

5

-4 -3 -2 -1 0 1 2 3 4 5






-4

-3

-2

-1

0

1

2

3

4

5

-4 -3 -2 -1 0 1 2 3 4 5






-4

-3

-2

-1

0

1

2

3

4

5

-4 -3 -2 -1 0 1 2 3 4 5


-4

-3

-2

-1

0

1

2

3

4

5

-4 -3 -2 -1 0 1 2 3 4 5

The line x + 2y = 4 divides the plane into two half-planes.

In one half-plane all the points satisfy x + 2y ≤ 4;in the other half-plane none do.

So once we have graphed the line,it only remains to determine which half-plane should be shaded.

-4

-3

-2

-1

0

1

2

3

4

5

-4 -3 -2 -1 0 1 2 3 4 5




-4

-3

-2

-1

0

1

2

3

4

5

-4 -3 -2 -1 0 1 2 3 4 5




There are two methods for doing that.

The first method is the test point method:test any point which is not on the line itself.

Any of the six points we looked at would do.

For example, (2, 3) does not satisfy the inequality;

therefore (2, 3) is in the wrong half-plane;

-4

-3

-2

-1

0

1

2

3

4

5

-4 -3 -2 -1 0 1 2 3 4 5

therefore we shade theother half-plane.






-4

-3

-2

-1

0

1

2

3

4

5

-4 -3 -2 -1 0 1 2 3 4 5







-4

-3

-2

-1

0

1

2

3

4

5

-4 -3 -2 -1 0 1 2 3 4 5


The other method for determining which half-plane to shadeis to solve the inequality for one of the variables.

x + 2y ≤ 4x ≤ 4− 2y

This means that x is less than equal to the other side:

the variable x gets smaller towards the left,

-4

-3

-2

-1

0

1

2

3

4

5

-4 -3 -2 -1 0 1 2 3 4 5

right

left

so we should shade the leftside of the line x +2y = 4.


x + 2y ≤ 4x ≤ 4− 2y



-4

-3

-2

-1

0

1

2

3

4

5

-4 -3 -2 -1 0 1 2 3 4 5

right

left



x + 2y ≤ 4x ≤ 4− 2y



-4

-3

-2

-1

0

1

2

3

4

5

-4 -3 -2 -1 0 1 2 3 4 5

right

left



x + 2y ≤ 4x ≤ 4− 2y



-4

-3

-2

-1

0

1

2

3

4

5

-4 -3 -2 -1 0 1 2 3 4 5

right

left



x + 2y ≤ 4x ≤ 4− 2y



-4

-3

-2

-1

0

1

2

3

4

5

-4 -3 -2 -1 0 1 2 3 4 5

right

left


Remember the algebra about inequalities:Multiplication by a negative reverses the inequality!

For example,−2x + 4y ≥ 6

− 2x ≥ 6− 4y

x ≤ −3 + 2y .



− 2x ≥ 6− 4y

x ≤ −3 + 2y .



− 2x ≥ 6− 4y

x ≤ −3 + 2y .



− 2x ≥ 6− 4y

x ≤ −3 + 2y .

Now let’s complicate things.

Let’s graph the system of inequalities

x + 2y ≤ 4 Iy ≤ 7− 3x II.

We’ve already done I, so let’s look at II.

The line y = 7− 3x has slope −3 and y -intercept at (0, 7).

-1

0

1

2

3

4

5

6

7

8

-1 0 1 2 3 4 5

We can use (0, 0) as a testpoint: it makes the in-equality true, so we shadethe lower left half-plane.

Now let’s complicate things.

Let’s graph the system of inequalities

x + 2y ≤ 4 Iy ≤ 7− 3x II.

We’ve already done I, so let’s look at II.

The line y = 7− 3x has slope −3 and y -intercept at (0, 7).

-1

0

1

2

3

4

5

6

7

8

-1 0 1 2 3 4 5

We can use (0, 0) as a testpoint: it makes the in-equality true, so we shadethe lower left half-plane.

Inequality I:

-1

0

1

2

3

4

5

6

7

8

-1 0 1 2 3 4 5

Inequalities I and II together:

-1

0

1

2

3

4

5

6

7

8

-1 0 1 2 3 4 5

Final Answer:

-1

0

1

2

3

4

5

6

7

8

-1 0 1 2 3 4 5

The final answer is the region that satisfies all the inequalitiesfrom the original system.

There are two methods for putting the separate inequalitiestogether.

1. Use a different color for each—then the final answer is thedarkest region;

2. use each inequality to successively eliminate part of theplane—the the final answer is what is left over.

The final answer is the region that satisfies all the inequalitiesfrom the original system.

There are two methods for putting the separate inequalitiestogether.

1. Use a different color for each—then the final answer is thedarkest region;

2. use each inequality to successively eliminate part of theplane—the the final answer is what is left over.

Here’s the last problem done using successive elimination.Inequality I (red means eliminated):

-1

0

1

2

3

4

5

6

7

8

-1 0 1 2 3 4 5

Inequality II (the final answer is white):

-1

0

1

2

3

4

5

6

7

8

-1 0 1 2 3 4 5

We’re interested in using these graphs to solve LP problemsso our systems will usually include the natural constraints.

Example:x ≥ 0y ≥ 0

x + 2y ≤ 4 Iy ≤ 7− 3x II

The addition of the natural constraints produces a slight change:

since x and y must both be nonnegative,the answer includes only points from the first quadrant.

We’re interested in using these graphs to solve LP problemsso our systems will usually include the natural constraints.

Example:x ≥ 0y ≥ 0

x + 2y ≤ 4 Iy ≤ 7− 3x II

The addition of the natural constraints produces a slight change:

since x and y must both be nonnegative,the answer includes only points from the first quadrant.

-1

0

1

2

3

4

5

6

7

8

-1 0 1 2 3 4 5

I

II

In the context of LP problems, this is called the feasible set.

-1

0

1

2

3

4

5

6

7

8

-1 0 1 2 3 4 5

I

II

In the context of LP problems, this is called the feasible set.

We need to know a little bit more about the feasible set:

We also need to know the coordinates of the corners.

-1

0

1

2

3

4

-1 0 1 2 3 4

I

II

A B

C

D

A is obviously (0, 0).B and D are also easy:they’re intercepts of thelines.

-1

0

1

2

3

4

-1 0 1 2 3 4

I

II

A B

C

D

B is the x-interceptof II: y = 7− 3x .

So

0 = 7− 3x7/3 = x ,

and B = (7/3, 0).

Similarly, D is the y -intercept of I: x + 2y = 4.

0 + 2y = 4y = 2D = (0, 2)

-1

0

1

2

3

4

-1 0 1 2 3 4

I

II

A B

C

D


So

0 = 7− 3x7/3 = x ,

and B = (7/3, 0).


0 + 2y = 4y = 2D = (0, 2)

-1

0

1

2

3

4

-1 0 1 2 3 4

I

II

A B

C

D


So

0 = 7− 3x7/3 = x ,

and B = (7/3, 0).


0 + 2y = 4y = 2D = (0, 2)

-1

0

1

2

3

4

-1 0 1 2 3 4

I

II

A B

C

D


So

0 = 7− 3x7/3 = x ,

and B = (7/3, 0).


0 + 2y = 4y = 2D = (0, 2)

-1

0

1

2

3

4

-1 0 1 2 3 4

I

II

A B

C

D

C is the only hard one.It is formed by the intersec-tion of lines I and II, so wemust solve a system of lin-ear equations to find it:

x + 2y = 4y = 7− 3x .

By substitution,

x + 2(7− 3x) = 4−5x + 14 = 4

−5x = −10x = 2y = 7− 3(2) = 1.

-1

0

1

2

3

4

-1 0 1 2 3 4

I

II

A B

C

D


x + 2y = 4y = 7− 3x .

By substitution,

x + 2(7− 3x) = 4−5x + 14 = 4

−5x = −10x = 2

y = 7− 3(2) = 1.

-1

0

1

2

3

4

-1 0 1 2 3 4

I

II

A B

C

D


x + 2y = 4y = 7− 3x .

By substitution,

x + 2(7− 3x) = 4−5x + 14 = 4

−5x = −10x = 2y = 7− 3(2) = 1.

So C = (2, 1). Now we have all the corners.

A (0, 0)B (7/3, 0)C (2, 1)D (0, 2)

This table will be crucial in actually solving LP problems.

Solving LP Problems

Now suppose our objective is to maximize the value of 3x + 2y onthe feasible set. Here is the previous feasible set together with thevalues of 3x + 2y indicated by color: green for high values and redfor low values.

Of all the points in thefeasible set, which is thegreenest?

Suppose we were lookingfor the minimum?

Solving LP Problems

Now suppose our objective is to maximize the value of 3x + 2y onthe feasible set. Here is the previous feasible set together with thevalues of 3x + 2y indicated by color: green for high values and redfor low values.


Suppose we were lookingfor the minimum?

Solving LP Problems

Now suppose the objective function is x − y .Remember: green for high values and red for low values.


Which is the reddest?

Solving LP Problems

Now suppose the objective function is x − y .Remember: green for high values and red for low values.


Which is the reddest?

It seems that both the max and min values are at corner points.

In fact this always true if the objective function is linear .

With a linear objective function,the color gradient is straight.

So to solve an LP problem,all you have to do is to compare values of the objective functionat the corner points of the feasible set:one of those points must be the answer.









Point 3x + 2y x − y

A (0, 0) 0 0B (7/3, 0) 7 7/3C (2, 1) 8 1D (0, 2) 4 −2

This problem contains an additional complication.

Example: Find the max and min of 3x + 2y subject to theconstraints

x ≥ 0 y ≥ 0I x − 2y ≥ −6 II x − 3y ≤ 2.

Graph:

-1

0

1

2

3

4

-1 0 1 2 3 4

I

IIA B

C

The corners B and C arejust intercepts.

A = (0, 0)B = (2, 0)C = (0, 3)

Note that the feasible setcontinues forever to the up-per right.



x ≥ 0 y ≥ 0I x − 2y ≥ −6 II x − 3y ≤ 2.

Graph:

-1

0

1

2

3

4

-1 0 1 2 3 4

I

IIA B

C


A = (0, 0)B = (2, 0)C = (0, 3)




x ≥ 0 y ≥ 0I x − 2y ≥ −6 II x − 3y ≤ 2.

Graph:

-1

0

1

2

3

4

-1 0 1 2 3 4

I

IIA B

C


A = (0, 0)B = (2, 0)C = (0, 3)


Here the objective function 3x + 2y is overlaid. See a problem?

There is no maximum, be-cause the objective func-tion just keeps gettingbigger toward the upperright.

There is a minimum at (0, 0).

How could we get those answers without using a computer tograph the objective function?













This feasible set is said to be unbounded.

-1

0

1

2

3

4

5

6

-1 0 1 2 3 4 5 6

I

IIA B

C

D

E

When the feasible setis unbounded we needadditional sample points,one from each unboundededge.

I’ll call D and E U-points.

They can be anywhere inside the unbounded edges.

This feasible set is said to be unbounded.

-1

0

1

2

3

4

5

6

-1 0 1 2 3 4 5 6

I

IIA B

C

D

E

When the feasible setis unbounded we needadditional sample points,one from each unboundededge.

I’ll call D and E U-points.

They can be anywhere inside the unbounded edges.

-1

0

1

2

3

4

5

6

-1 0 1 2 3 4 5 6

I

IIA B

C

D

E

Corners

A (0, 0)B (2, 0)C (0, 3)

U-points

DE

[h]

To get coordinates for D, notethat:

(a) D is on line I: x − 2y = −6;

(b) D is to the upper right of C .

We can use (b) to get one co-ordinate;then use (a) to get the other.

-1

0

1

2

3

4

5

6

-1 0 1 2 3 4 5 6

I

IIA B

C

D

E

Corners

A (0, 0)B (2, 0)C (0, 3)

U-points

DE

[h]

To get coordinates for D, notethat:


(b) D is to the upper right of C .We can use (b) to get one co-ordinate;then use (a) to get the other.

[h]

Corners

A (0, 0)B (2, 0)C (0, 3)

U-points

DE



By (b) we could use e.g. either x = 1or y = 4.

I’ll use the latter because it makesthe algebra easier.

By (a),x − 2(4) = −6

x = 2.

So for D we can use (2, 4).

[h]

Corners

A (0, 0)B (2, 0)C (0, 3)

U-points

DE





By (a),x − 2(4) = −6

x = 2.

So for D we can use (2, 4).

For E the method is the same.

[h]

Corners

A (0, 0)B (2, 0)C (0, 3)

U-points

D (2, 4)E

(a) E is on line II: x − 3y = 2;

(b) E is to the upper right of B.



By (a),x − 3(1) = 2

x = 5.

So for E we can use (5, 1).


[h]

Corners

A (0, 0)B (2, 0)C (0, 3)

U-points

D (2, 4)E





By (a),x − 3(1) = 2

x = 5.



[h]

Corners

A (0, 0)B (2, 0)C (0, 3)

U-points

D (2, 4)E





By (a),x − 3(1) = 2

x = 5.


Now complete the table.

Corners 3x + 2y

A (0, 0) 0B (2, 0) 6C (0, 3) 6

U-points

D (2, 4) 14E (5, 1) 17

The max value is at a U-point;the min value is at a corner.

The rule is:

1. if the best value is at a corner, that’s the answer;

2. if the best value is at a U-point, there is no answer.

Now complete the table.

Corners 3x + 2y

A (0, 0) 0B (2, 0) 6C (0, 3) 6

U-points

D (2, 4) 14E (5, 1) 17

The max value is at a U-point;the min value is at a corner.

The rule is:

1. if the best value is at a corner, that’s the answer;

2. if the best value is at a U-point, there is no answer.

Here’s the whole graph with all sample points and the objectivefunction overlaid.

Corners 3x + 2y

A (0, 0) 0B (2, 0) 6C (0, 3) 6

U-points

D (2, 4) 14E (5, 1) 17

The fact that the U-points take the highest values indicates thatthe values will increase forever.

Here’s the same feasible set with a different objective function,y − x .

Corners y − x

A (0, 0) 0B (2, 0) −2C (0, 3) 3

U-points

D (2, 4) 2E (5, 1) −4

Here the max is 3 but there is no min.

Here’s the same feasible set with a different objective function,y − x .

Corners y − x

A (0, 0) 0B (2, 0) −2C (0, 3) 3

U-points

D (2, 4) 2E (5, 1) −4

Here the max is 3 but there is no min.

Markov Chains

A Markov Chain is a system that can be in different states atdifferent times.

Example: Weather: Rainy, Sunny, Cloudy; times: days.

In this case the weather is observed every day, and the state of theweather is assigned to one of the three states and recorded.

As this continues a chain of successive states is recorded.

R → S → S → C → . . .

Note that in this case the weather one day might have someinfluence on the next day’s weather.

Markov Chains





R → S → S → C → . . .


Markov Chains





R → S → S → C → . . .


Markov Chains





R → S → S → C → . . .


Another way to describe a Markov Chain:it’s like a Bernoulli Process with two differences.

1. The trials can have more outcomes than just S or F .

2. The trials are not independent: one observation may affectthe next.

In a Bernoulli process the probabilities are always the same;

in a M.C. they may depend on the previous state.
















Transition DiagramsA good visual representation of a M.C. is by

a tree diagram with loops.Example: Suppose daily observations of the Dow Jones Average aremade and classified as either Rising, Falling, or Level.

As with a tree diagram, we can also indicate the conditionalprobabilities of moving along a particular edge.

For example, the .7 from R to L means:if today the DJ was rising,there is a 70% chance that tomorrow it will be level.

As with a tree diagram, we can also indicate the conditionalprobabilities of moving along a particular edge.

For example, the .7 from R to L means:if today the DJ was rising,there is a 70% chance that tomorrow it will be level.

We can trim off any edges with 0 probability.

This looks nice, but it can be hard to find the number you want.

We can trim off any edges with 0 probability.

This looks nice, but it can be hard to find the number you want.

Here’s another way to present the same information.

EndR F L

Sta

rt R 0 .3 .7F .4 .5 .1L .2 .6 .2

So for example, if today the DJ is falling, the probability that itwill rise tomorrow is 40%.

This table is called the transition matrix of the M.C.

It’s usually denoted by P.


EndR F L

Sta

rt R 0 .3 .7F .4 .5 .1L .2 .6 .2

So for example, if today the DJ is falling, the probability that itwill rise tomorrow is

40%.




EndR F L

Sta

rt R 0 .3 .7F .4 .5 .1L .2 .6 .2





EndR F L

Sta

rt R 0 .3 .7F .4 .5 .1L .2 .6 .2




P =

0 .3 .7.4 .5 .1.2 .6 .2

Things to remember:

1. rows correspond to starting states;

2. columns correspond to ending states;

3. both have the states in the same order (here R, F , L, left toright or top to bottom);

4. rows always sum to 1;

5. shorthand: probability of a transition from state i to state j ispij .

For example, the probability of a transition from F to L is p23,which is the entry in the second row and third column of P.

P =

0 .3 .7.4 .5 .1.2 .6 .2

Things to remember:







P =

0 .3 .7.4 .5 .1.2 .6 .2

Things to remember:







P =

0 .3 .7.4 .5 .1.2 .6 .2

Things to remember:







State 1 is Rise, State 2 is Fall, State 3 is Level.

P =

0 .3 .7.4 .5 .1.2 .6 .2

A lot of simple questions can be answered from P alone.

1. If tomorrow the DJ is level, what is the probability it will risethe day after?

20%

2. If the DJ fell today, what is most likely to happen tomorrow?Fall

3. If the DJ fell today, what is the probability of it falling thenext two days? 0.5 · 0.5 = 0.25

4. If the DJ fell today, what is the probability of it being levelthe next two days? 0.1 · 0.2 = 0.02

F.1−→ L

.2−→ L


P =

0 .3 .7.4 .5 .1.2 .6 .2


1. If tomorrow the DJ is level, what is the probability it will risethe day after? 20%

2. If the DJ fell today, what is most likely to happen tomorrow?

Fall



F.1−→ L

.2−→ L


P =

0 .3 .7.4 .5 .1.2 .6 .2




3. If the DJ fell today, what is the probability of it falling thenext two days?

0.5 · 0.5 = 0.25


F.1−→ L

.2−→ L


P =

0 .3 .7.4 .5 .1.2 .6 .2





4. If the DJ fell today, what is the probability of it being levelthe next two days?

0.1 · 0.2 = 0.02

F.1−→ L

.2−→ L


P =

0 .3 .7.4 .5 .1.2 .6 .2






F.1−→ L

.2−→ L


P =

0 .3 .7.4 .5 .1.2 .6 .2

Here’s a harder question.

If the DJ fell today, what is the probability it will be level two daysfrom now? In shorthand, p23(2) = ?

F?−→ ?

?−→ L

There are three cases:

F.4−→ R

.7−→ L F.5−→ F

.1−→ L F.1−→ L

.2−→ L

So the answer is 0.4 · 0.7 + 0.5 · 0.1 + 0.1 · 0.2.


P =

0 .3 .7.4 .5 .1.2 .6 .2



F?−→ ?

?−→ L


F.4−→ R

.7−→ L F.5−→ F

.1−→ L F.1−→ L

.2−→ L

So the answer is 0.4 · 0.7 + 0.5 · 0.1 + 0.1 · 0.2.


P =

0 .3 .7.4 .5 .1.2 .6 .2



F?−→ ?

?−→ L


F.4−→ R

.7−→ L F.5−→ F

.1−→ L F.1−→ L

.2−→ L

So the answer is 0.4 · 0.7 + 0.5 · 0.1 + 0.1 · 0.2.


P =

0 .3 .7.4 .5 .1.2 .6 .2



F?−→ ?

?−→ L


F.4−→ R

.7−→ L F.5−→ F

.1−→ L F.1−→ L

.2−→ L

So the answer is 0.4 · 0.7 + 0.5 · 0.1 + 0.1 · 0.2.


P =

0 .3 .7.4 .5 .1.2 .6 .2

Notice anything here?

0.4 · 0.7 + 0.5 · 0.1 + 0.1 · 0.2

It’s the product of the second row of P with the third column.

p23(2) =[.4 .5 .1

] .7.1.2


P =

0 .3 .7.4 .5 .1.2 .6 .2

Notice anything here?

0.4 · 0.7 + 0.5 · 0.1 + 0.1 · 0.2

It’s the product of the second row of P with the third column.

p23(2) =[.4 .5 .1

] .7.1.2

I’ll write Ri for the i th row of P, and Cj for the j th column of P.

We have just seen that

R2C3 = p23(2).

How about this?R3C1 = p31(2)

In general,RiCj = pij(2).

Because of this, a nice pattern arises when we compute P · P.

I’ll write Ri for the i th row of P, and Cj for the j th column of P.We have just seen that

R2C3 = p23(2).





R2C3 = p23(2).

How about this?R3C1 =

p31(2)




R2C3 = p23(2).





R2C3 = p23(2).





R2C3 = p23(2).




Using the cannon method, P · P becomes[C1 C2 C3

]

R1

R2

R3

p11(2) p12(2) p13(2)

p21(2) p22(2) p23(2)

p31(2) p32(2) p33(2)

That is, P2 is the matrix of two-step transition probabilities.This can be denoted by P(2).

P(2) = P2


]

R1

R2

R3

p11(2)

p12(2) p13(2)

p21(2) p22(2) p23(2)

p31(2) p32(2) p33(2)


P(2) = P2


]

R1

R2

R3

p11(2) p12(2)

p13(2)

p21(2) p22(2) p23(2)

p31(2) p32(2) p33(2)


P(2) = P2


]

R1

R2

R3

p11(2) p12(2) p13(2)

p21(2) p22(2) p23(2)

p31(2) p32(2) p33(2)


P(2) = P2


]

R1

R2

R3

p11(2) p12(2) p13(2)

p21(2)

p22(2) p23(2)

p31(2) p32(2) p33(2)


P(2) = P2


]

R1

R2

R3

p11(2) p12(2) p13(2)

p21(2) p22(2) p23(2)

p31(2) p32(2) p33(2)


P(2) = P2


]

R1

R2

R3

p11(2) p12(2) p13(2)

p21(2) p22(2) p23(2)

p31(2) p32(2) p33(2)


P(2) = P2

Summary

The probability of a transition from state i to state j in 2 steps is

pij(2) = the i , j entry in P2.

More generally,

The probability of a transition

from state ito state jin k steps is

pij(k) = the i , j entry in Pk .

Summary

The probability of a transition from state i to state j in 2 steps is

pij(2) = the i , j entry in P2.

More generally,




Example: Suppose IU is playing basketball against Purdue. State 1is “IU has just scored”; state 2 is “Purdue has just scored”. WhenIU scores, Purdue gets the ball, so in that case Purdue has a 70%chance of scoring next. When Purdue scores, IU gets the ball, andthen IU has an 80% of scoring next. Write out the transitionmatrix P and answer the following questions.

1. If Purdue has just scored, what is the probability that IUscores the next 2 baskets?

2. If Purdue has just scored, what is the probability that IUscores the basket after the next?

3. If Purdue has just scored, what is the probability that Purduescores the basket after the next?

4. If Purdue has just scored, what is the probability that IUscores the 4th basket from now?


Transition matrix:

IU Purdue

IU

0.3 0.7Purdue 0.8 0.2

P =

[0.3 0.70.8 0.2

]


Transition matrix:

IU Purdue

IU 0.3

0.7Purdue 0.8 0.2

P =

[0.3 0.70.8 0.2

]


Transition matrix:

IU Purdue

IU 0.3 0.7Purdue

0.8 0.2

P =

[0.3 0.70.8 0.2

]


Transition matrix:

IU Purdue

IU 0.3 0.7Purdue 0.8

0.2

P =

[0.3 0.70.8 0.2

]


Transition matrix:

IU Purdue

IU 0.3 0.7Purdue 0.8 0.2

P =

[0.3 0.70.8 0.2

]

States: IU, PU

P =

[0.3 0.70.8 0.2

]1: If Purdue has just scored, what is the probability that IU

scores the next 2 baskets?

Solution:PU

.8−→ IU.3−→ IU

p21 p11 = (.8)(.3) = .24

States: IU, PU

P =

[0.3 0.70.8 0.2


scores the next 2 baskets?

Solution:PU

.8−→ IU.3−→ IU

p21 p11 = (.8)(.3) = .24

States: IU, PU

P =

[0.3 0.70.8 0.2


scores the basket after the next?

Solution: We want p21(2), which is the 2, 1 entry in P2.

P2 =

[.65 .35.40 .60

]p21(2) = .40

Note that the rows in P2 also sum to 1.

States: IU, PU

P =

[0.3 0.70.8 0.2




P2 =

[.65 .35.40 .60

]p21(2) = .40


States: IU, PU

P =

[0.3 0.70.8 0.2




P2 =

[.65 .35.40 .60

]p21(2) = .40


States: IU, PU

P =

[0.3 0.70.8 0.2




P2 =

[.65 .35.40 .60

]p21(2) = .40


States: IU, PU

P =

[0.3 0.70.8 0.2

]

P2 =

[.65 .35.40 .60

]3: If Purdue has just scored, what is the probability that Purdue



p22(2) = .60

Alternatively, since this M.C. has only two states

p22(2) = 1− p21(2) = 1− .40 = .60,

using our previous answer.

States: IU, PU

P =

[0.3 0.70.8 0.2

]

P2 =

[.65 .35.40 .60




p22(2) = .60


p22(2) = 1− p21(2) = 1− .40 = .60,


States: IU, PU

P =

[0.3 0.70.8 0.2

]

P2 =

[.65 .35.40 .60




p22(2) = .60


p22(2) = 1− p21(2) = 1− .40 = .60,


States: IU, PU

P =

[0.3 0.70.8 0.2

]

P2 =

[.65 .35.40 .60


scores the 4th basket from now?


P4 = (P2)2

=

[.5625 .4375.5000 .5000

]p21(4) = .5

States: IU, PU

P =

[0.3 0.70.8 0.2

]

P2 =

[.65 .35.40 .60


scores the 4th basket from now?


P4 = (P2)2

=

[.5625 .4375.5000 .5000

]p21(4) = .5

State Vectors

Even when we’re not sure which state a M.C. is currently in,we can still describe the situation with a state vector.

If a M.C. has two states,and we have no idea which state it is in now,we can say that its state vector X is

X =[

0.5 0.5].

The first number is probability of being in state 1,the second is the probability of being in state 2.

Note that the numbers in the state vector are probabilities,which must sum to 1.

State Vectors



X =[

0.5 0.5].



State Vectors



X =[

0.5 0.5].



Example: Suppose a two state M.C. is twice as likely to now be instate 1 as to be in state 2. What is the state vector?

Solution:X =

[2x x

]Since the entries must sum to 1,

2x + x = 1

x = 1/3

X =[

2/31/3]


Solution:X =

[2x x


2x + x = 1

x = 1/3

X =[

2/31/3]


Solution:X =

[2x x

]

Since the entries must sum to 1,

2x + x = 1

x = 1/3

X =[

2/31/3]


Solution:X =

[2x x


2x + x = 1

x = 1/3

X =[

2/31/3]


Solution:X =

[2x x


2x + x = 1

x = 1/3

X =[

2/31/3]

Example: Suppose a 4 state M.C. now has a 20% probability ofbeing in state 1, equal probabilities of being in states 2 or 3, andthree times the probability of being in state 4 as in state 3. Whatis its current state vector?

Solution:X =

[0.2 x x 3x


0.2 + x + x + 3x = 1

5x = 0.8

x = 0.16

X =[

0.2 0.16 0.16 0.48]


Solution:X =

[0.2 x x 3x


0.2 + x + x + 3x = 1

5x = 0.8

x = 0.16

X =[

0.2 0.16 0.16 0.48]


Solution:X =

[0.2 x x 3x

]

Since the entries must sum to 1,

0.2 + x + x + 3x = 1

5x = 0.8

x = 0.16

X =[

0.2 0.16 0.16 0.48]


Solution:X =

[0.2 x x 3x


0.2 + x + x + 3x = 1

5x = 0.8

x = 0.16

X =[

0.2 0.16 0.16 0.48]


Solution:X =

[0.2 x x 3x


0.2 + x + x + 3x = 1

5x = 0.8

x = 0.16

X =[

0.2 0.16 0.16 0.48]

Any uncertainty about the current state is likely to be magnifiedfor future states, so we’ll need state vectors to describe futuresituations as well.

X0 = initial state vectorX1 = state vector after 1 stepX2 = state vector after 2 steps

...Xk = state vector after k steps

Recall our previous result.




In English:every factor of P takes you one step further into the future.

The same is true for the state vectors:

X1 = X0PX2 = X1P = X0P2

Xk+1 = XkP = X0Pk .

Example: Suppose a two state M.C. with transition matrix

P =

[0.4 0.60.2 0.8

]is equally likely initially to be in either state. What is theprobability it is in state 2 after two steps?

Solution:

X2 = X0P2 =[

0.5 0.5] [ 0.4 0.6

0.2 0.8

] [0.4 0.60.2 0.8

]

=[

0.3 0.7] [ 0.4 0.6

0.2 0.8

]=

[0.26 0.74

]


X1 = X0PX2 = X1P = X0P2

Xk+1 = XkP = X0Pk .


P =

[0.4 0.60.2 0.8


Solution:

X2 = X0P2 =[

0.5 0.5] [ 0.4 0.6

0.2 0.8

] [0.4 0.60.2 0.8

]

=[

0.3 0.7] [ 0.4 0.6

0.2 0.8

]=

[0.26 0.74

]


X1 = X0PX2 = X1P = X0P2

Xk+1 = XkP = X0Pk .


P =

[0.4 0.60.2 0.8


Solution:

X2 = X0P2 =[

0.5 0.5] [ 0.4 0.6

0.2 0.8

] [0.4 0.60.2 0.8

]

=

[0.3 0.7

] [ 0.4 0.60.2 0.8

]=

[0.26 0.74

]


X1 = X0PX2 = X1P = X0P2

Xk+1 = XkP = X0Pk .


P =

[0.4 0.60.2 0.8


Solution:

X2 = X0P2 =[

0.5 0.5] [ 0.4 0.6

0.2 0.8

] [0.4 0.60.2 0.8

]

=[

0.3 0.7] [ 0.4 0.6

0.2 0.8

]=

[0.26 0.74

]

Let’s continue this example.

P =

[0.4 0.60.2 0.8

]X0 =

[0.5 0.5

]X1 =

[0.3 0.7

]X2 =

[0.26 0.74

]X3 =

[0.252 0.748

]X32 =

[.2500000000000006 .7500000000000018

]X∞ =

[.25 .75

]


P =

[0.4 0.60.2 0.8

]X0 =

[0.5 0.5

]X1 =

[0.3 0.7

]X2 =

[0.26 0.74

]X3 =

[0.252 0.748

]X32 =

[.2500000000000006 .7500000000000018

]X∞ =

[.25 .75

]


P =

[0.4 0.60.2 0.8

]X0 =

[0.5 0.5

]X1 =

[0.3 0.7

]X2 =

[0.26 0.74

]X3 =

[0.252 0.748

]X32 =

[.2500000000000006 .7500000000000018

]X∞ =

[.25 .75

]

X∞ =[.25 .75

]What does X∞ mean?

You could say it contains the probabilities of being in each stateover the last ∞− 32 steps of the process.

We call X∞ the vector of stable probabilitiesand we denote it by W ,because this is Finite Math after all.

You can think of W ascontaining the long-term probabilities of each state,the fraction of time the M.C. spends in each state in the long run.

X∞ =[.25 .75





X∞ =[.25 .75





X∞ =[.25 .75





How can we calculate W without using a computer?

Notice that W has two important characteristics:

1. it is a state vector, so its entries sum to 1;

2. it is stable, meaning WP = W .

The last property is because WP = X∞P represents the statevector after ∞+ 1 steps, which should be the same as X∞.

Each of these properties can be expressed in matrix equation form



















1. W is a state vector, so its entries sum to 1.

Think of W as

W =[

w1 w2 w3 . . . wn

].

Then property 1 is equivalent to

w1 + w2 + w3 + · · ·+ wn = 1

[1 1 1 . . . 1

]

w1

w2

w3...

wn

= 1


Think of W as

W =[

w1 w2 w3 . . . wn

].


w1 + w2 + w3 + · · ·+ wn = 1

[1 1 1 . . . 1

]

w1

w2

w3...

wn

= 1


Think of W as

W =[

w1 w2 w3 . . . wn

].


w1 + w2 + w3 + · · ·+ wn = 1

[1 1 1 . . . 1

]

w1

w2

w3...

wn

= 1

2. W is stable, meaning WP = W .

We’ll just do a little elementary algebra:

WP = W

WP −W = 0

(where 0 is the zero matrix)

W (P − I) = 0



WP = W

WP −W = 0


W (

P − I) = 0



WP = W

WP −W = 0


W (P − I) = 0

W (P − I) = 0

Now there’s a slight problem:

In chapters 5 and 6 we learned about matrix methods for solvingsystems of equations.

In all those methods we were solving for a column vector,on the right.

For example, AX = B.

But here we are solving for W ,which is a row vector on the left.

Is everything we learned previously useless here?

W (P − I) = 0

Now there’s a slight problem:

In chapters 5 and 6 we learned about matrix methods for solvingsystems of equations.

In all those methods we were solving for a column vector,on the right.

For example, AX = B.

But here we are solving for W ,which is a row vector on the left.

Is everything we learned previously useless here?

There’s simple trick to save us:

turn everything sideways.

This is called transposing a matrix.

Example:

A =

[1 2 34 5 6

]

At =

1 42 53 6




Example:

A =

[1 2 34 5 6

]

At =

1 42 53 6




Example:

A =

[1 2 34 5 6

]

At =

1 42 53 6

When you transpose matrices, you also have to reverse the order inwhich they are multiplied:

You’re switching the positions of the lower left and upper rightcannons. 7

89

[

1 2 34 5 6

]becomes 1 4

2 53 6

[

7 8 9]

So we’ll just transpose the equation W (P − I) = 0:

(P − I)tW t = 0

Notice that the other equation can written in a similar way.

[1 1 1 . . . 1

]

w1

w2

w3...

wn

= 1

[1 1 1 . . . 1

]W t = 1

So we’ll just transpose the equation W (P − I) = 0:

(P − I)tW t = 0

Notice that the other equation can written in a similar way.

[1 1 1 . . . 1

]

w1

w2

w3...

wn

= 1

[1 1 1 . . . 1

]W t = 1

[1 1 1 . . . 1

]W t = 1

(P − I)tW t = 0

One great feature of matrix equations is that they’re stackable,like modular furniture.

1 1 1 . . . 1

(P − I)t

W t =

100...0

1 1 1 . . . 1

(P − I)t

W t =

100...0

Now we can apply the method of chapter 5:convert to augmented matrix form

1 1 1 . . . 1 10

(P − I)t 00

and then row reduce.

Method to Calculate W

To the find the vector W of stable probabilities for a M.C. withtranstion matrix P, form the augmented matrix

1 1 1 . . . 1 10

(P − I)t 00

and row reduce.

At the end you will have W t,i.e. W written as a column vector.

Example: Find W for the M.C. in the previous example, with

P =

[0.4 0.60.2 0.8

]

Solution:

P − I =

[0.4 0.60.2 0.8

]−[

1 00 1

]

=

[−0.6 0.6

0.2 −0.2

]

(P − I)t =

[−0.6 0.2

0.6 −0.2

]


P =

[0.4 0.60.2 0.8

]

Solution:

P − I =

[0.4 0.60.2 0.8

]−[

1 00 1

]

=

[−0.6 0.6

0.2 −0.2

]

(P − I)t =

[−0.6 0.2

0.6 −0.2

]


P =

[0.4 0.60.2 0.8

]

Solution:

P − I =

[0.4 0.60.2 0.8

]−[

1 00 1

]

=

[−0.6 0.6

0.2 −0.2

]

(P − I)t =

[−0.6 0.2

0.6 −0.2

]

(P − I)t =

[−0.6 0.2

0.6 −0.2

]Therefore the augmented matrix we need is 1 1 1

−0.6 0.2 00.6 −0.2 0

.

Now we row reduce.

0.6R1+→ R2 :

1 1 10 0.8 0.6

0.6 −0.2 0

− 0.6R1+→ R3 :

1 1 10 0.8 0.60 −0.8 −0.6

(P − I)t =

[−0.6 0.2

0.6 −0.2


−0.6 0.2 00.6 −0.2 0

.Now we row reduce.

0.6R1+→ R2 :

1 1 10 0.8 0.6

0.6 −0.2 0

− 0.6R1+→ R3 :

1 1 10 0.8 0.60 −0.8 −0.6

(P − I)t =

[−0.6 0.2

0.6 −0.2


−0.6 0.2 00.6 −0.2 0

.Now we row reduce.

0.6R1+→ R2 :

1 1 10 0.8 0.6

0.6 −0.2 0

− 0.6R1+→ R3 :

1 1 10 0.8 0.60 −0.8 −0.6

1 1 10 0.8 0.60 −0.8 −0.6

Notice that row 3 is an exact multiple of row 2.Therefore we can cancel one of them.[

1 1 10 0.8 0.6

]

5/4·→ R2 :

[1 1 10 1 0.75

]

− 1R2+→ R1 :

[1 0 0.250 1 0.75

]So that

W =[

0.25 0.75].

1 1 10 0.8 0.60 −0.8 −0.6


1 1 10 0.8 0.6

]

5/4·→ R2 :

[1 1 10 1 0.75

]

− 1R2+→ R1 :

[1 0 0.250 1 0.75

]So that

W =[

0.25 0.75].

1 1 10 0.8 0.60 −0.8 −0.6


1 1 10 0.8 0.6

]

5/4·→ R2 :

[1 1 10 1 0.75

]

− 1R2+→ R1 :

[1 0 0.250 1 0.75

]

So thatW =

[0.25 0.75

].

1 1 10 0.8 0.60 −0.8 −0.6


1 1 10 0.8 0.6

]

5/4·→ R2 :

[1 1 10 1 0.75

]

− 1R2+→ R1 :

[1 0 0.250 1 0.75

]So that

W =[

0.25 0.75].

Mysteriously, some students do not enjoy calculating stable vectors.

If you are repulsed by my method,have a look at the method shown in the book (section 8.3).

If you’re good at algebra you can solve for W byconverting WP = W into ordinary equations and throwing in

w1 + w2 + w3 + · · ·+ wn = 1

to get a system of equations you can then solve to get the the ws.

Regular Markov ChainsI’ve misled you.There are M.C.s for which X∞ does not exist.

If this M.C. starts in state 1, then

X0 =[

1 0]

X1 =[

0 1]

X2 =[

1 0]

Xeven =[

1 0]

Xodd =[

0 1].



X0 =[

1 0]

X1 =[

0 1]

X2 =[

1 0]

Xeven =[

1 0]

Xodd =[

0 1].



X0 =[

1 0]

X1 =[

0 1]

X2 =[

1 0]

Xeven =[

1 0]

Xodd =[

0 1].



X0 =[

1 0]

X1 =[

0 1]

X2 =[

1 0]

Xeven =[

1 0]

Xodd =[

0 1].

Similarly with this M.C.

The problem is:not enough connections between the states.

Similarly with this M.C.

The problem is:not enough connections between the states.

The most connections possible: every state is connected to everyother state.

You can recognize that immediately when you look at thetransition matrix.

P =

.2 .3 .5.1 .2 .7.4 .5 .1

P =

.2 0 .8.1 .9 0.4 .5 .1

On the left: fully connected.On the right: some connections missing.



P =

.2 .3 .5.1 .2 .7.4 .5 .1

P =

.2 0 .8.1 .9 0.4 .5 .1




P =

.2 .3 .5.1 .2 .7.4 .5 .1

P =

.2 0 .8.1 .9 0.4 .5 .1




P =

.2 .3 .5.1 .2 .7.4 .5 .1

P =

.2 0 .8.1 .9 0.4 .5 .1


Do we need every connection to be presentin order for X∞ to exist?

No, but they must eventually be present.

In other words,if we wait long enough all transitions become possible.

A regular Markov chain is for which there is some number k suchthat P(k) has all positive entries.









Example: The M.C. with transition matrix

P =

.2 0 .8.1 .9 0.4 .5 .1

has some connections missing, but what if we are patient?

P(2) = P2 =

.36 .40 .24.11 .81 .08.17 .50 .33

All two step transitions are possible.This M.C. is regular.


P =

.2 0 .8.1 .9 0.4 .5 .1


P(2) = P2 =

.36 .40 .24.11 .81 .08.17 .50 .33



P =

.2 0 .8.1 .9 0.4 .5 .1


P(2) = P2 =

.36 .40 .24.11 .81 .08.17 .50 .33


Determining whether a M.C. is regular

Suppose a Markov chain has transition matrix P.

1. Raise P to higher and higher powers.

2. If the zeros eventually disappear then the M.C. is regular.

3. If notice a repeating cycle containing zeros then it is notregular.

Two things make this easy :

1. don’t do any arithmetic: just note 0s and +s;

2. use repeated squarings to look farther faster.

Determining whether a M.C. is regular

Suppose a Markov chain has transition matrix P.

1. Raise P to higher and higher powers.

2. If the zeros eventually disappear then the M.C. is regular.

3. If notice a repeating cycle containing zeros then it is notregular.

Two things make this easy :

1. don’t do any arithmetic: just note 0s and +s;

2. use repeated squarings to look farther faster.

Example: Is the M.C. with transition matrix

P =

.2 0 .80 .9 0.4 0 .1

regular or not?

Solution: First: we’ll ignore the numbers and just keep track of 0sand +s.

P =

+ 0 +0 + 0+ 0 +


P =

.2 0 .80 .9 0.4 0 .1

regular or not?


P =

+ 0 +0 + 0+ 0 +

P =

0 + 0+ 0 +0 + 0

Now we’ll compute P2 = P(2). Keep in mind the following simple

rules.+ + = + 0 + 0 = 00 (∗) = 0 + + (∗) = +

P2 =

+ 0 +0 + 0+ 0 +

That’s an improvement: 4 zeros instead of 5.

What’s next?

P4 = (P2)2 =

+ 0 +0 + 0+ 0 +

P =

0 + 0+ 0 +0 + 0


rules.+ + = + 0 + 0 = 00 (∗) = 0 + + (∗) = +

P2 =

+ 0 +0 + 0+ 0 +


What’s next?

P4 = (P2)2 =

+ 0 +0 + 0+ 0 +

P =

0 + 0+ 0 +0 + 0


rules.+ + = + 0 + 0 = 00 (∗) = 0 + + (∗) = +

P2 =

+ 0 +0 + 0+ 0 +


What’s next?

P

4 = (P2)2 =

+ 0 +0 + 0+ 0 +

P =

0 + 0+ 0 +0 + 0


rules.+ + = + 0 + 0 = 00 (∗) = 0 + + (∗) = +

P2 =

+ 0 +0 + 0+ 0 +


What’s next?

P4 = (P2)2 =

+ 0 +0 + 0+ 0 +

P

0 + 0+ 0 +0 + 0

P2 =

+ 0 +0 + 0+ 0 +

P4 =

+ 0 +0 + 0+ 0 +

Now we have stopped making progress:

we’ve entered a cycle.

Since there are still zeros, we’ll never get rid of them.

The M.C. is not regular.

P

0 + 0+ 0 +0 + 0

P2 =

+ 0 +0 + 0+ 0 +

P4 =

+ 0 +0 + 0+ 0 +

Now we have stopped making progress:

we’ve entered a cycle.

Since there are still zeros, we’ll never get rid of them.

The M.C. is not regular.


P =

1 0 01 0 00 .2 .8

regular or not?


P =

+ 0 0+ 0 00 + +


P =

1 0 01 0 00 .2 .8

regular or not?


P =

+ 0 0+ 0 00 + +

P =

+ 0 0+ 0 00 + +

Now start repeatedly squaring.

P2 =

+ 0 0+ 0 0+ + +

P4 =

+ 0 0+ 0 0+ + +

Not regular.

P =

+ 0 0+ 0 00 + +


P2 =

+ 0 0+ 0 0+ + +

P4 =

+ 0 0+ 0 0+ + +

Not regular.

P =

+ 0 0+ 0 00 + +


P2 =

+ 0 0+ 0 0+ + +

P4 =

+ 0 0+ 0 0+ + +

Not regular.


P =

0 0 10 0 1.8 .2 0

regular or not?


P =

0 0 +0 0 ++ + 0


P =

0 0 10 0 1.8 .2 0

regular or not?


P =

0 0 +0 0 ++ + 0

P =

0 0 +0 0 ++ + 0


P2 =

+ + 0+ + 00 0 +

P4 =

+ + 0+ + 00 0 +

Not regular.

P =

0 0 +0 0 ++ + 0


P2 =

+ + 0+ + 00 0 +

P4 =

+ + 0+ + 00 0 +

Not regular.

P =

0 0 +0 0 ++ + 0


P2 =

+ + 0+ + 00 0 +

P4 =

+ + 0+ + 00 0 +

Not regular.


P =

0 1 00 0 1.8 .2 0

regular or not?


P =

0 + 00 0 ++ + 0


P =

0 1 00 0 1.8 .2 0

regular or not?


P =

0 + 00 0 ++ + 0

P =

0 + 00 0 ++ + 0


P2 =

0 0 ++ + 00 + +

P4 =

0 + ++ + ++ + +

P8 =

+ + ++ + ++ + +

Regular.

P =

0 + 00 0 ++ + 0


P2 =

0 0 ++ + 00 + +

P4 =

0 + ++ + ++ + +

P8 =

+ + ++ + ++ + +

Regular.

P =

0 + 00 0 ++ + 0


P2 =

0 0 ++ + 00 + +

P4 =

0 + ++ + ++ + +

P8 =

+ + ++ + ++ + +

Regular.

P =

0 + 00 0 ++ + 0


P2 =

0 0 ++ + 00 + +

P4 =

0 + ++ + ++ + +

P8 =

+ + ++ + ++ + +

Regular.

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