COMM 161Intro to Mathematical Analysis for

Management

Agenda• Section 1: Matrices/Linear Equations (15 min)• Section 2: Matrix Algebra/Applications (15

min)• Section 3: Input Output Models (15 min)• Section 4: Probability and Markov

Systems(20 min)• Section 5: Rates of Change & Derivatives (20

min)• Section 6: Techniques of Differentiation (20

min)• Section 7: Applications of Derivatives (20

min)• Section 8: Functions of Several Variables (20

min)• Section 9: 3D Optimization (20 min)

COMM 161Section 1: Systems of Linear Equations

and Matrices

Section 1: Systems of Linear Equations and

Matrices• Three Ways of Solving a System of

Equations:– Graphically– Algebraically via Substitution– Algebraically via Elimination

• We will use the theory of matrices to solve systems of equations

Section 1: Systems of Linear Equations and

Matrices• A Matrix is a rectangular array of

numbers– System:

• 3x-y = 2• x+y = 10

3 1 21 1 10

Coefficients Constants

Section 1: Systems of Linear Equations and

Matrices• Different Row Operations for Solving a

System:– Interchange any two equations

Section 1: Systems of Linear Equations and

Matrices• Different Row Operations for Solving a

System:– Multiply a row by any non-zero constant

Section 1: Systems of Linear Equations and

Matrices• Different Row Operations for Solving a

System:– Replace an equation by the sum of that

equation and a constant multiple of any other equation

Section 1: Systems of Linear Equations and

Matrices• Reduced Row Echelon Form

– The first non-zero entry in each row is one– The columns of the pivots are clear

• Pivoting – Using a coefficient to clear a column– There is only one ‘one’ per row

1 0 0 90 0 1 40 1 0 2

1 0 0 80 1 0 40 0 0 0

Section 1: Systems of Linear Equations and

Matrices• Guass-Jordan Elimination Method

– Write the augmented matrix – Interchange rows, if necessary, to obtain a

nonzero first entry. Pivot on this entry.– Interchange rows, if necessary, to obtain a

nonzero second entry in the second row. Pivot on this entry.

– Continue until in reduced row form.

Section 1: Systems of Linear Equations and

Matrices• Possible Outcomes for a system:

– Unique Solution

– No Solution

– Infinitely Many Solutions

1 0 0 90 0 1 40 1 0 2

1 1 2 30 1 1 10 0 0 1

1 0 0 80 1 0 40 0 0 0

Reduced Row Form

Final Row doesn’t make sense, 0+0+0 does not equal 1

The last row is true in all cases, therefore there is an infinite amount of solutions

Section 1: Systems of Linear Equations and

Matrices• Applying this information:

– The lazy scholar makes its cheesecakes using three different types of cream cheese: A, B, C. The chocolate cake uses 1 pound of A and 2 pounds of B. The vanilla cake uses 2 pounds of A and 3 pounds of C. The strawberry cake uses 3 pounds of B and 4 pounds of C. Lazy scholar has a massive abundance of cream cheese and wants to use 34 pounds of A, 38 pounds of B and 60 pounds of C, how many cheesecakes can be made?

Section 1: Systems of Linear Equations and

MatricesChocolate Vanilla Strawberry TotalA 1 (x) 2 (y) 0 (z) 34B 2 (x) 0 (y) 3 (z) 38C 0 (x) 3 (y) 4 (z) 60

Section 1: Systems of Linear Equations and

Matrices

Lazy Scholar can make 10 chocolate cheesecakes, 12 vanilla cheesecakes and 6 strawberry cheesecakes.

COMM 161Section 2: Matrix Algebra and

Applications

Section 2: Matrix Algebra and Applications

• Matrix Dimensions– A matrix is an ordered rectangular array of

real numbers with m rows and n columns.

• Matrix Entries– Denoted by aij

• ith row, and jth column • a14 = 0 or a31 = -2

2 4 1 06 7 8 32 0 1 9

A

So, this is an example of a 3x4 matrix or 3 rows by 4 columns

Section 2: Matrix Algebra and Applications

• Addition and Subtraction of Matrices– Matrices must be the same size– The sum of A+B is found by adding

corresponding entries in the two matrices– The difference of A-B is found by

subtracting the corresponding entries in B and A

813732420

A

192763031

B

Section 2: Matrix Algebra and Applications

0 1 2 3 4 02 3 3 6 7 73 2 1 9 8 1

A B

1 5 45 9 14

5 10 7

0 1 2 3 4 02 ( 3) 3 6 7 73 2 1 9 8 ( 1)

A B

1 1 41 3 01 8 9

Section 2: Matrix Algebra and Applications

• Scalar Multiplication– Any matrix A can be multiplied by c, which

is a real number

813732420

A

5(0) 5(2) 5(4)5 5( 2) 5(3) 5(7)

5(3) 5(1) 5(8)A

0 10 2010 15 35

15 5 40

Section 2: Matrix Algebra and Applications

• Transpose of a Matrix– Basically, you flip the rows with the

columns and the columns with the rows.

963852741

TA

987654321

A

Section 2: Matrix Algebra and Applications

• Matrix Multiplication – Size of the Product– If A is a matrix of size m x n and B is a

matrix of size n x p then the product of AB will be a matrix of size m x p

– I.E. Matrix A has a size of 3x2– Matrix B has a size of 2x5

• The product of Matrix AB will be 3x5– Note, The column size of A must equal the

row size of B– Matrix A Size 3x4 and Matrix B 3x4

• The Product cannot be computed

3231

2221

1211

cccccc

C

3231

2221

1211

bbbbbb

B

333231

232221

131211

aaaaaaaaa

A

Section 2: Matrix Algebra and Applications

For matrices A, B, and C, let AB = C.

Then

And so on…

31132112111111 bababac

32132212121112 bababac

31232122112121 bababac

Section 2: Matrix Algebra and Applications

• The identity matrix of size n is given by:

1000001000001000001000001

5I

Diagonal of 1’s, zeros every other entry

Section 2: Matrix Algebra and Applications

For matrices A, B, and C, the following hold:

I is the identity matrix, O is the zero matrix.

Associative law: (AB)C = A(BC)

(A + B) + C = A + (B + C)Distributive law: A(B + C) = AB + AC

(A+B)C = AC + BC Zero matrix: OA = AO = OIdentity matrix: IA = AI = A

Section 2: Matrix Algebra and Applications

• Finding the inverse of a matrix– Given the square matrix A.– Adjoin the identity matrix I (of the same size) to

form the augmented matrix: [A | I]– Use row operations to reduce the matrix to the

form: [I | B] (if possible)– Matrix B is the inverse of A

Ex. Find the inverse of A

112331

111A

Section 2: Matrix Algebra and Applications

1 1 1 1 0 00 2 2 1 1 00 3 1 2 0 1

2 1R R

( 2)3 1R R

1 1 1 1 0 01 3 3 0 1 0

2 1 1 0 0 1

1 1 1 1 0 00 1 11/ 2 1/ 2 00 3 1 2 0 1

Step 1

Step 3

Step 2

122R

Section 2: Matrix Algebra and Applications

( 3)3 2R R

132R

1 1 1 1 0 00 1 1 1/ 2 1/ 2 00 0 2 7 / 2 3/ 2 1

1 1 1 1 0 00 1 1 1/ 2 1/ 2 00 0 1 7 / 4 3/ 4 1/ 2

1 1 0 11/ 4 3/ 4 1/ 20 1 0 5 / 4 1/ 4 1/ 20 0 1 7 / 4 3/ 4 1/ 2

( 1)1 3R R

2 3R R

Step 5

Step 4

Step 6

Section 2: Matrix Algebra and Applications

1 2R R1 0 0 3/ 2 1/ 2 00 1 0 5 / 4 1/ 4 1/ 20 0 1 7 / 4 3/ 4 1/ 2

13/ 2 1/ 2 05 / 4 1/ 4 1/ 27 / 4 3/ 4 1/ 2

A

Step 7

Section 2: Matrix Algebra and Applications

D is called a determinant. If D = 0 then the matrix is singular (has no inverse).

Ex.

acbd

DAA

dcba

A

1 and exists then

0,bc-adD If

11

2143

A

Section 2: Matrix Algebra and Applications

3(2) 4(1) 2 D

1 2 411 32

A

1 21/ 2 3/ 2

2143

A

Section 2: Matrix Algebra and Applications

• If AX = B is a linear system of equations (number of equations = number of variables) and A-1 exists, then X = A-1B is the unique solution of the system.

42643

yxyx

AX = B

46

2143yx

AX = BA-1AX = A-1B(A-1A=I) IX = A-1BFinally, X = A-1B

Section 2: Matrix Algebra and Applications

3 41 2

A

3 4 61 2 4

xy

1 1 2

1/ 2 3/ 2A

1 2 3 4 1 2 61/ 2 3/ 2 1 2 1/ 2 3/ 2 4

xy

23

xy

So (–2 , 3) is the solution

Multiply by the inverse

COMM 161Section 3: Input-Output Models

Section 3: Input-Output Models

In an input – output model, the matrix equation giving the net output of goods and services needed to satisfy external demand is:

X AX D

The solution to the equation is:

Which gives the amount of goods and services that must be produced to satisfy external demand.

1X I A D

Technology Matrix

Production vector

External Demand

Section 3: Input-Output Models

Ex. A simple economy consists of manufacturing (M) and service (S) and has the following input-output model (units are in millions of dollars) : M

SM S

0.4 0.10.25 0.5

a.Determine the amount of manufactured products consumed for $300 million worth of service.

(0.1)(300) = 30 So $30 million manufactured products are

consumed

Section 3: Input-Output Models

b. Find the gross output of goods needed to satisfy a consumer demand for $90 million worth of manufactured products and $60 million worth of service.

1X I A D

.6 .1.25 .5

I A

1 1.82 .36

( ).91 2.18

I A

1.82 .36 90.91 2.18 60

185.4212.7

$185.4 million worth of manufactured products and $212.7 million worth of service.

Section 3: Input-Output Models

• If A is the technology matrix, then the ijth entry of

(I – A) –1 is the change in the number of units Sector i must produce in order to meet a one-unit increase in external demand for Sector j products.

Change in production = (I – A) –1D+

Where D+ is the change in external demand.

Post-midterm

• I apologize in advance for the boring slides

• Tips when practicing:– Communication and thinking– Applications!– General optimization

• Buckle down, let’s start

Probability and Markov Systems

Some Key Terms

- Sample space -Disjoint- Event (Mutually exclusive)- Complement -Probability

distribution- Union -Relative frequency- Intersection -Trials/sample size

Probability and Markov Systems

Theoretical Probability

P(E) = Number of favorable outcomes = n(E) Total number of outcomes n(S)

Probability Rules• All probabilities are between 0 and 1 inclusive

0 <= P(E) <= 1 • The sum of all the probabilities in the sample space

is 1• The probability of an event which cannot occur is 0.• The probability of any event which is not in the

sample space is zero. • The probability of an event which must occur is 1.• The probability of the sample space is 1. • The probability of an event not occurring is one

minus the probability of it occurring.P(E') = 1 - P(E)

Addition rule-General-Mutually exclusive events

Conditional probability-Independent events-Dependent events

Markov Systems-States-Time-step-State transition diagram-Transition matrix-State transition probabilities-Initial distribution vector-Steady-state vector

Average Rate of Change

Introduction to Derivatives-Average rate of change-Instantaneous rate of change-Differentiation-Tangent line

Techniques of Differentiation

-Product rule-Quotient rule-Chain rule-Logarithmic and exponential functions -Implicit differentiation

Quick Tips for Differentiation

• First derivative– Rate of change, velocity, marginal revenue

• Second derivative– Acceleration, change in the ___ rate

• Revenues = Price x Quantity• Profit = Revenues – Cost• Formula sheet! Use chapter summaries

Applications of the Derivative

• Relative extrema– Maxima and minima

• Absolute extrema– Maximum and minimum

• Critical points– Stationary (maximum or minimum)– Endpoints– Singular

• Fence, volume

• Concavity– Up or down– Point of inflection

• Point of diminishing return• Optimization question

1) Find the first derivative2) Critical points3) First derivative test4) Solve for optimized value

• Elasticity

• Inventory Management

DON’T PANIC

Functions of Several Variables

• Distance formula• Graph of a function of several variables

– Paraboloid– Level curves– Saddle point

• Partial derivatives– Marginal costs– Mixed partial derivative

• Optimization of function of several variables1) Solve for partial derivatives2) Find critical points3) Second derivative test (Hessian)

• Lagrange multipliers• Regression

– Residuals– Best-fit

Exam Tips• Use your class notes and work through

problems from class• Go to your Professor’s office hours • Understand theory but practise,

practise, practise!• Before the exam- sleep, whole grains,

fruit• Dress in layers, bring a watch• Questions?

Thanks for coming today!