Applications to Markov Chains

The Markov chains described in this section are used as mathematical models of a wide variety of situations in biology, business, chemistry, engineering, physics, and elsewhere. In each case, the model is used to describe an experiment or measurement that is preformed many times in the same way, where the outcome of the trial of the experiment will be one of several specified possible outcomes, and where the outcome of one trial depends only on the immediately preceding trial.

A vector with nonnegative entries that add up to 1 is called a probability vector. A stochastic matrix is a square matrix whose columns are probability vectors. A Markov chain is a sequence of probability vectors x0, x1, x2,.... together with a stochastic matrix P, such that x1=Px0, x2=Px1....Then the Markov chain described by the first order difference equation xi+1=Pxk for k=0,1,2.....

When a Markov chain of vectors in R^n describes a system or a sequence of experiments, the entries in xk, the probabilities that the system is in each of n possible states, or the probabilities that the outcome of the experiment is one of n possible outcomes. For this reason, xk is often called a state vector.

Example One: Examine a model for population movement between a city and its suburbs. The annual migration between these two parts of the metropolitan region was governed by the migration matrix M.

From: 

City Suburbs      

M = [.95        .03 ]                

[.05        .97 ]       

That is each year 5% of the population moved to the suburbs, and 3% of the suburban population moves to the city. The columns of M are probability vectors, so M is a stochastic matrix. Suppose the 2014 population of the region is 600,000 in the city and 400,000 in the suburbs. Then the initial distribution population in the region is given by x0. What is the distribution population in 2015? in 2016? 

After one year, the population vector 

[600,000

400,000]

changed to 

[.95    .03  x  600,000] = [582,000]

[.05    .97  x  400,000] = [418,000]

If we divide by both sides of the this equation by the total population of 1 million, and use the fact that kMx= M(kx), we find that 

[.95  .03  x .600 = [.582]

[.05  .97  x .400 = [.418]

The vector x1 = [.582

                           .418]

gives the population distribution in 2015. that is 58.2% of the region lived in the city and 41.8% lived in the suburbs. Similarly the population distribution in 2016 is described by a vector x2, where

x2 = Mx1 = [.95  .03  x  .582] = [.565]

                     [.05  .97  x .418]   =[.435]

gives the population distribution in 2016, that is 56.5% of the region lived in the city and 43.5% lived in the suburbs. 

If P is a stochastic matrix, then the steady-state vector (or equilibrium vector) for P is a probability vector q such that pq=q it can be shown that every stochastic matrix has a steady-state vector. 

Example two: The probability vector q = [.375

                                                               .625] is a steady state vector for the population migration matrix M, because 

Mq = [.95  .03  x  .375] = [.35625 + .01875] = [.375] q

          [.05  .97  x  .625] = [.01875 + .60625] = [.625] q

If the total populaiton in example one, is 1 million, then q in this example would correspond to having 375,000 persons in the city and 625,000 in the suburbs. At the end of one year, the migration out of the city would be (.05)(375,000) = 18,750 persons, and the migration into the city from the suburbs would be (0.3)(625,000) = 18,750 persons. As a result, the populaiton in the city would remain the same. Similarly, the suburban populaiton would be stable. 

Theorem 18: If P is an mxn regular stochastic matrix, then P has a unique steady-state vector q. Further, xo is any initial state and xk+1 = Pxk, for k = 0,1,2,....then the Markov chain {xk} converges to q as k goes to infinity. 

Examine Problem: On any given day a student is either healthy or ill. Of the students who are healthy today, 95% will be healthy tomorrow. Of the students who are ill today, 55% will still be ill tomorrow. 

a.) What is the stochastic matrix for this situation?

b.) Suppose 20% of the students are ill on Monday. What faction or percentage of the students are likely to be ill on Tuesday or Wednesday? 

c.) If a student is well today, what is the probability that he or she will be well two days from now? 

Solving matrixes using cofactor expansion:

Co-factor Expansion: 

Formula: Given A= [aij], the (i, j) cofactor of A is the number Cij given by Cij=(-1)^(i+j) det Aij

then det A= a11C11+a12C12+......+a1nC1n. This formula is called a cofactor expansion across the first row of A. 

Theorem One: The determinant of an m x n matrix A can by computed by a cofactor expansion across any row or down any column. The expansion across the ith row using the cofactors in the above equation is det A = a11C11+a12C12+....+a1nC1n. The cofactor expansion down the jth column is det A= a1jC1j+a2jC2j+.....+anjCnj. 

The plus or minus sign is the (i,j)-cofactor depends on the position of a1j in the matrix, regardless of the sign of a1j itself. The factor (-1)^(i+j) determines the following checkerboard pattern of signs: 

Example One: Use a cofactor expansion to solve the matrix. 

Theorem One is helpful for computing the determinant of a matrix that contains many zeros. For example, if a row is mostly zeros, then the cofactor expansion across that row has many terms that are zero, and the cofactors in those terms need not to be calculated. The same approach works with a column that contains many zeros. 

Examples: Find the cofactor expansion for each matrix. 

Theorem Two: If A is a triangular matrix, then det A is the product of the entries on the main diagonal of A. 

Example: Find the cofactor expansion

Note: Today's standards, a 25 x 25 matrix is small. Yet it would be impossible to calculate a 25 x 25 determinant by cofactor expansion. In general, a cofactor expansion requires more than n! multiplications, and 25! is approximately 1.5 x 10^25. If a computer performs one trillion multiplications per second, it would have to run for more than 500,000 years to compute a 25 x 25 determinant by this method. 

Examine Problem: Find the cofactor expansion and the determinant of the following matrix equation.

Linear Algebra 2- Properties of Matrix Multiplication (2.1)

Matrix Multiplication: 
Theorem 2: Let A be an m x m matrix, and let B and C have sizes for which the indicated sums and products are defined.
a.) A(BC) = (AB)C                                                (Associative Law of Multiplication)
corresponds to composition of linear transformations.
b.) A(B+C) = AB+AC                                           (Left Distributive Law)
c.) (B+C)A = BA+CA                                           (Right Distributive Law)
d.) r(AB) = (rA)B = A(rB)
for any scalar r.
e.) I_mA = A = AI_m                                             (Identity for Matrix Multiplication)

Example: Product 

Example: Associative Law of Multiplication: 

Example: Identity for Matrix Multiplication:

Warnings: 
1. In general, AB does not equal BA.
2. The cancellation laws do not fold for matrix multiplication. that is, if AB = AC, then it is not true in general that B = C.
3. If a product AB is the zero matrix, you cannot conclude in general that either A = 0 or B = 0.

Powers of a Matrix: 
If A is an n x n matrix and if k is a positive integer, then A^k denotes the product of k copies of A.
                                                    A^k = A x A x A.....
If A is nonzero and if x is in R^n, then A^k is the result of left multiplying x by A in the amount of k times. If k = 0 then A^0x should be x itself, which will be the identity matrix.

Example: 

 











Example: 

If you still need some help with matrix multiplication here is a good link to help you out.
https://www.mathsisfun.com/algebra/matrix-multiplying.html
Also i have found a really good video on Powers of a Matrix and how to use the powers as an implication of multiply matrix's. Its called power of Matrix's and you can find it on Youtube.

An example problem could be: Find all properties from theorem 2 (a-e) and also find the power matrix's for A, B, and C.

Solving One-Variable Linear Equations

Solving One Variable Equations: 

"Linear equations are equations with a plain old variable like "x" rather than something more complicated like "x^2"

You've probably already solved linear equations; you just didn't know it. 

When solving a one variable equation, you need to "undo" whatever has been done to the variable. 

 A solution: a solution to an equation is a value or set of values that makes the equation true. 

For Example in the Equation: 

x + 2 = 3

the solution is x = 1, because 1 + 2 = 3. 

Lets Review: Additive Inverse: 

An additive number is something you add to a number to make it equal 0. 

2 + (-2) = 0

5 + -5 = 0

Multiplicative Inverse: 

A multiplicative inverse is something you multiple by another number to equal 1. 

2 X 1/2 = 1

You do this in order to get the variable by itself; in technical terms you are "isolating the variable."

For Example: x + 6 = -3

You first want to get the x on the left side by itself. So you will subtract the 6 to the right side. So now the equation looks like x = -3 - 6. So now we solve for x. -3 - 6 = -9. So x = -9.

Whatever you do to an equation,
do the   S A M E   thing
to   B O T H   sides of that equation!

Example: x - 3 = -5

Since you want to get x by itself, so you want to add 3 to both sides. 

x - 3 = -5

+3 = +3 

x = -2

Then the solution is x = -2. 

Lets solve this equation: 3x + 6 + 4x = 13

First we need to combine like terms: those terms are the 3x + 4x = 7x. 

So now the equation look like this: 7x + 6 = 13

So now to get the variable by itself we need to subtract the 6 from both sides. 

So we get 7x = 13 - 6

7x = 7

now we divide by 7 to get x by itself to both sides. 

So x = 1. 

Now we need to check our work

3 (1) + 6 + 4 (1) = 13

3 + 6 + 4 = 13

13 = 13. 

Therefore the solution is x = 1. 

Lets solve this equation: 3 - 5x = 18.

First we need to move the 3 to the other side to get x by itself, so lets subtract the 3 from both sides.

-5x = 18 - 3

-5x = 15. 

Now we need to divide by the -5 to get x by itself. 

x = 15 / -5

x = -3.

Now we need to check our work so we plug x = -3 back into the equation. 

3 - 5(-3) = 18

3 - (-15) = 18

18 = 18. 

Therefore the solution is x = -3. 

Oder of Operations

Order of Operations: 

What is the answer to this ? 4 + 6 x 2 =

Did you get 20?

Can you see how the answer could be 16?

There are two possible ways to figure out this problem. 

          Do the addition first :                                                 Do the multiplication first: 

                4 + 6 x 2 =                                                                                  4 + 6 x 2 =

                    10 x 2 =20                                                                                  4 + 12 = 16

Which is it? 

Both answers can't be right or we would always be arguing to the answers of math problems. The nice thing about math is there is always just ONE answer. 

So a long time ago mathematicians decided to make a set of rules for what to do first in a math problem. 

The rules are called the Order of Operations!

which is abbreviated by PEMDAS.

The P in PEMDAS stands for "parenthesis"

Parenthesis in math are used to group important things together, so you always do them first. 

For example: 

 Do inside the parenthesis first:

Here's another Example: 

 In this problem we have to do paranthesis first then exponents then division.

The classic sentence that math geeks have been using for years to remember the Order of Operations (PEMDAS) is: 

Please Excuse My Dear Aunt Sally

Can you think of a sentence to go along with this acronym?

Murder Mystery- Geometry Game

After a lesson on shapes, angles and measurements i will put my students to the  test with a fun interactive game called Murder Mystery. This game will get the students up out of their seats to find clues and find who the suspect is. This game is kind of like the board game clue, students have to search through clues to find the murder.

Geometry Game- Murder Mystery: 

The Scene: The police are called to a high school, lying on the floor in the library is a dead student. As police search the library they find 5 clues written down by witness. They have sent the clues to you to decipher. They also provide a list of those present at the time of the murder. (This list is all the students in the class,  with specific details on each of the students). There are 32 suspects, each clue will eliminate half the number of suspects remaining. When all clues have been solved the suspected person will be revealed. 

Clue One: Get into Shape:

The answer to each number is a letter A=1, B=2, etc. 

1. Total number of sides of all four squares.

2. An octagon has ? sides. 

3. A square based pyramid has ? vertices. 

4. A pentagon has ? sides. 

5. A parallelogram has ? sides.

There are many more questions about 20 but I'm just giving you an example of questions you can ask. 

Clue Two: Look From a New Angle: 

I will provide the students with different kinds of angles and they will have to figure out the angle measure of each line. By the same system as above a = 10 degree, b = 20 degrees, and so on. 

For example: 

The rest of the clues also have to do with things the students learned from this particular lesson, they will have a clue on compass/directions, and also coordinating points on a plane. The last clue i will give them a story and they will have to find the error that is wrong in that story to find the suspect. 

I got this idea from another teacher but put my own spin on it, I will use my actually students as the suspects and I will let them choose if they wanna work together or not. None of the students will know who the suspect is until the game is over. This gets them up and moving around and also has them use their ipad's for clue number two. The story I also provide them with will also be available on their ipad's so they can see if they can spot what doesn't belong in the story. I have never done this game before but am looking forward to trying it in my classroom one day. I think this game can be for any grade and for any subject. The teacher just needs to have fun creating the questions, and the students need to have fun trying to figure out the clues. 

Welcome to my Technology Mathematical Educational Blog

This is a blog that I have already created to enhance students knowledge about specific topics in Discrete Mathematics (also know as Calc 4). I will be incorporating technology into my classroom also so many of my posts will be ways to incorporate technology into the mathematical classroom. I will be sharing different lesson plans, activities and my thoughts on certain math topics. Can't wait to hear feedback from my fellow classmates about incorporating technology into the classroom. At the rate technology keeps improving, we as teachers need to find new ways to keep up with the students and make learning more interesting to them.