Many types of applications involve a finite set of state {s1, s2,…, sn} of a population • for instance, residents of a city may live downtown or in the suburbs soft drink

Every month, 15% of the subscriber of company A changes to use the service of company B, and 10% of company B’s subscribers changes to use the service of A.. How many subscribers will

Group:

Students’ list:

- Hồ Lê Khánh Dy 2052431 –

- Nguyễn Đình Hoàng – 2052480

- Nghiêm Đức Tài – 20523412

- Nguyễn Phúc Th nh 2052728 ị –

- Trần Bảo Tín – 2052748

- Đỗ Diệp Phương Trâm - 2052286

Semester: HK202

Lecturer: Phan Thị Khánh Vân

Submission day: 30/5/2021

LIST OF CONTENTS

1 Markov chain

2 Determinants

3 System of linear equations

state is represented by a number p ij , where 0 ≤ p ij ≤ 1

B Definition:

• The matrix P = (

p 11 p 12 ⋯ p 1n

p 21

⋮ p 22 ⋮ ⋯ ⋯ p 2n ⋮

p n1 p n2 ⋯ p nn

) is called the matrix of transition probabilities

• At each transition, each member in a given state must either stay in that state or change to another state It means that:

∑ 𝑝 𝑖𝑗 𝑛 𝑖=1

= 1, ∀ 𝑗 = 1 𝑛

• The nth state vector of a Markov chain for which P is the matrix of transition

probabilities and X is the initial state vector is: 0

𝑋 𝑛 = 𝑃𝑋 𝑛−1 = 𝑃 2 𝑋 𝑛−2 = = 𝑃 𝑛 𝑋 0

C Practical problems:

a Two competing companies offer mobile phone service to a city with 100 000

households Suppose that each citizen uses one of these services Every month, 15%

of the subscriber of company A changes to use the service of company B, and 10% of company B’s subscribers changes to use the service of A Company A now has 60

000 subscribers and Company B has 40 000 subscribers How many subscribers will

each company have after 2 months?

b Consider an experiment of mating rabbits We watch the evolution of a particular

gene that appears in two types, G or g A rabbit has a pair of genes, either GG (dominant), Gg (hybrid the order is irrelevant, so gG is the same as Gg) or gg –

(recessive) In mating two rabbits, the offspring inherits a gene from each of its parents with equal probability Thus, if we mate a dominant (GG) with a hybrid (Gg), the offspring is dominant with probability 1/2 or hybrid with probability 1/2 Start with a rabbit of given character (GG, Gg, or gg) and mate it with a hybrid The offspring produced is again mated with a hybrid, and the process is repeated through

a number of generations, always mating with a hybrid (i) Write down the transition probabilities of the Markov chain thus defined (ii) Assume that we start with a hybrid rabbit Let µn be the probability distribution of the character of the rabbit of the n-th generation In other words, µn(GG),µn(Gg),µn(gg) are the probabilities that the n-th

generation rabbit is GG, Gg, or gg, respectively Compute µ1,µ2,µ3

D Solution:

a 𝑊𝑒 ℎ𝑎𝑣𝑒: 𝑋 0 = (60000 40000) , 𝑃 = ( 0. 0. 85 15 0.1 0.9 )

𝑋 2 = 𝑃 2 𝑋 0 = (0 0. 85 15 0.1 0.9 ) 2 ∗ (60000 40000) = ( 51250 48750)

So : There are 51250 subscribers from company A after 2 months

There are 48750 subscribers from company B after 2 months

b (i) From the given information, we have the transition probabilities:

Gg gg

GG 0.5 0.5 0 GG

0.25 0.5 0.25 Gg

Or we can write it in the following form:

𝑃 = ( 0.25 0.5 0.25 0.5 0.5 0

(ii) We start with 𝑋 0 being distributed as 𝜇 0 = ( 𝜇 0 (1), 𝜇 0 (2), 𝜇 0 ( 3)) = (0,1,0),and, letting

𝑚𝑢 𝑛 = ( 𝜇 0 (n), 𝜇 0 (n), 𝜇 0 (n)) be the distribution of 𝑋 𝑛 , we have:

μ = μ ∗ 𝑃 𝑛 𝑜 𝑛

So that:

𝜇 = 𝜇 ∗ 𝑃 = 1 0 (0 1 0) ∗ ( 0.25 0.5 0.25 0.5 0.5 0

0 0.5 0.5 ) = (0.25 0.5 0.25)

𝜇 2 = 𝜇 ∗ 𝑃 0 2 = (0 1 0) ∗ ( 0.25 0.5 0.25 0.5 0.5 0

2

= (0.25 0.5 0.25)

𝜇 3 = 𝜇 ∗ 𝑃 0 3 = (0 1 0) ∗ ( 0.25 0.5 0.25 0.5 0.5 0

3

= (0.25 0.5 0.25)

2 Determinant:

A Defintion:

In mathematics, the determinant is a scalar value that is a function of the entries of a square matrix

A n and the determinant of a matrix A is denoted det(A), det A, or |A| n

B Classification and calculation:

- General matrix:

1) n = 1, A = a => det(A) = a 11 11 2) n = 2, A = a a 11 12

a 21 a 22

=> det(A) = (-1) 1+1 a 11 |M 11 | + (-1) 1+2 a |M 12 12 | = a a 11 22 a – 12 a 21

(expand the determinant in row 1) 3) n = 3, A = a a a 11 12 13

a a a 21 22 23

a a a 31 32 33

=> det(A) = (-1) 1+1 a 11 |M 11 | + (-1) 1+2 a |M 12 12 | + (-1) 1+3 a 13 |M 13 |

= a det a a 11 22 23 – a 12 det a 21 a 23 + a 13 .det a 21 a 22

a a 32 33 a a 31 33 a 31 a 32 = a 11 a 22 a 33 a – 11 a 23 a 32 a – 12 a 21 a 33 + a a a + a a 12 23 31 13 21 a a 32 – 13 a a 22 31 = a 11 a a 22 33 + a 12 a a 23 31 + a 13 a a – a 21 32 11 a 23 a 32 a – 12 a 21 a 33 a – 13 a 22 a 31 (expand the determinant in row 1)

*Especially: We can calculate the determinant of A by the fastest way like this: 3 + Rewrite the elements in the matrix in rows and columns respectively

a a a 11 12 13

a a a 21 22 23

a a a 31 32 33 + Add the first 2 columns of the matrix to the right of the previously written elements

a a a a a 11 12 13 11 12

a 21 a 22 a 23 a 21 a 22

a a a a a 31 32 33 31 32 + det(A) = ∑ (product of elements on each blue diagonal) – ∑ (product of elements on each red

diagonal)

a a a a a 11 12 13 11 12

a a a a a 21 22 23 21 22

Recommandé pour toi

a a a a a 31 32 33 31 32

=> det(A) = a 11 a a 22 33 + a a 12 23 a + a 31 13 a a 21 32 – a a a a 11 23 32 – 12 a a a 21 33 – 13 a a 22 31

4) n = 4, A = a a a a 11 12 13 14

a a a a 21 22 23 24

a a a a 31 32 33 34

a a a a 41 42 43 44

=> det(A) = (-1) 1+1 a 11 |M 11 | + (-1) 1+2 a a

12 |M 12 | + (-1) 1+3

13 |M 13 | + (-1) 1+4 a 14 |M 14 | = (-1) 1+1 a 11 det a a a 22 23 24 + (-1) 1+2 a 12 det a 21 a 23 a 24

a a a 32 33 34 a a a 31 33 34

a a a 42 43 44 a a a 41 43 44

+(-1) 1+3 a det a a a + (-1) det a a a

a a a 31 32 34 a a a 31 32 33

a a a 41 42 44 a a a 41 42 43 = a 11 a a 22 33 a + a a a 44 11 23 34 a + a a 42 11 24 a a a 32 43 – 11 a 24 a a a 33 42 – 11 a 22 a a 34 43 – a 11 a a 23 32 a 44 – a a a a 12 21 33 44 – a a a a 12 23 34 41 – a 12 a a a 24 31 43 + a 12 a a a 24 33 41 + a 12 a a a 21 34 43 +

a a a a 12 23 31 44 + a 13 a a a 21 32 44 + a 13 a a a 22 34 41 + a a a 13 24 31 a 42 – a a a 13 24 32 a – a a 41 13 21 a a – 34 42

a a a a 13 22 31 44 – a 14 a a a 21 32 43 – a 14 a a a 22 33 41 – a 14 a a a 23 31 42 + a 14 a a a 23 32 41 + a 14 a a a 21 33 42 +

a a a a 14 22 31 43

5) n ≥ 5, we should turn the matrix A into the upper triangular matrix A’ n n and det (A n ) = det(A’ n )

= product of elements on the main diagonal of A’ n

Suite du document ci-dessous

DE THI GIUA KI (MIDTERM)

FORM ĐỀ THI ĐSTT CUỐI KÌ - Sample Final Test

16

12

a a a 11 12 13 … a 1n a a a 1 2 …… a x

a a a 21 22 23 … a 2n 0 b b 1 …… b y

A n = a a a 31 32 33 … a 3n => A’ n = 0 0 c …… c z

a a a n1 n2 n3 … a nn 0 0 0 ……… m

=> det(A n ) = det(A’ n ) = a.b.c…m

- Diagonal matrix:

a 0 0 ……… 0

0 b 0 ……… 0

A n = 0 0 c ……… 0 => det(A ) = n a.b.c…m

………

0 0 0 ……… m

- Identity matrix:

1 0 0 ……… 0

0 1 0 ……… 0

I n = 0 0 1 ……… 0 => det(I n ) = 1

………

0 0 0 ……… 1

- Triangular matrix:

a a a 1 2 …… a x a 0 0 ……… 0

0 b b 1 …… b y b 1 b 0 ……… 0

A n = 0 0 c …… c z or A n = c c 2 1 c ……… 0

0 0 0……… m m m m x y z …… m

=> det (A n ) = a.b.c…m

C Properties of determinant:

+ When we swap 2 rows or swap 2 columns, the determinant is changed sign

Example: det 1 2 = 1.4 – 2.3 = -2 => det 3 4 = 3.2 1.4 = 2 –

3 4 1 2 det 1 2 = 1.4 2.3 = -2 => det 2 1 = 2.3 4.1 = 2 – –

3 4 4 3 + If matrix A has 2 proportional rows or 2 proportional columns then det(A) = 0

Example: det 1 4 = 8 – 4 = 0

2 8 + If matrix A has all elements on a row = 0 or on a column = 0, then det(A)=0

0 0 0 0 0 0 0 0

Example: det 1 2 3 1 2 =

4 5 6 4 5 6 4 5

= 0.2.6 + 0.3.4 + 0.1.5 4.2.0 5.3.0 6.1.0 = 0 – – – + The common factor of a row or the common factor of a column can be taken out of the determinant

Example: 6 8 10 3 4 5 1 4 5

det 9 5 7 = 2.det 9 5 7 = 2.3.det 3 5 7

15 2 4 15 2 4 5 2 4

+ det(A ) = det(A n T n )

Example: det 6 7 = det 6 8 = 6.9 – 7.8 = -2

8 9 7 9 + det(A B ) = det(A n n n ).det(B n )

det 1 2 5 6 = det 1 2 det 5 6 = (4 3.2).(5.8 7.6) = 4

3 4 7 8 3 4 7 8 + det(α.A ) = α n n det(A n )

Example: det 2 3 7 = det 6 14 = 2 det 3 7 = 4.(3.2 2 – 8.7) = -200

8 2 16 4 8 2 + det(A ) = (det A) n n

Example: det 3 5 = (3.9 2 – 7.5) = 64 2

7 9

Let A, B and C be defined complex numbers the complex plane as in The vectors from C A and from C B are given by to to

z 1 = (x 1 −x 3 ) + i(y 1 − y ) 3 z1=(x1−x3) + i(y1−y3)

z 2 = (x 2 −x 3 ) + i(y 2 − y ) 3

so the area of the triangle is:

A= ½ Z1.Z2 A= ½ | ( Im ( (x 1 −x 3 )− i(y 1 −y 3 )) ((x 2 −x 3 )− i(y 2 −y 3 )) ) |

= 1/ 2|(x1−x3)(y2−y3)−(y1−y3)(x2−x3)|

=1/2|x1y2−y1x2+x2y3−y2x3+x3y1−y3x1|

Problem2:

Find the area of the triangle??

Evaluate that determinant We will expand on column 1

-2

(-2)

1

6

-1

-1

-1

= -2 ( 5 + 1 ) - 1 ( 2 + 1 ) + 6 ( 2 - 5 ) = -2 ( 6 ) - 1 ( 3 ) + 6 ( -3 ) = -12 - 3 - 18 = -33

So the area of the triangle is ½ detA =33/2= 16.5

Problem2: Calculate the matrix: 1 1 5 3

A = 2 6 8 9

2 5 6 7

1 2 5 6

Solution: 1 1 5 3 1 1 5 3 1 1 5 3 1 1 5 3

2 6 8 9 => 0 4 -2 3 => 0 4 -2 3 => 0 4 -2 3

2 5 6 7 0 3 -1 4 0 0 -5/2 -5/4 0 0 -5/2 -5/4

1 2 5 6 0 1 0 3 0 0 1/2 9/4 0 0 0 2

 det(A)= 1.4.(-5/2).2= -20

3.System of linear equations:

A Theory:

• In mathematics, a system of linear equations (or linear system) is a collection of linear equations involving the same set of variables

• A solution to a linear system is an assignment of numbers to the variables such that all the equations are simultaneously satisfied

B Definition:

• A general linear system of equations m in the n unknowns can be written as:

a 11 x 1 + a 12 x 2 + + a x + + a 1j j 1n x n = b 1

a i1 x 1 + a i2 x 2 + + a x ij j + + a in x n = b i (1)

a m1 x 1 + a m2 x 2 + + a x + + a mj j mn x n = b m

• Matrix is called the augmented matrix for the system (1), which is obtained by adjoining column to matrix as the last column B A

𝐴 = (𝑎 |𝑏 𝐵 𝑖𝑗 𝑖 ) 𝑚∗(𝑛+1)

• The system (1) is called a homogeneous if B = 0 m×1 and a nonhomogeneous if B ≠

0 m×1

• A non-homogeneous linear system A.X = b , A Rm×n If: ∈

- r(A) < r(A|b): there is no solution

- r(A) = r(A|b) = n: the solution is unique

- r(A) = r(A|b) = r < n: there is an infinite number of solutions

• A homogeneous system of linear system with b = 0 If :

- r(A) = n : there is only trivial solution X = 0

- r(A) < n : there is infinitely many solutions or non-trivial solutions

C Practical problems:

a Problem 1: Balance the chemical equation :

NH 3 + O 2 → NO + H 2 O (Not balanced) Solution :

- To balance the equation, we insert unknowns, multiplying the reactants and the products to get an equation of the form:

(x ) 1 NH 3 + (x )O 2 2 → (x 3 )NO + (x ) H O 4 2

- Next, by applying Law of Conservation of Matter, we compare the number of nitrogen (N), hydrogen (H) and oxygen (O) atoms of the reactants with the number of the products We obtain three equations :

N : x = x ; H : 3x = 2x ; O : 2x = x + x ; 1 3 1 4 2 3 4

- It is important to note that we made use of the subscripts because they count the number of atoms of a particular element Rewriting these equations in standard form, we see that we have a homogenous linear system in four unknowns, that is :

x 1 , x , x and x 2 3 4

x 1 - x = 0 3

3x 1 - 2x = 0 4

2x 2 - x - x = 0 3 4

- Writing this equations or system in matrix form, we have the augmented matrix : ( 1 0 −1 0 0 3 0 0 −2 0

- First, we subtract row 1 multiplied by 3 from row 2: R2 = R2 − 3R1 : ( 1 0 −1 0 0 0 0 3 −2 0

- Swap rows 2 and 3, then divide row 2 by 2: R2 = R2/2 and row 3 by 3: R3 = R3/3 After that, add row 3 to row 1: R1 = R1 + R3 We obtain :

( 1 0 0 1 −1/2 −1/2 0 0 −2/3 0

- Add row 3 multiplied by 1/2 to row 2: R2= R2 + R3/2 We have : ( 1 0 0 −2/3 0 0 1 0 −5/6 0

- Let x = x We get : 4 4

x 1 = (2/3).x 4

x 2 = (5/6).x 4

x 3 = (2/3).x 4

x 4 = x 4

- To balance the equation without fractions, we choose x = 6 4

- Then, our balance equation is :

4NH 3 + 5O 2 → 4NO + 6H 2 O

b Problem 2 : A dietitian is planning a meal that supplies certain quantities of vitamin

C, calcium, and magnesium Three foods will be used, their servings measured in milligrams The nutrients supplied by these foods and the dietary requirements are given in the table below Determine the servings (mg) of Food 1,2 and 3 necessary to

meet the dietary requirements

Solution :

- From the table, we archieve a non-homogenous linear system in three unknowns Food 1, Food 2, Food 3 denoted as f , f , f respectively : 1 2 3

30f 1 + 45f + 15f = 1785 2 3

20f 1 + 45f + 20f = 1642.5 2 3

30f 1 + 60f + 35f = 2472.5 2 3