1314 01 09 matrix algebra&marcov chain

Matrix Algebra

Asep HP Kesumajana

GL-5001 Pemodelan Geologi

GL-5001 - Pemodelan Geologi

4. Matrix Algebra & Markov Chain

2

MATRICES A m x n matrix is a rectangular array of numbers with m rows and

n columns. The array is enclosed in curved brackets and is usually denoted

by an upper case letter.

a 2 x 4 matrix.

The element in the ith row, jth column of a matrix A is labelled as aij.

the elements of a 3 X 3 matrix A:

often write A = (aij)

17214352

A

333231

232221

131211

aaaaaaaaa

A



3

example of a data matrix

08.0...77.232.1333.45.....................37.0...43.374.1597.4442.0...95.201.1870.43

...

130...

21

5232322 OPOFeOAlSiO

Specimen

SpecimenSpecimen



4

Equality of matrices A = B implies that aij = bij for every i and j, all the elements in corresponding positions of

A, B are equal

611524213

A

611524213

B



5

Addition of matrices

A + B = (aij + bij) Example:

cA = (caij)

611524213

A

620472311

B

1231196504

BA

Multiplication by a scalar



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Pre-multiplication by a row vector Each column of the matrix is multiplied by the vector as if it were

itself a vector, for example:

The result is a row vector.

Each row of the matrix is treated as a vector and used to multiply the column vector, for example

The result is a column vector

Post-multiplication by a column vector

5327821354

532

1525

231

141732



7

Multiplication of matrices by matrices

AB<>BA

612237104241

A

341253

B

2932443523161519

AB



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Diagonal Matrix

Non-zero elements occur only in the leading diagonal Pre-multiplication of a matrix A by a diagonal matrix multiplies the

elements of a row by the same number

Post-multiplication results in the all the elements in a column of A being multiplied by the same number:

1210161263

654321

200040003

A

6832169

1246

200040003

321143612

A



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Identity matrix This is a diagonal matrix whose elements in the leading diagonal

are all equal to unity. For example a 3 X 3 identity matrix would be

Behave like number 1 in ordinary aritmatic

100010001



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Inverse matrix The inverse of matrix A is written as A-1 and is

defined to be the matrix such that

AA-1 = A-1A = I, the identity matrix

Division of matrices by matrices is not defined, but we can multiply by inverse matrices.



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Transpose of a matrix The transpose of a matrix A, denoted by AT,

is obtained by writing the rows of A as columns. For example

572143

A

517423

TA



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Symmetric matrix The first row of a symmetric matrix is the same as the,

first column, and similarly for the other rows and columns, for example

943415352

A



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Orthogonal matrices

A matrix A is said to be orthogonal if ATA = 1, that is, the rows are orthogonal to its columns. For example

They are used in the rigid rotation of axes in such techniques as principal components analysis.

1001

cossinsincos

cossinsincos



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Translation, Rotation, and Rescaling Translation:

Tx translation in x direction Ty translation in y direction

equation: x’ = x + Tx y’ = y + Ty

matrix:

1010001

1,,1,','

yx TTyxyx



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Translation, Rotation, and Rescaling Rotation with center point at (0,0):

(x,y) the point to be rotated (x’,y’) the result Rotation angle (α)

equation: x’ = x cos α – y sin α y’ = y cos α + x sin α

Matrix:

1000cossin0sincos

1,,1,','

yxyx



16

Translation, Rotation, and Rescaling Rotation with center point at (xr,yr):

(x,y) the point to be rotated (x’,y’) the result Rotation angle (α)

equation: x’ = x cos α – y sin α + (1- cos α)xr + yr sin α

y’ = y cos α + x sin α + (1- cos α)yr – xr sin α simplified:

x’ = xr + (x – xr) cos α – (y – yr) sin α

y’ = yr + (y – yr) cos α + (x – xr) sin α Matrix:

1sin)cos1(sin)cos1(0cossin0sincos

1,,1,','

rrrr xyyxyxyx



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Translation, Rotation, and Rescaling Rescale with center point at (0,0):

Sx scale in x direction Sy scale in y direction

equation: x’ = x Sx y’ = y Sy

matrix:

1000000

1,,1,',' y

x

SS

yxyx



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Translation, Rotation, and Rescaling Rescale with center point at (xs,ys):

Sx scale in x direction Sy scale in y direction

equation: x’ = x Sx + (1 – Sx)xs

y’ = y Sy + (1 – Sy)ys

Simplified: x’ = xs + (x – xs) * Sx

y’ = ys + ( y – ys)* Sy

matrix:

1110000

1,,1,','

sysx

y

x

ySxSS

Syxyx



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Translation, Rotation, and Rescaling Combination with (0,0) as a center point Rotation Translation Rotation & Translation

Rotation & Translation scale Rotation, Translation & scale

For (x,y)

1000cossin0sincos

1000000

1000cossin0sincos

yx

yx

y

x

SSSS

SS

10cossin0sincos

1010001

1000cossin0sincos

TyTxSSSS

TyTxSSSS

yx

yx

yx

yx

10cossin0sincos

)1,,(1,','TyTx

SSSS

yxyx yx

yx



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Translation, Rotation, and Rescaling Combination with (0,0) as a center point Atau menggunakan pers:

x’ = Sx (x cos α – y sin α) + Tx

y’ = Sy (y cos α + x sin α) + Ty



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Translation, Rotation, and Rescaling Combination with (xp,yp) as a center point rotation skale

1sincossincos0cossin0sincos

11sin)cos1(1sin)cos1(0cossin0sincos

1110000

1sin)cos1(sin)cos1(0cossin0sincos

ppypppxp

yx

yx

pypypypxpxpx

yx

yx

pypx

y

x

pppp

xySyyxSxSSSS

ySxSySxSySxSSSSS

ySxSS

S

xyyx



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Translation, Rotation, and Rescaling Combination with (xp,yp) as a center point rotation & Scale Translation

equation: x’ = xp + Sx ([x – xp] cos α – [y – yp] sin α) + Tx

y’ = yp + Sy ([y – yp] cos α + [x – xp] sin α) + Ty


1010001


yppypxppxp

yx

yx

yxppypppxp

yx

yx

TxySyTyxSxSSSS

TTxySyyxSxSSSS



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Skew / Shear Shear in x direction

Shear in y direction

10001001

)1,,(1,',' xSHyxyx

10001001

)1,,(1,','SHy

yxyx

0

0

0

0



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Reflection Reflection to X axis

Reflection to Y axis

Reflection to point (0,0)

100010001

)1,,(1,',' yxyx

100010001

)1,,(1,',' yxyx

100010001

)1,,(1,',' yxyx

- 5

0

0-4

-5

-4



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Reflection Reflection to line (y=x)

Reflection to line (y=-x)

100001010

)1,,(1,',' yxyx

100001010

)1,,(1,',' yxyx

0

0

0

-6

Markov Chain

Asep HP Kesumajana

GL-5001 Pemodelan Geologi



27

Markov Chain Type of analysis: Time/space series Probabilistic model that exhibits a special

type of dependence: depend only on the present state of the system

and not on Preceding states each state in the future depends only on the

present state of the process the present in the future does not dependent on

the past.



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Limitations of Markov Models Markov models are limited by two major

assumptions: Transitions (probabilities) of changing from one state to

another are constant. Markov model is used only when a constant failure rate and repair rate assumption is justified.

The transition probabilities are determined only by the present state and not by the system's history. Future states of the system are independent of all but the current state of the system.



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Uses of markov Chain used in prediction of sequence of data of:

Artificial music. Spread of epidemics. Traffic on highways. Occurrence of accidents. Growth and decay of living organisms. Emission of particles from radioactive sources. Number of people waiting in a line (queue). Arrival of telephone calls at a particular telephone exchange. Geology: stratigraphic problem, biostratigraphic

problem etc.

Markov models are the only accurate method for modeling complex situations.

The complex proofs related to these models can be found in many reliability engineering handbooks and related publications



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Notation: A Markov chain is a sequence X1, X2, X3, ... of random

variables. called the state space, with Xn is the state at time n. If the conditional distribution of Xn+1 on past states is a

function of Xn alone

A Markov chain is characterized by the conditional distribution

which is called the transition probability of the process. This is sometimes called the "one-step" transition probability.



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Notation (cont) The probability of a transition in two, three, or more steps is

derived from the one-step transition probability and the Markov property:

Likewise,

These formulas generalize to arbitrary future times n+k by

multiplying the transition probabilities and integrating k times. The marginal distribution P(Xn) is the distribution over states at

time n. The initial distribution is P(X0). The evolution of the process through one time step is described by



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Example 1: Television Viewers

This week: 40% watched station X 60% watched station Y. each week: 15% of the X viewers switched to station Y 5% of the Y viewers switched to station X.



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Example 1:

Pk: 2 dimension column vector proposed to people watching station X and Y during k week

k=0: P0 = [0.40,0.60]T

k=1: watched station X:

(1.00 - 0.15) * 0.40 + 0.05 * 0.60 = 0.37 or 0.85 * 0.40 + 0.05 * 0.60 = 0.37

watched station Y: 0.15 * 0.40 + (1.00-0.05) * 0.60 = 0.63 or 0.15 * 0.40 + 0.95 * 0.60 = 0.63

P1 = [0.37,0.63]T



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Example 1:

if using 2 dimension matrix (A):

The matrix A called the transition matrix (or Markov matrix) we see that P1=AP0

From:

X Y

To: X 0.85 0.05

Y 0.15 0.95

63.037.0

60.040.0

95.015.005.085.0

1P



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Example 1:

K=2:

or

so: P2 = AP1 = A(AP0) = A2P0, P3 = AP2 = A(A2P0) = A3P0, and soon

in general: Pk = AkP0

try with k=10

654.0346.0

63.037.0

95.015.005.085.0

2P

654.0346.0

60.040.0

95.015.005.085.0

95.015.005.085.0

2P



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Example 2: Lithologic problem There is 3 lithologic type in 1 borhole say: SH, SS and LS. The

lithologic over laying by other lithology are following: LS

over laying by LS = 10%, over laying by SH = 60% and over laying by SS = 30%

SS over laying by LS = 30%, over laying by SH = 60% and over laying by SS = 10%

SH over laying by LS = 30%, over laying by SH = 40% and over laying by SS = 30%



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Find probability of LS, SS, SH on the 3rd level folowing LS on the 1st level:



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probability of LS on 3rd level following LS on 1st level:

LS, LS, LS probability: P(LS,LS,LS) = 0.1 * 0.1 = 0.01 LS, SS, LS probability: P(LS,SS,LS) = 0.3 * 0.3 = 0.09 LS, SH, LS probability: P(LS,SH,LS) = 0.6 * 0.3 = 0.18

P(LS on 3rd level following LS on 1st level): 0.01+0.09+0.18=0.28



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probability of SS on 3rd level following LS on 1st level:

LS, LS, SS probability: P(LS,LS,SS) = 0.1 * 0.3 = 0.03 LS, SS, SS probability: P(LS,SS,SS) = 0.3 * 0.1 = 0.03 LS, SH, SS probability: P(LS,SH,SS) = 0.6 * 0.3 = 0.18

P(SS on 3rd level following LS on 1st level): 0.03+0.03+0.18=0.24



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probability of SH on 3rd level following LS on 1st level:

LS, LS, SH probability: P(LS,LS,SH) = 0.1 * 0.6 = 0.06 LS, SS, SH probability: P(LS,SS,SH) = 0.3 * 0.6 = 0.18 LS, SH, SH probability: P(LS,SH,SH) = 0.6 * 0.4 = 0.24

P(SS on 3rd level following LS on 1st level): 0.06+0.18+0.24=0.48

Third level LS: SS: SH:

First level LS: 0.28 0.24 0.48



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Find probability of LS, SS, SH on the 3rd level following SS on the 1st level:



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probability of LS, SS, SH on 3rd level following SS on 1st level:

P(LS on 3rd level following SS on 1st level): 0.03+0.03+0.18=0.24

P(SS on 3rd level following SS on 1st level): 0.09+0.01+0.18=0.28

P(SH on 3rd level following SS on 1st level): 0.18+0.06+0.24=0.48

Third level

LS: SS: SH:

First level SS: 0.24 0.28 0.48



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Find probability of LS, SS, SH on the 3rd level following SH on the 1st level:



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probability of LS, SS, SH on 3rd level following SH on 1st level:

P(LS on 3rd level following SH on 1st level): 0.03+0.09+0.12=0.24 P(SS on 3rd level following SH on 1st level): 0.09+0.03+0.12=0.24 P(SS on 3rd level following SH on 1st level): 0.18+0.18+0.16=0.52

Third level

LS: SS: SH:

First level SH: 0.24 0.24 0.52



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for all lithologic:

3rd level (initial/state) LS SS SH

1st level (initial/ state)

LS 0.28 0.24 0.48 SS 0.24 0.28 0.48 SH 0.24 0.24 0.52



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involves three states of starting lithology: LS, SS, or SH any given lithology in the starting state, will have

different pobability in the next layer upward or downward.

The information (probability) about the transition that occurs at the next lithology can be summarized in the following transformation matrix (probability matrix).

In many cases, the probability matrix can be calculated from the frequency matrix estimated from the observation of the actual physical sequence and boundries.

The matrix describing the frequencies from one lithology to another lithologies is called frequency matrix.



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Transformation matrix (probability matrix) A:

If matrix AxA=A2

Overlaying lithology LS SS SH

Starting lithology

LS 10 30 60 SS 30 10 60 SH 30 30 40

3rd level (initial/state) LS SS SH

1st level (initial/ state)

LS 0.28 0.24 0.48 SS 0.24 0.28 0.48 SH 0.24 0.24 0.52



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initial state distribution matrix (1st order Markov Chain): The 1st lithology is shale, what is the probability for each lithology in the 3rd level up 1st level, 30% probability is LS, 30% probability is SS and 40% probability is SH can be summarized in the initial state distribution matrix (1st

order Markov Chain):

LS SS SH P0 LS 0.10 0.30 0.60



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for the third level: P1=P0xA

P1:

P1 is the state distribution following two transitions (3rd level, 2nd order Markov Chain) from the initial lithology of shale (1st level).

we can Predict that 3rd level (2nd order Markov Chain) : 24% probability being LS, 24% probability being SS, and 52% probability being SH again.

0.10 0.30 0.60 0.30 0.30 0.40 x 0.30 0.10 0.60 = 0.24 0.24 0.52

0.30 0.30 0.40



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Iterate to find P2 (3rd order Markov Chain) the probability at forth level of occurrences. P2 is also called the state distribution following three

transitions. P2 = P1 x A = P0 x A x A = P0 x A2

P2 = P1 x A:

0.10 0.30 0.60

0.24 0.24 0.52 x 0.30 0.10 0.60 = 0.25 0.25 0.50 0.30 0.30 0.40



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Iterate to find P2 (cont) P2 = P1 x A = P0 x A x A = P0 x A2

AND

0.10 0.30 0.60 0.10 0.30 0.60 0.28 0.24 0.48 0.30 0.10 0.60 x 0.30 0.10 0.60 = 0.24 0.28 0.48 0.30 0.30 0.40 0.30 0.30 0.40 0.24 0.24 0.52

0.28 0.24 0.48 0.30 0.30 0.40 x 0.24 0.28 0.48 = 0.25 0.25 0.50

0.24 0.24 0.52



52

Iterate to find P2 (cont)

P2 is the state distribution following three transitions (4th level, 3rd order Markov Chain) from the initial lithology of shale (1st level).

we can Predict that 4th level (3rd order Markov Chain) has about: 25% probability being LS, 25% probability being SS, and 50% probability being SH again.

In General: Pn = Pn-1 x A = P0 x An



53

How about P3 and P4 ? For P3:

Its find that P2 = P3 its mean that P2 = C This means that if the situation ever becomes one that can

be described by the state distribution: P = [0.25 0.25 0.50]

then at no time after that will it change. P is called the stable-state vector for this Markov process

0.28 0.24 0.48

0.25 0.25 0.50 x 0.24 0.28 0.48 = 0.25 0.25 0.50

0.24 0.24 0.52



54

Example 3: Lithologic problem Predict the 2 unit lithology in upper and

bottom of given stratigraphic column.

LS

SSSH

LSSHSSLS

COSHSS

SSSHSSSH

CO

LS

SH

SS

SH

SH

SS

SH

SS

SH

SH



55

Downward prediction Frequency matrix

S0 first order Markov Chain (probalility matrix)

SH SS LS COSH 7 8 1 1 17SS 5 7 2 1 15LS 4 0 0 0 4CO 2 0 0 0 2

18 15 3 2 38

SH SS LS COSH 0.412 0.471 0.059 0.059SS 0.333 0.467 0.133 0.067LS 1 0 0 0CO 1 0 0 0

1 LS2 SH3 SH4 SS5 SH6 SS7 SS8 LS9 SH

10 SS11 LS12 SH13 SH14 SH15 CO16 SH17 SS18 SH19 SH20 SS21 SH22 SS23 SH24 SS25 SS26 CO27 SH28 SH29 LS30 SH31 SH32 SH33 SS34 SS35 SS36 SS37 SS38 SS39 SH40 ?41 ? GL-5001 - Pemodelan Geologi


56

Downward prediction (cont)

S1 second order Markov Chain (probalility matrix)

S2 third order Markov Chain (probalility matrix)

1 LS2 SH3 SH4 SS5 SH6 SS7 SS8 LS9 SH

10 SS11 LS12 SH13 SH14 SH15 CO16 SH17 SS18 SH19 SH20 SS21 SH22 SS23 SH24 SS25 SS26 CO27 SH28 SH29 LS30 SH31 SH32 SH33 SS34 SS35 SS36 SS37 SS38 SS39 SH40 ?41 ?

SH SS LS COSH 0.427 0.435 0.084 0.054SS 0.461 0.403 0.086 0.050LS 0.444 0.444 0.056 0.056CO 0.444 0.444 0.056 0.056


37 SS38 SS39 SH40 SS41 ?

37 SS38 SS39 SH40 SS41 SH



57

Upward prediction

Frequency matrix

S0 first order Markov Chain (probalility matrix)

41 ?40 ?39 LS38 SH37 SH36 SS35 SH34 SS33 SS32 LS31 SH30 SS29 LS28 SH27 SH26 SH25 CO24 SH23 SS22 SH21 SH20 SS19 SH18 SS17 SH16 SS15 SS14 CO13 SH12 SH11 LS10 SH9 SH8 SH7 SS6 SS5 SS4 SS3 SS2 SS1 SH

SH SS LS COSH 0.39 0.28 0.222 0.111SS 0.53 0.47 0 0LS 0.33 0.67 0 0CO 0.50 0.50 0 0

SH SS LS COSH 7 5 4 2 18SS 8 7 0 0 15LS 1 2 0 0 3CO 1 1 0 0 2

17 15 4 2 38



58

Upward prediction (cont)

S1 second order Markov Chain (probalility matrix)

S2 third order Markov Chain (probalility matrix)

41 ?40 ?39 LS38 SH37 SH36 SS35 SH34 SS33 SS32 LS31 SH30 SS29 LS28 SH27 SH26 SH25 CO24 SH23 SS22 SH21 SH20 SS19 SH18 SS17 SH16 SS15 SS14 CO13 SH12 SH11 LS10 SH9 SH8 SH7 SS6 SS5 SS4 SS3 SS2 SS1 SH



41 ?40 SH39 LS38 SH37 SH

41 SH40 SH39 LS38 SH37 SH

1314 01 09 matrix algebra&marcov chain

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