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MA1506

Mathematics II

Chapter 5

Matrices and their uses

Chew T S MA1506-14 Chapter 5 1

This chapter consists of two parts

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In part one, we shall study


Matrix operations

Some special matrices

Inverse matrix and unique solution of AX=B

Determinant and inverse matrix

Leontief Input-Output model

PART ONE

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5.1 What is a Matrix?


rewritten as

2x1 matrix2x2 Matrix

2 7 3x y+ =

4 8 11x y+ =

2 7 3

4 8 11

x

y

=

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3x3 Matrix

m x n Matrix: m rows, n columns

Entry at i-th row j-th column

5.1 What is a Matrix?


11 1

1

n

m mn

a a

a a

We may write

11 12 13

21 22 23

31 32 33

a a a

a a aa a a

( )ijA a=

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5.2 Matrix operations


Matrix addition

Scalar multiplication

Matrix multiplication

Matrix transposition

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Matrix Addition


m x nmatrices

Term by term addition


( )ijA a=( )ijB b=

( )ij ijA B a b+ = +

1 2 7 3 8 5

4 8 6 9 10 17

+ =

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Scalar multiplication


m x nmatrix

Term by term multiplication

real or complex number


( )ijA a=c

( )ijcA ca=1 2 3 6

34 8 12 24

=

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Matrix Multiplication


Not term by term but row to column

m x nmatrix n x pmatrix


( )ijA a= ( )ijB b=

m x pmatrixAB C=( )ijC c=

1 1 2 2ij i j i j in njc a b a b a b= + + +

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Example


21

32

11

654

321

++++

++++=

)2(63514)1(62514

)2(33211)1(32211

= 78

12


Animation slide

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In general

Non commutative


AB BA

2 7 1 3 16 14 8 2 1 20 4

AB = =

1 3 2 7 14 312 1 4 8 0 6

BA = =

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Matrix transposition


m x nmatrixswap rows with columns


( )ijA a=

( )T jiA a= n x mmatrix

1 2 4

6 8 9

T

=

1 6

2 8

4 9

1 7 9

6 8 2

4 10 12

T =

1 6 4

7 8 10

9 2 12

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( )T

TA A=

( )T T TA B A B+ = +

( )T TcA cA=

( )T T TAB B A=

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Symmetric matrix


An n x nmatrix is symmetricif

5.3 Special matrices

TA A=

1 7 9

7 8 2

9 2 12

0 0 4

0 8 0

4 0 1

1 0

0 1

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Anti-Symmetric matrix


An n x nmatrix is anti-symmetricor skewsymmetricif


TA A=

0 7 9

7 0 2

9 2 0

0 22 0

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Properties of Symmetric matrices


IfA is symmetricand Bis any square matrix

T

B B+

TBAB

is symmetric

is symmetric

The above property will be used in T8 Q4


( ) ( )T T T T T T B B B B B B+ = + = +

Hence

( ) ( )T T T T T T T BAB B A B BAB= =

Hence

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Properties of anti-Symmetric matrices


IfA is anti-symmetricand Bis any square matrix

TBAB

TB B is anti-symmetric

is anti-symmetric

The above property will be used in T8 Q4


( ) ( ) ( )T T T T T T B B B B B B = =

Hence

Hence

( ) ( )T T T T T T T BAB B A B BAB= =

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Identity matrix


n x nidentity matrix

sometimes denoted by

In


1 0 00 1 0

0 0 0 1

I=

AI IA A= =

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Vectors as special matrices

Matrices containing only one column are often

called column vectors or vectors

Matrices containing only one row are often

called row vectors or vectors


1

2

1

2

3

[ ]1 2 [ ]1 2 3


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Vectors


=1

0

0

1

=

1

0

0

0

1

0

0

0

1

== =

(0,1)

(1,0)

(1,0,0) (0,1,0)

(0,0,1)


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a

b

1 0

0 1a b

= +

1 0 0

0 1 0

0 0 1

a

b a b c

c

= + +

=

=


ai bj+

ai bj ck + +

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Rotation


Rotation (anti-clockwise)

through angle

Let


cos sin( )

sin cosR

=

cos sinRi i j = +

sin cosRj i j = +

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Rotation


Rotation (anti-clockwise)

through angle

R transforms


cos sinRi i j = +

sin cosRj i j = + sin cosi j +

cos sini j +

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cos sin( )

sin cos

R

=

[ ]

15061506 cos sin

( ) sin cos

cos(1506 ) sin(1506 )

sin(1506 ) cos(1506 )

R

=

=


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Orthogonal matrix


Ann x nmatrix, Bis orthogonal if

is orthogonal

TBB I=

cos sin

sin cos

cos sin cos sin

sin cos sin cos

2 2

2 2

cos sin 0

0 sin cosI

+= =

+


Check

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cos sin

sin cos

( )cos ,sin

( )sin ,cos

( )cos , sin

( )sin , cos

An nn matrix A isorthogonal iff its columns

are perpendicular unitvectors.

An nn matrix A isorthogonal iff its rows areperpendicular unitvectors.

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Involutory matrix ( a matrix that is its own inverse)

AA I=21

a b

A aab

=

Example

Then

AA I=

= 1 2

or

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Shear angle and parallel to x-axis


Let S=

1 tan 1

0 1 0

Si

=

1

0

i

= =

1 tan 0

0 1 1Sj

=


Then

So

tan tan

1

i j

= = +

1 tan

0 1

Si i= tansj i j= +

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Si i= tansj i j= +

tan i j +

multiplication multiplication

j becomes ,turn angle clockwise,measurefrom

tan i j +

j

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(S transforms)


Si i= tansj i j= +

tan i j +

S maps

We may say that S maps i ito

mapsj to

tan i j +

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Si =1

0

Sj=1 tan0 1

S

=

tan

1

1 tan1 2

0 1

= +

1 2 tan

1 0 2 1

+ = +

1 2tan

2

+ =

1

2

1 2tan

2

+

or 1 1 tan 1 1 2tan

2 0 1 2 2S

+ = =

1 (1 2 ) 1 22

S S i j Si Sj = + = +

30Chew T S MA1506-14 Chapter 5


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Si =1

0

Sj=1 tan0 1

S

=

tan

1

1 tan( 1) 2

0 1

= +

1 2 tan

( 1) 0 2 1

+ = +

1 2 tan

2

+ =

1

2

1 2 tan

2

+

or 1 1 tan 1 1 2 tan

2 0 1 2 2S

+ = =

1 (( 1) 2 ) ( 1) 22

S S i j Si Sj = + = +

31Chew T S MA1506-14 Chapter 5


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Question 7 (a) [5 marks] (2007 Exam )In two dimensions, we perform a shear

[shearing angle 45 degrees] by means offorces parallel to an axis which makes anangle of 30 degrees with respect to thepositive x-axis. Find the final location of

the point which was originally at (x, y)coordinates (0, 1).

(0,1)

3045

Rotate 30

Shear45

Rotate 30

parallel to x axis

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cos sin( )

sin cosR

=

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Summary We have learnt

symmetric matrix A

anti symmetric matrix A

identity matrix

orthogonal matrix B

shear matrix

vector

rotation matrix

TA A=

TA A= I

TBB I=

1

2

[ ]1 21 tan

( ) 0 1S

= cos sin

( )sin cos

R

=


Involutory matrix A 2A I=

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The inverse matrix of A is denoted by1

A

WARNING: We only define inverse for

n n matrices, i.e. square matrices.

Let A=11 1

1

n

n nn

a a

a a

be an n n matrix

5.4 Inverse matrix and unique solution


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B=

1

2

n

b

b

b

X=

1

2

n

x

x

x

Consider a linear system AX=B

11 1

1

n

n nn

a a

a a

1

2

n

x

x

x

1

2

n

b

b

b

i.e., =



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which is equivalent to

11 1 12 2 1 1... n na x a x a x b+ + + =

21 1 22 2 2 2... n na x a x a x b+ + + =..............................................

1 1 2 2 ...n n nn n na x a x a x b+ + + =



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Theorem 1The linear system of equations AX=B

has a unique solution if and only if A has an

inverse 1A

How to find 1A if1

A

exists?

We can perform elementary rowoperations to A to get 1A


Remark: B can be any n-dim vector

Proof omitted


See Appendix

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5.5 Determinants and inverse


In this section, we introduce determinants,

which help us to determine whether ann n matrix has an inverse or not.

Consider11 12

21 22

a aa a

1

2

x

x

1

2

b

b

=

Then

11 1 12 2 1

a x a x b+ =

21 1 22 2 2a x a x b+ =

(1)

(2)

5 5 D t i t d i

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Hence 22 12(1) (2)a a get

11 22 12 21 1 1 22 12 2( )a a a a x b a a b = Similarly

21 12 22 11 2 1 21 11 2( )a a a a x b a a b =

Therefore the linear system of equations

has unique solution if and only if



11 22 12 21 0a a a a

5 5 Determinants and inverse

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We define the determinant of

11 12

21 22

a aa a

to be 11 22 12 21a a a a

We write

det11 12

21 22

a a

a a

11 22 12 21a a a a

or 11 1221 22

a aa a

11 22 12 21a a a a

=

=




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Let A=11 12 13

21 22 23

31 32 33

a a a

a a a

a a a

Then we define the determinant of A,

(denoted by det(A) or )A

to bedet(A)

11a2322

32 33

aa

a a

12a2321

31 33

aa

a a

13a+ 21 22

31 32

a a

a a

=

How to get this? see next slide




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called cofactorexpansion

det A



cofactorcofactorcofactor

Animation slide


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Cofactor expansion can be done about any

row or column



2x2 matrix


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Examples



Animation slide


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Cofactor expansion can be used for any

n x ndeterminant



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2ndmethod (cofactor expansion) finding Inverse


Work out the matrix of cofactor of each entry

Take transpose

Divide by determinant


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=

1

=

1

=

1

det

=1

Let

Then


Inverse of 2x2 matrix


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Important Properties of Determinants


size of M

det( ) (det )(det ) det( )ST S T TS = =

det detT

M M=

det( ) detn

cM c M =


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If A is upper (lower) triangular, thendet(A)=product of all diagonal entries of A

1 2 4

0 3 5 1 3 6 18

0 0 6

= =

4 0 0

1 5 0 4 5 6 1202 3 6

= =



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Theorem 2


An nxn matrix A has an inverse if and only if det(A)0

Proof omitted

1 1 1

0 2 3

0 0 3

A

=

1 1 1

0 0 30 0 2

B

=

det 1 2 3 6A= =

det 1 0 2 0B= =

A has an inverse

B does not have aninverse


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Theorem3


Let A be an nxn matrix. Then AX=B has a unique solution

if and only if det(A)0Proof omitted

1

2

3

1 1 1 4

0 2 3 5

0 0 3 6

x

x

x

=

has a unique solution

1 1 1

det 0 2 3 1 2 3 6 0

0 0 3

= =

since

Remark: B can be any n-dim vector


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Theorem 4


Let A be nxn matrix. Then AX=0

has nontrivial (nonzero) solutionsif and only if det(A)= 0.

Proof omitted

Note that A0=0.Hence zero vector 0= is always a solution of AX=0

0

0

0

Zero vector


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1

2

3

1 1 1 0

0 2 3 0

0 0 3 0

x

x

x

=

has a unique solution

Then the unique solution should be1

2

3

0

0

0

x

x

x

=


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Remark 2:AX=0 has either unique solution

(detA0)or infinitely many solutions

(detA=0)

Remark 3:If detA =0 , then AX=B, where B 0,

has two cases: infinitely many solns or no

solns


5 6 L i f I O M d l

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5.6 Leontief Input-Output Model


we use this model to analyze economics of

interdependent sectors

Example ---Oil and Transportation industries

(1) Total output $y of Transportation industry requires

(i) $0.50y gasoline from the oil industry

(ii) $0.20y transportation of equipment from thetransportationindustry


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We will look at a single oil company and a single

transportation company as a closed system


(2) The total output $x of Oil industry requires

(i) $0.12x transportation of gasoline from thetransportationindustry

(ii) $0.32x oil-based fuels for processing fromthe oil industry


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(3)Suppose that the demand from the outsidesector of the economy (all consumers outsideof oil and transportation) is:

$15 billion for oil

$1.2 billion for transportation


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$x b = the total output from oil company=oil demand

$y b = the total output from transportation company

=transportation demand


Internal demand External

demandFrom oilcompany Fromtransportationcompany

Oil demand

x0.32x 0.50y d1= $15b

Transportation

demand y0.12x 0.20y d2= $1.2b

.


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Setting up the demand equations

The total output of each company will equal thesum of the internal and external demands:

Expressed as a matrix equation:


X MX D= +

1

2

0.32 0.50

0.12 0.20

x x y d

y x y d

= + +

= + +

1

20.32 0.500.12 0.20

dx xdy y

= +


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Solving the demand equations

Solve forX

:

In our example:


IX MX D =

called technologymatrix

X MX D= +1( )X I M D=

1

2

0.32 0.50

0.12 0.20

dx x

dy y

= +

( )I M X D =

0.32 0.50

0.12 0.20M

=


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( ) 1 1.65 1.03

0.25 1.40I M

=

0.68 0.50

0.12 0.80I M =


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The solution

Putting it all together:

In order to meet the demand the companies need

to produce$26.0 billion of oil

$5.4 billion of transportation


1

2

151.2

dDd = =

( )1 1.65 1.03 15 26.0

0.25 1.40 1.2 5.4X I M D

= = =


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0.32 0.50 15

0.12 0.20 1.2

x x y

y x y

= + +

= + +Can you solve the above without using matrices ?

Sure, you can

However if it involves many variables, then solving

it by matrices is easier.

End of part I

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Chapter 5

PART TWO


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In part II, we shall study

Eigenvalues and eigenvectors---very important

concepts


Diagonalization of matrix

Weather forecasting model

Trace of a matrix

5 7 Eigenvalues and eigenvectors

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parallel

=

=

Vectors are in differentdirections

Vectors are in samedirection

Consider

5.7 Eigenvalues and eigenvectors

Vectors are in different

directions


Rewrite5.7 Eigenvalues and eigenvectors

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Let T=

T

Then

= =T

So we are interested in the last case

2 221 1

=

T multiplying21

by

is the same as multiplying2

1

by 2

Rewrite


Eigenvector T i t i5.7 Eigenvalues and eigenvectors

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Eigenvector


After multiplication (transformation),

does not change the direction

eigenvalue eigenvector

u

Tu

u

Matrix multiplication

is the same as scalar multiplication

which is much easier

T is nxn matrix

suppose We may use

u uorTu u=


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How to find Eigenvalues and Eigenvectors


We want to find nonzero eigenvectors

Tu u= How to find eigenvalue and eigenvector u

First note that 0 0T =Hence zero vector 0 is always an eigenvector


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we want nonzero vector

By Thm 4 in section 5.5, we have

Hence we can find nonzero eigenvaluesusing above equation

T-isnot meaningful

How to find nonzeroeigenvectors

First note that

Tu u Iu = =

( ) 0T I u =

det( ) 0T I =

Example5.7 Eigenvalues and eigenvectors

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Example


Find eigenvalues of

Cont.5.7 Eigenvalues and eigenvectors

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Cont.


( ) 0T I u =

Now we shall find eigenvectorassociated to eigenvalue =2 u

=

Tu u=

1 2 2 0( 2 ) 2 2 2 0T I

= =

1 2

2 2T

=

2, 3=


Cont.

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two equations same

1 equation 2 unknowns

They are many solutions, i.e. many eiqenvectors

Cont.


Cont.

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choose

Eigenvectors associated to eigenvalue 2 are ofthe form

1

12 2

=

0

get eigenvector

need only to choose one

=

All eigenvectors are

parallel i.e. all arelinearly dependent

Cont.

1=

1

12


Cont.

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eigenvector associated to = -3

two eqs same

Now find

1 3 2 0( ( 3) )

2 2 3 0

T I

+ = =

+

Cont.


Cont.

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choose get eigenvector

1

2 2

=

eigenvectors associated to =-3 is of the form

0

need only to choose one

2 0 + =

=

Cont.

1=1

2

Example5.7 Eigenvalues and eigenvectors

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Example


1

01

i

i

=

0i =

0i = Two eqs same

( ) 0T I

=

det( ) 0T I =T=

consider i=

Cont5.7 Eigenvalues and eigenvectors

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Cont.


eigenvector

0i =0i =

Two eqs same

1

i i

=

1i

get complex eigenvectorchoose 1=

use 0i =

cont.5.7 Eigenvalues and eigenvectors

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cont.


get eigenvector

10

1

i

i

=

0i =

0i + = Two eqs same

1

i i

=

choose get complexeigenvector

Now consider second eigenvalue

get

use 0i + =

i=

1

i

i=


Recallcont.cos sin

( )R

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When rotating through 90 degrees, everyrealvector should change direction,so

has NO real eigenvector

1

i

represents rotation through 90 degrees

are its eigenvectors which are complex

Recall

So the matrix we just mentioned

cont. ( )sin cos

R

=

1

i

5.8 Diagonalization

5.8 Diagonalization

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We say a matrix A is diagonalizable if

Matrix of eigenvectorsDiagonal matrix ofeigenvalues

ANS:If A is nxn matrix , and it has n non paralleleigenvectors, then A can be diagonalized.


1A PDP

=

5.8 Diagonalization

When a given matrix A can be diagonalized ?

Example

5.8 Diagonalization

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Example


has eigenvalues 2 and -3and corresponding eigenvectors

From the first example in section 5.7, we know

Then2 0

0 3

1e

2e

= =

1 2

2 2

1

12

1

2

= =

1 1

12

2

5.8 Diagonalization

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Next we can verify that

Chew T S MA1506-14 Chapter 587

1 2

2 2

P

2 0

0 3

1P

=

Example

5.8 Diagonalization

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p


has only one eigenvalue since

It has only one eigenvector

Not possible to diagonalize

1 tan

0 1

21 tandet (1 ) 00 1

= =

1

0

1 tan

0 1

Findingn

M5.8 Diagonalization

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11 1

2

0

0

n

n n

nM PD P P P

= =

Finding M


S

5.8 Diagonalization

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[ ]

1506

1506cos sin

( )sin cos

cos(1506 ) sin(1506 )

sin(1506 ) cos(1506 )

R

=

=

Some special cases Notused this method

Involutory matrix A 2A I=

(1)

(2)

Then

=

5 9 Model weather forecasting

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5.9 Model weather forecasting

Today Tomorrow ProbabilityRainy Rainy

Sunny

Sunny Rainy

Sunny


60%

70%

30%

40%

We can write the above information inthe following matrix form

5..9 Model weather forecasting

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Current : R S Next:R

columns add to 1


called transit ion matrix

0.6 0.3

0.4 0.7

R R S RM

R S S S

= = S


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Today is sunny, will it be rainy 2 days later?

S

S

R

S

R

S

R

0.7

0.7

0.3

0.3 0.4

0.6

0.39

0.610.12

0.18

0.49

0.21


0.6 0.3

0.4 0.7

R R S RM

R S S S

= =

Observe that


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In fact

2 0.6 0.3 0.6 0.3

0.4 0.7 0.4 0.7

M =

0.6 0.6 0.3 0.4 0.3 0.7 0.6 0.3

0.4 0.6 0.7 0.4 0.7 0.7 0.4 0.3

+ + = + +

Observe that

2 22

2 2

R R S RM

R S S S

=

2

2

0.48 0.39

0.52 0.61

S R

S S

= =

Summary of the previous slide5..9 Model weather forecasting

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Today is sunny, will it besunny 2 days later?


0.6 0.3

0.4 0.7

R R S RM

R S S S

= =

Today is rainy, will it besunny 2 days later?

Probability=0.52 Probability=0.61

2 22

2 2

0.48 0.39

0.52 0.61

R R S RM

R S S S

= =

Today is rainy, will it berainy 2 days later?

Probability=0.48

Today is sunny, will it berainy 2 days later?

Probability=0.39

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T d i R i il l i t b i 30 d l t ?


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Eigenvaluesof


0.6 0.3

0.4 0.7M

=

are

Corresponding eigenvectors are

1 0.3 = 2 1 =

1

1

1

4

3

Today is Rainy, will i t be rainy 30 days later?

Find

30

M Should use30 30 1

M PD P

=

1 1

4 3

7 7


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When use this method?suggest: n>4

41

3

P =

1 7 7

3 3

7 7

P

=

30 630 0.3 0 2 10 0

0 1 0 1D

=

630 30 1

4 31 1

2 10 0 7 74

3 31 0 13 7 7

M PD P

=

3 3

7 7

4 4

7 7

=

5.10 Trace of a Matrix

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Let Mbe a square matrix.

The trace of M, denoted Tr (M),is the sum of thediagonal entries

( ) ( )Tr MN Tr NM=5.10 Trace of a Matrix

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( ) ( )Tr MN Tr NM =

1

( ) ( ) ( )Tr M Tr PDP Tr D

= =Given matrix


For a diagonalizable matrix M,Tr(M)= Tr(D)=sum of its eigenvalues.

Example 1


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Use this to check your calculations ofeigenvalues and to find the remainingeigenvalue


0.6 0.3

0.4 0.7M

=

eigenvalue 1 0.3 =

Tr(M)=0.6+0.7=1.3

Hence the 2ndeigenvalue is2 1 =

a p e

Tr(M)= Tr(D)=sum of its eigenvalues.

Example 25.10 Trace of a Matrix

4 4 matrixA = entries= {1 2 3 16}

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entries= {1,2,3...16}

All rows, all columns,all diagonalshave the same sum

T A BExample 3


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( ) ( )( ) ( )

Tr T Tr A BTr A Tr B

= += +

T A B = +Suppose that

Find1 2 5 6 A=3 4 7 8

T =

[ ] [ ]1 1( ) ( ) ( ) 5 13Tr B Tr T Tr A

= =

( )Tr B

Solution Tr is linear

We need thisproperty inTutorial 9 Q3

R k


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Remark:

det = det(1

)= det(1) = det

= product of eigenvalues

0.6 0.3

0.4 0.7M

=

1 0.3 = 2 1 =eigenvalues

EndChapter 5

DetM=(0.3)(1)=0.3

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cont. cossin ( sin )

d d

dt dt

= =

Appendix

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( )dt dt

is constant

Composing two shears Appendix

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S:shear degrees parallel to x axis

S: shear degrees parallel to x axis

Still a shear but note that

Rotation in 3D Appendix

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Rotate 90 degrees (anticlockwise) about z-axisRotate 90 degrees (anticlockwise) about x-axis

About z-axis About x-axis

Cont. Appendix

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Rotate 90 degrees

(anticlockwise) about z-axis

Rotate 90 degrees

(anticlockwise) about x-axis

D t i t f O th l M t i M

Appendix

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Determinant of Orthogonal Matrix


M

Examples

1P

Appendix

P P exits

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are parallel iff

11 22 12 21det P P P P P=

11 22 12 21det 0P P P P P= =

11 21

12 22

P P

P P= =iff

11

21

P

u P

=

12

22

Pv

P

=

12

22

Pv

P

= =

det 0P=

not parallel iff det 0P

So if not parallel then1

P

exits


Hence

11 12

21 22

P PP

P P

=

,u v

,u v

,u v

Row operations to find inverse

Appendix

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Row operations to find inverse


Refer to Textbook Section 3.3 for more details

Remark: 2

nd

method finding inverse see slide 46

Three row operations on a matrix

multiply a constant to a row

switch two rows add a multiple of another row

(see example)

AppendixCont.

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Example

Let A=

1 4 2

2 8 3

0 1 1

Suppose that we know that A has an inverse.

Now we shall do the following to get 1A


First writeAppendixCont.

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1 4 2 1 0 0

2 8 3 0 1 00 1 1 0 0 1

Try to get zero as many as possible

for the lower triangular part.

1 22R R+

1 4 2 1 0 0

0 0 7 2 1 00 1 1 0 0 1


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1 4 0 3 7 2 7 0

AppendixCont.

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3 12R R +1 4 0 3 7 2 7 0

0 1 0 2 7 1 7 1

0 0 1 2 7 1 7 0

2 14R R +1 0 0 11 7 2 7 4

0 1 0 2 7 1 7 10 0 1 2 7 1 7 0

When we reach this step,we get


1A 11 7 2 7 4

2 7 1 7 1

2 7 1 7 0

=

Consider linear system AX=B where

AppendixCont.

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Consider linear system AX B where

1 4 2

2 8 3

0 1 1

A= X=1

2

3

x

x

x

B= 232

1

Now the inverse of A exists. Hence we have


AX B=1 1A AX A B =

1IX A B

=1X A B=

Therefore

AppendixCont.

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Therefore

1

2

3

x

x

x

11 7 2 7 42 7 1 7 1

2 7 1 7 0

232

1

=

2

3

4

=

(11/ 7)( 2) (2 / 7)(32) ( 4)(1) 2 + + =

( 2 / 7)( 2) ( 1/ 7)(32) (1)(1) 3 + + =

(2 /7)( 2) (1/ 7)(32) (0)(1) 4 + + =

Endof

chewma1506-14 ch5

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