Matrices 2

MatricesMM1 module 3, lecture 2

David Godfrey

Slide number 2

Matrices lecture 2 objectives

• After this lecture you should have a clear understanding of:• Representing sets of points in matrix form;• What the unit square is;• Performing and classifying matrix transformations;• Performing combined matrix transformations;• The effect of an inverse matrix;• The relationship between the the determinant of a

transformation and the change in area of a shape;• The effect of a transformation using a singular matrix.

Slide number 3

Transformations and transitions

• This lecture considers the use of matrices to describe transformations and transitions.

• 2D and 3D transformations are widely used in computer graphics. We will only consider 2D transformations in this lecture.

• State transitions are often used to model large complex systems. Such systems can be described using a set of state variables that undergo changes in value only at discrete time points. These changes can often be represented using matrix equations.

• We will first consider transformations.

Slide number 4

Matrices and transformations

• A coordinate system can be used to uniquely identify points in m-dimensional space as column vectors (m 1 matrices).

• e.g. in 2-dimensional space the points A, B, and C are represented as:

A 11

B 23

C 1 1

A

B

C

Slide number 5

Multiple points

• The position of n points in space can be characterised by a grouping of n column vectors forming an m n matrix.

• e.g. in 2-dimensional space the triangle with vertices A, B, and C can be representedby the matrix:

1 2 11 3 1

A

B

C

Slide number 6

Matrix transformations…

• Matrices can also be used to represent transformations of points.

• e.g. the matrix

will transform A, B, and C to A’, B’, and C’.

2 11 1

A

B

C

A«

B«

C«

Slide number 7

…Matrix transformations…

• i.e.

A

B

C

A«

B«

C«

0

3

1

1

11

12'

5

1

3

2

11

12'

0

3

1

1

11

12'

C

B

A

Slide number 8


• Or, more succinctly in matrix form:

A

B

C

A«

B«

C«

050

313

131

121

11

12

Slide number 9


• Note: the determinant of the transformation matrix gives the scale of the area change.

• e.g. area [A, B, C] = 5, area [A´, B´, C´] = 15

A

B

C

A«

B«

C«

det 2 11 1

3

Slide number 10


• e.g. transform the shape with vertices

• by the matrix

• Note that the determinant gives the scale factor for the change in area

312

11det

,1

2 ,

2

1 ,

1

1 ,

1

2

DCBA

12

11

Slide number 11

…Matrix transformations

• All four vertices can be transformed using a single matrix equation

• Negative determinant reverses the ordering of vertices.

5033

1303

1211

2112

12

11

A

B

CA«

B«

C«

D

D«

Slide number 12

The unit square

• It is instructive to look at the effect of transformations on a simple shape such as the unit square.

• The matrix for the unit square is given by

O P

Q R

O P Q R

0 1 0 10 0 1 1

Slide number 13

a11 a12

a21 a22

O P Q R

0 1 0 10 0 1 1

O' P' Q' R'

0 a11 a12 a11 a12

0 a21 a22 a21 a22

Transforming the unit square

• Transforming the unit square gives information about the transformation matrix

• i.e. transforming the unit square results in [zeros, original matrix, sums of rows]

Slide number 14

Transformations: uniform scaling

• The general form of uniform scaling by k is:

• e.g. uniform scaling by 2 is given by

O P

Q R

O« P«

Q« R«

k 00 k

2 00 2

Slide number 15

Transformations: stretch…

• The general form of a stretch by k in the x–direction is:

• e.g. x–direction stretch by 2 is given by

k 00 1

2 00 1

O P

Q R

O« P«

Q« R«

Slide number 16

…Transformations: stretch

• The general form of a stretch by k in the y–direction is:

• e.g. y–direction stretch by -2 is given by

1 00 k

1 00 2

O P

Q R

O« P«

Q« R«

Slide number 17

Transformations: shear…

• The general form of a shear by k in the x–direction is:

• e.g. x–direction shear by 1.5 is given by

1 k0 1

1 1.50 1

O P

Q R

O« P«

Q« R«

Slide number 18

…Transformations: shear

• The general form of a shear by k in the y–direction is:

• e.g. y–direction shear by -1 is given by

1 0k 1

1 0 1 1

O P

Q R

O«

P«

Q«

R«

Slide number 19

Transformations: reflection

• The general forms of reflections in the x–axis and y–axis are:

• e.g. an x–axis reflection is given by

1 00 1

and 1 0

0 1

1 00 1

O P

Q R

O« P«

Q« R«

Slide number 20

Transformations: rotation…

• The general form of a rotation by k about the origin is:

cos k sin k sin k cos k

O«

R«

P«kQ«

cos(k)

sin(k)

kk

–sin(k)

cos(k)

Slide number 21

…Transformations: rotation

• e.g. the matrices representing rotations of 90° and –30° are:

cos 90 sin 90 sin 90 cos 90

0 11 0

cos 30 sin 30 sin 30 cos 30

3

2

1

2

1

2

3

2

Slide number 22

Combined transformations…

• If the matrices A and B represent two transformations, then the matrix product AB represents the combined transformation of first applying B and then applying A.

• i.e. let the matrix U represent the unit square.Then BU represents the transformation B applied to the unit square.Now A(BU) represents the transformation A applied to the unit square transformed by B.This sequence of transformations is the same as the combined transformation (AB) applied to U (matrix multiplication is associative)ABU = A(BU) = (AB)U

Slide number 23

…Combined transformations…

• e.g. let the matrices A and B represent x–shear by 1 and y–stretch by 2 transformations

2200

3210

1100

1010

20

21

20

21

20

01

10

11

2200

3210

2200

1010

10

11

2200

1010

1100

1010

20

01

20

01 ,

10

11

UAB

AB

BUA

BU

BA

Slide number 24

…Combined transformations

• Note that combined transformations are not commutative.i.e. AB ≠ BAe.g.

2200

2110 ,

2200

3210

20

11

10

11

20

01 ,

20

21

20

01

10

11

20

01 ,

10

11

BAUABU

BAAB

BA

Slide number 25

Inverse transformations…

• Let A be a transformation, so AU = V represents the transformation A applied to the unit square.

• Now if we premultiply the equation by A-1, the inverse of A, we get A-1AU = A-1V

• But A-1A = I , so U = A-1V. In other words A-1

reverses the transformation A.• A-1 (if it exists) is thus the inverse

transformation of A.

Slide number 26

…Inverse transformations

• e.g. y–shear by -1 is given by

A 1 0 1 1

AU 1 0 1 1

0 1 0 10 0 1 1

0 1 0 10 1 1 0

A 1AU 1 01 1

0 1 0 10 1 1 0

0 1 0 10 0 1 1

O P

Q R

O«

P«

Q«

R«

Slide number 27

Inverse transformations: singular…

• What if the transformation is singular?e.g.

A 1 2 1 2

, det A 0

AU 1 2 1 2

0 1 0 10 0 1 1

0 1 2 10 1 2 1

O P

Q R

O«

P«

Q«

R«

Slide number 28

…Inverse transformations: singular

• Singular transformations collapses the unit square into a straight line (or a single point).

• Singular transformations cannot be reversed because there isn’t a one-to-one mapping between the original and transformed spaces.

O P

Q R

O«

P«

Q«

R«

Slide number 29

Transition matrices

• Consider a system that can be characterised by a set of m variables, called state variables.

• Furthermore, these state variables are allowed to change value only at discrete time points, called transitions.

• If the system state variables at transition i are linearly related related to the state variables at transition (i – 1) the transition can be described using a transition matrix.

• Examples of such systems include Markov chains, where the transition matrix elements represent probabilities of changing from one state to another.

Slide number 30

Transition matrices: example 1…

• e.g. A railway has 600 wagons carrying goods from point A to point B. At the end of each week it finds that 30% of the wagons that started the week at A are at B and 20% of the wagons that have started at B are now at A.1) How many wagons are at A and B at the end of two

weeks if 300 wagons started at A and 300 started at B?

2) If there are 400 wagons at A and 200 at B at the end of a week, how many wagons were there at A and at B at the start of the week?

3) How many wagons would need to be at A and at B at the start of the week if there were to be the same numbers at the end of the week?

Slide number 31

…Transition matrices: example 1…

• The transition matrix describes the way the distribution of wagons changes from the start to the end of each week. It does not describe all the changes that may take place during the week.

• Think of the columns of the matrix representing “from” and the rows representing “to”

from A from Bto Ato B

a11 a12

a21 a22

Slide number 32


• A railway has 600 wagons carrying goods from point A to point B. At the end of each week it finds that 30% of the wagons that started the week at A are at B and 20% of the wagons that have started at B are now at A.

• Fill in the elements of the matrix remembering that if 0.3 of A's wagons end up at B then 0.7 remain at A (i.e. go from A to A).

from A from Bto Ato B

0.7 0.20.3 0.8

Slide number 33


1) How many wagons are at A and B at the end of two weeks if 300 wagons started at A and 300 started at B?

• Construct a vector representing the numbers of wagons:

• Calculate the numbers of wagons at the end of week 1:

• Calculate the numbers of wagons at the end of week 2:

• i.e. there are 255 wagons at A and 345 at B.

0.7 0.20.3 0.8

300300

270330

at Aat B

300300

0.7 0.20.3 0.8

270330

255345

Slide number 34


2) If there are 400 wagons at A and 200 at B at the end of a week, how many wagons were there at A and at B at the start of the week?

• If the transition matrix represents the forward process, the inverse represents the reverse process.

• Calculate the inverse of the transition matrix:

• Apply the inverse transformation:

• i.e. there were 560 wagons at A and 40 at B.

0.7 0.20.3 0.8

1

1

0.56 0.060.8 0.2 0.3 0.7

1.6 0.4 0.6 1.4

1.6 0.4 0.6 1.4

400200

56040

Slide number 35

…Transition matrices: example 1…3) How many wagons would need to be at A and at B at the

start of the week if there were to be the same numbers at the end of the week?

• This is called the steady state problem:

0

0

2.03.0

2.03.0

0

0

8.03.0

2.07.0

0

0

8.03.0

2.07.0

10

01 where

8.03.0

2.07.0

y

x

y

x

y

x

y

x

y

x

y

x

I

I

II

y

x

y

x

y

x

8.03.0

2.07.0such that find

Slide number 36



• Since the total number of wagons is 600, the steady state distribution is:

5

3 and

5

2or ,

3

2 gives

0

0

2.03.0

2.03.0

yx

y

yx

x

y

x

y

x

2

5600 240 at A

35600 360 at B

Slide number 37

…Transition matrices: example 1


• Check that the answer is the steady state solution:

360

240

360

240

8.03.0

2.07.0

Slide number 38

Transition matrices: example 2…

• e.g. A company has 2 warehouses with initial stock 20,000 and 10,000 in warehouse A and B, respectively. At the end of each week 70% of stock in A is sold and 2% transferred to B, while 85% of stock in B is sold and 1% transferred to A. How much stock is left in each warehouse at the end of 2 weeks?

• Construct the transition matrix:

• Note: we need an extra row and column to track movement between A, B, and outside.

from A from B from outto Ato B

to out

0.28 0.01 0.000.02 0.14 0.000.70 0.85 1.00

Slide number 39

…Transition matrices: example 2

• How much stock is left in each warehouse at the end of 2 weeks?• Construct a vector representing the stock:

• Calculate the stock at the end of week 1:

• Calculate the stock at the end of week 2:

500,22

800,1

700,5

0

000,10

000,20

00.185.070.0

00.014.002.0

00.001.028.0

0

000,10

000,20

outat

Bat

Aat

020,28

366

614,1

500,22

800,1

700,5

00.185.070.0

00.014.002.0

00.001.028.0

Slide number 40

Matrices lecture 2 objectives

• After this lecture you should have a clear understanding of:• Representing sets of points in matrix form;• What the unit square is;• Performing and classifying matrix transformations;• Performing combined matrix transformations;• The effect of an inverse matrix;• The relationship between the the determinant of a

transformation and the change in area of a shape;• The effect of a transformation using a singular matrix.

Matrices 2

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