6A Geometric transformations and matrix algebra 6B Linear transformations 6C Linear transformations and group theory 6D Rotations 6E Reflections 6F Dilations 6G Shears

syllabus

syllabus

rref

efer

erence

ence

Core topic:

Matrices and applications

In this

In this

cha

chapter

pter

6A Geometric transformations

and matrix algebra

6B

Linear transformations

6C Linear transformations and

group theory

6D Rotations

6E

Reflections

6F

Dilations

6G Shears

6

Transformations

Geometric transformations and matrix

algebra

In your junior mathematics studies you encountered the idea of translation, reflection, rotation and dilation and how these transformations changed the position, size and orientation of the original figure. However, at that stage your investigations were limited to identifying the type of transformation that had taken place, the position of the mirror line or centre of rotation, and perhaps the size of

the image figure.

However, now you have skills with matrices that will allow much greater detailed explanation of the position of images or conversely, the transformation necessary to map point (x, y) onto point (z, w).

The matrix algebra used is very straightforward and because we are lim-iting our discussion at this stage to 2-dimensional space, most of our mat-rices will be of order 2 × 2. Throughout this section you will be reminded of the properties of groups and how transformations involved in matrix algebra can be considered to be a group.

Transformations

A transformation t is an operation which maps each point of the Cartesian plane onto some other point on the plane.

Consider point P(x, y). Under a transformation t this point is mapped onto P′(x′, y′). The point P(x, y) is referred to as the original or pre-image point and P′(x′, y′) is the image.

This transformation can be written in its most general form as (x′, y′) =t(x, y).

t

P (x, y)

x y

P'(x', y')

Find the coordinates of the image points of A(2, −1) and B(3, 0) under the transformation defined by the equations:

x′ = 2x− 4xy+y2− 4 y′ = 6 − 2xy+x− 2y2

THINK WRITE

Think of x′ and y′ as functions of x

and y.

Substitute x= 2 and y =−1 into equations for x′ and y′.

For A(2, −1)

x′= 2(2) − 4(2)(−1) + (−1)2− 4 x′= 4 + 8 + 1 − 4

x′= 9

y′= 6 − 2(2)(−1) + (2) − 2(−1)2 x′= 6 + 4 + 2 − 2

x′= 10 1

1

Translations

The equations used in the previous example define a general transformation or map-ping. A translation is a specific transformation that involves a shift of each point in the same direction.

x′=x+a y′=y+b

Each x-coordinate is moved a units parallel to the

x-axis and each y-coordinate is moved b units parallel to the y-axis.

The image of P is written P′ and this translation can be expressed in matrix equation form as

= + where

1. is the vector holding the image coordinates (x′, y′) of point P′

THINK WRITE

Write the coordinates of the transformed image. The symbol → is used to denote ‘maps onto’.

A(2, −1) → A′(9, 10)

Substitute x= 3 and y= 0 for B. For B(3, 0)

x′= 2(3) − 4(3)(0) + (0)2− 4 x′= 6 − 0 + 0 − 4

x′= 2

y′= 6 − 2(3)(0) + (3) − 2(0)2 x′= 6 − 0 + 3 − 0

x′= 9

Write the coordinates of the transformed image. B(3, 0) → B′(2, 9)

Sketch each original point and its image. Notice that the transformation of A seems quite unconnected with the transformation of B.

2

3

B(3, 0)

A(2, –1)

x

y _A'(9,10)

B'(2, 9)

t t

0

P(x, y)

x y

t _b

a

P'(x', y')

x¢

y¢

x y

a b

x¢

2. represents the original coordinates (x, y) of point P

3. is the translation vector and represents information about the horizontal

and vertical displacement.

Note t (lower case) denotes the translation itself and T (upper case) denotes the matrix of the translation.

Therefore (x′, y′) =t(x, y) can be written in matrix form as

= +T

= +

=

x y

a b

x′ y′

x y

x y

a b

x+a y+b

Find the image of triangle PQR with vertices P(2, -3), Q(0, 1) and R(-1, -2) under the

translation vector . Sketch the original and image figures.

THINK WRITE

State the general translation matrix equation.

= +

First sketch the original points, with coordinates, as shown below. Substitute

x- and y-values for each point in turn.

For P(2, −3)

= +

=

= 5

1

–

1 x′

y′

x y

a b

2

x′ y′

2 3 –

5 1 –

2+5

3

– +–1

7 4 –

2

Note that the image has been moved 5 units to the right and 1 unit down but remains unchanged in shape, area, size and orientation. Such a transformation is said to be

congruent.

Successive translations

The translation above could have been achieved by a succession of translations that have the final effect of 5 across and 1 down. Any number of successive translations

could achieve this: and or the reverse order, and , and so on.

THINK WRITE

P(2, −3) maps to P′(7, −4). P(2, −3) → P′(7, −4) For Q(0, 1)

= +

=

=

Q(0, 1) maps to Q′(5, 0). Q(0, 1) → Q′(5, 0) For R(−1, −2)

= +

=

=

R(−1, −2) maps to R′(4, −3). R(−1, −2) → R′(4, −3)

Sketch the image points with the original.

3

x′ y′

0 1

5 1 –

0+5

1+–1

5 0

4

x′ y′

1 –

2 –

5 1 –

1

– +5

2

– +−1

4 3 –

5 6

Q(0, 1)

P(2, –3) R(–1, –2)

x y

R'(4, –3)

Q'(5, 0)

P'(7, – 4)

5 0

0 1 –

3 2

Show that the translation T1= followed by T2= maps the point P(2, -3) from the previous example to the same point P′(7, -4) as found in worked example 2, and that the order of the translation has no effect on the result.

THINK WRITE

Set up the general matrix equation.

= +T₁

Use T₁ followed by T₂. For P(2, −3)

= +

=

P(2, −3) → P′(5, −1)

is the image, P″ of image P′. = +T₂

= +

=

P′(5, −1) → P″(7, −4)

Therefore P(2, −3) → P′(5, −1) → P″(7, −4)

Use T₂ followed by T₁. = +

=

=

P(2, −3) → P′(4, −6)

= +

=

P′(4, −6) → P″(7, −4)

Therefore P(2, −3) → P′(4, −6) → P″(7, −4) 3

2

2 3

–

1 x′

y′

x y

2

x′ y′

2 3 –

3 2

5 1 –

3 x″

y″

x″ y″

x′ y′

5 1 –

2 3 –

7 4 –

4 x′

y′

2 3 –

2 3 –

2+2

3

– +–3

4 6 –

x″ y″

4 6 –

3 2

7 4 –

3

This example shows, but does not prove that a set of translations is commutative, that is the order of operation does not affect the final result.

Translation of a curve

THINK WRITE

Sketch the translated image in 2 stages.

5

P(2, –3)

x y

P''(7, –4) P'(4, – 6) t1

t₂

Find the equation of the curve y=x2 under the translation of T= .

Continued over page

THINK WRITE

Set up the general matrix equation. = +

To find the image of the curve we must express x and y as found in the original function in terms of x′ and y′.

= −

x = x′− 1

y = y′+ 4 Substitute for x and y in the original

function to obtain the function in terms of the image coordinates.

y′+ 4 = (x′− 1)2

Rearrange and expand this function. _y′=x′2_{− 2x′+ 1 − 4}

y′=x′2− 2x′− 3 Factorise to obtain the zeros for the

function.

x-axis intercepts occur when y= 0 0 = (x – 3)(x+ 1)

x-axis intercepts occur when

x – 3 = 0 or x +1 = 0

3 − x= 3 x=−1

Find the y-axis intercepts. y-axis intercepts occur when x= 0

y= 02− 2(0) − 3

y=−3

1 4

–

1 x′

y′

x y

1 4 –

2

x y

x′ y′

1 4 –

3

4

5

4

Geometric transformations

and matrix algebra

1 Find the image of each of the following points under the transformation defined by x′= 2xy− 3y+x2

y′=xy+ 4y−x

a (0, 0) b (2, −4) c (1, 1) d (−5, −2)

Sketch the original point and its image.

2 Find the image of each of the following points under the translation T= .

a (0, 0) b (2, −4) c (3, 5) d (−4, 1)

3 The vertices of a triangle are given by A(0, 0), B(3, 5) and C(7, 2). Find the image of the vertices under each of the following translations:

a b c d

THINK WRITE

Sketch the original and image functions.

Note that the turning point (0, 0) maps to (1, −4) which was the translation vector.

6 y

(3, 0) (–1, 0)

–3 –4

y = x 2

x

0

y' = x'2 – 2_{x' – 3}

remember

1. A translation T can be written as = +T in matrix equation form.

The matrix is the vector representing the coordinates of the point (x, y)

and represents the coordinates of the point (x′, y′) — the image of (x, y)

after translation.

2. A translation results in an image congruent to the original object.

3. A set of translations is commutative — the order of operation does not affect the final result.

x ′ y′

x y x

y x′ y′

remember

6A

W WORKEDORKED E Examplexample

1

W WORKEDORKED E Examplexample

2

2 5 –

4 2

4 0

0 2 –

4 The line y =−x + 4 undergoes a succession of translations defined by T₁= and

T₂= . Show that the order in which these translations take place has no effect on

the result.

5 The line y= 2x+ 3 undergoes a translation defined by T= . Find the equation of

the image curve and sketch the original curve and its image.

6 Find the equation of the image of each of the following curves under the following translations. Graph the original curve and its image using a graphics calculator.

a y=−x2 b y=x2− 4 c y=x2−x− 6

d x2+y2= 4 e y2+x2+ 6y= 0

7 Use similar methods to those in question 6 to find the translation vector that maps each of the pairs of points:

a (2, 4) → (0, 1) b (4, −1) → (3, 5) c (6, 2) → (2, −5)

Linear transformations

Have you ever wondered how programmers who develop computer games move and manoeuvre characters on a screen to get them to spin or shrink as they appear to move further away from the observer? The study of linear transformation forms the foundation for these changes of form and size — the warping of the plane on which the characters are mapped.

There are many different ways in which the original, or

pre-image, can be changed or moved so that it looks dif-ferent, or is in a different place.

A linear transformation l is a mapping of the pre-image P(x, y) onto the image P′(x′, y′) where:

x′=ax+by y′=cx+dy

for all real values of a, b, c, and d. In matrix form this system is written as:

=

=L where L= and is called the transformation matrix.

WORKED

Example

3

4 3

2 –

1 –

WORKED

Example

4

1 –

2

3 1 –

5 1 –

2 –

0

2 2

4 –

2

x′ y′

a b c d

x y

x′ y′

x y

This type of transformation leaves the origin unchanged and therefore differs from a translation. The transformation matrix can also be extracted from information about the original and image points. An example of this is shown in the following worked example.

Find the images of the vertices of a unit square ABCD under the transformation given by

L =

.

THINK WRITE

Set up the initial matrix equation where

the image of P is given as P′. =

Investigate the transformation of each point in turn.

(Recall that the symbol → is used to denote ‘maps onto’.)

For point D(0, 0)

= =

That is, D(0, 0) → D′(0, 0) For point A(0, 1)

= =

That is, A(0, 1) → A′(2, 1) For point B(1, 1)

= =

That is, B(1, 1) → B′(3, 0) For point C(1, 0)

= =

That is, C(1, 0) → C′(1, −1)

Plot the image on the same axes as the original.

x y

A(0, 1)

D(0, 0)

B (1, 1)

C (1, 0)

1 2

1

– 1

1

x′ y′

1 2

1

– 1

x y

2

x′ y′

1 2

1

– 1

0 0

0 0

x′ y′

1 2

1

– 1

0 1

2 1

x′ y′

1 2

1

– 1

1 1

3 0

x′ y′

1 2

1

– 1

1 0

1 1 –

3

C(1, 0) y

A(0, 1)

D(0, 0)

B(1, 1)

x A'(2, 1)

B'(3, 0)

C'(1, –1)

5

As hinted at in the introduction to this section, there are two ways to conceptualise a transformation. The more obvious way is to imagine that the points move to new pos-itions on the Cartesian plane. The other less obvious notion is that it is actually the Car-tesian plane on which the original points are plotted that undergoes distortions to yield the transformed image. Perhaps the former is more straightforward, but the end product will be the same.

Find the matrix of the linear transformation that maps A(1, 1) onto A′(2, -1) and B(2, -1) onto B′(1, -1).

THINK WRITE

Set up the initial matrix equation. =

State matrix equations for A, A′, B and B′.

For point A:

=

and for point B:

=

Multiply the matrices to arrive at 4 simultaneous equations for 4 unknowns, a, b, c and d.

From the equation for point A:

2 =a + b (1)

−1 =c + d (2)

From the equation for point B:

1 = 2a − b (3)

−1 = 2c − d (4)

Rearrange these equations and enter them into the graphics calculator as an augmented matrix and perform Gaussian elimination on it (using rref as described in chapter 5).

a +b = 2

c + d =−1

2a −b = 1

2c − d =−1

Enter this as:

Display shows [I|X]. a = 1, b = 1, c =− , d =−

Use these values to build L, the linear transformation matrix.

L =

1 x′

y′

a b

c d

x y

2

2 1 –

a b

c d

1 1

1 1 –

a b

c d

2 1 –

3

4

1 1 0 0 2

0 0 1 1 –1

2 –1 0 0 1

0 0 2 –1 –1

5 2₃--- 1

3

---6 1 1

2 3

---– –1₃

---6

Linear transformations

1 ii Which of the following transformations are linear? ii Write the transformation matrices for each of these.

a x′=x+y b x′=x− 1

y′= 2x+y y′=y+ 2

c x′= 2x – 3y d x′= +y

y′= 3x+ 2y

y′= 1 + e x′=x2

y′=y2

2 a Find the images of the points A (1, −2), B (2, 0) and C(−3, 1) under the following transformations:

i ii iii iv

b Sketch the original triangle from a and its 4 different images.

3 Find the image of the points (given below) under the transformation defined by:

x′=x− 2y y′=−2x+y

a A(2, −3) b B(−3, −1) c C(4, 1) Plot the original point and its image in each case.

4 Find the image of the pre-image points A(4, 1), B(−4, 1) and C (0, 5) under the trans-formation defined by:

x′=x+ 3y y′=−x− 2y

Plot the original and image points.

5 Find the matrix of the linear transformation which maps: a (1, 2) → (−3, 1) and (3, 0) → (1, 4)

b (−2, 3) → (0, 0) and (−2, 4) → (1, 1) c (2, −1) → (1, 1) and (2, 1) → (3, 6) d (3, 4) → (5, 0) and (−3, −2) → (−2, 4)

remember

1. A linear transformation can be represented by = or

= L where L= is the transformation matrix that maps

point (x, y) onto the image (x′, y′).

2. The transformed image can be found using = .

3. The transformed image is not congruent to the object. x′

y′

a b

c d

x y

x′ y′

x y

a b

c d

x y

a b c d

1 –

x′ y′

remember

6B

x

1

y

---WORKED

Example

5

2 1 1 1

1

– 0

0 –1

2 0 0 2

1 0 0 1

WORKED

Example

Linear transformations and group

theory

Earlier in your Mathematics C course of study you were introduced to group theory (chapter 4). You found that a system formed a group if the properties of closure and associativity applied and an identity element and inverse existed. These properties apply to many areas of mathematics — including linear transformations. In chapter 4 we investigated whether matrices, in general, formed a group; now we will study groups that perform linear transformations.

Closure

If l₁ is a linear transformation that maps (x, y) → (x′, y′) then (x′, y′) =l₁ (x, y)

If l₂ is a linear transformation that maps (x′, y′) → (x″, y″) then (x″, y″) =l₂ (x′, y′)

Therefore it follows that

(x″, y″) =l₂ [l₁(x, y)]

where l₂ is followed by l₁ and maps (x, y) → (x″, y″). This double transformation can be represented as a single, where l = l₂l₁. This is known as composition of transform-ations, where the order is significant.

From the Mathematics B course you would be familiar with the idea of composition of functions, where g(x) =h(f(x)) indicates that f(x) is the ‘inner’ function within the structure and general shape of h(x).

In matrix form

=L₂

=L₂

=L₂L₁

=L where L is a 2 × 2 matrix and L=L₂L₁

We can verify this result by considering the image of the point P(1, 2) after a linear transformation

L₁= followed by a linear transformation

L₂= . Show that following this double

transformation produces the point P″(3, 1). If we mapped P(1, 2) directly to P″(3, 1) in a single transformation, find the transformation matrix L. Is this transformation matrix L equivalent to L₁L₂ or L₂L₁?

x″ y″

x′ y′

L₁ x y

 

 

 

x y

x y

1 2 3 4 5

1 2 3 4 5 6 P

P" P' l₁

l₂ l

0 y

x

1 2 2 1

3 –3 3

Associativity

As seen with matrix operations, matrix multiplication is associative; that is, (L₁L₂)L₃=L₁(L₂L₃). Therefore linear transformations are associative; that is, (l₁l₂)l₃=l₁(l₂l₃).

Identity

Remember the identity element (IE) is one which leaves the original number unchanged. When dealing with linear transformation this means that matrix multi-plication has been performed which leaves the original point unchanged. This is the identity transformation and is denoted by l_i — and the matrix is I.

For a 2 × 2 matrix, I= .

Inverse transformations

An inverse transformation is one that maps the image back to the original point — where

(x, y) → (x′, y′) → (x, y).

This transformation is denoted by l–1. As with other inverses ll–1=l–1l As with other inverses ll–1=l_i

If l1 and l2 are 2 linear transformations such that L1= and L2= :

a find P′, the image of P (1, 3) under l₁

b find P≤, the image of P′ under transformation l₂

c find the single transformation of P such that l = l₂l₁

d verify that P≤ (as found in part b) is equal to LP

THINK WRITE

a Use matrix operation to find P′, the image of P (1, 3) under l_1..

= =

b Find P≤, the image of P′ under transformation l₂.

= =

c Find the single transformation of P such that l=l₂ l₁.

L = =

d Verify that P≤ (as found in part b) is equal to LP.

LP = = = P″

Therefore P(1, 3) → P″(0, 4)

1 1

2 –2

0 0

2 1

x′ y′

1 1

2 –2

1 3

4 4 –

x″ y″

0 0

2 1

4 4 –

0 4

0 0

2 1

1 1

2 –2

0 0

4 0

0 0

4 0

1 3

0 4

7

WORKED

Example

1 0 0 1

P(x, y)

x y

l–1 l

If L is the linear transformation matrix and A is the transformation matrix which returns the point to the original, then A=L–1.

As with general matrix terminology, the transformation l is non-singular; that is, l

has an inverse if it has a matrix l–1 that will map the image back to the original. Therefore, only linear transformations that have an inverse l–1 can be considered to form a group. If l is singular then the set of linear transformations does not form a group.

Abelian groups

If the composition of linear transformations is commutative, then the set will form an Abelian group. But in general, multiplication of linear transformations is not commuta-tive, that is l₁l₂≠l₂l₁.

a Find the image of the point P(2, 3) under l1 followed by l2 with

L₁= L₂=

b Verify that l₂= l₁–1 in 2 ways.

THINK WRITE

a Set matrices in =LP form. a = =

P′ is the point (13, −1). Now find P″ using P″=L₂P′.

= =

Since P″(x″, y″) = P(x, y), L₂ has mapped P′(x′, y′) back onto the original, therefore L₂ is the inverse linear transformation of L₁.

b Verify this by showing L₂L₁=I. b L₂L₁ = =

Verify by finding the inverse of L₁. L₁–1=

L₁–1=

L₁–1=

L₁–1= L₂

State the conclusion. Therefore L₂= L₁–1

2 3

1 –1

0.2 0.6

0.2 –0.4

1 x′

y′

x′ y′

2 3

1 –1

2 3

13 1 –

2 x″

y″

0.2 0.6

0.2 –0.4

13 1 –

2 3

1 0.2 0.6

0.2 –0.4

2 3

1 –1

1 0

0 1

2 1

ad–bc

--- d –b c

– a

1 2

– –3

--- –1 –3 1

– 2

0.2 0.6

0.2 –0.4

3

8

Images of curves — non-singular transformations

So far we have mainly considered only the images of individual points under linear

transformation where =L . Now consider the image of a curve — essentially

a set of points.

Determine whether the following linear transformation, l1 is singular or non-singular. x′= 2x + y

y′=−2x + 3y

THINK WRITE

State l₁ in matrix form.

L₁= for L =

Test to determine whether the determinant = 0.

|L₁|= ad − bc = 6 −−2 = 8

State your conclusion. Since det L₁≠ 0, L₁ is non-singular, that is, it has an inverse.

1

2 1

2

– 3

a b

c d

2

3

9

WORKED

Example

x′ y′

x y

Find the image of the line y = 2x − 3 under the linear transformation L = .

THINK WRITE

We need to express the original function in terms of the image points so

we need to find and substitute

image points for the original points x

and y.

= L

L–1 = L–1L

= L–1

Evaluate the inverse.

=

=

2 1

0 3

1

x y

x′ y′

x y

x′ y′

x y

x y

x′ y′

2

x y

1

6–0

--- 3 –1

0 2

x′ y′

1 2

--- 1

6

---–

0 1₃

---x′ y′

10

Images of curves — singular transformations

If a linear transformation L is singular, then L does not have an inverse and the method shown in worked example 10 cannot be used. We need to use a different approach as shown in the next worked example.

THINK WRITE

Express x′ and y′ in terms of the original points.

=

Therefore

x = x′− y′

y = y′

Simplify and rearrange the image equation.

Substitute for x and y in terms of the image points, into the original function:

y′= 2( x′− y′) − 3

= x′− y′− 3

y′= x′− 3

Some texts drop the ‘primes’ on x′ and y′

at this stage, but if they are left in it reminds us that the graph of this function is the image of the original.

y′= x′− 4

Graph the original and image functions.

3 x

y

1 2 ---x′–1₆---y′

0x′+1₃---y′

1 2 --- 1

6

---1 3

---4

1 3

--- 1

2 --- 1

6

---1 3

---2 3

---5 3₂--- 1

2

---6 x

y

3

–3

y = 2x – 3

– 41_– 2

y' = 3_–x'– 4

2 1 – 2

11– 2

0

Find the image of the circle x2+ y2= 1 under the linear transformation L = .

Continued over page

THINK WRITE

State the initial transformation in general matrix form.

= L

=

1 2

2 4

1 x′

y′

x y

1 2

2 4

x y

11

THINK WRITE

Find values for x′ and y′. =

Notice that the equation for y′ equals twice the equation for x′. Therefore this should be stated as the function of the image.

y′= 2x′

Sketch the original and image curves.

2 x′

y′

x+2y

2x+4y

3

4

x y

x2 _+y2 _{= 1}

y' = 2x' (0, 1)

(–1, 0) (1, 0)

(0, –1)

remember

1. (a) Linear transformations are closed.

(b) If (x′, y′) =l₁(x, y) where l₁ is a linear transformation that maps (x, y) → (x′, y′) and l₂ is a linear transformation that maps (x′, y′) → ( x″, y″) then (x″, y″) =l₂(x′, y′)

= l₂[l₁(x, y)] where l₂ is followed by l₁

2. Linear transformations are associative; that is, (l₁l₂)l₃=l₁(l₂l₃).

3. The identity transformation is denoted by l_i and is represented by the identity matrix I.

4. An inverse transformation is one that maps the image back to the original point where (x, y) → (x′, y′) → (x, y) and is denoted by l–1.

As with other inverses ll–1=l–1l

=l_i

Only linear transformations that have an inverse l–1 can be considered to form a group.

5. If linear transformations are commutative, then they will form an Abelian group. But in general, multiplication of transformations is not commutative, that is l₁l₂≠l₂l₁.

Linear transformations and

group theory

1 A linear transformation l₁ is defined as x′= 2x+ 5y y′=x+ 3y a What will the image of P(3, 5) be?

b Is this linear transformation singular? c Show that l₁–1(l₁P ) =P.

d Use this linear transformation to state the image of the following curves:

i y=x ii y= 3x+ 2 iii x2+y2= 2

2 Find the image of the circle x2+y2= 9 under each of the following transformations.

a b c

3 Find the image of the circle x2+y2= 9 under each of the following transformations.

a b c

4 Show that under any linear transformation the image of a straight line is itself a straight line.

5 a Sketch the following curves on separate axes. i y= 2x− 1

ii y=−x+ 4

b Find the image of each curve under the linear transformation c Sketch each image as well as the original curve.

6 Find the image of each of the following curves under the linear transformation .

a y=x2

b y= 2x+ 5

Rotations

A rotation is a transformation in which the plane rotates about a fixed point called the centre of rotation. This point is usually taken as the origin, the rotation in an anticlockwise direction is considered a positive rotation and, in a clockwise direction as a negative rotation.

Examine the diagram at right to note that the centre of rotation is the only point that doesn’t move.

6C

WORKED

Example

7, 8, 9

WORKED

Example

10 _{1 3}

1 4

1 3 1 0

1 – –1

3 2

WORKED

Example

11 _{2 0}

2 0

2 4 3 6

8 4 4 2

2 4 1 1

5 3 2 1

x y

0 A B

C

θ A'

In a rotation:

1. each original point rotates through the same angle of rotation.

2. the image is congruent to the original — the length, angle and area remain unchanged in the image. This is referred to as a congruent transformation.

3. r_q denotes rotation in a positive direction through an angle of θ and R_θ is the matrix of rotation.

With all the transformations that will be discussed we will generate matrices based on where the points (1, 0) and (0, 1) are mapped to on the plane, as a result of the trans-formation. These points are represented by columns 1 and 2 of the identity matrix:

↓ ↓

Special rotations

In this section we will discuss transformations involving rotations of 90°, 180°, 270° and 360°, as well as general rotations.

Rotation of 90°

Consider the figure below.

As the plane rotates through θ = 90° about the origin, point (1, 0) will map to point (0, 1) and point (0, 1) will map to point (−1, 0).

↓ ↓

Hence, the identity matrix, I, is altered to to achieve a rotation of 90° about the origin.

It is most important that you recognise the pattern that is displayed by the columns in the matrix and the coordinates of the image points. This concept forms the basis of the next section of work and totally eliminates ‘remembering’ formulas so that you will be able to understand what is happening to the points.

Hence R_90°= and is the matrix of rotation.

In general terms (x, y) → (−y, x)

=

x′=−y y′=x

1 0 0 1

x y

(1, 0) (0, 1)

(–1, 0)

(0, 1)

x y

90º 90º

0 –1 1 0

0 –1 1 0

x y

P(x, y) P'(x', y')

x′ y′

0 –1 1 0

As mentioned earlier, these rotation matrices should not be learned. They are quite similar and can be too readily confused. Sketch the original (1, 0) and (0, 1) points and then use their images to build the rotation matrices.

Rotation of 180°

In the diagrams below, notice that point (1, 0) is mapped onto point (−1, 0) and point (0,1) is mapped onto (0, −1).

Therefore R_180°= where (x, y) → (−x, −y).

Rotation of 270°

In the diagrams below, notice that point (1, 0) is mapped onto point (0, −1) and point (0, 1) is mapped onto point (1, 0).

Therefore R_270°= where (x, y) → (y, −x).

Rotation of 360°

R_360°= because R_360° essentially leaves the original unchanged (or mapped onto

itself).

General rotation of

θ

Consider the points (1, 0) and (0, 1) that are rotated through angle θ about the origin.

x y

(1, 0) 180º

(–1, 0)

y (0, 1)

(0, –1)

180

º

x x

y

180 º

P(x, y)

0

P'(x', y')

1

– 0

0 –1

y y

x (0, 1)

(1, 0) 0

270

º

x (1, 0) 270º

(0, –1)

0 x

y

P(x, y)

270

º

P'(x', y')

0 1 1 – 0

0 1 1 0

y

0

A(1, 0) B(0,1)

P sin θ –sinθ

cosθ cos θ

Q

1 1

θ θ

x A'

Careful examination of the diagram shows that point (1, 0) is mapped onto point (cos θ, sin θ) and point (0, 1) is mapped onto point (−sin θ, cos θ) where

cos θ=x (horizontal) and sin θ=y (vertical)

R_θ=

R_–θ, where θ is taken in a clockwise, negative rotation about the origin, and is shown in the diagram to the right.

R_−θ=

R_−θ= since cos (−θ) = cos θ and sin (−θ) =−sin θ

Both R_θ and R_−θ can be used to confirm the specific cases of R_90°, R_180° and R_270°.

R_90°= =

Remember that when you need to evaluate a trigonometric ratio:

1. sketch the angle concerned in the correct quadrant

2. write the coordinates or length of the sides on the right-angled triangle

3. in the unit circle, the cosine ratio involves only the x-coordinate and the sine ratio involves only the y-coordinate.

Verification of the other angle measures is left as a future exercise.

sinθ

cos

θ θ

cos θ –sin θ

sin θ cos θ y

P(x, y)

θ

– x

P'(x',y')

cos ( )–θ –sin ( )–θ sin ( )–θ cos ( )–θ

cos θ sin θ –sin θ cos θ

cos 90° –sin 90° sin 90° cos 90°

0 –1 1 0

Find the image of the point (2, -2) under a rotation of about the origin.

THINK WRITE

Write the general rotation matrix and sketch the original point (shown below).

R_θ=

Substitute for θ.

(Note: The small c is the symbol for circular or radian measure.)

R_θ= π 4

---c

1 cos θ –sin θ

sin θ cos θ

2 π

4 ---c

cos π 4

--- –sin π

4

---sin π 4

--- cos π

4

---12

THINK WRITE

Always use a sketch to develop the matrix. R_θ=

Set up the general matrix form for transformations.

=

=

Rationalise the denominator and simplify. =

=

=

Sketch the original and the image points.

3

√2 1

1

– 4 π

– 4 π

1

2

--- 1

2 ---–

1 2

--- 1

2

---4 x′

y′

1 2

--- 1

2 ---–

1 2

--- 1

2

---2 2 –

2

2 --- 2

2 ---+

2 2 --- 2

2 ---–

5

4

2

---0

4 2

--- 2

2 ---×

0

2 2 0

6

(2, –2) x y

– 4 π

(2 2, 0)√

Find the image of the line y=−x+ 4 under the rotation of about the origin.

THINK WRITE

Write the general R_θ matrix. R_θ=

Substitute for θ and evaluate using

the relevant triangle of ratios.

=

=

Set up the general transformation

matrix model, rearranged so that

is the subject.

=

=

Evaluate the inverse of R. =

=

Multiply out the matrices. x= x′+ y′

y=− x′+ y′

Substitute for x and y in the original function.

y= −x+ 4 becomes

− x′+ y′=− x′− y′+ 4

After applying the Distributive Law and rationalising the denominator, this expression can be simplified.

y′= +

y′= ( − 2)x′+ 4( − 1)

π

6

---c

1 cos θ –sin θ

sin θ cos θ

2 π

6 ---c

1 2

√ 3

– 6 π – 3 π R_π 6

---cos π 6

--- –sin π 6

---sin π 6

--- cos π 6 ---3 2 --- 1 2 ---– 1 2 --- 3 2 ---3 x y x′ y′ R_π 6 ---x y x y R_π 6 ---1 – x′

y′

4 ₃ 1

4 --- –1

4 ---– ---3 2 --- 1 2 ---1 2 ---– 3 2 ---x′ y′ 3 2 --- 1 2 ---1 2 ---– 3 2 ---x′ y′ 5 3 2 --- 1 2 ---1 2 --- 3 2 ---6 1 2 --- 3 2 --- 3 2 --- 1 2

---7 (1– 3)x′

3+1

--- 8 3+1

---3 3

13

Rotations

1 Construct matrices for the following anticlockwise rotations about the origin (the

angles are given in radians).

a b π c d 2π

2 Find the image of the following points under the given anticlockwise rotations about

the origin.

a (2, 1) θ=

c

b (0, 4) θ=πc

c (6, 3) θ= c d (1, −3) θ=−60°

e (2, 3) θ= 90° f (1, 1) θ= c

3 a Find the equation of the image of the line y=−3x+ 1 as a result of the following

rotations:

i θ= 45° ii θ= c iii θ=− c

b Sketch each original line and its image.

4 Find the equation of the image of the circle x2+y2= 1 after a rotation of

c

. What do

you notice? Can you explain why this is so?

THINK WRITE

Use your calculator only at the end to simplify surds for sketching purposes. 8

x y

y = –x + 4

y' = (√3 – 2)x' + 4 (√3 – 1)

remember

1. For general rotation θ in an anticlockwise direction about the origin

R_θ= .

2. Use the special right-angled triangles to obtain the trigonometric ratios. 3. Rotation is a congruent transformation.

cos θ –sin θ sin θ cos θ

remember

6D

π 2

--- 3π

2

---W WORKEDORKED

E Examplexample

12

π 3

---π 4

---π 6

---W WORKEDORKED

E Examplexample

13

π 2

--- π

2

---Work

SHEET

6.1

---Reflections

A reflection is a linear transformation in which every point of the original is reflected through a straight line called a mediator. This line can be thought of as a mirror. The diagram at right shows LABC reflected through the mediator m, at x= 1.

In a reflection:

1. corresponding points of the image and original figures are equidistant from and

perpendicular to the mediator

2. length, angle and area of the image and

original are unchanged, hence it is a congruent transformation 3. any points of the original on the mediator are left unchanged.

We usually let m denote the reflection transformation and M the reflection matrix.

=

=

=

=

x y

A

B C

x = 1 m A'

Reflection in the

x

-axis (where

y

0)

↓ ↓

Again, sketch the points (1, 0) and (0, 1) from the identity matrix I= .

Under a reflection in the x-axis, point (0, 1) will map to (0, −1) and point (1, 0) will map onto itself because it is on the mediator.

Therefore M_{y = 0}= .

Reflection in the

y

-axis (where

x

0)

If you sketch the original points (1, 0) and (0, 1) you will notice that if these points are reflected in the y-axis then point (1, 0) will map to (−1, 0) and point (0, 1), which is on the mediator, will map onto itself.

Therefore, M_{x = 0}= .

1 0 0 1

x y

(0, 1)

(1, 0)

(0, –1)

my = 0 0

1 0 0 –1

x y

(0, 1)

(1, 0) (–1, 0)

0 mx = 0

1 – 0

0 1

Find the image of point (3, 1) under reflection My= 0. Sketch the original and its image.

Continued over page

THINK WRITE

Sketch the diagram to construct your reflection matrix.

M_{y= 0}=

Write the initial transformation matrix statement.

= M_{y= 0} 1

x y

(0, 1)

(1, 0)

(0, –1)

my = 0 0

1 0

0 –1

2 x′

y′

x y

14

THINK WRITE

Substitute the necessary values and evaluate.

=

=

Sketch the original and image points. The image is the point (3, −1).

3 1 0

0 –1

3 1

3 1 –

4

x y

P(3, 1)

P'(3, –1)

0 my = 0

Find the image of y=x under reflection in the y-axis. Sketch the original and its image.

THINK WRITE

Sketch the diagram to construct your reflection matrix.

M_{x= 0}=

Write the initial transformation matrix statement and rearrange it to have the original points as the subject.

= M_{x= 0}

= M–1

Substitute for M_{x = 0} and evaluate the inverse.

=

= 1

x y

(0, 1)

(1, 0) (–1, 0)

0 mx = 0

1

– 0

0 1

2 x′

y′

x y

x y

x′ y′

3 x

y 1

1 –

--- 1 0

0 –1

x′ y′

1

– 0

0 1

x′ y′

15

WORKED

Example

Reflection in line

y

x

To find this reflection, sketch the situation as described. Remember to note the main points from the introduction to this section:

1. corresponding points of the image and original figures are equidistant from and perpendicular to the mediator 2. length, angle and area of the image and original are

unchanged, hence it is a congruent transformation

3. any points of the original on the mediator are left unchanged. We find that (1, 0) and (0, 1) map to each other, therefore

M_y=x= .

THINK WRITE

Multiply to give expressions for x and y. x =−x′ y = y′

Substitute for x and y into the original equation.

y′=−x′

Sketch the original and image graphs. Note the origin is left unchanged.

4

5

0 x

y

y = x

y' = –x' mx = 0

= =

x y

y = x

(1, 0) (0, 1)

0

0 1 1 0

Find the equation of the image y = x2 reflected in the line y = x.

THINK WRITE

Sketch the relevant diagram to establish the reflection matrix.

M_y=x=

Continued over page 1

= =

x y

y = x

(1, 0) (0, 1)

0

0 1 1 0

16

Reflection in the line

y

=

x

tan

q

This line might be more easily recognised as y=mx, where m is the gradient of the line which passes through the origin.

Remember that the gradient m=

and tangent ratio = .

Therefore the tangent and gradient ratios provide the

same information: .

Carefully examine these diagrams that illustrate reflection of the points (1, 0) and (0, 1) in the line

y=x tan θ.

THINK WRITE

Set up the initial matrix equation and

rearrange to have x and y as the subject. = M_y=x

= M_y=x

Find the inverse of M_{y = x}. =

=

Multiply matrices to determine x and y. x = y′ y = x′

Substitute for x and y in the original expression.

y = x2 becomes x′= y′2 y′ =

Sketch the original and image curves. Note that the points (1, 1) and (0, 0) are unchanged as they are on the mediator.

2

x′ y′

x y

x y

x′ y′

3 1

1 –

--- 0 –1

1

– 0

x′ y′

0 1 1 0

x′ y′

4

5

x′ ±

6 y

y = x y = x2

y = x

(1, 1)

x y' =√x'

y' = –√x' 0

–1

= =

x y

θ

y = x tanθ

θ

θ θ

A(1, 0) 1

A' (cos 2 , sin 2 )

0

B'(cos(90° – 2 )θ, – sin(90° – 2 θ) ) B(0, 1)

θ

90° – 2

θ

90° –

x y

θ

0

= =

θθ

my = x tan

y₂–y₁ x₂–x₁

---y₂–y₁ x₂–x₁

---Note the following from these diagrams. For the point (1, 0):

1. point A is reflected to a point equidistant from and perpendicular to the line 2. the angle from the x-axis to A′ is 2θ

3. the x-coordinate of the right-angled triangle is cos 2θ 4. the y-coordinate of this triangle is sin 2θ.

5. Hence point (1, 0) → (cos 2θ, sin 2θ). For the point (0, 1):

1. point B is reflected to a point equidistant from and perpendicular to the line 2. ∠MOB = 90° −θ therefore ∠MOB′= 90° −θ

3. therefore ∠XOB′= (90° −θ) −θ= 90° − 2θ 4. the x-coordinate = cos (90° − 2θ)

5. the y-coordinate =−sin (90° − 2θ) because the angle is in the fourth quadrant. 6. Hence point (0, 1) → [cos (90° − 2θ), −sin(90° − 2θ)].

7. Using trigonometric ratios, this simplifies to yield (sin 2θ, −cos 2θ). (Remember that sin 30° = cos 60°, etc.)

Using all this information from the reflection of points (1, 0) and (0, 1) in the line

y=x tan θ yields: M_{y = x tan θ}= cos 2θ sin 2θ . sin 2θ –cos 2θ

Find the matrix for the reflection in the line y = x. (Note that the sign applies only to the 3.)

Continued over page

THINK WRITE

Use a sketch to express as the tangent ratio of some angle.

tan =

State the general reflection matrix in the line

y=x tan θ, then substitute for θ.

M_{y=x tan θ} =

M_{y= x} =

3

1 3

1

2 √3

– 6 π

– 3 π

π 3

--- 3

2

π

3

---cos 2θ sin 2θ

sin 2θ –cos 2θ

3

cos 2π 3

--- sin 2π 3

---sin 2π 3

--- –cos 2π 3

---17

THINK WRITE

Evaluate these ratios using the following triangle. = 3 –1 2 √3 – 3 π 2 _—

3 π −1 2 --- 3 2 ---3 2 --- 1 2

---Find the image of the line y =−x − 1 as reflected in the line y = x.

THINK WRITE

Use the matrix from the previous example as M_{y = x}.

M_{y= x}=

Set up the initial matrix transformation and inverse statement.

= M_{y= x}

= M_{y= x}

Find the inverse and multiply the matrices.

=

=

x =− x′+ y′

y = x′+ y′

3 1 3 3 1 – 2 --- 3 2 ---3 2 --- 1 2

---2 x′

y′ 3

x y x y 3 x′ y′ 3 x y 1 1 4 ---– 3 4 ---– ---1 2

--- − 3

2

---− 3

2 --- −1

Reflections

1 Write the matrices for the following reflections:

a m_x=0 b m_y=0 c m_y=x

d m_y=2 e m_{y= x} f m_y=−x

2 Find the images of each of the following points under the reflection given below. Sketch each original and its image.

a y-axis b x-axis c y=−x d y= x

i (3, −1) ii (4, 2) iii (−1, −3) iv (−2, 4) v (3, 0) vi (−2, −1)

THINK WRITE

Substitute for x and y into the original equation. Make sure you carry through the minus sign from the function.

y= −x− 1 becomes

x′+ y′= x′− y′− 1

y′+ y′= x′− x′− 1

Simplify and rationalise the

denominators to find the equation of the image line.

The last line is included for ease of graphing only.

y′= x′− 1

y′= ( − 2)x′+ 1 −

=−0.27x′− 0.73

Sketch the original and its image.

4

3 2

--- 1

2

--- 1

2

--- 3

2

---1 2

--- 3

2

--- 1

2

--- 3

2

---5 1+ 3

2

--- 1– 3

2

---3 3

6

0 x

y

y' = –0.27x' – 0.73

y = –x – 1

y =√3x

remember

1. Reflection is a congruent transformation. 2. Reflection occurs through a mediator, m.

3. Reflection in the line y=x tan θ is represented by M= cos 2θ sin 2θ .

sin 2θ –cos 2θ

remember

6E

3

W WORKEDORKED E Examplexample

14

---3 Find the image of the following curves under each of the reflections given below. a y=x b y=x2 c y= 2x2+ 1 d y=−x2

i y-axis ii x-axis iii y=−x iv y= x (part a only)

Dilations

So far we have investigated 4 kinds of transformations. The translation shifted the figure on the plane, the general linear transformation produced an image that, on occasions, bore little resemblance to its original.

The rotation and reflection transformations are congruent transformations with the original

basically repositioned on the plane. A dilation

is a transformation in which point P and image

P′ are collinear from a fixed point, usually the

origin O, as shown in the figure at right.

The length OP′=kOP where k is referred to as

the dilation factor.

If k > 0, a dilation may be an enlargement (for k > 1) or a reduction (for 0 < k < 1).

If k < 0 then the image of the original has been mapped through the origin in a

reverse direction.

In this diagram, k=− , therefore the image

appears half the distance from the fixed point O and on the opposite side of O to the original points.

In a dilation:

1. length and area are not preserved; the shape will appear similar, but not congruent to

the original

2. the dilation d is denoted by the matrix D_{k, x} with the dilation factor of k given parallel

to the x-axis and the anchor line being the y-axis.

W WORKEDORKED

E Examplexample

15,16,17,18

3 3

---O

P

P'

O

A

A'

B

B'

C C'

k > 1

O C' C

B A

A'

B'

0 < k < 1

1 2

---O

A

C

B A'

B'

C'

Dilation parallel to the

x

-axis

The dilation matrix D_{k, x} of the points (1, 0) and (0, 1)

under the dilation d_{k, x}is given by D_{k, x}=

where

1. (0, 1) is left unchanged since it is on the anchor line 2. the x-coordinate is mapped k× 1 units away from the

anchor line.

This is shown graphically in the figure at right.

Dilation parallel to the x-axis can be thought of as pulling the plane away from the fixed point or anchor line — in this case, the y-axis.

Dilation parallel to the

y

-axis

This dilation pulls the plane away from the x-axis so all points on the x-axis are anchored.

The figure at right shows that

D_{k, y}=

where

1. (1, 0) is left unchanged since it is on the anchor line 2. the y-coordinate is mapped k× 1 units away from

the anchor line (the x-axis).

x y

(0, 1)

(1, 0) (k, 0)

0

k 0 0 1

0 x

y

(0, 1)

(1, 0) (0, k)

1 0 0 k

Find the coordinates of the image of point (4, 3) under the dilation factor of -2 parallel to the x-axis.

Continued over page

THINK WRITE

Sketch the dilation and construct the matrix from the sketch.

D_{–2, x}=

Write the general transformation matrix equation.

= D_{–2, x} 1

x y

(0, 1)

(1, 0) (–2, 0)

0

2

– 0

0 1

2 x′

y′

4 3

19

THINK WRITE

Multiply the matrices to produce the image coordinates.

=

=

Point (4, 3) maps to image point (−8, 3) under a

dilation of −2 parallel to the x-axis. 3

x y

P(4, 3) P'(–8, 3)

0

x′ y′

2

– 0

0 1

4 3

8 –

3

Find the equation of the image of y = 2x + 1 under the dilation d2, x.

THINK WRITE

Sketch the dilation and construct the matrix from the sketch.

D_{2, x}=

Set up the general transformation matrix equation and rearrange to

have as the subject.

= D_{2, x}

= D_2, –1_x

Find the inverse of D_{2, x} and substitute it into the equation.

=

=

Multiply the matrix equations. x = x′ y = y′

Substitute x and y in the original equation (y= 2x+ 1).

y = 2x + 1 y′= 2( x′) + 1 y′= x′+ 1 1

0 x

y

(0, 1)

(1, 0) (2, 0)

2 0

0 1

2

x y

x′ y′

x y

x y

x′ y′

3 1

2

--- 1 0

0 2

x′ y′

1 2 --- 0

0 1

x′ y′

4 1₂

---5

1 2

---20

THINK WRITE

Sketch the original and its image. Note that (0, 1) remains unchanged since it is on the anchor line of the

y-axis.

6

0 x

y

y = 2x + 1

y' = x' + 1

(0, 1)

Find the image of the circle x2+y2= 9 with a dilation factor of parallel to the y-axis.

Continued over page

THINK WRITE

Sketch the situation and use this to construct the dilation matrix.

=

=

Rearrange to put as the subject. = –1

Calculate the inverse of D and substitute it into the equation.

=

=

Multiply the matrices and write expressions for x and y.

x = x′ y = 3y′

Substitute x and y into the original equation and rearrange to fit the general equation of an ellipse.

x2+ y2= 9 becomes (x′)2+ (3y′)2= 9 x′2+ 9y′2= 9

1 3

---1

(0, 1_–)

3

x y

(0, 1)

(1, 0) 0

D1 3 ---,y

1 0

0 1

3

---x′

y′ D13---,y

x y

x y

x y

D1 3 ---,y

x′ y′

2 1 ₁

3

---

---1 3 --- 0

0 1

x′ y′

1 0

0 3

x′ y′

3

4

21

If you think about the original shape and its image as shown in this example you will understand that the dilation factor of , in effect, shrinks the original shape, parallel to the y-axis so the figure falls back towards the anchor line (the x-axis) and leaves all points on the x-axis unchanged.

Dilation about the origin,

d

The previous dilations have been mapped parallel to an axis, where that axis has provided the anchor line for the stretching of the plane. However, a dilation about the origin does not anchor to a line, but rather to a point — the origin. The diagram at right shows this stretch that results in both x and y coordinates being mapped a dilation factor of k from the origin. Therefore, if the original point is on the origin it will map onto itself. Only the dilation factor is given in the dilation matrix:

D_k=

THINK WRITE

This can be written as + = 1

since is the general equation of an

ellipse about the origin. Therefore

a = 3 So the length of the semi-major axis is 3.

b = 1 So the length of the semi-minor axis is 1.

Sketch the original and its image.

x2 9

--- y

2

1

---x2

a2 --- y

2

b2

---+ = 1

5

x y

x2 _{+ y}2 _{= 9}

— —

x' 2

+ y'2 = 1

9 1

0

1 3

---x y

(0, 1)

(1, 0) (0, k)

(k, 0) P (x, y)

P'(k_x, k_y)

0

Find the image of y = x2 under a dilation factor of −2 about the origin.

THINK WRITE

Sketch the situation to construct the matrix.

D_–2=

Set up the initial transformation matrix equation.

= D_–2

=

Evaluate the inverse and multiply. =

=

x =− x′

y =− y′

Substitute for x and y into the original equation and simplify.

y = x2 becomes − y′= (− x′)2

= x′2

=− x′2

Sketch the original and its image. The minus sign results in the image reversing its position with respect to the origin, and the factor of 2 results in the broader parabola.

1

x y

(1, 0)

(0, –2) (0, 1) (–2, 0)

0

2

– 0

0 –2

2 x′

y′

x y

x y

D_––₂1 x′ y