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Matrices of Geometric Transformations (DP IB Applications & Interpretation (AI)): Revision Note

Paul

Written by: Paul

Reviewed by: Dan Finlay

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DP Applications & Interpretation (AI): HLDP Applications & Interpretation (AI): HL

Matrices of geometric transformations

How do I find the matrix for a transformation?

  • You be asked to find the matrix transformation given an object and its image

    • You need to know three pairs of coordinates of the object and its image

  • You can just substitute coordinates of the object open parentheses x comma space y close parentheses and coordinates of the image open parentheses x apostrophe comma space y apostrophe close parentheses into the equation open parentheses table row a b row c d end table close parentheses open parentheses table row x row y end table close parentheses plus open parentheses table row e row f end table close parentheses equals open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses

    • Each pair of coordinates gives you two equations

    • You can solve the equations to find the unknown values

Transformations with translations

  • It is easiest if you know the images of open parentheses 0 comma space 0 close parentheses, open parentheses 1 comma space 0 close parentheses and open parentheses 0 comma space 1 close parentheses

    • The image of open parentheses 0 comma space 0 close parentheses is open parentheses e comma space f close parentheses

    • The image of open parentheses 1 comma space 0 close parentheses is open parentheses a plus e comma space c plus space f close parentheses

    • The image of open parentheses 0 comma space 1 close parentheses is open parentheses b plus e comma space d plus space f close parentheses

  • For example, suppose (0, 0) ⇾ (2, 5), (1, 0) ⇾ (3, -1) and (0, 1) ⇾ (5, 0)

    • open parentheses 2 comma space 5 close parentheses is the image of the origin so this is the translation vector

      • Subtract this from the image of the other two points

    • The matrix transformation is open parentheses table row 1 3 row cell negative 6 end cell cell negative 5 end cell end table close parentheses open parentheses table row x row y end table close parentheses plus open parentheses table row 2 row 5 end table close parentheses

Transformations with no translations

  • It is easier if you know there are no translations

    • This means that the origin does not change under the transformation

  • In these cases, you can quickly find the transformation matrix by seeing where open parentheses 1 comma space 0 close parentheses and open parentheses 0 comma space 1 close parentheses are mapped to

    • These coordinates are the columns of the matrices

      • open parentheses table row a b row c d end table close parentheses open parentheses table row 1 row 0 end table close parentheses equals open parentheses table row a row c end table close parentheses

      • open parentheses table row a b row c d end table close parentheses open parentheses table row 0 row 1 end table close parentheses equals open parentheses table row b row d end table close parentheses

  • For example, suppose (0, 0) ⇾ (0, 0), (1, 0) ⇾ (3, -1) and (0, 1) ⇾ (5, 0)

    • The transformation matrix is open parentheses table row 3 5 row cell negative 1 end cell 0 end table close parentheses

What are the matrices for common geometric transformations?

  • Most of the following transformation matrices are given in the formula booklet

    • The translation one is not given

Transformation

Matrix

Rotation anticlockwise (or counter-clockwise) through angle θ about the origin

open parentheses table row cell cos theta end cell cell negative sin theta end cell row cell sin theta end cell cell cos theta end cell end table close parentheses open parentheses table row x row y end table close parentheses

Rotation clockwise through angle θ about the origin

open parentheses table row cell cos theta end cell cell sin theta end cell row cell negative sin theta end cell cell cos theta end cell end table close parentheses open parentheses table row x row y end table close parentheses

Reflection in the line y equals left parenthesis tan theta right parenthesis x

open parentheses table row cell cos 2 theta end cell cell sin 2 theta end cell row cell sin 2 theta end cell cell negative cos 2 theta end cell end table close parentheses open parentheses table row x row y end table close parentheses

Enlargement with scale factor k, centre of enlargement at the origin (0, 0)

open parentheses table row k 0 row 0 k end table close parentheses open parentheses table row x row y end table close parentheses

Horizontal stretch (or stretch parallel to the x-axis) with scale factor k

open parentheses table row k 0 row 0 1 end table close parentheses open parentheses table row x row y end table close parentheses

Vertical stretch (or stretch parallel to the y-axis) with scale factor k

open parentheses table row 1 0 row 0 k end table close parentheses open parentheses table row x row y end table close parentheses

Translation by vector

open parentheses table row x row y end table close parentheses plus open parentheses table row p row q end table close parentheses

  • If you are given a line of reflection in the form y equals m x

    • then solve tan theta equals m

Examiner Tips and Tricks

It would be good practice for you to try to derive the matrices in the table above by considering what happens to the points open parentheses 1 comma space 0 close parentheses and open parentheses 0 comma space 1 close parentheses.

Worked Example

Triangle PQR has coordinates P(-1, 4), Q(5, 4) and R(2, -1).


The transformation T is a reflection in the line space y equals x square root of 3.

a) Find the matrix T that represents a reflection in the line space y equals x square root of 3.

3-6-1-ib-hl-ai-only-we2a-soltn

b) Find the position matrix, P’, representing the coordinates of the images of points P, Q and R under the transformation T.

3-6-1-ib-hl-ai-only-we2b-soltn

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    Author
    Reviewer
    Paul

    Author: Paul

    Expertise: Maths Content Creator

    Paul has taught mathematics for 20 years and has been an examiner for Edexcel for over a decade. GCSE, A level, pure, mechanics, statistics, discrete – if it’s in a Maths exam, Paul will know about it. Paul is a passionate fan of clear and colourful notes with fascinating diagrams.

    Dan Finlay

    Reviewer: Dan Finlay

    Expertise: Maths Subject Lead

    Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.