International A LevelFurther Maths: Further Pure 1EdexcelRevision NotesMatricesTransformations using MatricesCombinations of Matrix Transformations

Combinations of Matrix Transformations (Edexcel International A Level Further Maths)

Revision Note

Mark Curtis

Expertise

Maths

Combinations of Matrix Transformations

How do I find a single matrix that represents a combination of transformations?

  • A point open parentheses x comma space y close parentheses can be transformed twice

    • Firstly by the matrix bold P, then secondly by the matrix bold Q

    • This is called a combined (or composite or successive) transformation

  • A single matrix, bold M, representing the combined transformation can be found using matrix multiplication as follows: 

    • bold M equals bold QP 

      • The order matters: the first transformation is on the right in the multiplication

      • This order is the reverse of what you might expect! 

    • Be careful: bold PQ represents bold Q first, followed by bold P second

How do I find the inverse of a combined transformation?

  • The inverse of a product of matrices is the product of the inverses of the matrices in reverse order

    • open parentheses bold AB close parentheses to the power of negative 1 end exponent equals bold B to the power of negative 1 end exponent bold A to the power of negative 1 end exponent

  • Let bold M represent the transformation first by bold P, then second by bold Q

    • That meansbold M equals bold QP from above

  • Algebraically, bold M to the power of negative 1 end exponent equals open parentheses bold QP close parentheses to the power of negative 1 end exponent which gives bold M to the power of negative 1 end exponent equals bold P to the power of negative 1 end exponent bold Q to the power of negative 1 end exponent

    • This shows that the inverse, bold M to the power of negative 1 end exponent, first reverses bold Q , then reverses bold P

      • That is the order we would expect

Worked Example

Three transformations in the x-y plane are represented by the matrices below.

bold A equals open parentheses table row cell negative 1 end cell 0 row 0 cell negative 1 end cell end table close parentheses  represents a rotation of 180° about the origin
bold B equals open parentheses table row cell negative 1 end cell 0 row 0 1 end table close parentheses represents a reflection in the y-axis
bold C equals open parentheses table row 1 0 row 0 cell negative 1 end cell end table close parentheses represents a reflection in the x-axis

(a) Use matrix multiplication to prove that a reflection in the y-axis followed by a reflection in the x-axis is equivalent to a rotation of 180° about the origin.
 

The question requires transformation bold B followed by transformation bold C
This is the same as the matrix bold CB in that order (the first transformation appears on the right)

bold CB

Use matrix multiplication to find bold CB

bold CB space equals space open parentheses table row 1 0 row 0 cell negative 1 end cell end table close parentheses cross times open parentheses table row cell negative 1 end cell 0 row 0 1 end table close parentheses equals open parentheses table row cell open parentheses 1 cross times negative 1 space plus space 0 cross times 0 close parentheses end cell cell open parentheses 1 cross times 0 space plus space 0 cross times 1 close parentheses end cell row cell open parentheses 0 cross times negative 1 space plus space minus 1 cross times 0 close parentheses end cell cell open parentheses 0 cross times 0 space plus space minus 1 cross times 1 close parentheses end cell end table close parentheses

The question claims that this is equivalent to transformation  bold A
Simplify the working above and show that it is the same as matrix bold A

bold CB equals stretchy left parenthesis table row cell negative 1 end cell 0 row 0 cell negative 1 end cell end table stretchy right parenthesis equals bold A
Therefore a reflection in the y-axis followed by a reflection in the x-axis is equivalent to a rotation of 180° about the origin

You would not get the marks for multiplying BC (it must be CB)

(b) A different transformation is represented by bold CD where bold D to the power of negative 1 end exponent equals open parentheses table row 2 0 row 0 1 end table close parentheses.

Find and simplify the matrix representing the inverse of the transformation.

You need to find the inverse of bold CD
You need the rule that open parentheses bold CD close parentheses to the power of negative 1 end exponent equals bold D to the power of negative 1 end exponent bold C to the power of negative 1 end exponent
You can substitute in bold D to the power of negative 1 end exponent from the question

table row cell open parentheses bold CD close parentheses to the power of negative 1 end exponent end cell equals cell bold D to the power of negative 1 end exponent bold C to the power of negative 1 end exponent end cell row blank equals cell open parentheses table row 2 0 row 0 1 end table close parentheses bold C to the power of negative 1 end exponent end cell end table

You need to find bold C to the power of negative 1 end exponent
Use that bold M equals open parentheses table row a b row c d end table close parentheses space space rightwards double arrow space space bold M to the power of bold minus bold 1 end exponent equals fraction numerator 1 over denominator det space bold M end fraction open parentheses table row d cell negative b end cell row cell negative c end cell a end table close parentheses where det space bold M equals a d minus b c

table row cell bold C to the power of negative 1 end exponent end cell equals cell fraction numerator 1 over denominator 1 cross times open parentheses negative 1 close parentheses minus 0 cross times 0 end fraction open parentheses table row cell negative 1 end cell 0 row 0 1 end table close parentheses end cell row blank equals cell open parentheses table row 1 0 row 0 cell negative 1 end cell end table close parentheses end cell end table

(Note that a reflection in the x-axis is its own inverse!)
Substitute this into the working above and multiply the matrices

table row cell open parentheses bold CD close parentheses to the power of negative 1 end exponent end cell equals cell open parentheses table row 2 0 row 0 1 end table close parentheses open parentheses table row 1 0 row 0 cell negative 1 end cell end table close parentheses end cell row blank equals cell open parentheses table row 2 0 row 0 cell negative 1 end cell end table close parentheses end cell end table

table row blank blank cell open parentheses table row 2 0 row 0 cell negative 1 end cell end table close parentheses end cell end table

There are other ways to do this question, for example finding bold D first
If you saw that bold C to the power of bold minus bold 1 end exponent equals bold C (as it is a reflection in the x-axis), explain why clearly

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    Mark Curtis

    Author: Mark Curtis

    Mark graduated twice from the University of Oxford: once in 2009 with a First in Mathematics, then again in 2013 with a PhD (DPhil) in Mathematics. He has had nine successful years as a secondary school teacher, specialising in A-Level Further Maths and running extension classes for Oxbridge Maths applicants. Alongside his teaching, he has written five internal textbooks, introduced new spiralling school curriculums and trained other Maths teachers through outreach programmes.