GCSEFurther MathsAQARevision Notes5. Matrix TransformationsMatrix TransformationsTransformations with Matrices

Transformations with Matrices (AQA GCSE Further Maths): Revision Note

Exam code: 8365

Written by: Mark Curtis

Reviewed by: Dan Finlay

Updated on 27 June 2025

Transforming a point

How do I transform a point using a matrix?

A point (x, y) in a 2D plane can be transformed on to another point (x',y') by a matrix, M
- (x, y) is the object and (x',y') is the image
The coordinates of the image point can be found using matrix multiplication
To transform (x, y) by the matrix $(\begin{matrix} a & b \\ c & d \end{matrix})$
- Write (x, y) as a column vector, $(\begin{matrix} x \\ y \end{matrix})$
- Use matrix multiplication to work out $(\begin{matrix} a & b \\ c & d \end{matrix}) (\begin{matrix} x \\ y \end{matrix})$ , which gives $(\begin{matrix} x' \\ y' \end{matrix})$
- Write down the image point coordinates, (x', y')
In harder questions you may be given the image coordinates, (x', y') and asked to find the original coordinates
- introduce letters (e.g. x and y) for the original coordinates, (x, y)
- use the matrix M to set up and solve simultaneous equations to find x and y

Worked Example

A matrix, M, is given by $M = (\begin{matrix} 4 & 5 \\ 1 & - 2 \end{matrix})$ .

(a) Work out the coordinates of the image of the point $(2, 3)$ using the transformation represented by M.

Multiply the transformation matrix $M$ by the coordinates, written as a column vector

$(\begin{matrix} 4 & 5 \\ 1 & - 2 \end{matrix}) (\begin{matrix} 2 \\ 3 \end{matrix}) = (\begin{matrix} 4 \times 2 + 5 \times 3 \\ 1 \times 2 + - 2 \times 3 \end{matrix}) = (\begin{matrix} 23 \\ - 4 \end{matrix})$

Rewrite the answer as coordinates

$(23, - 4)$

(b) The image of another point, P, using the transformation represented by M is $(11, 6)$ .

Work out the coordinates of P.

You do not know the coordinates of P, so we can write it as $(x, y)$

Let $P = (x, y)$

Multiply the transformation matrix $M$ by the coordinates P, written as a column vector
This time, you know the image of the point after it is transformed, so can fill this in as the "answer"
$\begin{array}{rcl} (\begin{matrix} 4 & 5 \\ 1 & - 2 \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) & = & (\begin{matrix} 11 \\ 6 \end{matrix}) \\ (\begin{matrix} 4 x + 5 y \\ x - 2 y \end{matrix}) & = & (\begin{matrix} 11 \\ 6 \end{matrix}) \end{array}$

Equate the matching elements of the two matrices
$\begin{array}{rcl} 4 x + 5 y & = & 11 \\ x - 2 y & = & 6 \end{array}$

You now have a pair of simultaneous equations which can be solved using either elimination or substitution

Use substitution to rearrange the second equation to make $x$ the subject

$x = 2 y + 6$

Substitute this into the first equation, and solve to find $y$

$\begin{array}{rcl} 4 (2 y + 6) + 5 y & = & 11 \\ 8 y + 24 + 5 y & = & 11 \\ 13 y + 24 & = & 11 \\ 13 y & = & - 13 \\ y & = & - 1 \end{array}$

Substitute $y = - 1$ into the second equation to find $x$

$\begin{array}{rcl} x - 2 (- 1) & = & 6 \\ x + 2 & = & 6 \\ x & = & 4 \end{array}$

P is $(4, - 1)$

Rotation matrices

How do I find rotation matrices?

Imagine the unit square OABC
- It has a side-length of 1 unit
- O is the origin

The coordinates of A and C as column vectors are
- $A = (\begin{matrix} 1 \\ 0 \end{matrix})$ and $C = (\begin{matrix} 0 \\ 1 \end{matrix})$
Under a rotation about the origin, A moves to A' and C moves to C'
- The matrix, M representing this rotation is $M = (\begin{matrix} A' | & C' \end{matrix})$
- A' and C' are column vectors of their new positions
  - The points O and B are not needed, as we can draw the rotated square using just A' and C' (as O won't move)
For example:
- To find the matrix representing a rotation of 90° anticlockwise about the origin
  - A goes to $A' = (\begin{matrix} 0 \\ 1 \end{matrix})$ (on the positive y-axis)
  - C goes to $C' = (\begin{matrix} - 1 \\ 0 \end{matrix})$ (on the negative x-axis)
  - $M = (\begin{matrix} A' | & C' \end{matrix}) = (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})$
- To find the matrix representing a rotation of 180° about the origin
  - A goes to $A' = (\begin{matrix} - 1 \\ 0 \end{matrix})$ (on the negative x-axis)
  - C goes to $C' = (\begin{matrix} 0 \\ - 1 \end{matrix})$ (on the negative y-axis)
  - $M = (\begin{matrix} A' | & C' \end{matrix}) = (\begin{matrix} - 1 & 0 \\ 0 & - 1 \end{matrix})$
  - This is the same as $M = - I$ where I is the identity matrix

Worked Example

The matrix M represents a rotation of 270° anticlockwise about the origin.
Work out M.

A rotation of 270° anticlockwise is the same as a rotation of 90° clockwise

Consider how the points A and C on the unit square are transformed

The point A $(\begin{matrix} 1 \\ 0 \end{matrix})$ moves to A' $(\begin{matrix} 0 \\ - 1 \end{matrix})$

The point C $(\begin{matrix} 0 \\ 1 \end{matrix})$ moves to C' $(\begin{matrix} 1 \\ 0 \end{matrix})$

The transformation matrix is given by $M = (\begin{matrix} A' | & C' \end{matrix})$

$M = (\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix})$

Reflection matrices

How do I find reflection matrices?

Imagine the unit square OABC
- It has a side-length 1 unit
- O is the origin

The coordinates of A and C as column vectors are
- $A = (\begin{matrix} 1 \\ 0 \end{matrix})$ and $C = (\begin{matrix} 0 \\ 1 \end{matrix})$
Under a reflection about an axis (or y = ± x), A moves to A' and C moves to C'
- The matrix, M representing this reflection is $M = (\begin{matrix} A' | & C' \end{matrix})$
- A' and C' are column vectors of their new positions
  - The points O and B are not needed, as we can draw the reflected square using just A' and C' (as O won't move)
For example:
- To find the matrix representing a reflection about the x-axis
  - A stays where it is, so $A' = (\begin{matrix} 1 \\ 0 \end{matrix})$
  - C goes to $C' = (\begin{matrix} 0 \\ - 1 \end{matrix})$ (on the negative y-axis)
  - $M = (\begin{matrix} A' | & C' \end{matrix}) = (\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix})$
- To find the matrix representing a reflection in the line y = x
  - A goes to $A' = (\begin{matrix} 0 \\ 1 \end{matrix})$ (on the positive y-axis)
  - C goes to $C' = (\begin{matrix} 1 \\ 0 \end{matrix})$ (on the positive x-axis)
  - $M = (\begin{matrix} A' | & C' \end{matrix}) = (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})$
  - This is not the same as the identity matrix as the 1s are on the wrong diagonal

Worked Example

(a) The matrix M represents a reflection in the y-axis.
Work out M.
Consider how the points A and C on the unit square are transformed by a reflection in the y-axis

The point A $(\begin{matrix} 1 \\ 0 \end{matrix})$ moves to A' $(\begin{matrix} - 1 \\ 0 \end{matrix})$

The point C $(\begin{matrix} 0 \\ 1 \end{matrix})$ remains in the same place

The transformation matrix is given by $M = (\begin{matrix} A' | & C' \end{matrix})$

$M = (\begin{matrix} - 1 & 0 \\ 0 & 1 \end{matrix})$

(b)

The matrix N represents a reflection in the line $y = - x$ .

Work out N.

Consider how the points A and C on the unit square are transformed by a reflection in the line $y = - x$

The point A $(\begin{matrix} 1 \\ 0 \end{matrix})$ moves to A' $(\begin{matrix} 0 \\ - 1 \end{matrix})$

The point C $(\begin{matrix} 0 \\ 1 \end{matrix})$ moves to C' $(\begin{matrix} - 1 \\ 0 \end{matrix})$
The transformation matrix is given by $N = (\begin{matrix} A' | & C' \end{matrix})$

$N = (\begin{matrix} 0 & - 1 \\ - 1 & 0 \end{matrix})$

Enlargement matrices

How do I find enlargement matrices?

Imagine the unit square OABC
- It has a side-length of 1 unit
- O is the origin

The coordinates of A and C as column vectors are
- $A = (\begin{matrix} 1 \\ 0 \end{matrix})$ and $C = (\begin{matrix} 0 \\ 1 \end{matrix})$
Under an enlargement of scale factor k about the origin (including negative scale factors), A moves to A' and C moves to C'
- The matrix, M representing this enlargement is $M = (\begin{matrix} A' | & C' \end{matrix})$
- A' and C' are column vectors of their new positions
  - The points O and B are not needed, as we can draw the enlarged square using just A' and C' (as O won't move)
$A' = (\begin{matrix} k \\ 0 \end{matrix})$ and $C' = (\begin{matrix} 0 \\ k \end{matrix})$
- They are both just moving along the x and y axes respectively
So all enlargement matrices have the form $M = (\begin{matrix} k & 0 \\ 0 & k \end{matrix})$
- This is the same as $M = k I$ , where $I$ is the identity matrix
For example:
- The matrix representation of an enlargement of scale factor 3 about the origin is $(\begin{matrix} 3 & 0 \\ 0 & 3 \end{matrix})$
- The matrix representation of an enlargement of scale factor $- \frac{1}{2}$ about the origin is $(\begin{matrix} - 0.5 & 0 \\ 0 & - 0.5 \end{matrix})$

Worked Example

The matrix M representing a transformation is given by $(\begin{matrix} \frac{1}{4} & 0 \\ 0 & \frac{1}{4} \end{matrix})$ .

Describe geometrically the transformation represented by M.

The matrix $M$ can be written as a multiple of the identity matrix, $I$

$(\begin{matrix} \frac{1}{4} & 0 \\ 0 & \frac{1}{4} \end{matrix}) = \frac{1}{4} (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})$

So the unit square is being scaled by $\frac{1}{4}$

Enlargement by scale factor $\frac{1}{4}$ about the origin

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Transformations with Matrices (AQA GCSE Further Maths): Revision Note

Transforming a point

How do I transform a point using a matrix?

Worked Example

Rotation matrices

How do I find rotation matrices?

Worked Example

Reflection matrices

How do I find reflection matrices?

Worked Example

Enlargement matrices

How do I find enlargement matrices?

Worked Example

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Mark Curtis

Reviewer: Dan Finlay