Transforming a Point (Edexcel IGCSE Maths)

Spec point — 5G

Transforming a point

How do I transform a point using a matrix?

• A point (x,y) in a 2D plane can be transformed onto 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 $( \begin{gathered} a b \\ c d \end{gathered} )$

  • Write (x, y) as a column vector, $( \begin{gathered} x \\ y \end{gathered} )$

    • Note that this is the same as a $2\times1$ matrix

  • Use matrix multiplication to work out $( \begin{gathered} a b \\ c d \end{gathered} ) ( \begin{gathered} x \\ y \end{gathered} )$, which gives $( \begin{gathered} x ' \\ y ' \end{gathered} )$

    • $( \begin{gathered} x ' \\ y ' \end{gathered} ) = ( \begin{gathered} a b \\ c d \end{gathered} ) ( \begin{gathered} x \\ y \end{gathered} ) = ( \begin{gathered} a x + b y \\ c x + d y \end{gathered} )$

  • 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)

    • then use the matrix $M$ to set up and solve simultaneous equations to find x and y

  • Alternatively, you could use the inverse matrix $M^{-1}$ to undo the transformation

    • | M ( | x | y | ) | = ( | x ' | | y ' | | ) left right double arrow ( | x | y | ) = M ^ - 1 ( | x ' | | y ' | | ) | |
    • You can see this by solving the matrix equation for $( \begin{gathered} x \\ y \end{gathered} )$:

$\begin{gathered} M ( \begin{gathered} x \\ y \end{gathered} ) & = ( \begin{gathered} x ' \\ y ' \end{gathered} ) \\ M^{- 1} M ( \begin{gathered} x \\ y \end{gathered} ) & = M^{- 1} ( \begin{gathered} x ' \\ y ' \end{gathered} ) \\ I ( \begin{gathered} x \\ y \end{gathered} ) & = M^{- 1} ( \begin{gathered} x ' \\ y ' \end{gathered} ) \\ ( \begin{gathered} x \\ y \end{gathered} ) & = M^{- 1} ( \begin{gathered} x ' \\ y ' \end{gathered} ) \end{gathered}$

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

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

Answer:

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

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

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 $P$.

Answer:

Method 1

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

Let $P = ( x , y )$* *

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

• This time, you know the image of the point after it is transformed, so can fill this in as the "answer"

$\begin{gathered} ( \begin{gathered} 4 5 \\ 1 - 2 \end{gathered} ) ( \begin{gathered} x \\ y \end{gathered} ) & = ( \begin{gathered} 11 \\ 6 \end{gathered} ) \\ ( \begin{gathered} 4 x + 5 y \\ x - 2 y \end{gathered} ) & = ( \begin{gathered} 11 \\ 6 \end{gathered} ) \end{gathered}$

Equate the matching elements of the two matrices

$\begin{gathered} 4 x + 5 y & = 11 \\ x - 2 y & = 6 \end{gathered}$* *

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

Using substitution, start by rearranging the second equation to make $x$ the subject

$x=2y+6$

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

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

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

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

Final answer: P is $(4,-1)$

Method 2

Find the inverse of matrix $M$

• The inverse of $( \begin{gathered} a b \\ c d \end{gathered} )$ is $\frac{1}{a d - b c} ( \begin{gathered} d - b \\ - c & a \end{gathered} )$

$\begin{gathered} M^{- 1} & = \frac{1}{4 \times ( - 2 ) - 5 \times 1} ( \begin{gathered} - 2 & - 5 \\ - 1 & 4 \end{gathered} ) \\ = \frac{1}{- 13} ( \begin{gathered} - 2 & - 5 \\ - 1 & 4 \end{gathered} ) \\ = ( \begin{gathered} \frac{2}{13} & \frac{5}{13} \\ \frac{1}{13} & - \frac{4}{13} \end{gathered} ) \end{gathered}$

If $M ( \begin{gathered} x \\ y \end{gathered} ) = ( \begin{gathered} 11 \\ 6 \end{gathered} )$, then $( \begin{gathered} x \\ y \end{gathered} ) = M^{- 1} ( \begin{gathered} 11 \\ 6 \end{gathered} )$

$\begin{gathered} ( \begin{gathered} x \\ y \end{gathered} ) & = ( \begin{gathered} \frac{2}{13} & \frac{5}{13} \\ \frac{1}{13} & - \frac{4}{13} \end{gathered} ) ( \begin{gathered} 11 \\ 6 \end{gathered} ) \\ = ( \begin{gathered} \frac{2}{13} \times 11 + \frac{5}{13} \times 6 \\ \frac{1}{13} \times 11 + ( - \frac{4}{13} ) \times 6 \end{gathered} ) \\ = ( \begin{gathered} \frac{52}{13} \\ - \frac{13}{13} \end{gathered} ) \\ = ( \begin{gathered} 4 \\ - 1 \end{gathered} ) \end{gathered}$

Write as coordinates

Final answer: P is $(4,-1)$