Transforming a Point (Edexcel IGCSE Maths B): Revision Note

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

(a) Work out the coordinates of the image of the point $\left(2,3\right)$ using the transformation represented by $\mathbf{M}$.

Answer:

 Multiply the coordinates, written as a column vector, by the transformation matrix $\mathbf{M}$

$\left(\begin{array}{cc}4& 5\\ 1& -2\end{array}\right)\left(\begin{array}{c}2\\ 3\end{array}\right)=\left(\begin{array}{c}4\times 2+5\times 3\\ 1\times 2+-2\times 3\end{array}\right)=\left(\begin{array}{c}23\\ -4\end{array}\right)$ 

Rewrite the answer as coordinates

$\mathrm{(}\mathrm{23}\mathrm{,}\mathrm{-}\mathrm{4}\mathrm{)}$
  

(b) The image of another point, $P$, using the transformation represented by $\mathbf{M}$ is $\left(11,6\right)$. 

Work out the coordinates of $P$.

Answer:

Method 1

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

Let $P=\left(x,y\right)$ 

Multiply the coordinates of P, written as a column vector, by the transformation matrix $\mathbf{M}$

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

$\begin{array}{rcl}\left(\begin{array}{cc}4& 5\\ 1& -2\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)& =& \left(\begin{array}{c}11\\ 6\end{array}\right)\\ \left(\begin{array}{c}4x+5y\\ x-2y\end{array}\right)& =& \left(\begin{array}{c}11\\ 6\end{array}\right)\end{array}$ 

Equate the matching elements of the two matrices

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

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 $y$ 

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

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

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

P is $\mathrm{(}\mathrm{4}\mathrm{,}\mathrm{-}\mathrm{1}\mathrm{)}$

Method 2

Find the inverse of matrix $\mathbf{M}$

• The inverse of $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ is $\frac{1}{ad-bc}\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)$

$\begin{array}{rcl}{\mathbf{M}}^{\mathbf{-}\mathbf{1}}& =& \frac{1}{4\times \left(-2\right)-5\times 1}\left(\begin{array}{cc}-2& -5\\ -1& 4\end{array}\right)\\ & =& \frac{1}{-13}\left(\begin{array}{cc}-2& -5\\ -1& 4\end{array}\right)\\ & =& \left(\begin{array}{cc}\frac{2}{13}& \frac{5}{13}\\ \frac{1}{13}& -\frac{4}{13}\end{array}\right)\end{array}$

If $\mathbf{M}\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}11\\ 6\end{array}\right)$, then $\left(\begin{array}{c}x\\ y\end{array}\right)={\mathbf{M}}^{-1}\left(\begin{array}{c}11\\ 6\end{array}\right)$

$\begin{array}{rcl}\left(\begin{array}{c}x\\ y\end{array}\right)& =& \left(\begin{array}{cc}\frac{2}{13}& \frac{5}{13}\\ \frac{1}{13}& -\frac{4}{13}\end{array}\right)\left(\begin{array}{c}11\\ 6\end{array}\right)\\ & =& \left(\begin{array}{c}\frac{2}{13}\times 11+\frac{5}{13}\times 6\\ \frac{1}{13}\times 11+\left(-\frac{4}{13}\right)\times 6\end{array}\right)\\ & =& \left(\begin{array}{c}\frac{52}{13}\\ -\frac{13}{13}\end{array}\right)\\ & =& \left(\begin{array}{c}4\\ -1\end{array}\right)\end{array}$

Write as coordinates

P is $\mathrm{(}\mathrm{4}\mathrm{,}\mathrm{-}\mathrm{1}\mathrm{)}$