Combinations of Matrix Transformations (Edexcel International A Level (IAL) Further Maths: Further Pure 1): Revision Note

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

$\mathbf{A}=\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)$  represents a rotation of 180° about the origin
$\mathbf{B}=\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)$ represents a reflection in the y-axis
$\mathbf{C}=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$ 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 $\mathbf{B}$ followed by transformation $\mathbf{C}$
This is the same as the matrix $\mathbf{CB}$ in that order (the first transformation appears on the right)

$\mathbf{CB}$

Use matrix multiplication to find $\mathbf{CB}$

$\mathbf{CB}=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\times \left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}\left(1\times -1+0\times 0\right)& \left(1\times 0+0\times 1\right)\\ \left(0\times -1+-1\times 0\right)& \left(0\times 0+-1\times 1\right)\end{array}\right)$

The question claims that this is equivalent to transformation  $\mathbf{A}$
Simplify the working above and show that it is the same as matrix $\mathbf{A}$

$\mathbf{CB}=\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)=\mathbf{A}$
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 $\mathbf{CD}$ where ${\mathbf{D}}^{-1}=\left(\begin{array}{cc}2& 0\\ 0& 1\end{array}\right)$.

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

You need to find the inverse of $\mathbf{CD}$
You need the rule that ${\left(\mathbf{CD}\right)}^{-1}={\mathbf{D}}^{-1}{\mathbf{C}}^{-1}$
You can substitute in ${\mathbf{D}}^{-1}$ from the question

$\begin{array}{rcl}{\left(\mathbf{CD}\right)}^{-1}& =& {\mathbf{D}}^{-1}{\mathbf{C}}^{-1}\\ & =& \left(\begin{array}{cc}2& 0\\ 0& 1\end{array}\right){\mathbf{C}}^{-1}\end{array}$

You need to find ${\mathbf{C}}^{-1}$
Use that $\mathbf{M}=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\Rightarrow {\mathbf{M}}^{\mathbf{-}\mathbf{1}}=\frac{1}{\det \mathbf{M}}\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)$ where $\det \mathbf{M}=ad-bc$

$\begin{array}{rcl}{\mathbf{C}}^{-1}& =& \frac{1}{1\times \left(-1\right)-0\times 0}\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)\\ & =& \left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\end{array}$

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

$\begin{array}{rcl}{\left(\mathbf{CD}\right)}^{-1}& =& \left(\begin{array}{cc}2& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\\ & =& \left(\begin{array}{cc}2& 0\\ 0& -1\end{array}\right)\end{array}$

$\begin{array}{rcl}& & \left(\begin{array}{cc}2& 0\\ 0& -1\end{array}\right)\end{array}$

There are other ways to do this question, for example finding $\mathbf{D}$ first
If you saw that ${\mathbf{C}}^{\mathbf{-}\mathbf{1}}=\mathbf{C}$ (as it is a reflection in the x-axis), explain why clearly