Determinant of a Transformation Matrix (Edexcel IGCSE Maths B): Revision Note

A triangle has vertices with coordinates (1, -3), (7, -3) and (3, 1).

(a) Find the coordinates of the vertices of the image triangle when that triangle is transformed by the matrix $M=\left(\begin{array}{cc}-\sqrt{3}& -1\\ 1& -\sqrt{3}\end{array}\right)$.

Answer:

You can represent the triangle as a 2×3 matrix of vertices

• This will be the coordinates of the vertices as column vectors, but written in a single matrix

$\left(\begin{array}{ccc}1& 7& 3\\ -3& -3& 1\end{array}\right)$

To transform the triangle, multiply its matrix by the transformation matrix $\mathbf{M}$

$\begin{array}{rcl}& & \left(\begin{array}{cc}-\sqrt{3}& -1\\ 1& -\sqrt{3}\end{array}\right)\left(\begin{array}{ccc}1& 7& 3\\ -3& -3& 1\end{array}\right)\\ & =& \left(\begin{array}{ccc}\left(-\sqrt{3}\right)\times 1+\left(-1\right)\times \left(-3\right)& \left(-\sqrt{3}\right)\times 7+\left(-1\right)\times \left(-3\right)& \left(-\sqrt{3}\right)\times 3+\left(-1\right)\times 1\\ 1\times 1+\left(-\sqrt{3}\right)\times \left(-3\right)& 1\times 7+\left(-\sqrt{3}\right)\times \left(-3\right)& 1\times 3+\left(-\sqrt{3}\right)\times 1\end{array}\right)\\ & =& \left(\begin{array}{ccc}3-\sqrt{3}& 3-7\sqrt{3}& -1-3\sqrt{3}\\ 1+3\sqrt{3}& 7+3\sqrt{3}& 3-\sqrt{3}\end{array}\right)\end{array}$

Write the new vertices as coordinates

$\left(3-\sqrt{3},1+3\sqrt{3}\right)$, $\left(3-7\sqrt{3},7+3\sqrt{3}\right)$ and $\left(-1-3\sqrt{3},3-\sqrt{3}\right)$
 

(b) Find the area of the image triangle.

Answer:

The original and transformed triangles look like this:

Trying to calculate the area of the transformed triangle directly would be quite challenging, to say the least!

However the area of the original triangle is easy to calculate

• It has a base of 6, and a height of 4

• Use $\mathrm{Area}=\frac{1}{2}\times \mathrm{base}\times \mathrm{height}$

$\mathrm{area}\mathrm{of}\mathrm{original}\mathrm{triangle}=\frac{1}{2}\times 6\times 4=12$

Find the determinant of the transformation matrix $M=\left(\begin{array}{cc}-\sqrt{3}& -1\\ 1& -\sqrt{3}\end{array}\right)$

• The determinant of a 2×2 matrix $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ is $ad-bc$

$\begin{array}{rcl}\mathrm{determinant}\mathrm{of}M& =& \left(-\sqrt{3}\right)\times \left(-\sqrt{3}\right)-\left(-1\right)\times 1\\ & =& 3-\left(-1\right)\\ & =& 4\end{array}$

That is the area scale factor of the transformation

• So multiply the original area by 4 to find the transformed area

$12\times 4=48$

Area of the image triangle is 48