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FM June 2024 p12 q04
4167
The matrix M is given by \(\mathbf{M} = \begin{pmatrix} \frac{1}{2} & -\frac{1}{2}\sqrt{3} \\ \frac{1}{2}\sqrt{3} & \frac{1}{2} \end{pmatrix} \begin{pmatrix} 14 & 0 \\ 0 & 1 \end{pmatrix}\).
(a) The matrix M represents a sequence of two geometrical transformations in the x-y plane. Give full details of each transformation, and make clear the order in which they are applied. [4]
(b) Write \(\mathbf{M}^{-1}\) as the product of two matrices, neither of which is I. [2]
\((c) Find the equations of the invariant lines, through the origin, of the transformation represented by M. [5]\)
(d) The triangle ABC in the x-y plane is transformed by M onto triangle DEF. Given that the area of triangle DEF is 28 cm2, find the area of triangle ABC. [2]
Solution
(a) The matrix \(\mathbf{M}\) can be decomposed into two transformations: a stretch and a rotation. The first matrix \(\begin{pmatrix} 14 & 0 \\ 0 & 1 \end{pmatrix}\) represents a stretch parallel to the x-axis with a scale factor of 14. The second matrix \(\begin{pmatrix} \frac{1}{2} & -\frac{1}{2}\sqrt{3} \\ \frac{1}{2}\sqrt{3} & \frac{1}{2} \end{pmatrix}\) represents a rotation by \(\frac{\pi}{3}\) anticlockwise about the origin.
(b) To find \(\mathbf{M}^{-1}\), we take the inverse of each matrix in the product: \(\begin{pmatrix} 14 & 0 \\ 0 & 1 \end{pmatrix}^{-1} = \begin{pmatrix} \frac{1}{14} & 0 \\ 0 & 1 \end{pmatrix}\) and \(\begin{pmatrix} \frac{1}{2} & -\frac{1}{2}\sqrt{3} \\ \frac{1}{2}\sqrt{3} & \frac{1}{2} \end{pmatrix}^{-1} = \begin{pmatrix} \frac{1}{2} & \frac{1}{2}\sqrt{3} \\ -\frac{1}{2}\sqrt{3} & \frac{1}{2} \end{pmatrix}\).
(c) The invariant lines are found by solving \(\mathbf{M} \begin{pmatrix} x \\ y \end{pmatrix} = \lambda \begin{pmatrix} x \\ y \end{pmatrix}\). Solving gives the equations \(y = 2\sqrt{3}x\) and \(y = \frac{2}{3}\sqrt{3}x\).
(d) The area of triangle \(ABC\) is found using the determinant of \(\mathbf{M}\). Given \(|\mathbf{DEF}| = |\det \mathbf{M}| \times |\mathbf{ABC}|\), we have \(28 = 14 \times |\mathbf{ABC}|\), so \(|\mathbf{ABC}| = 2\) cm2.
