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FM November 2022 p12 q03
4255
The matrix M is given by M = \(\begin{pmatrix} 1 & 0 \\ 0 & k \end{pmatrix} \begin{pmatrix} 1 & 0 \\ k & 1 \end{pmatrix}\), where \(k\) is a constant and \(k \neq 0\) or 1.
(a) The matrix M represents a sequence of two geometrical transformations. State the type of each transformation, and make clear the order in which they are applied. [2]
(b) Write M-1 as the product of two matrices, neither of which is I. [2]
(c) Show that the invariant points of the transformation represented by M lie on the line \(y = \frac{k^2}{1-k}x\). [4]
(d) The triangle ABC in the x-y plane is transformed by M onto triangle DEF. Find the value of \(k\) for which the area of triangle DEF is equal to the area of triangle ABC. [2]
Solution
(a) The matrix \(\begin{pmatrix} 1 & 0 \\ k & 1 \end{pmatrix}\) represents a shear in the y-direction. The matrix \(\begin{pmatrix} 1 & 0 \\ 0 & k \end{pmatrix}\) represents a stretch parallel to the y-axis with scale factor \(k\). The transformations are applied in the order of shear followed by stretch.
(b) To find \(\mathbf{M}^{-1}\), we first find the inverse of each matrix: \(\begin{pmatrix} 1 & 0 \\ k & 1 \end{pmatrix}^{-1} = \begin{pmatrix} 1 & 0 \\ -k & 1 \end{pmatrix}\) and \(\begin{pmatrix} 1 & 0 \\ 0 & k \end{pmatrix}^{-1} = \begin{pmatrix} 1 & 0 \\ 0 & \frac{1}{k} \end{pmatrix}\). Thus, \(\mathbf{M}^{-1} = \begin{pmatrix} 1 & 0 \\ -k & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & \frac{1}{k} \end{pmatrix}\).
(c) The transformation \(\mathbf{M} = \begin{pmatrix} 1 & 0 \\ k^2 & k \end{pmatrix}\) transforms \(\begin{pmatrix} x \\ y \end{pmatrix}\) to \(\begin{pmatrix} x \\ k^2x + ky \end{pmatrix}\). Setting \(\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ k^2x + ky \end{pmatrix}\), we get \(k^2x + ky = y\), which simplifies to \(y = \frac{k^2}{1-k}x\).
(d) The area of triangle DEF is equal to the area of triangle ABC when \(|\det(\mathbf{M})| = 1\). The determinant \(\det(\mathbf{M}) = 1 \times k - 0 \times k^2 = k\). Thus, \(|k| = 1\), giving \(k = -1\).
