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FM Nov 2023 p12 q03
4187
The matrix \(\mathbf{M}\) is given by \(\mathbf{M} = \begin{pmatrix} k & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}\), where \(k\) is a constant and \(k \neq 0\) and \(k \neq 1\).
(a) The matrix \(\mathbf{M}\) represents a sequence of two geometrical transformations. State the type of each transformation, and make clear the order in which they are applied. [2]
The unit square in the \(x-y\) plane is transformed by \(\mathbf{M}\) onto parallelogram \(OPQR\).
(b) Find, in terms of \(k\), the area of parallelogram \(OPQR\) and the matrix which transforms \(OPQR\) onto the unit square. [3]
(c) Show that the line through the origin with gradient \(\frac{1}{k-1}\) is invariant under the transformation in the \(x-y\) plane represented by \(\mathbf{M}\). [3]
Solution
(a) The matrix \(\mathbf{M}\) is the product of two matrices: \(\begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}\) represents a shear parallel to the x-axis, and \(\begin{pmatrix} k & 0 \\ 0 & 1 \end{pmatrix}\) represents a stretch parallel to the x-axis by a factor of \(k\). The shear is applied first, followed by the stretch.
(b) The area of the parallelogram \(OPQR\) is given by the absolute value of the determinant of \(\mathbf{M}\). Calculating the determinant: \(\det(\mathbf{M}) = \det\left(\begin{pmatrix} k & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}\right) = k\). Thus, \(|OPQR| = |k|\).
The inverse matrix \(\mathbf{M}^{-1}\) is calculated as follows: \(\mathbf{M}^{-1} = \frac{1}{k} \begin{pmatrix} 1 & 0 \\ -1 & k \end{pmatrix}\).
(c) To show the line is invariant, consider the transformation of a point \(\begin{pmatrix} x \\ \frac{1}{k-1}x \end{pmatrix}\) under \(\mathbf{M}\):
\(\begin{pmatrix} k & 0 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ \frac{1}{k-1}x \end{pmatrix} = \begin{pmatrix} kx \\ x + \frac{1}{k-1}x \end{pmatrix} = \begin{pmatrix} kx \\ \frac{k}{k-1}x \end{pmatrix} = k \begin{pmatrix} x \\ \frac{1}{k-1}x \end{pmatrix}\).
This shows that the line \(y = \frac{1}{k-1}x\) is invariant under the transformation.
