AS June 2023 Paper 1 Q9
9 A transformation T of the plane is represented by the matrix \(\mathbf{M} = \begin{pmatrix} k + 1 & -1 \\ 1 & k \end{pmatrix}\), where \(k\) is a constant.
Show that, for all values of \(k\), T has no invariant lines through the origin. [6]
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix} k + 1 & -1 \\ 1 & k \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} kx + x - y \\ x + ky \end{pmatrix}\) | M1 | 1.1 |
| \(y = mx \Rightarrow x + ky = m(kx + x - y)\) | M1 | 3.1a |
| \(\Rightarrow x + kmx = m(kx + x - mx)\) \(\Rightarrow 1 + km = km + m - m^2\) | A1 | 2.1 |
| \(\Rightarrow m^2 - m + 1 = 0\) | A1 | 1.1 |
| discriminant \(= (-1)^2 - 4 = -3 \lt 0\) | M1 | 3.1a |
| so no real roots, and no invariant lines | A1 | 3.2a |
| [6] |
Notes
If a specific value for \(k\) used, allow max of 3 M marks (SC)
M1 A1: (2nd M1, 1st A1) or \(x + ky = m(kx + x - y) + c\)
\(\Rightarrow x + k(mx + c) = m(kx + x - mx - c) + c\)
\(\Rightarrow 1 + km = km + m - m^2\) and \(kc = c - mc\)
A1: (2nd) soi
M1: (3rd) oe or by solving to get \(m = \dfrac{1}{2} \pm \dfrac{\sqrt{3}}{2}\mathrm{i}\)
A1: (3rd) without wrong working
If invariant point (instead of line) only first M1 is available