A2 June 2022 Paper 1 Q4
4
(a) A transformation with associated matrix \(\begin{pmatrix} m & 2 & 1 \\ 0 & 1 & -2 \\ 2 & 0 & 3 \end{pmatrix}\), where \(m\) is a constant, maps the vertices of a cube to points that all lie in a plane.
Find \(m\). [3]
Find \(m\). [3]
(b) The transformations S and T of the plane have associated matrices \(\mathbf{M}\) and \(\mathbf{N}\) respectively, where \(\mathbf{M} = \begin{pmatrix} k & 1 \\ -3 & 4 \end{pmatrix}\) and the determinant of \(\mathbf{N}\) is \(3k + 1\). The transformation U is equivalent to the combined transformation consisting of S followed by T.
Given that U preserves orientation and has an area scale factor 2, find the possible values of \(k\). [4]
Given that U preserves orientation and has an area scale factor 2, find the possible values of \(k\). [4]
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{vmatrix} m & 2 & 1 \\ 0 & 1 & -2 \\ 2 & 0 & 3 \end{vmatrix} = 3m - 2 \times 4 + 1 \times (-2) = 3m - 10\) | M1 | 3.1a |
| so \(3m - 10 = 0 \Rightarrow m = \frac{10}{3}\) | A1 A1 | 1.1 1.1 |
| [3] |
Notes
M1: finding determinant
Alternative solution
| Scheme | Marks |
|---|---|
| \(\begin{pmatrix} m \\ 0 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} \times \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix} = \begin{pmatrix} m \\ 0 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ -6 \\ -5 \end{pmatrix}\) | M1 |
| \(= 3m - 10\ [= 0]\) \(\Rightarrow m = \dfrac{10}{3}\) | A1 A1 |
| [3] |
| Scheme | Marks | AO |
|---|---|---|
| \(\det\mathbf{M} = 4k + 3\) | B1 | 1.2 |
| \(\det(\mathbf{NM}) = \det\mathbf{N} \times \det\mathbf{M}\) | M1 | 3.1a |
| \(\Rightarrow (3k + 1)(4k + 3) = 2\) | M1 | 1.1 |
| \(k = -1\) or \(-\dfrac{1}{12}\) | A1 | 1.1 |
| [4] |
Notes
M1: Soi
M1: Equating to 2
Alternative solution
| Scheme | Marks |
|---|---|
| \(\det(NM) = (ak - 3b)(c + 4d) - (a + 4b)(ck - 3d)\) | B1 |
| \((ak - 3b)(c + 4d) - (a + 4b)(ck - 3d) = 2\) | M1 |
| \((3k + 1)(4k + 3) = 2\) | M1 |
| \(k = -1\) or \(-\dfrac{1}{12}\) | A1 |
| [4] |