AS June 2025 Paper 1 Q4
4
(a) The transformation T is represented by the matrix \(\mathbf{M} = \begin{pmatrix} 1 & -2 & 2 \\ 2 & 1 & 0 \\ 1 & 2 & -1 \end{pmatrix}\).
A shape \(\mathrm{S_1}\) is mapped to a shape \(\mathrm{S_2}\) by the transformation T.
Show that volume of \(\mathrm{S_1}\) is the same as the volume of \(\mathrm{S_2}\). [2]
A shape \(\mathrm{S_1}\) is mapped to a shape \(\mathrm{S_2}\) by the transformation T.
Show that volume of \(\mathrm{S_1}\) is the same as the volume of \(\mathrm{S_2}\). [2]
(b) Three planes have equations\[\begin{aligned} x - 2y + 2z &= \lambda, \\ 2x + y \phantom{{}+2z} &= 2, \\ x + 2y - z &= 0, \end{aligned}\]where \(\lambda\) is a constant.
(i) Explain why the three planes intersect at a point for any value of \(\lambda\). [2]
(ii) Use a matrix method to determine, in terms of \(\lambda\), the coordinates of this point. [4]
| Scheme | Marks | AO |
|---|---|---|
| \(\det\mathbf{M} = 1\) | M1 | 1.1 |
| \(\Rightarrow\) T preserves volumes as \(\det\mathbf{M}\) is the volume scale factor | A1 | 1.2 |
| [2] |
Notes
M1: calculating \(\det\mathbf{M}\) (BC)
A1: \(=1\) and conclusion (must mention scale factor or equivalent)
if ‘area’ instead of ‘volume’ then A0
| Scheme | Marks | AO |
|---|---|---|
| (i) Matrix of coeffs is \(\mathbf{M}\) | M1 | 1.1 |
| Det \(\mathbf{M} = 1 \Rightarrow \mathbf{M}\) is non-singular [so planes meet at a point] | A1 | 2.4 |
| [2] | ||
| (ii) \(\mathbf{M}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} \lambda \\ 2 \\ 0 \end{pmatrix}\ \left[\Rightarrow \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \mathbf{M}^{-1}\begin{pmatrix} \lambda \\ 2 \\ 0 \end{pmatrix}\right]\) | M1 | 1.1 |
| \(\mathbf{M}^{-1} = \begin{pmatrix} -1 & 2 & -2 \\ 2 & -3 & 4 \\ 3 & -4 & 5 \end{pmatrix}\) | B1 | 1.1 |
| \(\begin{pmatrix} -1 & 2 & -2 \\ 2 & -3 & 4 \\ 3 & -4 & 5 \end{pmatrix}\begin{pmatrix} \lambda \\ 2 \\ 0 \end{pmatrix}\) \(= \begin{pmatrix} -\lambda + 4 \\ 2\lambda - 6 \\ 3\lambda - 8 \end{pmatrix}\) | M1 | 1.1 |
| \(\Rightarrow\) Point of intersection is \((-\lambda + 4, 2\lambda - 6, 3\lambda - 8)\) | A1 | 1.1 |
| [4] |
Notes
(b)(i)
M1: soi
A1: or \(\det\mathbf{M} \ne 0\), or \(\mathbf{M}^{-1}\) exists
Allow SC2 for fully correct algebraic derivations of the point of intersection \((-\lambda + 4, 2\lambda - 6, 3\lambda - 8)\).
(b)(ii)
M1: translate to vector equation (soi)
B1: Correct inverse
M1: Attempt to multiply \(\mathbf{M}^{-1}\) by \(\begin{pmatrix} \lambda \\ 2 \\ 0 \end{pmatrix}\)
A1: accept in vector form
matrix working may be shown in part (b)(i)