FP1 June 2012 Q9
9. \[\mathbf{M} = \begin{pmatrix} 3 & 4 \\ 2 & -5 \end{pmatrix}\]
The transformation represented by \(\mathbf{M}\) maps the point \(S(2a - 7,\ a - 1)\), where \(a\) is a constant, onto the point \(S'(25,\ -14)\).
The point \(R\) has coordinates \((6,\ 0)\).
Given that \(O\) is the origin,
Triangle \(ORS\) is mapped onto triangle \(OR'S'\) by the transformation represented by \(\mathbf{M}\).
Given that \[\mathbf{A} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\]
The transformation represented by \(\mathbf{A}\) followed by the transformation represented by \(\mathbf{B}\) is equivalent to the transformation represented by \(\mathbf{M}\).
| Scheme | Marks |
|---|---|
| \(\det\mathbf{M} = 3(-5) - (4)(2) = -15 - 8 = \underline{-23}\) \(\underline{-23}\) | B1 |
| [1] |
| Scheme | Marks |
|---|---|
| Therefore, \(\begin{pmatrix} 3 & 4 \\ 2 & -5 \end{pmatrix}\begin{pmatrix} 2a - 7 \\ a - 1 \end{pmatrix} = \begin{pmatrix} 25 \\ -14 \end{pmatrix}\) Using the information in the question to form the matrix equation. Can be implied by any of the correct equations below. | M1 |
| Either, \(3(2a - 7) + 4(a - 1) = 25\) or \(2(2a - 7) - 5(a - 1) = -14\) or \(\begin{pmatrix} 3(2a - 7) + 4(a - 1) \\ 2(2a - 7) - 5(a - 1) \end{pmatrix} = \begin{pmatrix} 25 \\ -14 \end{pmatrix}\) Any one correct equation (unsimplified) inside or outside matrices | A1 |
| giving \(a = 5\) \(a = 5\) | A1 |
| [3] |
Notes
Alternative (Way 2)
| Scheme | Marks |
|---|---|
| \(\mathbf{M}: \begin{pmatrix} 2a - 7 \\ a - 1 \end{pmatrix} \to \begin{pmatrix} 25 \\ -14 \end{pmatrix}\) | |
| Therefore, \(\begin{pmatrix} 3 & 4 \\ 2 & -5 \end{pmatrix}\begin{pmatrix} 2a - 7 \\ a - 1 \end{pmatrix} = \begin{pmatrix} 25 \\ -14 \end{pmatrix}\) or \(\begin{pmatrix} 2a - 7 \\ a - 1 \end{pmatrix} = \begin{pmatrix} 3 & 4 \\ 2 & -5 \end{pmatrix}^{-1}\begin{pmatrix} 25 \\ -14 \end{pmatrix}\) Using the information in the question to form the matrix equation. Can be implied by any of the correct equations below. | M1 |
| \(\begin{pmatrix} 2a - 7 \\ a - 1 \end{pmatrix} = \dfrac{1}{(-23)}\begin{pmatrix} -5 & -4 \\ -2 & 3 \end{pmatrix}\begin{pmatrix} 25 \\ -14 \end{pmatrix} = \dfrac{1}{(-23)}\begin{pmatrix} -125 + 56 \\ -50 - 42 \end{pmatrix}\) | |
| Either, \((2a - 7) = 3\) or \((a - 1) = 4\) Any one correct equation. | A1 |
| giving \(a = 5\) \(a = 5\) | A1 |
| [3] |
| Scheme | Marks |
|---|---|
| \(\text{Area}(ORS) = \dfrac{1}{2}(6)(4);\ = \underline{12}\ (\text{units})^2\) M1: \(\dfrac{1}{2}(6)(\text{Their } a - 1)\) A1: 12 cao and cso | M1A1 |
| [2] |
Notes
Note A(6, 0) is sometimes misinterpreted as (0, 6) – this is the wrong triangle and scores M0 e.g. \(1/2 \times 6 \times 3 = 9\)
Alternative (Way 2 Determinant)
| Scheme | Marks |
|---|---|
| \(\textit{Area } ORS = \dfrac{1}{2}\begin{vmatrix} 6 & 3 & 0 & 6 \\ 0 & 4 & 0 & 0 \end{vmatrix}\) \(= \dfrac{1}{2}\left|(6 \times 4 - 3 \times 0 + 0 - 0 + 0 - 0)\right|\) Correct calculation | M1 |
| \(= 12\) | A1 |
| [2] |
| Scheme | Marks |
|---|---|
| \(\text{Area}(OR'S') = \pm 23 \times (12)\) \(\pm\det\mathbf{M} \times\) (their part (\(c\)) answer) | M1 |
| 276 (follow through provided area \(> 0\)) | A1ft |
| [2] |
Notes
A method not involving the determinant requires the coordinates of \(R'\) to be calculated ((18, 12)) and then a correct method for the area e.g. \((26 \times 25 - 7 \times 13 - 9 \times 12 - 7 \times 25)\) M1 = 276 A1
Alternative (Way 2 Determinant)
| Scheme | Marks |
|---|---|
| \(\textit{Area } ORS = \dfrac{1}{2}\begin{vmatrix} 18 & 25 & 0 & 18 \\ 12 & -14 & 0 & 12 \end{vmatrix}\) \(= \dfrac{1}{2}\left|(18 \times -14 - 12 \times 25 + 0 - 0 + 0 - 0)\right|\) Correct calculation | M1 |
| \(= 276\) | A1ft |
| [2] |
| Scheme | Marks |
|---|---|
| Rotation; \(90^\circ\) anti-clockwise (or \(270^\circ\) clockwise) about \((0,\ 0)\). B1: Rotation, Rotates, Rotate, Rotating (not turn) B1: \(90^\circ\) anti-clockwise (or \(270^\circ\) clockwise) about (around/from etc.) \((0,\ 0)\) | B1;B1 |
| [2] |
| Scheme | Marks |
|---|---|
| \(\mathbf{M} = \mathbf{BA}\) \(\mathbf{M} = \mathbf{BA}\), seen or implied. | M1 |
| \(\mathbf{A}^{-1} = \dfrac{1}{(0)(0) - (1)(-1)}\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix};\ = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\) \(\mathbf{A}^{-1} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\) | A1 |
| \(\mathbf{B} = \begin{pmatrix} 3 & 4 \\ 2 & -5 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\) Applies \(\mathbf{M}(\text{their } \mathbf{A}^{-1})\) | M1 |
| \(\mathbf{B} = \begin{pmatrix} -4 & 3 \\ 5 & 2 \end{pmatrix}\) | A1 |
| [4] | |
| 14 marks |
Notes
NB some candidates state \(\mathbf{M} = \mathbf{AB}\) and then calculate \(\mathbf{MA}^{-1}\) or state \(\mathbf{M} = \mathbf{BA}\) and then calculate \(\mathbf{A}^{-1}\mathbf{M}\). These could score M0A0 M1A1ft and M1A1M0A0 respectively.
Special case
| Scheme | Marks |
|---|---|
| \(\mathbf{M} = \mathbf{AB}\) \(\mathbf{M} = \mathbf{AB}\), seen or implied. | M0 |
| \(\mathbf{A}^{-1} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\) | A0 |
| \(\mathbf{B} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 3 & 4 \\ 2 & -5 \end{pmatrix} = \begin{pmatrix} 2 & -5 \\ -3 & -4 \end{pmatrix}\) Applies \((\text{their } \mathbf{A}^{-1})\mathbf{M}\) | M1A1ft |
Alternative (Way 2)
| Scheme | Marks |
|---|---|
| \(\mathbf{M} = \mathbf{BA}\) \(\mathbf{M} = \mathbf{BA}\), seen or implied. | M1 |
| \(\begin{pmatrix} 3 & 4 \\ 2 & -5 \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\) \(\begin{pmatrix} 3 & 4 \\ 2 & -5 \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\), with constants to be found. | A1 |
| \(\begin{pmatrix} 3 & 4 \\ 2 & -5 \end{pmatrix} = \begin{pmatrix} b & -a \\ d & -c \end{pmatrix}\) \(\begin{pmatrix} 3 & 4 \\ 2 & -5 \end{pmatrix} = \text{their } \begin{pmatrix} b & -a \\ d & -c \end{pmatrix}\) with at least two elements correct on RHS. | M1 |
| \(\mathbf{B} = \begin{pmatrix} -4 & 3 \\ 5 & 2 \end{pmatrix}\) Correct matrix for \(\mathbf{B}\) of \(\begin{pmatrix} -4 & 3 \\ 5 & 2 \end{pmatrix}\) or \(a = -4\), \(b = 3\), \(c = 5\), \(d = 2\) | A1 |
| [4] |