FP3 June 2017 Q6
6. The matrix \(\mathbf{M}\) is given by \[\mathbf{M} = \begin{pmatrix}1 & k & 0\\ 2 & -2 & 1\\ -4 & 1 & -1\end{pmatrix}, \quad k \in \mathbb{R},\ k \neq \frac{1}{2}\]
The straight line \(l_1\) is mapped onto the straight line \(l_2\) by the transformation represented by the matrix \[\begin{pmatrix}1 & 0 & 0\\ 2 & -2 & 1\\ -4 & 1 & -1\end{pmatrix}\]
Given that \(l_2\) has cartesian equation \[\frac{x - 1}{5} = \frac{y + 2}{2} = \frac{z - 3}{1}\]
| Scheme | Marks |
|---|---|
| \(\det\mathbf{M} = 1 \times (2 - 1) - k(-2 + 4)(+0) = 1 - 2k\) * or e.g. \(\det\mathbf{M} = (0) - 1(1 + 4k) - 1(-2 - 2k) = 1 - 2k\) * or rule of Sarrus: \(\det\mathbf{M} = 2 - 4k - 1 + 2k = 1 - 2k\) * Or e.g. \((1)\begin{vmatrix}-2 & 1\\ 1 & -1\end{vmatrix} - k\begin{vmatrix}2 & 1\\ -4 & -1\end{vmatrix} + 0\begin{vmatrix}2 & -2\\ -4 & 1\end{vmatrix}\) | M1A1* |
| (2) |
Notes
M1: Correct attempt at determinant (at least 2 ‘elements’ correct). May need to check as they might use a different row/column.
A1*: Obtains printed answer with no errors. If they use determinant notation as in the last example, then you must see at least one intermediate step before the printed answer e.g. minimally 1 - 2\(k\) + 0.
| Scheme | Marks |
|---|---|
| \(\left(\mathbf{M}^{\mathrm{T}}\right)\ \begin{pmatrix}1 & 2 & -4\\ k & -2 & 1\\ 0 & 1 & -1\end{pmatrix}\) or \((\text{minors})\ \begin{pmatrix}1 & 2 & -6\\ -k & -1 & 1 + 4k\\ k & 1 & -2 - 2k\end{pmatrix}\) or \((\text{cofactors})\ \begin{pmatrix}1 & -2 & -6\\ k & -1 & -1 - 4k\\ k & -1 & -2 - 2k\end{pmatrix}\) | B1 |
| \(\mathbf{M}^{-1} = \dfrac{1}{1 - 2k}\begin{pmatrix}1 & k & k\\ -2 & -1 & -1\\ -6 & -1 - 4k & -2 - 2k\end{pmatrix}\) | M1A1A1 |
| (4) |
Notes
M1: Full attempt at inverse ignoring determinant. Need to see all stages but allow numerical slips.
A1: 2 correct rows or 2 correct columns including reciprocal of determinant
A1: All correct including reciprocal of determinant
| Scheme | Marks |
|---|---|
| \(l_2: (1 + 5\lambda)\mathbf{i} + (-2 + 2\lambda)\mathbf{j} + (3 + \lambda)\mathbf{k}\) | M1A1 |
| \(\dfrac{1}{1}\begin{pmatrix}1 & 0 & 0\\ -2 & -1 & -1\\ -6 & -1 & -2\end{pmatrix}\begin{pmatrix}1 + 5\lambda\\ -2 + 2\lambda\\ 3 + \lambda\end{pmatrix} = \begin{pmatrix}1 + 5\lambda\\ -3 - 13\lambda\\ -10 - 34\lambda\end{pmatrix}\) or e.g. \(\dfrac{1}{1}\begin{pmatrix}1 & 0 & 0\\ -2 & -1 & -1\\ -6 & -1 & -2\end{pmatrix}\begin{pmatrix}1 & 5\\ -2 & 2\\ 3 & 1\end{pmatrix} = \begin{pmatrix}1 & 5\\ -3 & -13\\ -10 & -34\end{pmatrix}\) | M1A1 |
| \(\dfrac{x - 1}{5} = \dfrac{y + 3}{-13} = \dfrac{z + 10}{-34}\) oe \(\begin{matrix}a_1 + b_1\lambda\\ a_2 + b_2\lambda\\ a_3 + b_3\lambda\end{matrix} \to \dfrac{x - a_1}{b_1} = \dfrac{y - a_2}{b_2} = \dfrac{z - a_3}{b_3}\) | dM1A1 |
| (6) | |
| (12 marks) |
Notes
M1: Attempt \(l_2\) in parametric form
A1: Correct parametric form
M1: Puts \(k = 0\) in their \(\mathbf{M}^{-1}\) and multiplies this by their parametric form correctly. Or starts again to find the inverse and multiplies.
A1: Correct parametric form for \(l_1\) or correct matrix.
dM1: Attempts cartesian form from their parametric \(l_1\) correctly. Dependent on both previous M’s.
A1: A complete correct equation
If their \(\mathbf{M}^{-1}\) is incorrect in terms of \(k\) but by substituting \(k = 0\), a correct answer is obtained in (c) allow a full recovery.
(c) Way 2
| Scheme | Marks |
|---|---|
| \(\mathbf{i} - 2\mathbf{j} + 3\mathbf{k}\) and \(6\mathbf{i} + 4\mathbf{k}\) are on \(l_2\) | |
| \(\mathbf{M}^{-1}(\mathbf{i} - 2\mathbf{j} + 3\mathbf{k}) = \mathbf{i} - 3\mathbf{j} - 10\mathbf{k}\) \(\mathbf{M}^{-1}(6\mathbf{i} + 4\mathbf{k}) = 6\mathbf{i} - 16\mathbf{j} - 44\mathbf{k}\) | M1A1 |
| \(\begin{pmatrix}6 + 5\lambda\\ -16 - 13\lambda\\ -44 - 34\lambda\end{pmatrix}\) | M1A1 |
| \(\dfrac{x - 6}{5} = \dfrac{y + 16}{-13} = \dfrac{z + 44}{-34}\) oe \(\begin{matrix}a_1 + b_1\lambda\\ a_2 + b_2\lambda\\ a_3 + b_3\lambda\end{matrix} \to \dfrac{x - a_1}{b_1} = \dfrac{y - a_2}{b_2} = \dfrac{z - a_3}{b_3}\) | dM1A1 |
M1: Attempt two points on \(l_1\)
A1: Two correct points on \(l_1\)
M1: Uses their points to obtain parametric form for \(l_1\)
A1: Correct parametric form for \(l_1\) or correct position and direction.
dM1: Attempts cartesian form from their parametric \(l_1\) correctly. Dependent on both previous M’s.
A1: A complete correct equation
(c) Way 3
| Scheme | Marks |
|---|---|
| \(\begin{pmatrix}1 & 0 & 0\\ 2 & -2 & 1\\ -4 & 1 & -1\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix} = \begin{pmatrix}1\\ -2\\ 3\end{pmatrix} \Rightarrow \begin{pmatrix}x\\ y\\ z\end{pmatrix} = \begin{pmatrix}1\\ -3\\ -10\end{pmatrix}\) | M1A1 |
| \(\begin{pmatrix}1 & 0 & 0\\ 2 & -2 & 1\\ -4 & 1 & -1\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix} = \begin{pmatrix}5\\ 2\\ 1\end{pmatrix} \Rightarrow \begin{pmatrix}x\\ y\\ z\end{pmatrix} = \begin{pmatrix}5\\ -13\\ -34\end{pmatrix}\) | M1A1 |
| \(\dfrac{x - 1}{5} = \dfrac{y + 3}{-13} = \dfrac{z + 10}{-34}\) | dM1A1 |
M1: Solves \(\mathbf{M}\begin{pmatrix}x\\ y\\ z\end{pmatrix} = \begin{pmatrix}1\\ -2\\ 3\end{pmatrix} \Rightarrow \begin{pmatrix}x\\ y\\ z\end{pmatrix} = \ldots\)
A1: \(\mathbf{i} - 3\mathbf{j} - 10\mathbf{k}\). Correct vector or values for \(x\), \(y\) and \(z\)
M1: Solves \(\mathbf{M}\begin{pmatrix}x\\ y\\ z\end{pmatrix} = \begin{pmatrix}5\\ 2\\ 1\end{pmatrix} \Rightarrow \begin{pmatrix}x\\ y\\ z\end{pmatrix} = \ldots\)
A1: \(5\mathbf{i} - 13\mathbf{j} - 34\mathbf{k}\). Correct vector or values for \(x\), \(y\) and \(z\)
dM1: Attempts Cartesian form from their values correctly. Dependent on both previous M’s.
A1: A complete correct equation
(c) Way 4
| Scheme | Marks |
|---|---|
| \(l_2: (1 + 5\lambda)\mathbf{i} + (-2 + 2\lambda)\mathbf{j} + (3 + \lambda)\mathbf{k}\) | M1A1 |
| \(\begin{pmatrix}1 & 0 & 0\\ 2 & -2 & 1\\ -4 & 1 & -1\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix} = \begin{pmatrix}1 + 5\lambda\\ -2 + 2\lambda\\ 3 + \lambda\end{pmatrix} \Rightarrow \begin{aligned}x &= 1 + 5\lambda\\ y &= -3 - 13\lambda\\ z &= -10 - 34\lambda\end{aligned}\) M1: Uses \(\mathbf{Mx} = l_2\) in parametric form A1: Correct expressions for \(x\), \(y\) and \(z\) | M1A1 |
| \(\dfrac{x - 1}{5} = \dfrac{y + 3}{-13} = \dfrac{z + 10}{-34}\) | dM1A1 |
M1: Attempt \(l_2\) in parametric form correctly
A1: Correct
dM1: Attempts Cartesian form from their values correctly. Dependent on both previous M’s.
A1: A complete correct equation