| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2017 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | Transformation mapping problems |
| Difficulty | Challenging +1.2 This is a multi-part FP3 question involving standard matrix operations (determinant, inverse) and a 3D transformation problem. Parts (a) and (b) are routine Further Maths techniques. Part (c) requires applying the inverse transformation to find a pre-image line, which involves systematic calculation rather than novel insight. The question is moderately harder than average A-level due to being Further Maths content with 3D geometry, but follows predictable patterns for FP3 students. |
| Spec | 4.03j Determinant 3x3: calculation4.03o Inverse 3x3 matrix4.04a Line equations: 2D and 3D, cartesian and vector forms |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \(\det\mathbf{M}=1\times(2-1)-k(-2+4)(+0)=1-2k\) or \(\det\mathbf{M}=(0)-1(1+4k)-1(-2-2k)=1-2k\) or rule of Sarrus: \(\det\mathbf{M}=2-4k-1+2k=1-2k\) | M1A1* | M1: Correct attempt at determinant (at least 2 elements correct); A1: Obtains printed answer with no errors. If determinant notation used, must see intermediate step e.g. minimally \(1-2k+0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \((\mathbf{M}^T)=\begin{pmatrix}1&2&-4\\k&-2&1\\0&1&-1\end{pmatrix}\) or (minors) \(\begin{pmatrix}1&2&-6\\-k&-1&1+4k\\k&1&-2-2k\end{pmatrix}\) or (cofactors) \(\begin{pmatrix}1&-2&-6\\k&-1&-1-4k\\k&-1&-2-2k\end{pmatrix}\) | B1 | |
| \(\mathbf{M}^{-1}=\frac{1}{1-2k}\begin{pmatrix}1&k&k\\-2&-1&-1\\-6&-1-4k&-2-2k\end{pmatrix}\) | M1A1A1 | M1: Full attempt at inverse ignoring determinant (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 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \(l_2:(1+5\lambda)\mathbf{i}+(-2+2\lambda)\mathbf{j}+(3+\lambda)\mathbf{k}\) | M1A1 | M1: Attempt \(l_2\) in parametric form; A1: Correct parametric form |
| \(\frac{1}{1}\begin{pmatrix}1&0&0\\-2&-1&-1\\-6&-1&-2\end{pmatrix}\begin{pmatrix}1+5\lambda\\-2+2\lambda\\3+\lambda\end{pmatrix}=\begin{pmatrix}1+5\lambda\\-3-13\lambda\\-10-34\lambda\end{pmatrix}\) | M1A1 | M1: Puts \(k=0\) in \(\mathbf{M}^{-1}\) and multiplies by parametric form correctly; A1: Correct parametric form for \(l_1\) or correct matrix |
| \(\frac{x-1}{5}=\frac{y+3}{-13}=\frac{z+10}{-34}\) | dM1A1 | M1: Attempts Cartesian form from parametric \(l_1\) correctly. Dependent on both previous M's; A1: Complete correct equation |
# Question 6:
## Part (a):
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $\det\mathbf{M}=1\times(2-1)-k(-2+4)(+0)=1-2k$ or $\det\mathbf{M}=(0)-1(1+4k)-1(-2-2k)=1-2k$ or rule of Sarrus: $\det\mathbf{M}=2-4k-1+2k=1-2k$ | M1A1* | M1: Correct attempt at determinant (at least 2 elements correct); A1: Obtains printed answer with no errors. If determinant notation used, must see intermediate step e.g. minimally $1-2k+0$ |
## Part (b):
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $(\mathbf{M}^T)=\begin{pmatrix}1&2&-4\\k&-2&1\\0&1&-1\end{pmatrix}$ or (minors) $\begin{pmatrix}1&2&-6\\-k&-1&1+4k\\k&1&-2-2k\end{pmatrix}$ or (cofactors) $\begin{pmatrix}1&-2&-6\\k&-1&-1-4k\\k&-1&-2-2k\end{pmatrix}$ | B1 | |
| $\mathbf{M}^{-1}=\frac{1}{1-2k}\begin{pmatrix}1&k&k\\-2&-1&-1\\-6&-1-4k&-2-2k\end{pmatrix}$ | M1A1A1 | M1: Full attempt at inverse ignoring determinant (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 |
## Part (c):
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $l_2:(1+5\lambda)\mathbf{i}+(-2+2\lambda)\mathbf{j}+(3+\lambda)\mathbf{k}$ | M1A1 | M1: Attempt $l_2$ in parametric form; A1: Correct parametric form |
| $\frac{1}{1}\begin{pmatrix}1&0&0\\-2&-1&-1\\-6&-1&-2\end{pmatrix}\begin{pmatrix}1+5\lambda\\-2+2\lambda\\3+\lambda\end{pmatrix}=\begin{pmatrix}1+5\lambda\\-3-13\lambda\\-10-34\lambda\end{pmatrix}$ | M1A1 | M1: Puts $k=0$ in $\mathbf{M}^{-1}$ and multiplies by parametric form correctly; A1: Correct parametric form for $l_1$ or correct matrix |
| $\frac{x-1}{5}=\frac{y+3}{-13}=\frac{z+10}{-34}$ | dM1A1 | M1: Attempts Cartesian form from parametric $l_1$ correctly. Dependent on both previous M's; A1: Complete correct equation |6. The matrix $\mathbf { M }$ is given by
$$\mathbf { M } = \left( \begin{array} { r r r }
1 & k & 0 \\
2 & - 2 & 1 \\
- 4 & 1 & - 1
\end{array} \right) , k \in \mathbb { R } , k \neq \frac { 1 } { 2 }$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\operatorname { det } \mathbf { M } = 1 - 2 k$.
\item Find $\mathbf { M } ^ { - 1 }$ in terms of $k$.
The straight line $l _ { 1 }$ is mapped onto the straight line $l _ { 2 }$ by the transformation represented by the matrix
$$\left( \begin{array} { r r r }
1 & 0 & 0 \\
2 & - 2 & 1 \\
- 4 & 1 & - 1
\end{array} \right)$$
Given that $l _ { 2 }$ has cartesian equation
$$\frac { x - 1 } { 5 } = \frac { y + 2 } { 2 } = \frac { z - 3 } { 1 }$$
\item find a cartesian equation of the line $l _ { 1 }$
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP3 2017 Q6 [12]}}