| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2018 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Vector equation of a plane |
| Difficulty | Challenging +1.2 This is a multi-part Further Maths question requiring matrix inversion (standard cofactor method) and understanding that planes transform via the inverse transpose. While it involves several steps and Further Maths content, the techniques are routine: finding a 3×3 inverse follows a standard algorithm, and applying the transformation to the plane equation is methodical rather than requiring novel insight. The conceptual demand is moderate for FM students. |
| Spec | 4.03o Inverse 3x3 matrix4.03q Inverse transformations |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \( | \mathbf{M} | =4(2k)+5(k^2)(+0)\) |
| Minors: \(\begin{pmatrix}2k&k^2&-5k+6\\-5k&4k&-35\\0&0&8+5k\end{pmatrix}\) or cofactors: \(\begin{pmatrix}2k&-k^2&6-5k\\5k&4k&35\\0&0&8+5k\end{pmatrix}\) | B1 | A correct first step of minors or cofactors |
| \(\mathbf{M}^{-1}=\frac{1}{5k^2+8k}\begin{pmatrix}2k&5k&0\\-k^2&4k&0\\6-5k&35&8+5k\end{pmatrix}\) | M1B1A1 | M1: Fully recognisable attempt at the inverse including reciprocal of the determinant; B1: Any 2 correct rows or columns ignoring determinant (may be missing); A1: Fully correct inverse |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(\mathbf{M}^{-1}=-\frac{1}{3}\begin{pmatrix}-2&-5&0\\-1&-4&0\\11&35&3\end{pmatrix}\) | M1 | Substitutes \(k=-1\) |
| \(\Pi_2: x=s,\ y=t,\ z=2s-4\) | M1 | Attempts parametric form (\(s\neq 0, t\neq 0\)); any pair of letters (inc \(x\) and \(y\)) can be used as parameters |
| \(-\frac{1}{3}\begin{pmatrix}-2&-5&0\\-1&-4&0\\11&35&3\end{pmatrix}\begin{pmatrix}s\\t\\2s-4\end{pmatrix}\) | ddM1 | Attempts \(\mathbf{M}^{-1}\times\) their parametric form; depends on both M marks above |
| \(-\frac{1}{3}\begin{pmatrix}-2s-5t\\-s-4t\\11s+35t+6s-12\end{pmatrix}\) | A1 | Correct parametric form for \(\Pi_1\) with \(s, t\) |
| \(11x-5y+z=4\) | dddM1A1 | dddM1: Eliminates \(s\) and \(t\) to obtain a Cartesian equation; all 3 previous M marks needed; \(x=-2x-5y\) gets M0 here (unless parameters are now changed); A1: Correct equation (oe) |
# Question 5:
$\mathbf{M}=\begin{pmatrix}4&-5&0\\k&2&0\\-3&-5&k\end{pmatrix}$
## Part (a):
| Working/Answer | Marks | Notes |
|---|---|---|
| $|\mathbf{M}|=4(2k)+5(k^2)(+0)$ | B1 | Correct determinant in any form (quadratic may be unsimplified) |
| Minors: $\begin{pmatrix}2k&k^2&-5k+6\\-5k&4k&-35\\0&0&8+5k\end{pmatrix}$ or cofactors: $\begin{pmatrix}2k&-k^2&6-5k\\5k&4k&35\\0&0&8+5k\end{pmatrix}$ | B1 | A correct first step of minors or cofactors |
| $\mathbf{M}^{-1}=\frac{1}{5k^2+8k}\begin{pmatrix}2k&5k&0\\-k^2&4k&0\\6-5k&35&8+5k\end{pmatrix}$ | M1B1A1 | M1: Fully recognisable attempt at the inverse including reciprocal of the determinant; B1: Any 2 correct rows or columns ignoring determinant (may be missing); A1: Fully correct inverse |
## Part (b):
| Working/Answer | Marks | Notes |
|---|---|---|
| $\mathbf{M}^{-1}=-\frac{1}{3}\begin{pmatrix}-2&-5&0\\-1&-4&0\\11&35&3\end{pmatrix}$ | M1 | Substitutes $k=-1$ |
| $\Pi_2: x=s,\ y=t,\ z=2s-4$ | M1 | Attempts parametric form ($s\neq 0, t\neq 0$); any pair of letters (inc $x$ and $y$) can be used as parameters |
| $-\frac{1}{3}\begin{pmatrix}-2&-5&0\\-1&-4&0\\11&35&3\end{pmatrix}\begin{pmatrix}s\\t\\2s-4\end{pmatrix}$ | ddM1 | Attempts $\mathbf{M}^{-1}\times$ their parametric form; depends on both M marks above |
| $-\frac{1}{3}\begin{pmatrix}-2s-5t\\-s-4t\\11s+35t+6s-12\end{pmatrix}$ | A1 | Correct parametric form for $\Pi_1$ with $s, t$ |
| $11x-5y+z=4$ | dddM1A1 | dddM1: Eliminates $s$ and $t$ to obtain a Cartesian equation; all 3 previous M marks needed; $x=-2x-5y$ gets M0 here (unless parameters are now changed); A1: Correct equation (oe) |
**Total: 11 marks**5.
$$\mathbf { M } = \left( \begin{array} { r r r }
4 & - 5 & 0 \\
k & 2 & 0 \\
- 3 & - 5 & k
\end{array} \right) \text {, where } k \text { is a real constant, } k \neq 0 , k \neq - \frac { 8 } { 5 }$$
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $k$, the inverse of the matrix $\mathbf { M }$.
A transformation $T : \mathbb { R } ^ { 3 } \rightarrow \mathbb { R } ^ { 3 }$ is represented by the matrix
$$\left( \begin{array} { r r r }
4 & - 5 & 0 \\
- 1 & 2 & 0 \\
- 3 & - 5 & - 1
\end{array} \right)$$
The transformation $T$ maps the plane $\Pi _ { 1 }$ onto the plane $\Pi _ { 2 }$\\
Given that the plane $\Pi _ { 2 }$ has equation $2 x - z = 4$
\item find a cartesian equation of the plane $\Pi _ { 1 }$
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 2018 Q5 [11]}}