| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2011 |
| Session | June |
| Marks | 9 |
| Topic | Groups |
| Type | Matrix groups |
| Difficulty | Challenging +1.2 This is a straightforward abstract algebra question requiring verification of group axioms and basic subgroup analysis. Part (i) involves routine checking of closure, associativity, identity, and inverses for a simple matrix form. Part (ii) requires finding a cyclic subgroup of order 2 (standard) and showing no order-3 subgroup exists (using basic group theory). While this is Further Maths content making it harder than typical A-level, the techniques are mechanical and well-practiced for students at this level, with no novel insights required. |
| Spec | 8.03c Group definition: recall and use, show structure is/isn't a group8.03f Subgroups: definition and tests for proper subgroups |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Good attempt to multiply 2 matrices of the appropriate form: \(\begin{pmatrix} p & p \\ p & p \end{pmatrix}\begin{pmatrix} q & q \\ q & q \end{pmatrix}\) | M1 | |
| "Closure" noted or implied by correct product matrix \(= \begin{pmatrix} 2pq & 2pq \\ 2pq & 2pq \end{pmatrix} \in S\) | A1 | |
| Statement that \(\times_{M}\) known to be associative | B1 | |
| Alt. \([(p)(q)](r) = (p)[(q)(r)] = (4pqr)\) shown | B1 | |
| Identity is \(\begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix} (\in S)\) | B1 | |
| \(\begin{pmatrix} p & p \\ p & p \end{pmatrix}^{-1} = \begin{pmatrix} \frac{1}{4p} & -\frac{1}{4p} \\ -\frac{1}{4p} & \frac{1}{4p} \end{pmatrix} (\in S\) as \(p \neq 0)\) | B1 | |
| … and \((S, \times_{M})\) is a group since all four group axioms are satisfied | ||
| [5] | ||
| (ii) Attempt to look for a self-inverse element; i.e. solving \(p = \frac{1}{4p}\) | M1 | ft their \((p)^{-1}\) and |
| \(p = -\frac{1}{2}\) and noting that \(H = \{\mathbb{E}, \mathbb{A}\}\) where \(\mathbb{E} = \begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix}\), \(\mathbb{A} = \begin{pmatrix} -\frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{2} & -\frac{1}{2} \end{pmatrix}\) | A1 | |
| Looking for \(\{\mathbb{E}, \mathbb{B}, \mathbb{B}^3\}\) where \(\mathbb{B}^3 = \mathbb{E}\); i.e. solving \((4p)^3 = \frac{1}{2}\) | M1 | |
| Explaining carefully that \(p^3 = \frac{1}{8} \Leftrightarrow p = \frac{1}{2}\) and no such \(\mathbb{B}(\ne \mathbb{E})\) exists | A1 | |
| [4] |
**(i)** Good attempt to multiply 2 matrices of the appropriate form: $\begin{pmatrix} p & p \\ p & p \end{pmatrix}\begin{pmatrix} q & q \\ q & q \end{pmatrix}$ | M1 |
"Closure" noted or implied by correct product matrix $= \begin{pmatrix} 2pq & 2pq \\ 2pq & 2pq \end{pmatrix} \in S$ | A1 |
Statement that $\times_{M}$ known to be associative | B1 |
Alt. $[(p)(q)](r) = (p)[(q)(r)] = (4pqr)$ shown | B1 |
Identity is $\begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix} (\in S)$ | B1 |
$\begin{pmatrix} p & p \\ p & p \end{pmatrix}^{-1} = \begin{pmatrix} \frac{1}{4p} & -\frac{1}{4p} \\ -\frac{1}{4p} & \frac{1}{4p} \end{pmatrix} (\in S$ as $p \neq 0)$ | B1 |
… and $(S, \times_{M})$ is a group since all four group axioms are satisfied |
| [5] |
**(ii)** Attempt to look for a self-inverse element; i.e. solving $p = \frac{1}{4p}$ | M1 | ft their $(p)^{-1}$ and
$p = -\frac{1}{2}$ and noting that $H = \{\mathbb{E}, \mathbb{A}\}$ where $\mathbb{E} = \begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix}$, $\mathbb{A} = \begin{pmatrix} -\frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{2} & -\frac{1}{2} \end{pmatrix}$ | A1 |
Looking for $\{\mathbb{E}, \mathbb{B}, \mathbb{B}^3\}$ where $\mathbb{B}^3 = \mathbb{E}$; i.e. solving $(4p)^3 = \frac{1}{2}$ | M1 |
Explaining carefully that $p^3 = \frac{1}{8} \Leftrightarrow p = \frac{1}{2}$ and no such $\mathbb{B}(\ne \mathbb{E})$ exists | A1 |
| [4] |
---Consider the set $S$ of all matrices of the form $\begin{pmatrix} p & p \\ p & p \end{pmatrix}$, where $p$ is a non-zero rational number.
\begin{enumerate}[label=(\roman*)]
\item Show that $S$, under the operation of matrix multiplication, forms a group, $G$. [5]
\item Find a subgroup of $G$ of order 2 and show that $G$ contains no subgroups of order 3. [4]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2011 Q6 [9]}}