Pre-U Pre-U 9795/1 2020 Specimen — Question 8 5 marks

Exam BoardPre-U
ModulePre-U 9795/1 (Pre-U Further Mathematics Paper 1)
Year2020
SessionSpecimen
Marks5
TopicGroups
TypeMatrix groups
DifficultyChallenging +1.8 This is a challenging group theory question requiring verification of group axioms for a non-standard matrix set, identification of a specific subgroup, and proof of non-existence of order-3 subgroups using Lagrange's theorem or direct analysis. While the matrix form is simple, the abstract algebra reasoning and proof components place it well above average difficulty for A-level, though it's a relatively standard exercise within Further Maths group theory.
Spec4.03b Matrix operations: addition, multiplication, scalar8.03c Group definition: recall and use, show structure is/isn't a group8.03g Cyclic groups: meaning of the term
8 Consider the set \(S\) of all matrices of the form \(\left( \begin{array} { l l } p & p \\ p & p \end{array} \right)\), where p is a non-zero rational number.
  1. Show that \(S\), under the operation of matrix multiplication, forms a group, \(G\). (You may assume that matrix multiplication is associative.)
  2. Find a subgroup of \(G\) of order 2 and show that \(G\) contains no subgroups of order 3.
(a) 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 OR Alternative \([(p)(q)](r) = (p)[(q)(r)] = (4pqr)\) 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
Total: 5
(b) Attempt to look for a self-inverse element; i.e. solving \(p = \frac{1}{4p}\) using their \((p)^{-1}\) and E M1
\(p = -\frac{1}{2}\) and noting that \(H = \{\mathbf{E}, \mathbf{A}\}\) where \(\mathbf{E} = \begin{pmatrix}\frac{1}{2} & \frac{1}{2}\\ \frac{1}{2} & \frac{1}{2}\end{pmatrix}\), \(\mathbf{A} = \begin{pmatrix}-\frac{1}{2} & -\frac{1}{2}\\ -\frac{1}{2} & -\frac{1}{2}\end{pmatrix}\) A1
Looking for \(\{\mathbf{E}, \mathbf{B}, \mathbf{B}^2\}\) where \(\mathbf{B}^3 = \mathbf{E}\); 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 B \((\neq \mathbf{E})\) exists A1
Total: 4
**(a)** 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 **OR** Alternative $[(p)(q)](r) = (p)[(q)(r)] = (4pqr)$ 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

**Total: 5**

**(b)** Attempt to look for a self-inverse element; i.e. solving $p = \frac{1}{4p}$ using their $(p)^{-1}$ and **E** **M1**

$p = -\frac{1}{2}$ and noting that $H = \{\mathbf{E}, \mathbf{A}\}$ where $\mathbf{E} = \begin{pmatrix}\frac{1}{2} & \frac{1}{2}\\ \frac{1}{2} & \frac{1}{2}\end{pmatrix}$, $\mathbf{A} = \begin{pmatrix}-\frac{1}{2} & -\frac{1}{2}\\ -\frac{1}{2} & -\frac{1}{2}\end{pmatrix}$ **A1**

Looking for $\{\mathbf{E}, \mathbf{B}, \mathbf{B}^2\}$ where $\mathbf{B}^3 = \mathbf{E}$; 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 **B** $(\neq \mathbf{E})$ exists **A1**

**Total: 4**
8 Consider the set $S$ of all matrices of the form $\left( \begin{array} { l l } p & p \\ p & p \end{array} \right)$, where p is a non-zero rational number.
\begin{enumerate}[label=(\alph*)]
\item Show that $S$, under the operation of matrix multiplication, forms a group, $G$. (You may assume that matrix multiplication is associative.)
\item Find a subgroup of $G$ of order 2 and show that $G$ contains no subgroups of order 3.
\end{enumerate}

\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2020 Q8 [5]}}
This paper (4 questions)
View full paper
Q7 2 Q8 5 Q11 8 Q12 6