| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure (Further Additional Pure) |
| Year | 2017 |
| Session | Specimen |
| Marks | 13 |
| Topic | Groups |
| Type | Matrix groups |
| Difficulty | Challenging +1.8 This is a Further Maths group theory question requiring matrix multiplication, proof by contradiction, verification of group axioms, and finding a specific subgroup. While the individual components are accessible (matrix multiplication is routine, group axioms are standard), the abstract algebraic reasoning and multi-part structure with proof elements place it well above average difficulty. The subgroup identification requires insight into the structure of the group (essentially complex numbers under multiplication). |
| Spec | 4.01b Complex proofs: conjecture and demanding proofs4.03a Matrix language: terminology and notation4.03b Matrix operations: addition, multiplication, scalar8.03c Group definition: recall and use, show structure is/isn't a group8.03f Subgroups: definition and tests for proper subgroups |
The set $X$ consists of all $2 \times 2$ matrices of the form $\begin{pmatrix} x & -y \\ y & x \end{pmatrix}$, where $x$ and $y$ are real numbers which are not both zero.
\begin{enumerate}[label=(\roman*)]
\item \begin{enumerate}[label=(\alph*)]
\item The matrices $\begin{pmatrix} a & -b \\ b & a \end{pmatrix}$ and $\begin{pmatrix} c & -d \\ d & c \end{pmatrix}$ are both elements of $X$.
Show that $\begin{pmatrix} a & -b \\ b & a \end{pmatrix}\begin{pmatrix} c & -d \\ d & c \end{pmatrix} = \begin{pmatrix} p & -q \\ q & p \end{pmatrix}$ for some real numbers $p$ and $q$ to be found in terms of $a$, $b$, $c$ and $d$. [2]
\item Prove by contradiction that $p$ and $q$ are not both zero. [5]
\end{enumerate}
\item Prove that $X$, under matrix multiplication, forms a group $G$.
[You may use the result that matrix multiplication is associative.] [4]
\item Determine a subgroup of $G$ of order 17. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure 2017 Q8 [13]}}