| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2012 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Type | Matrix groups |
| Difficulty | Challenging +1.8 This is a Further Maths group theory question requiring understanding of matrix multiplication mod 6, commutativity verification, finding inverses, determining element orders, applying Lagrange's theorem to rule out subgroups, and analyzing group isomorphism based on element orders. While conceptually demanding and requiring multiple abstract algebra concepts, the calculations are relatively straightforward and the question provides significant scaffolding through its parts. |
| Spec | 4.03b Matrix operations: addition, multiplication, scalar8.03b Cayley tables: construct for finite sets under binary operation8.03c Group definition: recall and use, show structure is/isn't a group8.03e Order of elements: and order of groups8.03f Subgroups: definition and tests for proper subgroups8.03l Isomorphism: determine using informal methods |
| Answer | Marks | Guidance |
|---|---|---|
| \(\begin{pmatrix} 1 & 0 \\ n & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ m & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ n+m & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ m & n \end{pmatrix}\) | M1 | For multiplying 2 distinct matrices of the correct form both ways, or generalised form at least one way, |
| \(= \begin{pmatrix} 1 & 0 \\ m & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ n & 1 \end{pmatrix} \Rightarrow \text{commutative}\) | A1 | For stating or implying that addition is commutative and correct conclusion SR Use of numerical matrices must be generalised for any credit |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \((I =) \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\) | B1 | For correct identity |
| EITHER \(\begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}^{-1} = \begin{pmatrix} 1 & 0 \\ -2 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 4 & 1 \end{pmatrix}\) | M1 A1 | For using inverse property For correct inverse |
| OR \(\begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ n & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \Rightarrow 2 + n = 0 \Rightarrow \begin{pmatrix} 1 & 0 \\ 4 & 1 \end{pmatrix}\) | M1 A1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix}\) has order 2 | B1 | For correct order |
| 4 is not a factor of 6 | B1 | For correct reason (Award B0 for "Lagrange" only). Must be explicit about the '6' |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}\) OR \(\begin{pmatrix} 1 & 0 \\ 5 & 1 \end{pmatrix}\) has order 6, (or \(>\) 3) | B1* | For stating (that there is) an element of \(M\) with order 6 |
| OR \((M, \times)\) is cyclic, \(G\) is non-cyclic (having no element of order 6) OR \((M, \times)\) is commutative \(G\) is not commutative (being the non-cyclic group) \(\Rightarrow\) groups are not isomorphic | Award B1* for a relevant statement about \(M\) and \(G\) | |
| B1*dep | For correct conclusion and no false statements attached to conclusion | |
| [2] |
**(i)**
| $\begin{pmatrix} 1 & 0 \\ n & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ m & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ n+m & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ m & n \end{pmatrix}$ | M1 | For multiplying 2 distinct matrices of the correct form both ways, or generalised form at least one way, |
| $= \begin{pmatrix} 1 & 0 \\ m & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ n & 1 \end{pmatrix} \Rightarrow \text{commutative}$ | A1 | For stating or implying that addition is commutative and correct conclusion SR Use of numerical matrices must be generalised for any credit |
| | [2] | |
**(ii)**
| $(I =) \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ | B1 | For correct identity |
| EITHER $\begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}^{-1} = \begin{pmatrix} 1 & 0 \\ -2 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 4 & 1 \end{pmatrix}$ | M1 A1 | For using inverse property For correct inverse |
| OR $\begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ n & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \Rightarrow 2 + n = 0 \Rightarrow \begin{pmatrix} 1 & 0 \\ 4 & 1 \end{pmatrix}$ | M1 A1 | |
| | [3] | |
**(iii)**
| $\begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix}$ has order 2 | B1 | For correct order |
| 4 is not a factor of 6 | B1 | For correct reason (Award B0 for "Lagrange" only). Must be explicit about the '6' |
| | [2] | |
**(iv)**
| $\begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$ OR $\begin{pmatrix} 1 & 0 \\ 5 & 1 \end{pmatrix}$ has order 6, (or $>$ 3) | B1* | For stating (that there is) an element of $M$ with order 6 |
| OR $(M, \times)$ is cyclic, $G$ is non-cyclic (having no element of order 6) OR $(M, \times)$ is commutative $G$ is not commutative (being the non-cyclic group) $\Rightarrow$ groups are not isomorphic | | Award B1* for a relevant statement about $M$ and $G$ |
| | B1*dep | For correct conclusion and no false statements attached to conclusion |
| | [2] | |7 The set $M$ consists of the six matrices $\left( \begin{array} { l l } 1 & 0 \\ n & 1 \end{array} \right)$, where $n \in \{ 0,1,2,3,4,5 \}$. It is given that $M$ forms a group ( $M , \times$ ) under matrix multiplication, with numerical addition and multiplication both being carried out modulo 6 .\\
(i) Determine whether ( $M , \times$ ) is a commutative group, justifying your answer.\\
(ii) Write down the identity element of the group and find the inverse of $\left( \begin{array} { l l } 1 & 0 \\ 2 & 1 \end{array} \right)$.\\
(iii) State the order of $\left( \begin{array} { l l } 1 & 0 \\ 3 & 1 \end{array} \right)$ and give a reason why $( M , \times )$ has no subgroup of order 4.\\
(iv) The multiplicative group $G$ has order 6. All the elements of $G$, apart from the identity, have order 2 or 3 . Determine whether $G$ is isomorphic to ( $M , \times$ ), justifying your answer.
\hfill \mbox{\textit{OCR FP3 2012 Q7 [9]}}