| Exam Board | OCR MEI |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2015 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Type | Verify group axioms |
| Difficulty | Challenging +1.8 This is a comprehensive group theory question requiring verification of group axioms, closure proofs, modular arithmetic with matrices, construction of a Cayley table, and isomorphism determination. While it covers multiple concepts systematically, each part follows standard procedures taught in Further Maths. The modular arithmetic component and isomorphism check add complexity beyond routine exercises, but the question provides significant scaffolding through its structured parts. |
| Spec | 4.03a Matrix language: terminology and notation4.03b Matrix operations: addition, multiplication, scalar4.03c Matrix multiplication: properties (associative, not commutative)4.03n Inverse 2x2 matrix8.03c Group definition: recall and use, show structure is/isn't a group8.03d Latin square property: for group tables8.03e Order of elements: and order of groups8.03l Isomorphism: determine using informal methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| The set is closed: \(\begin{pmatrix}a & b\\0 & \frac{1}{a}\end{pmatrix}\begin{pmatrix}c & d\\0 & \frac{1}{c}\end{pmatrix} = \begin{pmatrix}ac & ad+\frac{b}{c}\\0 & \frac{1}{ac}\end{pmatrix}\) | M1 | Attempt to demonstrate closure |
| Two general and different matrices shown | A1 | Two general and different matrices |
| Correct product therefore shows closure | A1 | Correct product therefore shows closure |
| Identity is \(\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}\) or \(m(1,0)\) | B1 | |
| Each element has an inverse; Inverse of \(\begin{pmatrix}a & b\\0 & \frac{1}{a}\end{pmatrix}\) is \(\begin{pmatrix}\frac{1}{a} & -b\\0 & a\end{pmatrix}\) | M1, A1 | Attempt to demonstrate inverse; A general matrix and its inverse |
| Which is in M | A1 | |
| Total: 7 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| e.g. product: \(\begin{pmatrix}c & d\\0 & \frac{1}{c}\end{pmatrix}\times\begin{pmatrix}a & b\\0 & \frac{1}{a}\end{pmatrix} = \begin{pmatrix}ac & cb+\frac{d}{a}\\0 & \frac{1}{ac}\end{pmatrix}\) | M1 | Demonstration by multiplying 2 matrices each way round (one way might be quoted from (i)) |
| 2 correct (and different) products | A1 | 2 correct (and different) products |
| So No | A1 | Dep on previous 2 marks |
| Total: 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\begin{pmatrix}k & b\\0 & \frac{1}{k}\end{pmatrix}\begin{pmatrix}k & c\\0 & \frac{1}{k}\end{pmatrix} = \begin{pmatrix}k^2 & \ldots\\0 & \frac{1}{k^2}\end{pmatrix}\) | M1 | Multiplying 2 matrices in N (Allow matrices the same) |
| Sight of \(k^2\) | A1 | Sight of \(k^2\) |
| This is only in the set \(N_k\) if \(k^2 = k\) | A1 | |
| Given \(k \neq 0 \Rightarrow k = 1\) only | A1 | |
| Total: 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(m(1,1)^2 = \begin{pmatrix}1 & 1\\0 & 1\end{pmatrix}\begin{pmatrix}1 & 1\\0 & 1\end{pmatrix} = \begin{pmatrix}1 & 2\\0 & 1\end{pmatrix}\) | M1, A1 | |
| Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Cayley table for \(\{m(1,0), m(1,1), m(1,2), m(1,3)\}\): \(m(1,0)\) the identity, table entries as shown | B1 | \(m(1,0)\) the identity |
| Full correct Cayley table | B3 | \(-1\) each error |
| Total: 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Group table for R; OR any argument that states: \(a, b, c\) have order 2 | B3, B1 | \(-1\) each error |
| Reason for this | B1 | |
| But only one element of P has order 2 | B1 | |
| So No | B1 | Dependent on previous B3 |
| Total: 4 marks |
# Question 4:
## Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| The set is closed: $\begin{pmatrix}a & b\\0 & \frac{1}{a}\end{pmatrix}\begin{pmatrix}c & d\\0 & \frac{1}{c}\end{pmatrix} = \begin{pmatrix}ac & ad+\frac{b}{c}\\0 & \frac{1}{ac}\end{pmatrix}$ | M1 | Attempt to demonstrate closure |
| Two general and different matrices shown | A1 | Two general and different matrices |
| Correct product therefore shows closure | A1 | Correct product therefore shows closure |
| Identity is $\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}$ or $m(1,0)$ | B1 | |
| Each element has an inverse; Inverse of $\begin{pmatrix}a & b\\0 & \frac{1}{a}\end{pmatrix}$ is $\begin{pmatrix}\frac{1}{a} & -b\\0 & a\end{pmatrix}$ | M1, A1 | Attempt to demonstrate inverse; A general matrix and its inverse |
| Which is in M | A1 | |
| **Total: 7 marks** | | |
## Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| e.g. product: $\begin{pmatrix}c & d\\0 & \frac{1}{c}\end{pmatrix}\times\begin{pmatrix}a & b\\0 & \frac{1}{a}\end{pmatrix} = \begin{pmatrix}ac & cb+\frac{d}{a}\\0 & \frac{1}{ac}\end{pmatrix}$ | M1 | Demonstration by multiplying 2 matrices each way round (one way might be quoted from (i)) |
| 2 correct (and different) products | A1 | 2 correct (and different) products |
| So No | A1 | Dep on previous 2 marks |
| **Total: 3 marks** | | |
## Part (iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\begin{pmatrix}k & b\\0 & \frac{1}{k}\end{pmatrix}\begin{pmatrix}k & c\\0 & \frac{1}{k}\end{pmatrix} = \begin{pmatrix}k^2 & \ldots\\0 & \frac{1}{k^2}\end{pmatrix}$ | M1 | Multiplying 2 matrices in N (Allow matrices the same) |
| Sight of $k^2$ | A1 | Sight of $k^2$ |
| This is only in the set $N_k$ if $k^2 = k$ | A1 | |
| Given $k \neq 0 \Rightarrow k = 1$ only | A1 | |
| **Total: 4 marks** | | |
## Part (iv)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $m(1,1)^2 = \begin{pmatrix}1 & 1\\0 & 1\end{pmatrix}\begin{pmatrix}1 & 1\\0 & 1\end{pmatrix} = \begin{pmatrix}1 & 2\\0 & 1\end{pmatrix}$ | M1, A1 | |
| **Total: 2 marks** | | |
## Part (v)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Cayley table for $\{m(1,0), m(1,1), m(1,2), m(1,3)\}$: $m(1,0)$ the identity, table entries as shown | B1 | $m(1,0)$ the identity |
| Full correct Cayley table | B3 | $-1$ each error |
| **Total: 4 marks** | | |
## Part (vi)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Group table for R; OR any argument that states: $a, b, c$ have order 2 | B3, B1 | $-1$ each error |
| Reason for this | B1 | |
| But only one element of P has order 2 | B1 | |
| So No | B1 | Dependent on previous B3 |
| **Total: 4 marks** | | |4 M is the set of all $2 \times 2$ matrices $\mathrm { m } ( a , b )$ where $a$ and $b$ are rational numbers and
$$\mathrm { m } ( a , b ) = \left( \begin{array} { l l }
a & b \\
0 & \frac { 1 } { a }
\end{array} \right) , a \neq 0$$
(i) Show that under matrix multiplication M is a group. You may assume associativity of matrix multiplication.\\
(ii) Determine whether the group is commutative.
The set $\mathrm { N } _ { k }$ consists of all $2 \times 2$ matrices $\mathrm { m } ( k , b )$ where $k$ is a fixed positive integer and $b$ can take any integer value.\\
(iii) Prove that $\mathrm { N } _ { k }$ is closed under matrix multiplication if and only if $k = 1$.
Now consider the set P consisting of the matrices $\mathrm { m } ( 1,0 ) , \mathrm { m } ( 1,1 ) , \mathrm { m } ( 1,2 )$ and $\mathrm { m } ( 1,3 )$. The elements of P are combined using matrix multiplication but with arithmetic carried out modulo 4 .\\
(iv) Show that $( \mathrm { m } ( 1,1 ) ) ^ { 2 } = \mathrm { m } ( 1,2 )$.\\
(v) Construct the group combination table for P .
The group R consists of the set $\{ e , a , b , c \}$ combined under the operation *. The identity element is $e$, and elements $a , b$ and $c$ are such that
$$a ^ { * } a = b ^ { * } b = c ^ { * } c \quad \text { and } \quad a ^ { * } c = c ^ { * } a = b$$
(vi) Determine whether R is isomorphic to P .
Option 5: Markov chains
\section*{This question requires the use of a calculator with the ability to handle matrices.}
\hfill \mbox{\textit{OCR MEI FP3 2015 Q4 [24]}}