| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure (Further Additional Pure) |
| Session | Specimen |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Type | Verify group axioms |
| Difficulty | Challenging +1.8 This is a Further Maths group theory question requiring verification of all group axioms for a non-standard matrix set, including a proof by contradiction for closure and finding a specific-order subgroup. While the matrix multiplication is computational, proving closure properly, identifying the identity/inverses, and constructing a cyclic subgroup of order 17 requires solid abstract algebra understanding beyond typical A-level. |
| Spec | 1.01d Proof by contradiction4.03a 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 subgroups8.03i Properties of groups: structure of finite groups up to order 7 |
| Answer | Marks | Guidance |
|---|---|---|
| 8 | (i) | (a) |
| Answer | Marks |
|---|---|
| i.e. p(cid:32)ac(cid:16)bd and q(cid:32)ad(cid:14)bc | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| [2] | 1.1 | |
| 1.1 | May be clear from the product matrix | |
| 8 | (i) | (b) |
| Answer | Marks |
|---|---|
| both zero is false. | B1 |
| Answer | Marks |
|---|---|
| [5] | 1.1 |
| Answer | Marks |
|---|---|
| 2.4 | n |
| Answer | Marks |
|---|---|
| Contradiction clearly explained | OR |
| Answer | Marks | Guidance |
|---|---|---|
| 8 | (ii) | Closure follows from part (i) |
| Answer | Marks |
|---|---|
| since a2 (cid:32)b2 (cid:122)0 (as a, b not both zero) | B1 |
| Answer | Marks |
|---|---|
| [4] | 1.1 |
| Answer | Marks |
|---|---|
| 2.3 | n |
| Answer | Marks | Guidance |
|---|---|---|
| 8 | (iii) | The matrix witha(cid:32)cos (cid:11) 2(cid:83) (cid:12) , b(cid:32)sin (cid:11) 2(cid:83) (cid:12) |
| Answer | Marks |
|---|---|
| 17 | B1 |
| Answer | Marks |
|---|---|
| [2] | i |
| Answer | Marks |
|---|---|
| 2.4 | Suitable values of a and b |
Question 8:
8 | (i) | (a) | (cid:167)a (cid:16)b(cid:183) (cid:167)c (cid:16)d(cid:183) (cid:167)ac(cid:16)bd (cid:16)(ad(cid:14)bc)(cid:183)
(cid:168) (cid:184) (cid:168) (cid:184)(cid:32)(cid:168) (cid:184)
(cid:169)b a (cid:185) (cid:169)d c (cid:185) (cid:169)ad (cid:14)bc ac(cid:16)bd (cid:185)
i.e. p(cid:32)ac(cid:16)bd and q(cid:32)ad(cid:14)bc | M1
A1
[2] | 1.1
1.1 | May be clear from the product matrix
8 | (i) | (b) | Suppose that both p(cid:32)0 and q(cid:32)0, so that
ac(cid:16)bd (cid:32)0 and bc(cid:14)ad (cid:32)0
Then acd(cid:16)bd2 (cid:32)0 and
acd(cid:14)bc2 (cid:32)0
(cid:11) c2(cid:14)d2(cid:12)
Subtracting (cid:159)b (cid:32)0
Since c and d are not both zero c2 (cid:14)d2 (cid:122)0
p
(cid:159)b(cid:32)0
S
Similarly, ac(cid:16)bd (cid:32)0 and
(cid:11) c2(cid:14)d2(cid:12)
bc(cid:14)ad (cid:32)0(cid:159)a (cid:32)0
Since c and d are not both zero c2 (cid:14)d2 (cid:122)0
(cid:159)a(cid:32)0
This contradicts the condition that a, b are not
both zero, so the assumption that p and q are
both zero is false. | B1
M1
e
A1
A1
E1
[5] | 1.1
i
c
3.1a
2.2a
2.2a
2.4 | n
Explicit assumption that both are zero
Ceondition that a and b not both zero
and c and d not both zero need not be
mexplicitly stated
e.g. treating as simultaneous equations
(either a, b or c, d)
One case correctly concluded
Second case correctly concluded
Contradiction clearly explained | OR
B1 Suppose that both
p(cid:32)0 and q(cid:32)0so that
(cid:167)p (cid:16)q(cid:183)
det(cid:168) (cid:184)(cid:32) p2 (cid:14)q2 (cid:32)0
(cid:169)q p (cid:185)
M1 Then
(cid:170)(cid:167)a (cid:16)b(cid:183) (cid:167)c (cid:16)d(cid:183)(cid:186)
det(cid:171)(cid:168) (cid:184) (cid:168) (cid:184)(cid:187)(cid:32)0
(cid:172)(cid:169)b a (cid:185) (cid:169)d c (cid:185)(cid:188)
A1 so that c2 (cid:14)d2 (cid:32)0 and/or
a2 (cid:14)b2 (cid:32)0
A1 since neither {c and d} or {a and
b} can both be zero…
E1 This contradicts the condition that
a, b and c, d are not both zero, so the
assumption that p and q are both zero
is false.
8 | (ii) | Closure follows from part (i)
(Associativity is given)
Identity is when a(cid:32)1, b(cid:32)0 or noting
(cid:167)1 ((cid:16))0(cid:183)
(cid:168) (cid:184)(cid:143)X
(cid:169)0 1 (cid:185)
(cid:167)a (cid:16)b(cid:183)
Inverse of (cid:168) (cid:184) is
(cid:168) (cid:184)
(cid:169)b a (cid:185)
1 (cid:167) a (cid:16)((cid:16)b)(cid:183)
(cid:168) (cid:184)(cid:143)X
a2(cid:14)b2 (cid:169)(cid:16)b a (cid:185)
since a2 (cid:32)b2 (cid:122)0 (as a, b not both zero) | B1
B1
B1
E1
[4] | 1.1
1.1
2.1
2.3 | n
To earn this, they must show
appropriate form for A(cid:16)1 and note
e
why it exists
m
8 | (iii) | The matrix witha(cid:32)cos (cid:11) 2(cid:83) (cid:12) , b(cid:32)sin (cid:11) 2(cid:83) (cid:12)
17 17
generates …
the subgroup of rotations about O (through
multiples of 2(cid:83))
17 | B1
e
B1
[2] | i
c
3.1a
2.4 | Suitable values of a and b
Description of the subgroup8 The set $X$ consists of all $2 \times 2$ matrices of the form $\left( \begin{array} { r r } x & - y \\ y & x \end{array} \right)$, where $x$ and $y$ are real numbers which are not both zero.
\begin{enumerate}[label=(\roman*)]
\item (a) The matrices $\left( \begin{array} { c c } a & - b \\ b & a \end{array} \right)$ and $\left( \begin{array} { c c } c & - d \\ d & c \end{array} \right)$ are both elements of $X$.
Show that $\left( \begin{array} { c c } a & - b \\ b & a \end{array} \right) \left( \begin{array} { c c } c & - d \\ d & c \end{array} \right) = \left( \begin{array} { c c } p & - q \\ q & p \end{array} \right)$ for some real numbers $p$ and $q$ to be found in terms of $a , b , c$ and $d$.\\
(b) Prove by contradiction that $p$ and $q$ are not both zero.
\item Prove that $X$, under matrix multiplication, forms a group $G$. [You may use the result that matrix multiplication is associative.]
\item Determine a subgroup of $G$ of order 17.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure Q8 [13]}}