OCR Further Additional Pure AS 2018 June — Question 4 11 marks

Exam BoardOCR
ModuleFurther Additional Pure AS (Further Additional Pure AS)
Year2018
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGroups
TypeMatrix groups
DifficultyChallenging +1.2 This is a structured group theory question requiring matrix multiplication, systematic enumeration of group elements, and application of standard definitions (order, abelian, cyclic). While it involves Further Maths content (abstract algebra), the question guides students through each step methodically, making it more accessible than typical group theory problems. The main challenge is careful computation and recognizing the dihedral group structure, but no deep insight is required.
Spec8.03c Group definition: recall and use, show structure is/isn't a group8.03e Order of elements: and order of groups8.03g Cyclic groups: meaning of the term8.03h Generators: of cyclic and non-cyclic groups
4 The group \(G\) consists of a set of six matrices under matrix multiplication. Two of the elements of \(G\) are \(\mathbf { A } = \left( \begin{array} { l l } 0 & 1 \\ 1 & 0 \end{array} \right)\) and \(\mathbf { B } = \left( \begin{array} { l l } 1 & - 1 \\ 0 & - 1 \end{array} \right)\).
  1. Determine each of the following:
    • \(\mathbf { A } ^ { 2 }\)
    • \(\mathbf { B } ^ { 2 }\)
    • Determine all the elements of \(G\).
    • State the order of each non-identity element of \(G\).
    • State, with justification, whether \(G\) is
    • abelian
    • cyclic.
Question 4:
AnswerMarks Guidance
4(i) 1 0 1 0
A2 =   and B2 =  
AnswerMarks
0 1 0 1B1
B11.1a
1.1BC
BC
[2]
AnswerMarks
(ii)1 0
Clear statement that I =   is an element
0 1
0 1 1 1
AB =   BA =   = (AB)2
1 1 1 0
1 0
ABA or BAB =  
AnswerMarks
1 1B1
B1
B1
AnswerMarks
B12.2a
1.1a
1.1
AnswerMarks
1.1Anywhere (e.g. in a Cayley table)
BC
BC
[4]
AnswerMarks
(iii)A and B have order 2
AB and BA have order 3
AnswerMarks
ABA (or BAB) has order 2B1
B1
AnswerMarks
B12.1
1.1
AnswerMarks
1.1Stated or shown
[3]
AnswerMarks
(iv)G not abelian since (e.g.) AB  BA
G not cyclic since no element of order 6 (or not
AnswerMarks
abelian  not cyclic)E1
E12.4
2.4Must have been shown
Valid correct statement supported by
AnswerMarks
their elements and orders(not just claimed as a general
result for matrix multn.)
[2]
Question 4:
4 | (i) | 1 0 1 0
A2 =   and B2 =  
0 1 0 1 | B1
B1 | 1.1a
1.1 | BC
BC
[2]
(ii) | 1 0
Clear statement that I =   is an element
0 1
0 1 1 1
AB =   BA =   = (AB)2
1 1 1 0
1 0
ABA or BAB =  
1 1 | B1
B1
B1
B1 | 2.2a
1.1a
1.1
1.1 | Anywhere (e.g. in a Cayley table)
BC
BC
[4]
(iii) | A and B have order 2
AB and BA have order 3
ABA (or BAB) has order 2 | B1
B1
B1 | 2.1
1.1
1.1 | Stated or shown
[3]
(iv) | G not abelian since (e.g.) AB  BA
G not cyclic since no element of order 6 (or not
abelian  not cyclic) | E1
E1 | 2.4
2.4 | Must have been shown
Valid correct statement supported by
their elements and orders | (not just claimed as a general
result for matrix multn.)
[2]
4 The group $G$ consists of a set of six matrices under matrix multiplication. Two of the elements of $G$ are $\mathbf { A } = \left( \begin{array} { l l } 0 & 1 \\ 1 & 0 \end{array} \right)$ and $\mathbf { B } = \left( \begin{array} { l l } 1 & - 1 \\ 0 & - 1 \end{array} \right)$.\\
(i) Determine each of the following:

\begin{itemize}
  \item $\mathbf { A } ^ { 2 }$
  \item $\mathbf { B } ^ { 2 }$\\
(ii) Determine all the elements of $G$.\\
(iii) State the order of each non-identity element of $G$.\\
(iv) State, with justification, whether $G$ is
  \item abelian
  \item cyclic.
\end{itemize}

\hfill \mbox{\textit{OCR Further Additional Pure AS 2018 Q4 [11]}}
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Q1 8 Q2 9 Q3 3 Q4 11 Q5 8 Q6 8 Q7 13