Question
The function \(f:\mathbb{R} \times \mathbb{R} \to \mathbb{R} \times \mathbb{R}\) is defined by \(\boldsymbol{X} \mapsto \boldsymbol{AX}\) , where \(\boldsymbol{X} = \left[ \begin{array}{l}
x\\
y
\end{array} \right]\) and \(\boldsymbol{A} = \left[ \begin{array}{l}
a\\
c
\end{array} \right.\left. \begin{array}{l}
b\\
d
\end{array} \right]\) where \(a\) , \(b\) , \(c\) , \(d\) are all non-zero.
Consider the group \(\left\{ {S,{ + _m}} \right\}\) where \(S = \left\{ {0,1,2 \ldots m – 1} \right\}\) , \(m \in \mathbb{N}\) , \(m \ge 3\) and \({ + _m}\) denotes addition modulo \(m\) .
Show that \(f\) is a bijection if \(\boldsymbol{A}\) is non-singular.
Suppose now that \(\boldsymbol{A}\) is singular.
(i) Write down the relationship between \(a\) , \(b\) , \(c\) , \(d\) .
(ii) Deduce that the second row of \(\boldsymbol{A}\) is a multiple of the first row of \(\boldsymbol{A}\) .
(iii) Hence show that \(f\) is not a bijection.
Show that \(\left\{ {S,{ + _m}} \right\}\) is cyclic for all m .
Given that \(m\) is prime,
(i) explain why all elements except the identity are generators of \(\left\{ {S,{ + _m}} \right\}\) ;
(ii) find the inverse of \(x\) , where x is any element of \(\left\{ {S,{ + _m}} \right\}\) apart from the identity;
(iii) determine the number of sets of two distinct elements where each element is the inverse of the other.
Suppose now that \(m = ab\) where \(a\) , \(b\) are unequal prime numbers. Show that \(\left\{ {S,{ + _m}} \right\}\) has two proper subgroups and identify them.
Answer/Explanation
Markscheme
recognizing that the function needs to be injective and surjective R1
Note: Award R1 if this is seen anywhere in the solution.
injective:
let \(\boldsymbol{U}, \boldsymbol{V} \in ^\circ \times ^\circ \) be 2-D column vectors such that \(\boldsymbol{AU} = \boldsymbol{AV}\) M1
\({\boldsymbol{A}^{ – 1}}\boldsymbol{AU} = {\boldsymbol{A}^{ – 1}}\boldsymbol{AV}\) M1
\(\boldsymbol{U} = \boldsymbol{V}\) A1
this shows that \(f\) is injective
surjective:
let \(W \in ^\circ \times ^\circ \) M1
then there exists \(\boldsymbol{Z} = {\boldsymbol{A}^{ – 1}}\boldsymbol{W} \in ^\circ \times ^\circ \) such that \(\boldsymbol{AZ} = \boldsymbol{W}\) M1A1
this shows that \(f\) is surjective
therefore \(f\) is a bijection AG
[7 marks]
(i) the relationship is \(ad = bc\) A1
(ii) it follows that \(\frac{c}{a} = \frac{d}{b} = \lambda \) so that \((c,d) = \lambda (a,b)\) A1
(iii) EITHER
let \(\boldsymbol{W} = \left[ \begin{array}{l}
p\\
q
\end{array} \right]\) be a 2-D vector
then \(\boldsymbol{AW} = \left[ \begin{array}{l}
a\\
\lambda a
\end{array} \right.\left. \begin{array}{l}
b\\
\lambda b
\end{array} \right]\left[ \begin{array}{l}
p\\
q
\end{array} \right]\) M1
\( = \left[ \begin{array}{l}
ap + bq\\
\lambda (ap + bq)
\end{array} \right]\) A1
the image always satisfies \(y = \lambda x\) so \(f\) is not surjective and therefore not a bijection R1
OR
consider
\(\left[ {\begin{array}{*{20}{c}}
a&b \\
{\lambda a}&{\lambda b}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
b \\
0
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{ab} \\
{\lambda ab}
\end{array}} \right]\)
\(\left[ {\begin{array}{*{20}{c}}
a&b \\
{\lambda a}&{\lambda b}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
0 \\
a
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{ab} \\
{\lambda ab}
\end{array}} \right]\)
this shows that \(f\) is not injective and therefore not a bijection R1
[5 marks]
the identity element is \(0\) R1
consider, for \(1 \le r \le m\) ,
using \(1\) as a generator M1
\(1\) combined with itself \(r\) times gives \(r\) and as \(r\) increases from \(1\) to m, the group is generated ending with \(0\) when \(r = m\) A1
it is therefore cyclic AG
[3 marks]
(i) by Lagrange the order of each element must be a factor of \(m\) and if \(m\) is prime, its only factors are \(1\) and \(m\) R1
since 0 is the only element of order \(1\), all other elements are of order \(m\) and are therefore generators R1
(ii) since \(x{ + _m}(m – x) = 0\) (M1)
the inverse of x is \((m – x)\) A1
(iii) consider
M1A1
there are \(\frac{1}{2}(m – 1)\) inverse pairs A1 N1
Note: Award M1 for an attempt to list the inverse pairs, A1 for completing it correctly and A1 for the final answer.
[7 marks]
since \(a\), \(b\) are unequal primes the only factors of \(m\) are \(a\) and \(b\)
there are therefore only subgroups of order \(a\) and \(b\) R1
they are
\(\left\{ {0,a,2a, \ldots ,(b – 1)a} \right\}\) A1
\(\left\{ {0,b,2b, \ldots ,(a – 1)b} \right\}\) A1
[3 marks]
Question
\(S\) is defined as the set of all \(2 \times 2\) non-singular matrices. \(A\) and \(B\) are two elements of the set \(S\).
(i) Show that \({({A^T})^{ – 1}} = {({A^{ – 1}})^T}\).
(ii) Show that \({(AB)^T} = {B^T}{A^T}\).
A relation \(R\) is defined on \(S\) such that \(A\) is related to \(B\) if and only if there exists an element \(X\) of \(S\) such that \(XA{X^T} = B\). Show that \(R\) is an equivalence relation.
Answer/Explanation
Markscheme
(i) \(A = \left( {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right)\)
\({A^T} = \left( {\begin{array}{*{20}{c}} a&c \\ b&d \end{array}} \right)\) M1
\({({A^T})^{ – 1}} = \frac{1}{{ad – bc}}\left( {\begin{array}{*{20}{c}} d&{ – c} \\ { – b}&a \end{array}} \right)\;\;\;\)(which exists because \(ad – bc \ne 0\)) A1
\({A^{ – 1}} = \frac{1}{{ad – bc}}\left( {\begin{array}{*{20}{c}} d&{ – b} \\ { – c}&a \end{array}} \right)\) M1
\({({A^{ – 1}})^T} = \frac{1}{{ad – bc}}\left( {\begin{array}{*{20}{c}} d&{ – c} \\ { – b}&a \end{array}} \right)\) A1
hence \({({A^T})^{ – 1}} = {({A^{ – 1}})^T}\) as required AG
(ii) \(A = \left( {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right)\;\;\;B = \left( {\begin{array}{*{20}{c}} e&f \\ g&h \end{array}} \right)\)
\(AB = \left( {\begin{array}{*{20}{c}} {ae + bg}&{af + bh} \\ {ce + dg}&{cf + dh} \end{array}} \right)\) M1
\({(AB)^T} = \left( {\begin{array}{*{20}{c}} {ae + bg}&{ce + dg} \\ {af + bh}&{cf + dh} \end{array}} \right)\) A1
\({B^T} = \left( {\begin{array}{*{20}{c}} e&g \\ f&h \end{array}} \right)\;\;\;{A^T} = \left( {\begin{array}{*{20}{c}} a&c \\ b&d \end{array}} \right)\) M1
\({B^T}{A^T} = \left( {\begin{array}{*{20}{c}} e&g \\ f&h \end{array}} \right)\left( {\begin{array}{*{20}{c}} a&c \\ b&d \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {ae + bg}&{ce + dg} \\ {af + bh}&{cf + dh} \end{array}} \right)\) A1
hence \({(AB)^T} = {B^T}{A^T}\) AG
\(R\) is reflexive since \(I \in S\) and \(IA{I^T} = A\) A1
\(XA{X^T} = B \Rightarrow A = {X^{ – 1}}B{({X^T})^{ – 1}}\) M1A1
\( \Rightarrow A = {X^{ – 1}}B{({X^{ – 1}})^T}\) from a (i) A1
which is of the correct form, hence symmetric AG
\(ARB \Rightarrow XA{X^T} = B\) and \(BRC = YB{Y^T} = C\) M1
Note: Allow use of \(X\) rather than \(Y\) in this line.
\( \Rightarrow YXA{X^T}{Y^T} = YB{Y^T} = C\) M1A1
\( \Rightarrow (YX)A{(YX)^T} = C\) from a (ii) A1
this is of the correct form, hence transitive
hence \(R\) is an equivalence relation AG