Question 4:
4 | (b) | (i) |  1 0   1 0 
A  A =  (m, n ≥ 0)
m n m 1 n 1
 1 0
=   S since two non-negative
m+n 1
integers add to give a non-negative integer so S
is closed (under ) | M1
A1
[2] | 2.4
2.2a | Considering 2 different, general,
independent elements of S
multiplied together.
Some justification must be given
4 | (b) | (ii) |  1 0 
A −n 1 = (where n > 0)
− n 1
S since –n is a negative integer so S is not a
subgroup since the inverse axiom is not met | M1
A1
[2] | 2.4
2.2a | Forming the correct inverse of
any non-identity element of S.
Can be general or specific.
Some justification must be given.
Must be from correct reasoning.
4 | (c) | (i) |  − 1 0−   − 1 0−   1 0 
 =
0 1 0 1 0 1
   1 0   − 1 0−   
, ,  cao
0 1 0 1 | M1
A1
[2] | 3.1a
2.5 |  − 1 0− 
Identifying in any
0 1
form
Must be notationally correct (ie
elements separated by comma in
curly braces) but subset alone is
sufficient. Either order.
Could be defined indirectly; eg
 − 1 0− 
{A, A2} where A = .
0 1
4 | (c) | (ii) | (Subgroup of order 2 must be isomorphic to the
(cyclic) group of order 2 so) if M is the non-
identity element, M = M–1 (or M2 = I or
equivalent).
This, together with d e t M = 1 means that
a b  d −b
M−1= =
   
c d −c a 
a=d and b=c=0.
But a2 = 1 since ad – bc = 1 => a = d = –1 is
only possibility for g as a  1 (M is not the
identity) | B1
B1
[2] | 2.4
2.4 | Or M–1 is in the subgroup
(closure) and can’t be I so must
be M. Or M2 is in the subgroup
and can’t be M so must be I.
Must see form for inverse.
Or, instead, complete argument
involving square of matrix (must
see form for square and correct
conclusions based on this
equalling identity and det
equalling 1) | a − = 1 ( s o d b c b c = − 1 ) a a d n d
 2  2  a b a + + ( b c b a2  ) d
=    c d  ( c a  + + ) d d  b c 
+ −  ( ) 1 a a d + ( ) b a d 
= 
+ ( ) c a d  d + − ( ) 1 a d
  1 0
=  e ith   e + = r ( 0 a d
0 1  
 + − = ( ) 1 a a d − = 1 1 # )
o   + r ( 0 a d b  = = 0 c = a d 1
a 2 = n d 1 a n d a d 2  = = 1 a = d − 1
s  = in c e 1 a d = 1 g iv e s th e id e n tity )
4 | (d) | The inverse property is not satisfied.
 1 0   1 0 
eg is in the set but as d e t = 0 it
1 0 1 0
does not have an inverse in the set. | B1
B1
[2] | 2.4
2.4 | Any singular real 2 by 2 matrix.
Stating only that it has no inverse
is insufficient; some justification
is required.