Question 4:
4 | (a) | (i) | 2n + 1 + 2m + 1 = 2(n + m + 1) so not closed
[0 ∉ G so] no identity.
Since no identity the inverse property cannot be
satisfied. | B1
B1
B1
[3] | 2.1
2.2a
2.2a | Must show some working
Could be seen with next B1
Cannot gain this B1 without
previous B1
4 | (a) | (ii) | (a+b 2)(c+d 2)=ac+2bd+(bc+ad) 2 so
closed
a = 1, b = 0 ⇒ 1 [∈ G] so identity exists
1 1
eg =1− 2∉G so inverse property not
2+ 2 2
satisfied | B1
B1
B1
[3] | 2.1
2.2a
2.2a | Must show some working
1 must be seen
Single numerical counter example
is sufficient or
−1
0 ∉ 𝐺𝐺 | Elements must be general
and distinct
For need to
𝑎𝑎 𝑏𝑏√2
justify2 ans2wer 2eg 2 is
𝑎𝑎 −2𝑏𝑏 +𝑎𝑎 −2𝑏𝑏
not always in G 𝑎𝑎
2 2
𝑎𝑎 −2𝑏𝑏
4 | (a) | (iii) | a, b∈R⇒ab∈R so closed
so identity exists
1 ∈ ℝ 𝑎𝑎𝑛𝑛 𝑎𝑎so 𝑎𝑎 i×nv1er=se p1r×op𝑎𝑎er=ty 𝑎𝑎no∈t sℝatisfied
−1 | B1
B1
B1
[3] | 2.1
2.2a
2.2a | Need justification
1 must be seen
4 | (b) | (i) | 0−1 ∉ ℝ
 0 
 
 0 −1
1 −1 −i
 
2−i −1
 0 −i
 
−i 0  | B1
B1
B1
[3] | 3.1a
1.1
1.1
4 | (b) | (ii) | They are not isomorphic because M contains only one
element of order 2 while N is known to contain at least
3. | B1 | 2.4 | Or other valid reason (eg M is
cyclic while N is not since it
requires more than 1 element to
generate it)
[1]
 1 −2 −2a  a−2b−2c 
    
1 −2 1 −2 b = 1 −2a+b−2c
3   3 
 −2 −2 1c  −2a−2b+c
e.f = 0 ⇒ a + b + c = 0
−b−c−2b−2c −b−c
   
∴A.f = 1 2(b+c)+b−2c = b
3   
2(b+c)−2b+c  c 
a
 
∴A.f = b =f so f is an e-vec of A
 
c | M1
M1
M1
A1
[4] | 3.1a
3.1a
1.1
3.2a | Finding vector A.f
Using perpendicularity condition
to find a relationship between a, b
and c
Eliminating a, b or c consistently
in all 3 components to derive A.f
in two unknowns or eliminating b
or c in x, a or c in y and a or b in
z.
Completing substitution and
correct conclusion | soi
Alternative method:
 1   0 
   
f =λ −1 +µ 1
   
 0   −1 | M1 | Expressing a general f in terms of
two non-parallel vectors which
are both perpendicular to e | Showing that a specific
perpendicular vector is an
e-vec SC2 or M1SC1
3.1a
 1 −2 −2  1   0 
1     
∴A.f = −2 1 −2 λ −1 +µ 1 =
3       
 −2 −2 1       0    −1    
 1 −2 −2 1   1 −2 −2 0 
1λ  −2 1 −2  −1  +1µ  −2 1 −2  1 
3         3        
−2 −2 1 0  −2 −2 1−1 | M1 | 1.1 | 1.1 | Opening brackets
 3   0 
= 1λ  −3  +1µ  3 
3     3    
 0  −3 | M1 | 3.1a | Multiplying vectors into matrix
Alternative method:
 1   0 
   
f =λ −1 +µ 1
   
 0   −1
M1
Expressing a general f in terms of
two non-parallel vectors which
are both perpendicular to e
Showing that a specific
perpendicular vector is an
e-vec SC2 or M1SC1
 1   0 
   
∴A.f =λ −1 +µ 1 =f so f is an e-vec of A
   
   
 0  −1 | A1 | 3.2a | Completing and correct
conclusion
Alternative Method 2:
To find e-vals put
1−λ −2 −2
3 3 3
 −2 1−λ −2  =0⇒λ3−λ2 −λ+1=0
3 3 3
 
−2 −2 1−λ
 
3 3 3
(λ = –1 gives e) so consider λ = 1:
 1 −2 −2
a a a
 3 3 3
If f =  b  then we need −2 1 −2  b  =  b 
  3 3 3    
c  −2 −2 1  c c
 3 3 3 
 1 −2 −2a a   1 3 −1 − 3 2 − 3 2  a
or 1 −2 1 −2  b  =  b  or  −2 1−1 −2  b  =0
3     3 3 3  
−2 −2 1c c  −2 − 2 1 −1  c
 3 3 3 
⟹But𝑎𝑎 +𝑏𝑏+𝑐𝑐=0 so e.f =0
𝑎𝑎 1
f𝑓𝑓 m=us�t 𝑏𝑏be� p𝑎𝑎e𝑛𝑛r𝑎𝑎pe 𝑒𝑒nd=ic u�l1ar� to e 𝑎𝑎 +𝑏𝑏+𝑐𝑐= 0⟹
𝑐𝑐 1
⟹ | [4]
M1
M1
M1
A1 | 3.1a
1.1
3.1a
3.2a
[4] | For attempt at ch eqn eg
seen
𝑎𝑎𝑒𝑒𝑡𝑡|𝑨𝑨−𝜆𝜆𝑰𝑰|