Marking Scheme of F.6 Pure Mathematics Quiz
(Matrices, system of linear equations, inequalities)
1.
(a) (I + A)(I  A + A 2) = I + A3 = I
(∵ A3 = 0)
1A
(b) I + PAP 1 = 2P  PIP1 + PAP 1 = 2P  P(I + A)P1 = 2P
 det[P(I + A)P1] = det(2P)
 detP  det(I + A)  det(P1) = 23detP
 det(I + A) = 8 (∵ detP = 1)
1M
1M
1A
By (a),
(I + A)(I  A + A 2) = I  det(I + A)(I  A + A 2) = detI  det(I + A)det(I  A + A 2) = 1
 8det(I  A + A 2) = 1
1
8
 det(I  A + A2) =
1M
1A
 n  1
Apply AM > GM on 1, 2, …, n, we get n! < 
 .
 2 
n
2.
Apply AM > GM on
1
1
1
,
, …,
, we get
1 2 2  3
n  (n  1)
1
1
1

 ... 
1 2 2  3
n  (n  1)
>
n
n
1
1  2  ...  n 2  (n  1)
1 
1 1   1 1 
1
        ...   

1 2   2 3 
 n n 1
>
n
1
>
n 1
n
1
,
(n!) (n  1)
2
2
2
n
1
(n!) (n  1)
2
hence n! > n  1
n 1
2
.
1
1
1
3.
(a) (i)
2
1
6
1 
 0  (2   )(1   )  6  0  (   4)(   1)  0
   4 or  1
1A
1A
 2 1  x1 
 x1 
 3 1  x1   0
(ii) 
    1   
    
 6 1  y1
 y1
 6 2  y1  0
1M
 x1   1 
One possible solution is     
 y1   3
1A
 2 1  x2 
 x2 
 2 1   x2   0
Similarly, 
    4   
    
 6 1  y2 
 y2 
 6 3  y2   0
 x2   1
One possible solution is     
 y2   2
1A
(b) Choose x = 1 and y = 0, we have     0
Choose x = 0 and y = 1, we have     0
(c)
 x
 x1 
 x2 
 x   x1  x2 
 x   x1
            
   
 y
 y1
 y2 
 y  y1  y2 
 y  y1
As
(d)
2M
x1
x2
y1
y2
x2    
   ... (**)
y2    
2M
 0 , (**) has unique solution , .
2M
f (  )  det( A  I )  0 has roots 3 and 2.
Since f (  ) is quadratic equation in  with leading coefficient 1,
f ( )  (  3)(  2)  2  5  6 .
Thus, f ( A)  A2  5 A  6 I.
2M
 x
 x1 
 x2 
For any x, y R , by (c), put           .
 y
 y1 
 y2 
1M
 x
 x1 
 x2 
f ( A)    ( A2  5 A  6)    ( A2  5 A  6) 
 y
 y1 
 y2 
 x1 
 x2 
  ( A  2 I )( A  3I )   ( A  3I )( A  2 I ) 
 y1 
 y2 
 0
 
 0
 0 0
By (b), f ( A)  

 0 0
1M
4.
(a) (i)
2
1

5
2
 1 p  2

q
pq
p2  2 p  2

1 2


p  2  ~ 0 1
0 p
p 2  p  1

1
q
pq
p2  2 p  2  q


p

2
p  p  2 
1
1M
q
1 2

~ 0 1
pq
 0 0  pq  2 p  q  2

1 

p  --- (*)
p  2 
1
Hence result follows.
(ii) (pq + 2p  q + 2)z = p + 2
(p + 1)(2  q)z = p + 2
1
(1) For unique solution, p  1 and q  2.
1
(2) For distinct solutions (i.e. infinitely many solutions), p = 2 and q = 2
2A
(b) If
(c) (i)
p = 2 , q = 2 , then by (a)(ii)(2), (E)
has infinitely many solutions
Using (*), y + (2 + 2)z = 2 and x + 2(2)  2z = 1
Hence the solution set is {(5 + 2z , 2 , z) : z  R}
Put (x , y , z) = (5 + 2z , 2 , z) into x2  10y2  9z2 = 0 .
(5 + 2z)2  10(2)2  9z2 = 0  5z2  20z + 15 = 0
z = 1 or 3
1M
1A
1M
Hence (x , y , z) = (7 , 2 , 1) or (11 , 2 , 3)
Yes. For q = 1 and p  1, by (a)(ii)(1), (E) has a unique solution.
 [ corresponding to (E)]  0
Now [ corresponding to (F)] = [ corresponding to (E)]  0
Hence (F) has a unique solution.
1A
1M
1
(ii) Suppose (x0 , y0 , z0) is the unique solution to (E) , then (x0 , z0 , y0) is a
Solution to (F) , given that (E) and (F) have common solution, by
(c)(i)
1M
(x0 , y0 , z0) = (x0 , z0 , y0)  y0 = z0 --- (#)
From (*),
p2
p2
p2
z0 =
=
=
( p  1)( 2  q)
( p  1)( 2  1)
p 1
y0 = p  (p + q)z0 = p  (p + 1)z0 = p  (p + 2) = 2
4
p2
By (#),
= 2 => p = 
3
p 1
END
1M+1A