Question 4:
4 | (a) | 1−λ 7 8
det(P−λI)= −6 12−λ 12
−2 4 8−λ
=(1−λ)((12−λ)(8−λ)−12×4)
−7(−6(8−λ)−12×−2))+8(−6×4−(12−λ)×−2)
=(1−λ)(96−20λ+λ2 −48)
−7(−48+6λ+24)+8(−24+24−2λ)
=(1−λ)(λ2 −20λ+48)−7(6λ−24)+8(−2λ)
=λ2 −20λ+48−λ3+20λ2 −48λ−42λ+168−16λ
=−λ3+21λ2 −126λ+216=0 www | M1
M1
A1
[3] | 1.1
1.1
1.1 | Formation of appropriate determinant.
May be implied by 2nd M1.
Attempt to expand determinant.
Allow one arithmetic or algebraic
error.
AG. Must expand all brackets
(except those with just ± or ±1
outside) before reaching AG. | Must have “= 0”.
4 | (b) | (i) |  1 7 8 1  3
    
−6 12 12 −2 = −6
    
 −2 4 8 2  6
 1
 
=3 −2
(so it is an e-vector) and the e-value is 3
 
 2 | M1
A1 | 1.1
1.1 | Finding Pe. Must specify “Pe =”.
and equating to λe and comparing.
Accept “λ = 3” as declaration. | Verification must not start by
assuming λ = 3.
Do not accept embedded answer.
Alternative method:
1−λ 7 8 1  3−λ
    
−6 12−λ 12 −2 = −6+2λ soi
    
 −2 4 8−λ  2  6−2λ  | M1 | 1.1 | Finding (P – λI)e. in terms of λ.
Could be implied by three equations:
3 – λ = 0, –6 + 2λ = 0, 6 – 2λ = 0.
 1 7 8 1  1
    
or −6 12 12 −2 =λ −2
    
 −2 4 8 2  2
giving three correct equations in λ
from x, y and z components.
0
 
= 0 if λ = 3 (so it is an e-vector) so e-value is 3
 
0 | A1 | 1.1 | or declaring λ = 3 as solution to all
three equations. . | Verification must not start by
assuming λ = 3. | Verification must not start by
assuming λ = 3.
Alternative method:
1−λ 7 8 1  3−λ
    
−6 12−λ 12 −2 = −6+2λ soi
    
 −2 4 8−λ  2  6−2λ 
M1
1.1
Finding (P – λI)e. in terms of λ.
Could be implied by three equations:
3 – λ = 0, –6 + 2λ = 0, 6 – 2λ = 0.
A1
1.1
or declaring λ = 3 as solution to all
three equations. .
4 | (b) | (ii) |  1 7 8 x x  6x 
      
−6 12 12 y =6 y or 6y
      
 −2 4 8 5 5 6×5
Any two of:
x + 7y + 40 = 6x
–6x + 12y + 60 = 6y
–2x + 4y + 40 = 30
correct
⇒ x = 15 & y = 5 (so the eigenvector with an
eigenvalue of 6 with z = 5 is 15i + 5j + 5k). | M1
M1
A1
[3] | 3.1a
1.1
3.2a | Using Pv = 6v (or (P – 6I)v = 0)) and
z = 5. Can be implied by two correct
equations in x and y below. Can be
awarded if z replaced by 5 later, even
after errors.
Multiplying the vector into the matrix
and equating 2 of the components to
find 2 correct simultaneous equations
in x and y soi by answers. Need not
be simplified.
Allow embedded. Not 3i + j + k but
ISW once x = 15 and y = 5 seen. | −5 7 8x 0
    
or −6 6 12 y = 0
    
 −2 4 25 0
–5x + 7y = –40 (–5x + 7y + 40 = 0)
–x + y = –10 (–6x + 6y + 60 = 0)
–x + 2y = –5 (–2x + 4y + 10 = 0)
[2]
4 | (c) | (ii) | 12 11 1  21 14 7 
   
 3 4 −7  −  7 14 −14 
   
1  6 6 3   7 7 14 
∴E −1 =  
9 15 0 0  
   
+ 0 15 0
   
   
  0 0 15 
 6 −3 −6  2 −1 −2
1   3 3 3
= 9    − − 4 1 − 5 1 7 4    or    − − 9 1 4 − 9 5 1 7 9 4   
9 9 9 | B1
[1] | 1.1 | Must see this expression oe. eg could
3 2 1  2 12 11 1 
   
see 1 2 −2 for 3 4 −7 .
   
1 1 2    6 6 3  
Condone uncancelled fractions.
[4]