Question 2:
2 | (a) | 10−λ 12 −8
det(A−λI)= −1 2−λ 4
3 6 2−λ
=(10−λ) ( (2−λ)2−4×6)
−12 (−(2−λ)−4×3 )−8 (−1×6−3(2−λ) )
=(10−λ)(λ2−4λ+4−24)−12(−2+λ−12)
−8(−6−6+3λ)
=(10−λ)(λ2−4λ−20)−12(λ−14)
−8(−12+3λ)
=10λ2−40λ−200−λ3+4λ2+20λ−12λ+168
+96−24λ
=−λ3+14λ2−56λ+64
So char eqn is –λ3 + 14λ2 – 56λ + 64 = 0 | M1
M1
A1 | 1.1a
1.1
1.1 | DR
Formation of appropriate
determinant
Attempt to expand determinant.
Condone one minor slip
AG Some intermediate working
must be shown.
Must be exactly the equation
given. | May be implied
2 | (a) | Alternative method: | B1 | AG. Calculation of determinant
and trace. Clear calculation must
be seen; just tr(A) = 14 is B0.
tr(A)=10+2+2=14
det(A)==10(4−24)−12 (−2−12 )−8 (−6−6 )
=−200+168+96=64
M1 | Attempt to find either tr(A2) or
the trace of the minor matrix.
Condone one minor slip.
64 96 −48
tr ( A 2 ) =tr  0 16 24  =64+16+4=84
 
30 60 4
2 4 10 −8 10 12
+ +
or 6 2 3 2 −1 2
=(4−24)+(20+24)+(20+12)=56
So characteristic equation is | A1 | AG. From correct working only.
Must be exactly the equation
given.
−λ3+14λ2− 1( 142−84 ) λ+64=0
2
∴−λ3+14λ2−56λ+64=0
[3]
B1
AG. Calculation of determinant
and trace. Clear calculation must
be seen; just tr(A) = 14 is B0.
M1
Attempt to find either tr(A2) or
the trace of the minor matrix.
Condone one minor slip.
A1
AG. From correct working only.
Must be exactly the equation
given.
2 | (b) | C-H ⇒ –A3 + 14A2 – 56A + 64I = 0
⇒–A2 + 14A – 56I + 64A–1 = 0
⇒A −1= 1 (A2−14A+56I)
64
 2 
10 12 −8 10 12 −8
 
   
 −1 2 4 −14 −1 2 4 
   
−1= 1   3 6 2  3 6 2 
A  
64 1 0 0 
+56   
0 1 0
   
 
0 0 1
 
−20 −72 64
A −1= 1  14 44 −32 
64 
 −12 −24 32 | B1
M1
A1FT
M1
A1
[5] | 1.1
1.1
1.1
1.1
1.1 | Correct statement of C-H theorem
using char eqn. Condone 0 or O
for 0 but must be an equation
Multiplying throughout by A–1 to
leave A2, A and A–1 terms. OA–1
must be 0 (or 0 or O).
Correct rearrangement of quoted
characteristic eqn to find a matrix
expression for A–1 (or kA–1)
Clear evidence of substitution of
A into their matrix equation for
kA–1.
140 168 −112
 
NB 14A= −14 28 56
 
 42 84 28
Must be derived from correct
working.
−10 −36 32
1  
or 7 22 −16 or
32 
 
 −6 −12 16
−0.3125 −1.125 1
 
0.21875 0.6875 −0.5
 
 
−0.1875 −0.375 0.5 | 64 96 −48
A 2 =  0 16 24 
 
30 60 4
−84 −168 112
 
−14A+56I= 14 28 −56
 
 −42 −84 28
Correct answer from no working or
other method: 0/5
 5 9 
− − 1
 16 8 
 7 11 1
or −
 
32 16 2
 
3 3 1
− − 
 16 8 2
2 | (c) | –23 + 14×22 – 56×2 + 64
= –8 + 56 – 112 + 64 = 0 ⇒ (λ – 2) is a factor
–λ3 + 14λ2 – 56λ + 64
= λ2(2 – λ) – 12λ(2 – λ) + 32(2 – λ)
= (2 – λ)(λ2 – 12λ+ 32) = (2 – λ)(4 – λ)(8 – λ)
So eigenvalues are 2, 4 and 8
2 0 0
 
D= 0 4 0
 
0 0 8 | B1
M1
A1
A1FT
[4] | 3.1a
1.1
3.1a
1.1 | For one linear factor soi
Attempt to factorise (could also
be by eg symbolic division or
comparing coefficients)
soi
FT their solutions in correct order | (λ – 2), (λ – 4) or (λ – 8)
If M0 then SC2 for correct answer.
If M0 then SC1 for incorrect order
of correct diagonal elements.