Question 5:
5 | (a) | a 0  −
( a ) ( 3 ) 0   = − − =
2 3  −
e-vals are 3 and a and no others.
a 0x  ax  3x
=3:   = = 
2 3y 2x+3y 3y
0
=3: e =  www
1 1
a 0x  ax  ax
=a:   = = 
2 3y 2x+3y ay
a 3  − 
a :  = e = www
2 2
2 1 ( a 3 ) 2 2 2 c o s 14  e .e = =  − +
1 2
(a2 – 6a + 13 = 8 so)
a = 1 or a = 5 www | M1
A1
M1
A1
M1
A1
M1
A1 | 3.1a
1.1
1.1
1.1
1.1
1.1
2.2a
1.1 | Forming characteristic equation.
Consideration of Ae = e
(or (A – λI)e = 0) for one correct
e-value.
(ie ax = 3x or 2x + 3y = 3y)
Or any non-zero (possibly
algebraic) multiple.
Consideration of Ae = e
(or (A – λI)e = 0) for other
correct e-value.
(ie (ax = ax and) 2x + 3y = ay)
Or any non-zero (possibly
algebraic) multiple.
Forming numerical equation in a
by using the dot product with
reasonable attempt at e . M0 if
2
e .e =0. Ignore minor
1 2
inconsistency in angle between
their e-vecs (ie acute/obtuse).
For both and no others. | o r
 a − 3 0   x   ( a − 3 ) x   0 
= =
2 0 y 2 x 0
 0 
If eg then condone omission
s
of condition on s (ie s 0).
0 0 x
or   
2 3−ay
 0  0
=  =  
2x+(3−a)y 0
See above note on multiplier.
 a − 3   1 
21 o r 2
a − 3
Could argue via gradients;
direction of e is along y-axis so
1
‘gradient’ of e must be 1;
2
2
=  1 (= 1 insufficient for
a − 3
M1).
Not from wrong e-vecs,
1  2  3−a
eg   or   or  
0 a−3  2 
Alternative method for final M1A1M1A1:
p  
q 1 p 2 q 2 c o s 14  e =  e .e = =  +
2 q 1 2 | M1 | 1.1
Forming dot product of their e-
vecs with general form for e to
2
derive numerical equation in two
unknowns, the components of e .
2
 1 
=> p2 = q2 => p = q => e = oe
2  1 | A1 | 1.1 | Must be two possible e-vecs. | Must be two possible e-vecs.
 a 0   1   a   1 
 = = a
2 3  1 2  3  1 |  a 0   1   a   1 
 = = a
2 3  1 2  3  1 | M1 | 2.2a | Using e-vec property with e and
2
a as the e-val. |  0 0−   1 
o r 
2  3 a 1
 0   0 
= =
2  ( 3 − a ) 0
a = 1 or a = 5 www | A1 | 1.1 | For both and no others. |  1 
Not from e =
1 0
[8]
5 | (b) | (i) | 2 ( a 3 ) 3 a 0   − + + =
 P 2 − ( a + 3 ) P + 3 a I = O
 P 2 − ( a + 3 ) P = − 3 a I
(So need − 3 a = 1  ) a = − 13
So r = ( − ( a + 3 ) = ) − 83 oe | M1
M1
A1
A1 | 3.1a
3.1a
2.2a
2.2a | Correctly expanding
characteristic equation.
Using C-H theorem and
expressing either equation in
form with which direct
comparison can be made (either
can be implied by correct
equations in a and r but not only
by correct answers).
C-H must be used.
r=−22. C-H must be used
3 | Must be an equation but can be
recovered.
P 2 + r P − I = O
If M1M0 and C-H apparently
used in derivation of a and r then
SCB1 if a and r correctly found.
Alternative method for final M1A1M1A1:
p  
q 1 p 2 q 2 c o s 14  e =  e .e = =  +
2 q 1 2
M1
1.1
Using e-vec property with e and
2
a as the e-val.
Alternative method: | B1 | Use of C-H. Can be implied by
substitution of both e-vals into
scalar equation | Must be 1 not I. Can be
recovered
P 2 + r P = I oe (must be the characteristic
equation) => 2 + r = 1 oe
8
3 2 + r  3 = 1  3 r = − 8  r = −
3 | M1 | Substituting numerical e-val into
char equation to derive value of r.
8
a 2 + r  a = 1  a 2 − a = 1
3 | 8
a 2 + r  a = 1  a 2 − a = 1
3 | M1 | Substituting algebraic e-val into
char equation with their value of r
1
3a 2−8a−3=0a=− or a=3
3 | A1 | (3a + 1)(a – 3)
a = 3 must be explicitly rejected. | (3a + 1)(a – 3)
1
But a  3 (given) so a=−
3 | a = 3 must be explicitly rejected.
[4]
5 | (b) | (ii) | P 4 = ( I − r P ) 2 = I − 2 r P + r 2 P 2
or P 4 = ( I + 83 P ) 2 = I + 1 63 P + 6 49 P 2
= I − 2 r P + r 2 ( I − r P ) = (1 + r 2 ) I − ( 2 r + r 3 ) P
o r = I + 1 63 P + 6 49( ( I + 8 P )
3
= ( 1 + 6 49 ) I + 1 63 + 5 1 2 ) P
2 7
P 4 = 7 39 I + 6 5 6 P sos= 73, t = 656 cao
2 7 9 27 | M1
M1
A1
[3] | 3.1a
2.2a
1.1 | Squaring and expanding
expression for P2 with r algebraic
or their value. Need I2 = I and
IP = P soi.
Substituting for P2 and collecting
I and P terms and no others.
NB Other approaches are possible
(eg squaring I = P2 + rP or
multiplying throughout by P2).
1st M1 for a useful operation and
substitution, 2nd M1 for further
algebra leading to correct form.
Allow embedded answer | Or multiplying by P to find P3
and eliminating P2.
P 3 = P − r P 2 = P − r ( I − r P ) or
P 3 = P + 83 P 2 = P + 83 ( I + 83 P )
Or multiplying by P again and
eliminating P2 again.
P4 =(1+r2)P2−rP
=(1+r2)(I−rP)−rP=etc
If M1M0 because correct form
not reached then SCB1 for
correct values of s and t properly
obtained
B1
Use of C-H. Can be implied by
substitution of both e-vals into
scalar equation
Must be 1 not I. Can be
recovered
PMT
Need to get in touch?
If you ever have any questions about OCR qualifications or services (including administration, logistics and teaching) please feel free to get in
touch with our customer support centre.
Call us on
01223 553998
Alternatively, you can email us on
support@ocr.org.uk
For more information visit
ocr.org.uk/qualifications/resource-finder
ocr.org.uk
Twitter/ocrexams
/ocrexams
/company/ocr
/ocrexams
OCR is part of Cambridge University Press & Assessment, a department of the University of Cambridge.
For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored. © OCR
2023 Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee. Registered in England. Registered office
The Triangle Building, Shaftesbury Road, Cambridge, CB2 8EA.
Registered company number 3484466. OCR is an exempt charity.
OCR operates academic and vocational qualifications regulated by Ofqual, Qualifications Wales and CCEA as listed in their
qualifications registers including A Levels, GCSEs, Cambridge Technicals and Cambridge Nationals.
OCR provides resources to help you deliver our qualifications. These resources do not represent any particular teaching method
we expect you to use. We update our resources regularly and aim to make sure content is accurate but please check the OCR
website so that you have the most up-to-date version. OCR cannot be held responsible for any errors or omissions in these
resources.
Though we make every effort to check our resources, there may be contradictions between published support and the
specification, so it is important that you always use information in the latest specification. We indicate any specification changes
within the document itself, change the version number and provide a summary of the changes. If you do notice a discrepancy
between the specification and a resource, please contact us.
Whether you already offer OCR qualifications, are new to OCR or are thinking about switching, you can request more
information using our Expression of Interest form.
Please get in touch if you want to discuss the accessibility of resources we offer to support you in delivering our qualifications.