General Certificate of Education January 2007 Advanced Level Examination

MATHEMATICS Unit Further Pure 4 Wednesday 31 January 2007

MFP4

9.00 am to 10.30 am

For this paper you must have: * an 8-page answer book * the blue AQA booklet of formulae and statistical tables. You may use a graphics calculator.

Time allowed: 1 hour 30 minutes Instructions Use blue or black ink or ball-point pen. Pencil should only be used for drawing. Write the information required on the front of your answer book. The Examining Body for this paper is AQA. The Paper Reference is MFP4. Answer all questions. Show all necessary working; otherwise marks for method may be lost. * *

* *

Information The maximum mark for this paper is 75. The marks for questions are shown in brackets. * *

Advice Unless stated otherwise, you may quote formulae, without proof, from the booklet. *

P89990/Jan07/MFP4 6/6/

MFP4

2

Answer all questions.

1 Show that the system of equations x þ 2y
z ¼ 0 3x
y þ 4z ¼ 7 8x þ y þ 7z ¼ 30 is inconsistent.

2

(4 marks)

(a) Show that ða
bÞ is a factor of the determinant


a b c


D ¼

b þ c c þ a a þ b


bc ca ab


(2 marks)

(b) Factorise D completely into linear factors.

(5 marks)

3 The points P, Q and R have position vectors p, q and r respectively relative to an origin O, where 2 3 2 3 2 3 1
3 9 4 5 4 5 4 p¼ 1 ,q¼ 4 and r ¼ 2 5 4 20 4 (a)

(i) Determine p  q .

(2 marks)

(ii) Find the area of triangle OPQ .

(3 marks)

(b) Use the scalar triple product to show that p, q and r are linearly dependent, and interpret this result geometrically. (3 marks)

P89990/Jan07/MFP4

3

2

1 4 4 The matrices MA ¼ 0 0 A and B respectively.

3 2 3 0 0 1 0 0 0
1 5 and MB ¼ 4 0
1 0 5 represent the transformations 1 0 0 0 1

(a) Give a full geometrical description of each of A and B.

(5 marks)

(b) Transformation C is obtained by carrying out A followed by B. (i) Find MC , the matrix of C.

(2 marks)

(ii) Hence give a full geometrical description of the single transformation C. (2 marks)

5

(a) Find, to the nearest 0.1°, the acute angle between the planes with equations r . ð3i
4j þ kÞ ¼ 2 and r . ð2i þ 12j
kÞ ¼ 38 (b) Write down cartesian equations for these two planes. (c)

x
a y
b z
c ¼ ¼ , cartesian equations for the line of l m n intersection of the two planes. (5 marks)

(a) Find the eigenvalues and corresponding eigenvectors of the matrix   1 2 X¼ 5 4 (b)

(2 marks)

(i) Find, in the form

(ii) Determine the direction cosines of this line.

6

(4 marks)

(2 marks)

(6 marks)

(i) Write down a diagonal matrix D, and a suitable matrix U, such that X ¼ UDU
1 (ii) Write down also the matrix U
1 .

(2 marks) (1 mark)

(iii) Use your results from parts (b)(i) and (b)(ii) to determine the matrix X5 in the   a b form , where a, b, c and d are integers. (3 marks) c d

Turn over for the next question

P89990/Jan07/MFP4

s

Turn over

4




1 2 7 The transformation S is a shear with matrix M ¼ . Points ðx, yÞ are mapped
2 3 under S to image points ðx 0 , y 0 Þ such that 

x0 y0



  x ¼M y

(a) Find the equation of the line of invariant points of S.

(2 marks)

(b) Show that all lines of the form y ¼ x þ c , where c is a constant, are invariant lines of S. (3 marks) (c) Evaluate det M, and state the property of shears which is indicated by this result. (2 marks) (d) Calculate, to the nearest degree, the acute angle between the line y ¼
x and its image under S. (3 marks)

2

3 4
1 2 8 The matrix P ¼ 4 1 1 3 5 , where a is constant.
2 0 a (a)

(i) Determine det P as a linear expression in a . (ii) Evaluate det P in the case when a ¼ 3 . (iii) Find the value of a for which P is singular.

(2 marks) (1 mark) (2 marks)

(b) The 3  3 matrix Q is such that PQ ¼ 25I . Without finding Q: (i) write down an expression for P
1 in terms of Q;

(1 mark)

(ii) find the value of the constant k such that ðPQÞ
1 ¼ kI ;

(2 marks)

(iii) determine the numerical value of det Q in the case when a ¼ 3 .

(4 marks)

END OF QUESTIONS

Copyright Ó 2007 AQA and its licensors. All rights reserved.

P89990/Jan07/MFP4

MFP4 - AQA GCE Mark Scheme 2007 January series

Key to mark scheme and abbreviations used in marking M m or dM A B E or ft or F CAO CSO AWFW AWRT ACF AG SC OE A2,1 –x EE NMS PI SCA

mark is for method mark is dependent on one or more M marks and is for method mark is dependent on M or m marks and is for accuracy mark is independent of M or m marks and is for method and accuracy mark is for explanation follow through from previous incorrect result correct answer only correct solution only anything which falls within anything which rounds to any correct form answer given special case or equivalent 2 or 1 (or 0) accuracy marks deduct x marks for each error no method shown possibly implied substantially correct approach

MC MR RA FW ISW FIW BOD WR FB NOS G c sf dp

mis-copy mis-read required accuracy further work ignore subsequent work from incorrect work given benefit of doubt work replaced by candidate formulae book not on scheme graph candidate significant figure(s) decimal place(s)

No Method Shown Where the question specifically requires a particular method to be used, we must usually see evidence of use of this method for any marks to be awarded. However, there are situations in some units where part marks would be appropriate, particularly when similar techniques are involved. Your Principal Examiner will alert you to these and details will be provided on the mark scheme. Where the answer can be reasonably obtained without showing working and it is very unlikely that the correct answer can be obtained by using an incorrect method, we must award full marks. However, the obvious penalty to candidates showing no working is that incorrect answers, however close, earn no marks. Where a question asks the candidate to state or write down a result, no method need be shown for full marks. Where the permitted calculator has functions which reasonably allow the solution of the question directly, the correct answer without working earns full marks, unless it is given to less than the degree of accuracy accepted in the mark scheme, when it gains no marks. Otherwise we require evidence of a correct method for any marks to be awarded.

3

MFP4 - AQA GCE Mark Scheme 2007 January series

MFP4 Q 1

Solution 0 1 2

1 2 −1 −1 0 3 −1 4 7 → 0 −1 7 7 8 1 7 30 0 −15 15 30 1 2 −1 0 → 0 −1 1 1 0 −1 1 2

Marks M1

Comments R2′ = R2 – 3R1 R3′ = R3 – 8R1 Penalise numerical errors once only, at this stage

A1 A1

Or ∆ = – 7 – 3 + 64 – 8 – 4 – 42 = 0

and ∆x or ∆y or ∆z = 0 shown also Explaining this ⇒ inconsistency Or Solving (1) & (2), say, to get x = λ, y = 1 – λ, z = 2 – λ

Inconsistency noted/explained ft provided working is clear

B1F

4

(M1) (A1) (A1) (B1)

(4)

So showing ∆ = 0 and thinking this is it scores M1A1A0B0

(4) 4

Checking to show inconsistency

(M1) (A1) (A1) (B1)

Substg. in (3) ⇒ 15 = 30 Total 2(a)

Total

a −b b c ∆ = b−a c+a a+b c(b − a) ca ab

M1

1 b c = (a – b) − 1 c + a a + b −c ca ab

A1

C1′ = C1 – C2

2

Or Setting b = a ⇒ C1 = C2 ⇒ ∆ = 0 ⇒ (a – b) a factor of ∆

(M1) (A1)

(2)

Or ∆=(a – b)(c3 +a2b+ab2 –abc –b2 c–a2 c )

(M1) (A1)

(2)

4

Factor theorem

Must be completely correct

MFP4 - AQA GCE Mark Scheme 2007 January series

MFP4 (cont) Q 2(b)

Solution 1 b−c c = (a – b) −1 c − b a + b −c a(c − b) ab

Marks

Total

Comments

M1

C2′ = C2 – C3

A1

2nd linear factor extracted

0 0 a+b+c e.g. ∆= (a – b)(b – c) −1 −1 a+b −c −a ab

M1

Genuine attempt at both remaining linear factors: e.g. R1′ = R1 + R2

and then expanding final det. ∆ = – (a + b + c)(a – b)(b – c)(c – a)

A1 A1

3rd factor All correct

1 1 c = (a – b)(b – c) − 1 −1 a + b −c −a ab

Or By cyclic symmetry, (b – c) and (c – a) are also factors

Final linear factor & checking sign of a coefficient. Or Expanding the determinant fully ∆= Multiplying out (a – b)(b – c)(c – a)(a + b + c)

(M1) (A1) (A1) (M1) (A1) (M1) (A1)

(5)

(M1)

(A1) = Fully correct working to show the two things are identically equal & checking for (A1) sign Total i j k ⎡ 4 ⎤ M1 A1 3(a)(i) p × q = 1 1 4 = ⎢ −32 ⎥ ⎢ ⎥ −3 4 20 ⎣⎢ 7 ⎦⎥ (ii) A = 1 | p × q | 2 = 1 42 + 322 + 7 2 2 = 33 2

5

No fudging, or jumping straight to the answer allowed (5) 7

2

M1 For attempt at | p × q |

B1 A1F

1 1 4 ⎡ 4 ⎤ ⎡9 ⎤ (b) p × q . r = ⎢ −32 ⎥ . ⎢ 2 ⎥ or −3 4 20 ⎢ ⎥ ⎢ ⎥ 9 2 4 ⎣⎢ 7 ⎦⎥ ⎣⎢ 4 ⎦⎥ = 36 – 64 + 28 = 0

3

M1 Give when “= 0” reached

A1

(⇒ Lin Dep) O, P, Q, R Or p, q, r co-planar

ft

B1 Total

5

3 8

MFP4 - AQA GCE Mark Scheme 2007 January series

MFP4 (cont) Q Solution 4(a) A is a Rotation thro’ 90o about Ox B is a Reflection in y = 0 (i.e. x–z plane) (b)(i)

⎡1 0 0 ⎤ MC = MB MA = ⎢⎢ 0 0 1 ⎥⎥ ⎣⎢ 0 1 0 ⎥⎦

(ii) C is a Reflection in y = z

Marks M1 A1 A1 M1 A1

Total

M1 A1

2

M1 A1

2

Comments

5

Give M1 for any series of reflections

N.B. In (i): ⎡1 0 0 ⎤ MA MB = ⎢⎢ 0 0 −1⎥⎥ scores M0 ⎢⎣ 0 −1 0 ⎥⎦ but ft “Reflection in y = – z” in (ii) 9 5(a)

scalar product Use of cos θ = product of moduli Numerator = ± 43 Denominator = 26 . 149 θ = 46.3o

(b) 3x – 4y + z = 2 and 2x + 12y – z = 38 (c)(i) (3i – 4j + k) × (2i + 12j – k) = – 8i + 5j + 44k p.v. of any point on line

Must be (3i – 4j + k) and (2i + 12j – k)

B1 B1 A1

Dr. = 5.099... × 3.742... = 0.6908... 4

B1 B1

2

M1 A1 M1 1

e.g. (0, 5, 22), (8, 0, – 22), (4, 2 , 0)

A1

x − xc y − yc z − zc = = −8 5 44

B1F

2

Or Adding ⇒ 5x + 8y = 40 (e.g.) x −8 y x y −5 Or = = λ = = µ −8 5 −8 5 x = 8 – 8λ , x = – 8µ y = 5λ y = 5 + 5µ ⇒ z = 44λ – 22 ⇒ z = 44λ + 22 x − xc y − yc z − zc = = 5 44 −8 (ii)

M1

82 + 52 + 442 = 45 −8 1 44 d.c.s are , and 45 9 45

5

ft

(M1) (dM1) (A1)

Eliminating one variable Parametrisation attempted

(M1)

Substg. to find third variable

(A1) B1F

(5)

B1F

2

Total

ft

13

6

ft

MFP4 - AQA GCE Mark Scheme 2007 January series

MFP4 (cont) Q Solution 6(a) Char. Eqn. is λ2 – 5λ – 6 = 0 Solving ⇒ λ = – 1 or 6 Substg. either λ back

Marks B1 M1 A1 M1

⎡1⎤

Comments

λ = – 1 ⇒ x + y = 0 ⇒ evecs. α ⎢ ⎥ −1

A1

⎡2⎤ λ = 6 ⇒ 5x – 2y = 0 ⇒ evecs. β ⎢ ⎥ ⎣5⎦

A1

6

Any non-zero β

⎡ −1 0 ⎤ D= ⎢ ⎥ ⎣ 0 6⎦

B1F B1F

2

ft evals. ft evecs. (must correspond to their evals.)

B1F

1

ft their U (provided non-singular)

⎣

(b)(i)

Total

⎦

⎡ 1 2⎤ U= ⎢ ⎥ ⎣ −1 5 ⎦

–1 1 ⎡5 −2 ⎤ (ii) U = 7 ⎢1 1 ⎥ ⎣ ⎦ (iii) X5 = U D5 U – 1 ⎡ 1 2 ⎤ ⎡ −1 0 ⎤ = 1 ⎢ 7 ⎣ −1 5 ⎥⎦ ⎢⎣ 0 65 ⎥⎦

Any non-zero α

M1 ⎡5 −2 ⎤ ⎢1 1 ⎥ ⎣ ⎦

⎡ 2221 2222 ⎤ = ⎢ ⎥ ⎣5555 5554 ⎦

Use of correct D5 (ft) N.B. 65 = 7776

B1F A1

3 12

7(a) Setting x′ = x and y′ = y x = – x + 2y and y = – 2x + 3y gives y = x (b) ⎡ −1 2 ⎤ ⎡ x ⎤ ⎡ x + 2c ⎤ ⎢ −2 3 ⎥ ⎢ x + c ⎥ = ⎢ x + 3c ⎥ ⎦ ⎣ ⎦⎣ ⎣ ⎦

A1

(d)

2

M1A1

And y′ = x′ + c also

(c)

Or via evals/evecs

M1

det M = 1 ⇒ Areas of shapes invariant ⎡ −1 2 ⎤ ⎡ a ⎤ ⎡ −3a ⎤ ⎢ −2 3 ⎥ ⎢ −a ⎥ = ⎢ −5a ⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦

B1

3

B1 B1

2

Explanation

M1

⇒ Image of y = – x under S is y = 5 x 3 5 Angle is 135o – tan – 1 3 = 76o

A1 B1F

3

N.B. Final angle can be gained via scalar product: (i − j) ⋅ (−3i − 5j) cos θ = 2 34 ⇒ θ = cos – 1 (1/ 17 ) = 76° Total

10

7

ft

MFP4 - AQA GCE Mark Scheme 2007 January series

MFP4 (Cont) Q Solution det P = 4 a + 6 + 4 + a = 5a + 10 8(a)(i)

Marks M1 A1

Total 2

(ii) When a = 3, det P = 25

B1F

1

ft

(iii) Setting their det P = 0 ⇒ a = – 2

M1 A1F

2

ft

B1

1

M1 A1

2

(b)(i) P – 1 = 1 Q 25 (ii) (PQ) – 1 = (25 I) – 1 = 1 I 25 Or (PQ) – 1 = Q – 1 P – 1

Ignore (PQ) – 1 = P – 1 Q – 1 if they can make it work

(M1) (A1)

= Q–1. 1 Q = 1 I 25 25 (iii) det PQ = det (25 I) = 253 or 15625 det PQ = det P . det Q ⇒ 253 = 25 det Q ⇒ det Q = 252 or 625

Comments

(2)

M1 A1 M1 A1

Total TOTAL

8

Used 4 12 75

AQA – Further pure 4 – Jan 2007 – Answers Question 1:

Exam report

= x + 2 y − z 0  4z 7 3 x − y += 8 x + y + 7 z = 30 

l1= 3 x − y + 4 z 7 l2  l2 ⇔ 11x + 11 = z 37 l2 + l3  l3 = 7 z 14 2l2 + l1  7x +

7 3 x − y + 4 z =  37  ⇔ x + z = The system is inconsistent. 11  2  x + z =

Although intended as a straightforward starter, this question caused as much difficulty as any on the paper. Around a third of candidates thought that it was sufficient to show that the determinant of the coefficient matrix was zero – ie for nonuniqueness of solution(s) – while many others relied on faulty arithmetic to support their claim of inconsistency.

Question 2:

a

Exam report

b

a −b

c

b

c

b + c c + a a + b =b − a c + a a + b (C1 ' =− C1 C2 ) ∆= bc ca ab c(b − a ) ca ab 1

b

c

1

b

c

∆ = (a − b) −1 c + a a + b = (a − b) 0 a + b + c a + b + c −c ca ab −c ca ab 1 b c −b 1 b c 0 ∆ = (a − b)(a + b + c) 0 1 1 = (a − b)(a + b + c) 0 1 −c ca ab −c ca ab − ca (a − b)(a + b + c)(ab − ca + c(c − b)) = (a − b)(a + b + c)(a (b − c) − c(b − c)) ∆= ∆= (a − b)(a + b + c)(b − c)(a − c)

Question 3:

This was the type of question previously handled very badly, but this time most candidates appeared well-drilled in how to manipulate determinants. The greatest obstacles to a completely successful conclusion to the question lay in the widespread instinct to expand the remaining determinant at too early a stage, leaving many candidates unable to factorise a quadratic expression in two variables, for instance; and in the more minor error of assuming that the final expression was fully cyclically symmetric in a, b, c, but failing to check that there was a ‘minus sign’ difference between what they were expecting and what they had been given.

Exam report

i

j

k 1 4 1 4 1 1 a) i) p = ×q 1 1 = i− j+ k 4 −3 20 −3 4 4 20 −3 4 20 p×q = 4i − 32 j + 7k 33 1 1 2 p= ×q 4 + (−32) 2 += 72 2 2 2 1 1 4 −3 4 4 20 −3 20 b) p.(q × r ) =−3 4 20 = − +4 2 4 9 4 9 2 9 2 4 ii ) Area of OPQ=

p.(q × r ) = 16 − 40 + 12 + 180 − 24 − 144 = 0 The vector p, q and r are coplanar or The points O, P Q and R are coplanar

This question was handled very well indeed, apart from minor slips in the arithmetic. The geometrical interpretation of linear dependence sought was that p, q and r were co-planar. Many candidates eased their uncertainty as to what was required by saying it in several different ways.

Question 4:

Exam report

0 0 1 0 0  1      = −1  0 Cos (90) − Sin(90)  . MA 0 0 =  0 1 0   0 Sin(90) Cos (90)      0 This is a rotation 90 about the x − axis. 1 0 0   = M B  0 −1 0  0 0 1   This transformation maps the base vectors as such i → i j → − j This is a reflection in the xOz plane (or y = 0 plane) k → k b) The matrix corresponding to the transformation A followed by B 1 0 01 0 0  1 0 0      is M C =M B M A =  0 −1 0   0 0 −1 = 0 0 1 0 0 10 1 0  0 1 0      ii ) C is mapping the vectors of the base as such i

→

i

→ k This is a reflection in the plane y = z. k → j j

This question relied on the use of information given in the formulae booklet, and the majority of candidates coped very well with it. Many fell at the obvious hurdle in part (b), by supposing that “A followed by B” was represented by the matrix M A M B , rather than M B M A . Candidates who failed to attempt to describe the three transformations by a single, rather than multiple (ie composite), description were not awarded marks. There was no penalty for candidates who, for instance, called a plane a line, so long as they gave the right equation.

Question 5:

Exam report

a )Normal vectors to the plane are 3  2      n1 =  −4  and n 2 = 12  1   −1     Let's work out the angle between these vectors 3  2     n1.n 2 = −4  . 12  =6 − 48 − 1 =−43 1   −1    n1 =

9 + 16 + 1=

Cosθ =

n1.n 2 n1 n 2

26 and n 2 =

4 + 144 + 1=

149

−43 = this gives θ 133.69o 26 149

The acute angle between the plane is 180-133.69 = 46.3o 3    b) r.  −4  = 2 ⇔ 3 x − 4 y + z = 2 1    2    r. 12=  38 ⇔ 2 x + 12 y − z = 38  −1   c) i ) A direction vector of the line of intersection is u =n1 × n 2 i j k 3 1 3 −4 −4 1 3 −4 1 = u= i− k+ k 12 −1 2 −1 2 12 2 12 −1 u = −8i + 5k + 44k 2 3 x − 4 y + z = A point belonging to both plane satisfies  38 2 x + 12 y − z = 2 3 x − 4 y + z = ⇔ 5 x + 8 y = 40

for example (8, 0, − 22)

Cartesian equations of the line of intersection is : ii ) u = (−8) 2 + (5) 2 + (44) 2 = 45 The direction cosines are :

−8 5 1 44 , = , 45 45 9 45

x − 8 y z + 22 = = −8 5 44

This was the longest question on the paper, and contained a mixture of ideas from across the unit’s topics. Most candidates coped very favourably with this variety and, minor arithmetical slips apart, marks were high. The biggest obstacle to success came in part (b)(i), where candidates needed to find the coordinates of any one point on the line of intersection. Setting x, y or z equal to zero and finding a value for the other two variables from the resulting simultaneous equations was the most common approach, and this worked well for most candidates.

Question 6:

Exam report

a ) Let's work out det(X-λ I )=0 ⇔

1− λ 5

2 =0 ⇔ (1 − λ )(4 − λ ) − 10 =0 4−λ

⇔ λ 2 − 5λ − 6 = 0 ⇔ (λ − 6)(λ + 1) = 0 The eigenvalues are λ = 6 and λ = −1 x b)i ) To find the eigenvectors, we solve (X-λ I )   = 0  y x  2 2 x  −1, we have (X + I )   = For λ = 0 ⇔   = 0  y 5 5 y  0 2 x + 2 y = ⇔ ⇔ x+ y = 0 5 x + 5 y 0

1  An eigenvector is    −1

x  −5 2   x  For λ = 6, we have (X − 6I )   = 0 ⇔ 0   =  y  5 −2   y  0 −5 x + 2 y =  2 ⇔ ⇔ 5x = 2y An eigenvector is   0 5 x − 2 y = 5  1 2  −1 0  The matrices D=    and U =   −1 5   0 6 1 2 1  5 −2  = 5 + 2 = 7 and U −1 =  ii ) Det (U) =  −1 5 7 1 1  1  1 2   (−1)5 0   5 −2  5 −1 U = = iii ) X5 UD     7  −1 5   0 65   1 1  X

5

1  1 2   −5 2  1  −5 + 2 × 65 2 + 2 × 65   2221 2222  = =        7  −1 5   65 65  7  5 + 5 × 65 −2 + 5 × 65   5555 5554 

This question was handled very confidently by most candidates, and apart from a small number of candidates who clearly deduced the 5 matrix X from a calculator, despite the question’s injunction to use the previous results, marks of 10, 11 or 12 were very common indeed.

Question 7:

Exam report

 x   −1 2   x   − x + 2 y   x '  = a) M    =    =     y   −2 3   y   −2 x + 3 y   y '  All the points on the line are invariant = so x ' x= and y ' y  x =− x + 2 y 2 x − 2 y =0 this gives:  ⇔ −2 x + 3 y 0 y = 2 x − 2 y = The line of invariant points has equation y = x.  x   − x + 2( x + c)   x + 2c   x '  b) M= =     =     x + c   −2 x + 3( x + c)   x + 3c   y '  It is obvious that y ' = x '+ c .For all c, the line y = x + c is invariant. c) det(M ) =

−1 2 =−3 + 4 =1 −2 3

Through a shear, the area is invariant. 1  d ) The line with equation y = − x has direction vector u1 =    −1 The image of the line y = − x:  x   −1 2   x   −3 x   x '  M=    =   =     − x   −2 3   − x   −5 x   y '  5 5 5 x '− 3 y ' = y ' = −5 x = × −3 x = x '. 0 3 3 The image of the line y = − x is the line 5 x − 3 y = 0 3 A direction vector of this line is u 2 =   5 Let's work out the angle between these lines 1   3  u1.u 2 =  .   =3 − 5 =−2 , u1 = 1 + 1 = 2  −1  5  and u 2 = 9 + 25 = 34 Cosθ =

u1.u 2 = u1 u 2

−2 so θ 104.04o = 2 34

The acute angle between the lines is 180o − 104o.04 = 76o to the nearest degree.

This question was slightly tricky in some respects. However, the right approach cut through the apparent problems remarkably quickly, and many candidates were well up to the task. In part (a), it was essential to note that x′ = x and y′=y. in order to make rapid progress, although many candidates took the alternate route and found a single (repeated) eigenvalue of 1 for the matrix before proceeding successfully by that means. In part (b), x  it was important to work with M   from the  x + c outset, and then note that y’= x’+c also. In part (d), x a similar start with M   again cut out −  x unnecessary work.

Question 8:

Exam report

4 −1 2 a ) i ) det(P ) = 1 1 3 = 4a + (a + 6) + 2(0 + 2) −2 0 a det(P= ) 5a + 10 ii ) det(P ) when= a 3 is 15 + 10 = 25 iii ) P is singular when det(P )=0. This happens when a = −2 This was the last question on the paper because, since it involves a 3 × 3 matrix, it was potentially the most timeconsuming question. However, candidates found the structure of the question very helpful.

b) PQ = 25I i ) Multiply both sides by P −1 : P −1PQ = 25P −1I 1 Q = P −1 25 1 1 Q= I 25 25 iii ) det(PQ) = det(P) det(Q) ii ) ( PQ ) = Q −1P −1 =Q −1 × −1

det(25I ) = 25 × det(Q) 253 = 25det(Q)

Grade Mark

so det(Q) = 252 = 625

Grade boundaries Max 75

A 61

B 53

C 45

D 38

E 31