Cambridge A Level Mathematics - Further 9231 — 2012 Oct/Nov Paper 1 · Variant 2

*7150461845* UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level FURTHER MATHEMATICS 9231/12 Paper 1 October/November 2012 3 hours Additional Materials: Answer Booklet/Paper Graph Paper List of Formulae (MF10) READ THESE INSTRUCTIONS FIRST If you have been given an Answer Booklet, follow the instructions on the front cover of the Booklet. Write your Centre number, candidate number and name on all the work you hand in. Write in dark blue or black pen. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. The use of a calculator is expected, where appropriate. Results obtained solely from a graphic calculator, without supporting working or reasoning, will not receive credit. You are reminded of the need for clear presentation in your answers. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 4 printed pages. JC12 11_9231_12/FP © UCLES 2012 [Turn over

2 1 Find the cartesian equation corresponding to the polar equation r = (√2) sec(θ −1 4π). [3] Sketch the the graph of r = (√2) sec(θ −1 4π), for −1 4π < θ < 3 4π, indicating clearly the polar coordinates of the intersection with the initial line. [2] 2 The curve C has equation y = 2x 1 2 for 0 ≤x ≤4. Find (i) the mean value of y with respect to x for 0 ≤x ≤4, [3] (ii) the y-coordinate of the centroid of the region enclosed by C, the line x = 4 and the x-axis. [3] 3 Find the general solution of the differential equation d2x dt2 + 4dx dt + 13x = 26t2 + 3t + 13. [6] 4 Let f(r) = r(r + 1)(r + 2). Show that f(r) −f(r −1) = 3r(r + 1). [1] Hence show that n ∑ r=1 r(r + 1) = 1 3n(n + 1)(n + 2). [2] Using the standard result for n ∑ r=1 r, deduce that n ∑ r=1 r2 = 1 6n(n + 1)(2n + 1). [2] Find the sum of the series 12 + 2 × 22 + 32 + 2 × 42 + 52 + 2 × 62 + . . . + 2(n −1)2 + n2, where n is odd. [3] 5 Let In denote ã ∞ 0 xne−2x dx. Show that In = 1 2nIn−1, for n ≥1. [2] Prove by mathematical induction that, for all positive integers n, In = n! 2n+1 . [6] 6 Use de Moivre’s theorem to show that cos 4θ = 8 cos4θ −8 cos2θ + 1. [3] Without using a calculator, verify that cos 4θ = −cos 3θ for each of the values θ = 1 7π, 3 7π, 5 7π, π. [2] Using the result cos 3θ = 4 cos3θ −3 cos θ, show that the roots of the equation 8c4 + 4c3 −8c2 −3c + 1 = 0 are cos 1 7π, cos 3 7π, cos 5 7π, −1. [2] Deduce that cos 1 7π + cos 3 7π + cos 5 7π = 1 2. [2] © UCLES 2012 9231/12/O/N/12

3 7 The curve C has equation y = λx + x x −2, where λ is a non-zero constant. Find the equations of the asymptotes of C. [3] Show that C has no turning points if λ < 0. [3] Sketch C in the case λ = −1, stating the coordinates of the intersections with the axes. [3] 8 The curve C has parametric equations x = 1 3t3 −ln t, y = 4 3t 3 2, for 1 ≤t ≤3. Find the arc length of C. [6] Find also the area of the surface generated when C is rotated through 2π radians about the x-axis. [4] 9 The plane Π has equation r = 2i + 3j −k + λ(i −2j + 2k) + µ(3i + j −2k). The line l, which does not lie in Π, has equation r = 3i + 6j + 12k + t(8i + 5j −8k). Show that l is parallel to Π. [4] Find the position vector of the point at which the line with equation r = 5i −4j + 7k + s(2i −j + k) meets Π. [4] Find the perpendicular distance from the point with position vector 9i + 11j + 2k to l. [4] 10 Write down the eigenvalues of the matrix A, where A =  1 4 −16 0 2 3 0 0 3 . [1] Find corresponding eigenvectors. [4] Let n be a positive integer. Write down a matrix P and a diagonal matrix D such that An = PDP−1. [2] Find P−1 and An. [5] Hence find lim n→∞3−nAn. [1] © UCLES 2012 9231/12/O/N/12 [Turn over

4 11 Answer only one of the following two alternatives. EITHER The roots of the equation x4 −3x2 + 5x −2 = 0 are α, β, γ, δ, and αn + βn + γ n + δ n is denoted by Sn. Show that Sn+4 −3Sn+2 + 5Sn+1 −2Sn = 0. [2] Find the values of (i) S2 and S4, [3] (ii) S3 and S5. [6] Hence find the value of α2(β3 + γ 3 + δ 3) + β2(γ 3 + δ 3 + α3) + γ 2(δ 3 + α3 + β3) + δ2(α3 + β3 + γ 3). [3] OR The linear transformation T : >4 →>3 is represented by the matrix M, where M =  2 1 −1 4 3 4 6 1 −1 2 8 −7 . The range space of T is R. In any order, (i) show that the dimension of R is 2, (ii) find a basis for R and obtain a cartesian equation for R, (iii) find a basis for the null space of T. [9] The vector 8 7 k ! belongs to R. Find the value of k and, with this value of k, find the general solution of Mx = 8 7 k !. [5] Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. University of Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge. © UCLES 2012 9231/12/O/N/12

CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the October/November 2012 series 9231 FURTHER MATHEMATICS 9231/12 Paper 1, maximum raw mark 100 This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the October/November 2012 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.

Page 2 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2012 9231 12 © Cambridge International Examinations 2012 Mark Scheme Notes Marks are of the following three types: M Method mark, awarded for a valid method applied to the problem. Method marks are not lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. Correct application of a formula without the formula being quoted obviously earns the M mark and in some cases an M mark can be implied from a correct answer. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated method mark is earned (or implied). B Mark for a correct result or statement independent of method marks. • When a part of a question has two or more "method" steps, the M marks are generally independent unless the scheme specifically says otherwise; and similarly when there are several B marks allocated. The notation DM or DB (or dep*) is used to indicate that a particular M or B mark is dependent on an earlier M or B (asterisked) mark in the scheme. When two or more steps are run together by the candidate, the earlier marks are implied and full credit is given. • The symbol implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A or B marks are given for correct work only. A and B marks are not given for fortuitously "correct" answers or results obtained from incorrect working. • Note: B2 or A2 means that the candidate can earn 2 or 0. B2/1/0 means that the candidate can earn anything from 0 to 2. The marks indicated in the scheme may not be subdivided. If there is genuine doubt whether a candidate has earned a mark, allow the candidate the benefit of the doubt. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. • Wrong or missing units in an answer should not lead to the loss of a mark unless the scheme specifically indicates otherwise. • For a numerical answer, allow the A or B mark if a value is obtained which is correct to 3 s.f., or which would be correct to 3 s.f. if rounded (1 d.p. in the case of an angle). As stated above, an A or B mark is not given if a correct numerical answer arises fortuitously from incorrect working. For Mechanics questions, allow A or B marks for correct answers which arise from taking g equal to 9.8 or 9.81 instead of 10.

Page 3 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2012 9231 12 © Cambridge International Examinations 2012 The following abbreviations may be used in a mark scheme or used on the scripts: AEF Any Equivalent Form (of answer is equally acceptable) AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid) BOD Benefit of Doubt (allowed when the validity of a solution may not be absolutely clear) CAO Correct Answer Only (emphasising that no "follow through" from a previous error is allowed) CWO Correct Working Only – often written by a ‘fortuitous' answer ISW Ignore Subsequent Working MR Misread PA Premature Approximation (resulting in basically correct work that is insufficiently accurate) SOS See Other Solution (the candidate makes a better attempt at the same question) SR Special Ruling (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the light of a particular circumstance) Penalties MR –1 A penalty of MR –1 is deducted from A or B marks when the data of a question or part question are genuinely misread and the object and difficulty of the question remain unaltered. In this case all A and B marks then become "follow through " marks. MR is not applied when the candidate misreads his own figures – this is regarded as an error in accuracy. An MR–2 penalty may be applied in particular cases if agreed at the coordination meeting. PA –1 This is deducted from A or B marks in the case of premature approximation. The PA –1 penalty is usually discussed at the meeting.

Page 4 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2012 9231 12 © Cambridge International Examinations 2012 Qu No Commentary Solution Marks Part Mark Total 1 Re-writes equation and uses compound angle formula. Changes to cartesian. Sketches graph. 2 4 cos =       −π θ r 2 2 sin 2 cos =       + θ θ r 2 2 sin cos = + ⇒ = + ⇒ y x r r θ θ or x y − = 2 . Straight line at – 4 π to the initial line. Point (2,0) clearly indicated. M1 A1 A1 B1 B1 3 2 [5] 2 (i) (ii) Uses formula for mean value. Integrates Uses formula for y- coordinate. Integrates. 4 d 2 4 0 2 1 x x ∫ = 3 8 3 1 4 0 2 3 =         x ∫ ∫ 4 0 2 1 4 0 d 2 d 4 2 1 x x x x [ ] 2 3 32 3 16 3 4 4 0 2 3 4 0 2 = × =         = x x M1 M1A1 M1 M1A1 3 3 [6] 3 Solves AQE. Finds CF. Form for PI and differentiates. Compares cefficients and solves. 3 2 0 13 4 2 ± − = ⇒ = + + m m m i CF: ) 3 sin 3 cos ( e 2 t B t A t + − PI: p x q pt x r qt pt x 2 2 2 = ⇒ + = ⇒ + + = && & 2 26 13 = ⇒ = p p 1 3 13 8 − = ⇒ = + q q p 1 13 13 4 2 = ⇒ = + + r r q p GS: 1 2 ) 3 sin 3 cos ( e 2 2 + − + + = − t t t B t A x t M1 A1 M1 M1A1 A1 6 [6]

Page 5 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2012 9231 12 © Cambridge International Examinations 2012 Qu No Commentary Solution Marks Part Mark Total 4 Verifies given result. Uses method of differences to sum first series. Subtracts ∑ = n r r 1 to obtain sum of second series. Splits series into two series. Applies sum of squares formula to obtain result. )1 2 )( 1 ( )1 ( )1 ( ) 2 )( 1 ( + − + + = + − − + + r r r r r r r r r r )1 ( 3 + = r r (AG) ∑ = = + n r r r 1 )1 ( [ ] [ ] [ ] { } 0 ( f )1( f ... ) 2 ( f )1 ( f )1 ( f ) ( f 3 1 − + + − − − + − − n n n n ) 2 )( 1 ( 3 1 + + = n n n (AG) (Award B1 if ‘not hence’.) ∑ ∑ ∑ = = = − + = n r n r n r r r r r 1 1 1 2 )1 ( = 2 )1 ( 3 ) 2 )( 1 ( + − + + n n n n n )1 2 )( 1 ( 6 1 )3 4 2 )( 1 ( 6 1 + + = − + + = n n n n n n (AG) ) )1 ( ... 4 2 ( ) ... 2 1( 2 2 2 2 2 2 − + + + + + + + n n = 6 2 1 2 1 4 6 )1 2 )( 1 ( n n n n n n       +       − + + + =…= )1 ( 2 1 2 + n n B1 M1 A1 M1 A1 M1 M1A1 1 2 2 3 [8] 5 Integrates by parts to obtain reduction formula. (States proposition.) Proves base case. Shows 1 P P + ⇒ k k . States conclusion. ∫ ∫ ∞ − ∞ − ∞ − +       − = 0 2x - 1 0 2 0 2 d 2 e 2 e d e x nx x x x n x n x n 1 2 − = nI n (AG) Pn : 1 2 ! + = n n n I n = 1 2 1 e 2 1 d e 0 0 2x - 2x - 0 =    − = = ∞ ∞∫ x I 2 1 2 !1 4 1 2 1 2 1 = = × = I ∴P1 true. Pk : 1 2 ! + = k k k I for some integer k. 2 1 1 2 )! 1 ( 2 ! 2 1 + + + + = × + = ∴ k k k k k k I 1 P P + ⇒ ∴ k k Hence by PMI Pn is true for all positive integers n. M1 A1 B1 B1 B1 M1A1 A1 2 6 [8]

Page 6 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2012 9231 12 © Cambridge International Examinations 2012 Qu No Commentary Solution Marks Part Mark Total 6 Proves initial result. Verifies any two cases (B1) Verifies remaining two cases (B1) Shows roots of equation to be as given. States sum of roots of equation. θ θ θ θ 4 sin i 4 cos ) sin i (cos 4 + = + = 4 3 2 2 3 4 (is) c(is) 4 (is) 6c (is) c 4 c + + + + 2 2 2 2 4 ) c 1( ) c 1( c 6 4 cos − + − − = c θ 1 cos 8 cos 8 2 4 + − = θ θ (AG) π π π π 7 3 cos 7 4 cos 7 4 cos − =       − − = π π π π π 7 9 cos 7 2 cos 7 2 cos 7 12 cos − =       + − =      − = π π π π 7 15 cos 7 1 cos 7 6 cos 7 20 cos − = − = = π π 3 cos )1 ( 1 4 cos − = − − = = 1 cos 8 cos 8 2 4 + − θ θ = – ( θ θ cos 3 cos 4 3 − ) ⇒ 0 1 3 8 4 8 2 3 4 = + − − + c c c c (*) ⇒ 1 - , 7 5 cos , 7 3 cos , 7 1 cos π π π are the roots. (AG) Sum of roots of (*) are 2 1 8 4 − = − ⇒ Result (AG) M1 M1 A1 B1 B1 M1 A1 M1A1 3 2 2 2 [9] 7 Finds asymptotes to C. Differentiates and equates to 0. Sketch of graph. Deduct 1 mark for poor forms at infinity. Deduct 1 mark if intersections with axes not shown. Vertical asymptote x = 2. 2 2 1 − + + = x x y λ 1 + = ⇒ x y λ is oblique asymptote. 0 ) 2 ( 2 2 = − − = ′ − x y λ for turning points. ⇒ > − = 0 ) 2 ( 2 2 x λ no turning points if 0 < λ . Or 0 2 4 4 0 2 = − + − ⇒ = ′ λ λ λ x x y Uses discriminant to show ⇒ < 0 8λ no T.P.s. Axes and asymptotes. LH branch. RH branch. (Indicating intersections with axes at (0,0) and (3,0).) B1 M1A1 M1A1 A1 (M1A1) (A1) B1 B1B1 3 3 3 [9]

Page 7 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2012 9231 12 © Cambridge International Examinations 2012 Qu No Commentary Solution Marks Part Mark Total 8 Differentiates. Squares and adds. Uses arc length formula. Integrates. Obtains result. Uses surface area formula. Integrates. Inserts limits. Obtains result. t t x 1 2 − = & 2 1 2t y = & 2 2 2 2 1 4 1 d d       + = +       − = t t t t t t s ∫       + = 3 1 2 d 1 t t t s 3 1 3 ln 3       + = t t 3 ln 3 26 3 1 3 ln 9 + = − + = (= 9.77) ∫               + = 3 1 2 2 3 d 1 3 4 2 t t t t S π t t t d 3 8 3 1 2 1 2 7 ∫         + = π 3 1 2 3 2 9 3 2 9 2 3 8         + = t t π [ ]           + − + = 3 2 9 2 3 2 3 18 3 8π       − = 27 64 3 3 160 π (= 283 or π 0. 90 = ) B1 M1A1 M1 A1 A1 M1 A1 M1 A1 6 4 [10]

Page 8 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2012 9231 12 © Cambridge International Examinations 2012 Qu No Commentary Solution Marks Part Mark Total 9 Finds vector normal to Π. Dot product of this with general point on l1. Deduces result. Cartesian equation of Π. Substitutes general point of l2. Finds value of parameter. Finds p.v. of intersection. Finds distance from point to known point on l. Finds distance along l from known point to foot of perpendicular from given point to l. F.t. on non- hypotenuse side (must be real). Writes a set of three equations in three unknowns for the intersection of l with Π. Solves the set of equations. Finds p.v. of intersection. k j i k j i n 7 8 2 2 1 3 2 2 1 + + = − − = 138 7 8 2 8 12 5 6 8 3 =           ⋅           − + + t t t or 0 7 8 2 8 5 8 =           ⋅           − Independent of t ⇒ parallel, or ⇒ parallel. Π: 21 7 8 2 = + + z y x Sub. s x 2 5 + = , s y − − = 4 , s z + = 7 2 − = ⇒s and line meets Π at point with p.v. i – 2j + 5k Take (9,11,2) as A, (3,6,12) as B and let C be foot of perpendicular from A to l. 161 10 5 6 AB 2 2 2 = + + = 153 153 153 ) k 8 j 5 i8 )( k 10 j 5 i6 )( 8 5 6 ( 1 BC 2 2 2 = = − + − + + = 2 2 or 8 153 161 AC = − = (= 2.83) Alternatively: µ λ µ λ µ λ 2 2 1 7 2 3 4 3 2 2 5 − + − = + + − = − − + + = + s s s 2 − = ⇒s , 2 = ⇒λ , 1 − = µ and line meets Π at point with p.v. i – 2j + 5k M1A1 A1 A1 B1 M1 A1 A1 B1 M1 A1 A1 (B1) (M1A1) (A1) 4 4 4 [12]

Page 9 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2012 9231 12 © Cambridge International Examinations 2012 Qu No Commentary Solution Marks Part Mark Total 9 Finds vector BA . Finds distance from point to known point on l. Finds distance along l from known point to foot of perpendicular from given point to l. F.t. on non-hypotenuse side (must be real). Finds vector . AC Uses AC perpendicular to l to find t. Finds length AC. BA = 6i + 5j – 10j 8 5 8 10 5 6 8 5 8 1 2 2 − − + + k j i = 8 153 1224 = or 2 2 ( = 2.83) Alternatively: (A) Take (9,11,2) as A, (3,6,12) as B and let C be foot of perpendicular from A to l. AB = 161 10 5 6 2 2 2 = + + BC = ) 8 5 8 )( 10 5 6 ( 8 5 8 1 2 2 2 k j i k j i − + − + + + 153 153 153 = = AC = 8 153 161 = − or 2 2 ( = 2.83) (B) AC           − − − = t t t 8 10 5 5 6 8 1 0 8 5 8 . 8 10 5 5 6 8 = ⇒ =           −           − − − t t t t AC = 8 2 0 2 2 2 2 = + + B1 M1A1 A1 (B1) (M1) (A1) (A1 ) (B1) (M1A1) (A1) (4) (4) (4) [12]

Page 10 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2012 9231 12 © Cambridge International Examinations 2012 Qu No Commentary Solution Marks Part Mark Total 10 States eigenvalues. Finds eigenvectors. States P and D. Finds inverse of P. Finds An. States required limit. Eigenvalues are 1, 2, 3. 1 = λ e1 =           = − 0 0 28 3 1 0 16 4 0 k j i ~           0 0 1 2 = λ e2 =           = − − 0 1 4 1 0 0 16 4 1 k j i 3 = λ e3 =          − = − − − 2 6 4 3 1 0 16 4 2 k j i ~          − 1 3 2 P =           − 1 0 0 3 1 0 2 4 1 D =           n n 3 0 0 0 2 0 0 0 1 Det P = 1 ⇒ Adj P = P-1 =           − − 1 0 0 3 1 0 14 4 1 An = PDP-1           − − = n n n 3 0 0 2.3 2 0 14 4 1 P or P n           − + n n n n 3 0 0 3 2 0 3.2 2.4 1 1           + − − − + − = + n n n n n n n 3 0 0 ] 3 2.3 [ 2 0 ] 3.2 2. 12 14 [ ] 2.4 4 [ 1 1           − → − 1 0 0 3 0 0 2 0 0 3 n n A as n ∞ → . B1 M1A1 A1 A1 B1 B1 M1A1 M1A1 A1 B1 1 4 2 5 1 [13]

Page 11 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2012 9231 12 © Cambridge International Examinations 2012 Qu No Commentary Solution Marks Part Mark Total 11 (i) (ii) EITHER Substitute α into equation. Multiply by n α . Obtain result. Uses ( ) ∑ ∑ ∑ − = αβ α α 2 2 2 Finds 4 S from formula. αβγδ αβγ ∑ = −1 S Finds 3 S from formula. Finds 5 S from formula. α is a root 0 2 5 3 2 4 = − + − ⇒ α α α 0 2 5 3 1 2 4 = − + − ⇒ + + + n n n n α α α α Repeat for δ γ β , , and sum ⇒ 0 2 5 3 1 2 4 = − + − + + + n n n n S S S S (AG) 6 ) 3 ( 2 0 2 = − × − = S 26 4 2 0 5 6 3 4 = × + × − × = S 2 5 2 5 1 = − − = − S 15 2 5 2 4 5 0 3 3 − = × + × − × = S 75 0 2 6 5 ) 15 ( 3 5 − = × + × − − × = S ∑ − = 5 3 2 3 2 S S S β α 15 ) 75 ( ) 15 ( 6 − = − − − × = M1 A1 B1 M1A1 M1A1 M1A1 M1A1 M1 M1A1 2 3 6 3 [14]

Page 12 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2012 9231 12 © Cambridge International Examinations 2012 Qu No Commentary Solution Marks Part Mark Total 11 (i) (ii) (iii) OR Reduces M to echelon form. Uses dimension theorem. States basis for R. Finds cartesian equation for R. Finds basis for null space. Evaluates k. Finds a particular solution. Finds general solution.           − − 7 - 1 4 8 2 1 6 4 3 1 1 2 →…           − → 0 2 - 4 0 0 0 3 1 0 1 1 2 Dim(M) = 4 – 2 = 2 Basis for R is                           − 2 4 1 , 1 3 2 . (OE) µ λ + = 2 x µ λ 4 3 + = y 0 2 = + − ⇒ z y x µ λ 2 + − = z 0 4 2 = + − + t z y x 0 2 3 = − + t z y λ = t and µ = z µ λ 3 2 − = ⇒y and µ λ 2 3 + − = x ⇒Basis of null space is                                    − 0 1 3 - 2 , 1 0 2 3 (OE) 9 0 7 8 2 − = ⇒ = + − × k k           − =           −           − 9 7 8 2 4 1 2 1 3 2 5 (OE) (via equations) x =             − +            − +             − 0 1 3 2 1 0 2 3 0 0 2 5 µ λ . M1A1 A1 B1 M1A1 M1 A1A1 B1 M1A1 M1A1 9 5 [14]