Examiners commentaries 2014

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1 Examiners commentaries 204 Examiners commentaries 204 MT73 Algebra Important note This commentary reflects the examination and assessment arrangements for this course in the academic year The format and structure of the examination may change in future years, and any such changes will be publicised on the virtual learning environment (VLE). Information about the subject guide Unless otherwise stated, all cross-references will be to the latest version of the subject guide (203). You should always attempt to use the most recent edition of any Essential reading textbook, even if the commentary and/or online reading list and/or subject guide refers to an earlier edition. If different editions of Essential reading are listed, please check the VLE for reading supplements if none are available, please use the contents list and index of the new edition to find the relevant section. General remarks Learning outcomes At the end of this course and having completed the essential reading and activities you should be able to: use the concepts, terminology, methods and conventions covered in the course to solve mathematical problems in this subject solve unseen mathematical problems involving understanding of these concepts and application of these methods see how algebra can be used to solve problems in economics and related subjects demonstrate knowledge and understanding of the underlying principles of algebra. Approach the examination as a test of your understanding Success in this paper is achieved by knowing and understanding the material covered in each chapter very well, starting at the beginning with the basic material and building on it to apply it to, and understand, the material covered in the later chapters. When you study it is important to understand the theory behind each process, to understand how and why it works, rather than just be able to carry out a method mechanically.

2 MT73 Algebra Key steps to improvement Try to gain an overview of the entire syllabus and to study the full extent of the topics described in the subject guide. A good way to review the material is to look at the learning outcomes at the end of each chapter. Since the whole syllabus is examinable, any topic could potentially appear in any part of the five questions. The basic tool you have learned in linear algebra for solving a system of linear equations is the method of Gaussian elimination. Being able to interpret information about the solutions of the system, or about the subspaces associated with the matrix from the reduced row echelon form is key to obtaining the correct answers to many of the questions on the examination paper. Ensure that you have worked through all the exercises in the subject guide and all the exercises in the relevant parts of the Essential reading text. Think about how the theory applies to them. The more practice and understanding you have, the greater the ease with which you can answer questions during the examination. It is important that you understand any notation you are using and that you use it correctly. Mathematics is a precise subject. The definitions and notations all have precise meanings and you should know them. You should read a question carefully and answer exactly what is asked; don t leave out any parts (or you will miss some marks) and don t do more than what is required (or you will use time for something which will not gain any marks). There may be different methods to answer a question, and unless a specific method is mentioned, any correct method is acceptable. However, there may be some advantage to using one method, and some methods may take less time than others. In this subject, often a good knowledge of the theory will provide a shorter, more efficient method to answer a question. Stop and think about what you are being asked and how to best approach the question. Write your answers clearly. All answers should be justified by showing work It is essential that you always include the entire solution. This means two things. First, you should not simply write down the answer in the examination script, but you should explain the method by which it is obtained. Secondly, you should include rough working (even if it is messy!). The Examiners want you to get the right answers, of course, but it is more important that you prove you know what you are doing: that is what is really being examined. If you have not completely solved a problem, you may still be awarded marks for a partial, incomplete, or slightly wrong, solution; but, if you have written down a wrong answer and nothing else, no marks can be awarded. On the other hand, if you merely write down a correct answer which requires some working out (and is not merely a deduction) without showing how you derived it, you will not receive full credit. Most of the marks are assigned for demonstrating that you know and understand how to carry out a given method. So it is certainly in your interest to include all your workings. Check your work 2 It is important to check your work as you proceed, particularly where this is easy to do. Stop and ask yourself, Is my answer sensible?. In many cases lack of accuracy caused problems which could have been avoided by checking. For example, when finding an eigenvector there should be a row of zeros in the reduced row echelon form of the appropriate matrix, so that you get a non-zero eigenvector. If not, either the eigenvalue

3 Examiners commentaries 204 or the row reduction is incorrect, and should be redone. Eigenvectors can be checked using the matrix. Inverse matrices can be checked by multiplying the matrix times its inverse to get the identity matrix. If you find that you have made a mistake, go back and correct it. If you can t find your error, then at least say that you have made one, and try to carry on. Expectations of the examination paper Every examination paper is different. Previous years papers and the sample examination paper in the subject guide are only indicative. You should not assume that the same questions and topics will appear in next year s examination or in the same order or arrangement. Each year, the Examiners want to test that candidates know and understand the material in the syllabus and, in setting an examination paper, they try to test whether the candidate does indeed know the methods and definitions, understands them, and is able to use them (and not merely whether they vaguely remember them or have learned solutions to past papers). Every year there are some questions which are likely to seem unfamiliar, or different from previous years questions. You should expect to be surprised by some of the questions. Of course, you will only be examined on material in the syllabus, so all questions can be answered using the course material. All exam questions will include some basic material, because later material depends on it. There will be enough, routine, familiar content in the examination so that a student who has achieved competence in the unit will pass, but, of course, for a high mark, more is expected: you will have to demonstrate an ability to solve new and unfamiliar problems. Essential reading The Essential textbook for this course is Linear Algebra: Concepts and Methods by Martin Anthony and Michele Harvey (Cambridge University Press). Whenever a section of the subject guide is referred to in the Specific comments on questions that follow, you should also look at the corresponding sections of the Essential textbook for detailed explanations and examples. Calculators You are reminded that calculators are not permitted in the examination for this course, under any circumstances. The Examiners know this, and so they set questions that do not require a calculator. It is a good idea to prepare for this by not using your calculator as you study and revise. 3

4 MT73 Algebra Question spotting Many candidates are disappointed to find that their examination performance is poorer than they expected. This can be due to a number of different reasons and the Examiners commentaries suggest ways of addressing common problems and improving your performance. We want to draw your attention to one particular failing question spotting, that is, confining your examination preparation to a few question topics which have come up in past papers for the course. This can have very serious consequences. We recognise that candidates may not cover all topics in the syllabus in the same depth, but you need to be aware that Examiners are free to set questions on any aspect of the syllabus. This means that you need to study enough of the syllabus to enable you to answer the required number of examination questions. The syllabus can be found in the Course information sheet in the section of the VLE dedicated to this course. You should read the syllabus very carefully and ensure that you cover sufficient material in preparation for the examination. Examiners will vary the topics and questions from year to year and may well set questions that have not appeared in past papers every topic on the syllabus is a legitimate examination target. So although past papers can be helpful in revision, you cannot assume that topics or specific questions that have come up in past examinations will occur again. If you rely on a question spotting strategy, it is likely you will find yourself in difficulties when you sit the examination paper. We strongly advise you not to adopt this strategy. 4

5 Examiners commentaries 204 Examiners commentaries 204 MT73 Algebra Important note This commentary reflects the examination and assessment arrangements for this course in the academic year The format and structure of the examination may change in future years, and any such changes will be publicised on the virtual learning environment (VLE). Information about the subject guide Unless otherwise stated, all cross-references will be to the latest version of the subject guide (203). You should always attempt to use the most recent edition of any Essential reading textbook, even if the commentary and/or online reading list and/or subject guide refers to an earlier edition. If different editions of Essential reading are listed, please check the VLE for reading supplements if none are available, please use the contents list and index of the new edition to find the relevant section. Comments on specific questions Zone A Candidates should answer all FIVE questions. All questions carry equal marks. Question Let A = and b = (a) Find the reduced row echelon form of the augmented matrix (A b). Hence find the general solution of the system of linear equations Ax = b (expressing your solution in vector form). (b) Let c, c 2,..., c 5 denote the column vectors of the matrix A. (i) Can c 4 be expressed as a linear combination of c and c 2? Justify your answer. Write down a linear combination if one exists. (ii) With reference to the reduced row echelon form of A, explain why the set B = {c, c 2, c 5 } is a basis for R 3. Write down [b] B, the coordinates of the vector b in the basis B. (iii) Define what it means to say that a set of vectors {v, v 2,..., v n } is linearly dependent. Show that the set of column vectors {c, c 2, c 3 } is linearly dependent. (iv) Explain why {c, c 2 } is a basis of the subspace S = Lin{c, c 2, c 3, c 4 } of R 3. Deduce that the subspace S is a plane in R 3, and find the Cartesian equation of this plane. 5

6 MT73 Algebra Reading for this question This question examines basic skills from several parts of the subject guide. Part (a) tests your ability to solve systems of linear equations using Gaussian elimination, which is covered in Chapter 3. Then part (b) tests your ability to interpret information from the reduced row echelon form of the coefficient matrix. To answer the questions concerning basis and linear independence requires your knowledge of the basic material of Chapters 7 and 8, in particular Sections Approaching the question Let A = b = (a) To find the general solution of the system of linear equations Ax = b, put the augmented matrix into reduced row echelon form. (A b) = R R R 2 3R R3+R R 3+R R R R R3 R 2 2R Set the non-leading variables equal to arbitrary parameters; say, x 3 = s and x 4 = t. Then the general solution is: x 2 4s 4t x 2 3 9s 3t x = x 3 = s = 0 + s + t 0 = p + sv + tv 2, s, t R. x 4 t 0 0 x You can easily check your solution by calculating Ap = b, Av = 0, Av 2 = 0. (b) (i) There are several ways to do this question. The most straightforward is just to solve, 9 4 = a 3 + b 0. Equating the components, the second line immediately gives a = 4, and then the first line gives b = 3. Then you need to check that the third equation is also satisfied, = 4( ) + 3(). So, c 4 = 4c + 3c 2. Alternatively, the quickest way is to notice the relationship between c, c 2, c 4 as shown by looking at the corresponding columns of the RREF of A. If you denote these as ĉ, ĉ 2, ĉ 4, then it is clear that ĉ 4 = 4ĉ + 3ĉ 2, and the same relationship applies to the columns of A. Alternatively, you can see this relationship by using the null space vector v 2 = ( 4, 3, 0,, 0) T and the basic fact that if A is an m n matrix and x = (x, x 2,..., x n ) T, then Ax = x c + x 2 c x n c n, 6

7 Examiners commentaries 204 where c, c 2,..., c n are the columns of A. This yields the equality Av 2 = 4c 3c 2 + 0c 3 + c 4 + 0c 5 = 0 from which you can again conclude that c 4 = 4c + 3c 2. (ii) The column vectors c, c 2, c 5 correspond to the columns with leading ones in the RREF of A, which indicates that they are linearly independent. Since R 3 has dimension three, three linearly independent vectors in R 3 are a basis. Using the dimension of R 3 to conclude that the three linearly independent vectors are a basis is an important part of this argument. Alternatively, you could deduce from the RREF of A that the reduced row echelon form of a matrix consisting of the three column vectors, c, c 2, c 5 would be the identity matrix, and therefore {c, c 2, c 5 } is a basis of R 3. From the solution, since Ap = b, you can write b = 2c + 3c 2 + c 5, using the same basic fact as above. Therefore the coordinate matrix of b in this basis is [b] B = 2 3 Think about what it is you needed to know about a basis of R 3 in order to answer this part of the question. The material is all contained in Chapter 8, in particular Section 8.3. (iii) The set {v, v 2,..., v n } is linearly dependent if and only if the vector equation a v + a 2 v a n v n = 0 has a non-trivial solution; that is, a solution with not all coefficients a i = 0. While it is correct to say that the vectors are linearly dependent if and only if some vector in the set is a linear combination of the others, this is a consequence of the definition. Looking at the solution of Ax = b, the vector v = ( 4, 9,, 0, 0) T is a solution of the homogeneous system. Since Av = 0, you can deduce that B 4c 9c 2 + c 3 = 0. This is a non-trivial linear combination showing that the vectors are linearly dependent. Alternatively, you could show that the determinant of the matrix whose columns are c, c 2, c 3 is equal to zero, and conclude that the equation a c + a 2 c 2 + a 3 c 3 = 0 has a non-trivial solution. (iv) The subspace Lin{c, c 2 } contains the vector c 4 by (i) since c 4 = 4c + 3c 2 is a linear combination of c, c 2. By (iii), c 3 = 4c + 9c 2, so c 3 is also a linear combination of c, c 2. Therefore, S = Lin{c, c 2, c 3, c 4 } = Lin{c, c 2 }. Since the vectors c, c 2 are linearly independent (neither is a scalar multiple of the other, or this follows from (ii)), the set {c, c 2 } is a basis of S. You do need to state why the set {c, c 2 } spans S and why it is linearly independent. Then you can conclude it is a basis of S. The subspace S has a basis consisting of two vectors in R 3, and a two-dimensional subspace of R 3 is a plane through the origin. To find a Cartesian equation, you can evaluate the determinant (which is most easily expanded using column three) 3 x 0 y = x 2y + z = 0. z You should understand why and how this works, as shown in Section 8.3 of the subject guide.. 7

8 MT73 Algebra Alternatively, you can use the parametric equation x = sc + tc 2, s, t R and eliminate s and t from the three resulting component equations to obtain a Cartesian equation in x, y, z. The plane has equation x 2y + z = 0. Check that this is correct; the vectors c, c 2, c 3, c 4 should all satisfy this equation. Question 2 (a) Let A be an m n matrix. Define what is meant by the range of A, R(A). Prove that R(A) is a subspace of R m. (b) Let c, c 2, c 3 denote the columns of the matrix (i) Find rank(a) A = (ii) Find a basis for the null space of A, N(A). (iii) Find a basis for R(A). (iv) Let d = c c 2 c 3. Before attempting to solve the system, explain why you know that Ax = d is consistent. Write down a general solution of the system of equations Ax = d. (v) State the rank-nullity (dimension) theorem for matrices carefully defining each term. Verify the theorem for the matrix A. (c) Let A be an m n matrix and let B be an n p matrix. Show that Deduce that rank(ab) rank(a). R(AB) R(A). Reading for this question This question, which is mainly about the range of A is testing material from Chapters 5, 7 and 8 of the subject guide. The range, R(A), is first introduced in Section 5.4, where it is also shown that a system of equations Ax = b is consistent if and only if b is a linear combination of the columns of A. In Section you will find the proof that R(A) is a subspace of R m, and later in Section you find it is equivalent to the column space of A, denoted CS(A). These are just two different ways of looking at the same subspace associated with the matrix A. For the remainder of the question you need to know about basis and dimension, Chapter 8. In particular, the rank nullity theorem is Section 8.5. Part (b)(iv) tests the basic fact from Section 5.3 that the general solution of a system of linear equations is composed of a particular solution plus the general solution of the homogeneous system. Part (c) requires knowledge of how to show a subspace is contained in another subspace. An example of this is shown in Section Approaching the question (a) Let A be an m n matrix. The range of A, R(A), is the set R(A) = {Ax x R n }. Since A is m n, Ax R m, so this is a subset of R m. 8

9 Examiners commentaries 204 To show it is a subspace, first note that it is non-empty. Since A0 = 0, 0 R(A) so it is non-empty. Next you need to show that it is closed under addition and scalar multiplication. For addition, let x, y R(A), so that Av = x and Aw = y, for some v, w R n. Then x + y = Av + Aw = A(v + w) with v + w R n = x + y R(A). For scalar multiplication, let x R(A) and α R. Then αx = α(av) = A(αv) with αv R n = αx R(A). This proves that R(A) is a subspace of R m. (b) Let c, c 2, c 3 denote the columns of the matrix A. To answer the questions, put the matrix A into reduced row echelon form A = (i) rank(a) = 2. (ii) To find a basis of N(A), find the solutions of Ax = 0. If x = (x, y, z) T, set the non-leading variable z = t, where t R is an arbitrary parameter. Then a general solution is x = x y z = 7t 2t t = t 7 2 Therefore a basis of N(A) is {v}, where v = (7, 2, ) T. = tv, t R. (iii) A basis of the range of A, R(A), is given by the columns of A which correspond to the leading ones in the reduced row echelon form of A, so a basis of R(A) is 5 2, (iv) If d = c c 2 c 3, then Ax = d is consistent because d is a linear combination of the columns of A; that is, d is in the column space of A, CS(A), and CS(A) = R(A). The range of A consists of all vectors b for which Ax = b is consistent. To write down a general solution of Ax = d, use the fact that a general solution of a system of linear equations is the sum of a particular solution and the general solution of Ax = 0, which you have already found in part (ii). You can find a particular solution using the fact that for any m n matrix A and vector x = (x, x 2,..., x n ) T, Ax = x c + x 2 c x n c n, where c, c 2,..., c n are the columns of A. Since d = c c 2 c 3, you can deduce that a particular solution p such that Ap = d is p = (,, ) T. Then a general solution of the system of equations Ax = d is x = p + tv = + t 7 2, t R. Of course, you could also find a solution by finding d = ( 7, 2,, 2) T and solving the system Ax = d by Gaussian elimination, which would take considerably more time. 9

10 MT73 Algebra (v) The rank-nullity (dimension) theorem for an m n matrix A, states that rank(a) + nullity(a) = n, where: n is the number of columns of A, rank(a) is the dimension of R(A), and nullity(a) is the dimension of N(A). Here notation is important. R(A) and N(A) denote the subspaces, not their dimensions. The rank-nullity theorem is about the dimensions of the subspaces. So you could also state the theorem as dim(r(a)) + dim(n(a)) = n. For the matrix A, you know from parts (ii) and (iii) that dim(n(a)) = and dim(r(a)) = rank(a) = 2. Since A has 3 columns, this gives rank(a) + nullity(a) = 2 + = 3 = n. (c) Let A be an m n matrix and let B be an n p matrix. Then AB is an m p matrix. To show that R(AB) R(A), you need to show that any x R(AB) is also contained in R(A). (Both ranges are contained in R m.) Let x R(AB). Then there is a vector v R p such that (AB)v = x. Since B is n p, the vector Bv R n and x = (AB)v = A(Bv) = x R(A). Since any vector in R(AB) is also in R(A), this shows that R(AB) R(A). Therefore, dim(r(ab)) dim(r(a)); that is, rank(ab) rank(a). Question 3 (a) A student takes out a loan of $P at a monthly interest rate of %. Each month she makes a repayment of $50. If the loan is reduced monthly with her repayments and y t is the loan balance outstanding at the end of month t, explain why y t satisfies the difference equation y t+ =.0y t 50. Solve the difference equation. Use your answer to find an expression for the maximum amount P she can borrow if the loan must be repaid in two years. (b) In any year t, the fixed total population of 800 inhabitants of Island is divided into those living by the lake, and those living in the forest. Initially 200 inhabitants live by the lake. If x t denotes the number of inhabitants living by the lake in year t, and y t denotes the number living in the forest in year t, then the yearly population movements are given by the following system of difference equations x t = 0.7 x t y t y t = 0.3 x t y t. (i) Each year, what percentage of the population living by the lake moves to the forest? 0 (ii) Express the equations in matrix form, as x t = Ax t. Find the eigenvalues of A and find a corresponding eigenvector for each eigenvalue. Find an invertible matrix P and a diagonal matrix D such that P AP = D.

11 Examiners commentaries 204 (iii) Using the diagonalisation in part (ii), find expressions for x t and y t in terms of t. Use your answers to determine the long term population distribution; that is, find what happens to x t as t, writing down the eventual population in each location. Reading for this question Part (a) of this question concerns a financial application resulting in a first order difference equation, as covered in Sections of the subject guide. Part (b) requires a knowledge of diagonalisation (Chapter 0) and its application to solving systems of first-order linear difference equations, as shown in Section.2. In particular, the system in this question is a Markov chain, Section.2.4 of the subject guide. Approaching the question (a) The loan balance outstanding in month t + is the loan balance outstanding in month t, plus the interest for one month on this balance, which is (0.0)y t, minus a monthly repayment of $50. Therefore The solution of this equation is given by y t+ = y t + 0.0y t 50 =.0y t 50, t N. y t = y + (y 0 y )(.0) t, where y is the time-independent solution. Here, y = Therefore, the solution is 50 ( (.0)) = 50.0 = 5000, and y 0 = P. y t = (P 5000)(.0) t, t N. Alternatively, you can obtain the same solution by considering y t as a geometric sum. y 0 = P y = P (.0) 50 y 2 = (P (.0) 50)(.0) 50 = P (.0) 2 50(.0) 50 Continuing, y t = P (.0) t 50 ( + (.0) + + (.0) t ). Evaluating the geometric sum, y t = P (.0) t 50((.0)t ) (.0 ), t N y t = P (.0) t 5000(.0) t which gives the same expression for y t as above. Alternatively, you can solve this equation using the same method as for second-order difference equations as shown in Sections First solve the homogeneous difference equation y t+ (.0)y t = 0.

12 MT73 Algebra This has auxiliary equation z (.0) = 0, so the general solution of the homogeneous equation is y t = A(.0) t. Then find a particular (constant) solution y by substituting into the original equation, y = (.0)y 50 = y = 50 (.0) = Putting this together, the general solution of the equation y t+ =.0y t 50 is y t = A(.0) t It now remains to evaluate A using the initial conditon y 0 = P, y 0 = A = P = A = P 5000 = y t = (P 5000)(.0) t , t N. If the loan must be repaid in two years, then y 24 = 0. Solving this for P, y 24 = (P 5000)(.0) 24 = 0 P (.0) 24 = 5000((.0) 24 ) = P = 5000((.0)24 ) (.0) 24 = (.0) 24. (b) Yearly population movements on Island are given by the system of equations { xt = 0.7 x t y t y t = 0.3 x t y t, t N (i) In each year, 30% of those living by the lake move to the forest. (ii) In matrix form the equations are x t = Ax t : ( xt y t ) = ( ) ( xt y t ). This is a Markov process, so λ = is one of the eigenvalues. To diagonalise A, first find the other eigenvalue from the characteristic equation. A λi = 0: 0.7 λ λ = λ + λ2 0.5 = λ 2.2λ = (λ )(λ 0.2) = 0. So the eigenvalues are λ = and λ 2 = 0.2. To find a corresponding eigenvector for each eigenvalue, solve (A λi)v = 0. For λ = : A I = /3 0 0 = v = 5. 3 For λ = 0.2: A 0.2I = = v 2 =. Then if P = 5 3 and D = 0, P AP = D

13 Examiners commentaries 204 (iii) You can solve this system using the diagonalisation either by a change of variables (Section.2.2) or using powers of the matrix A (Section.2.3). Using matrix powers, you have x t = Ax t x t = A t x 0. Then, as shown in Section., A t = P D t P, so the solution of the system of difference equations is x t = A t x 0 = P D t P x 0. Initially there are 200 inhabitants by the lake, so you can take x 0 = (200, 600) T, or you can work this question using the distribution vector x 0 = (0.25, 0.75) T, as long as you interpret your solution correctly. Then x t is given by: 5 t x t = 3 0 (0.2) t = 3 0 (0.2) t xt 5 x t = = (0.2) t. 3 y t The sequences are { xt = (0.2) t y t = (0.2) t or x t = (0.2)t y t = (0.2)t 5 As t, x t 00. The eventual population distribution is 500 inhabitants 3 living by the lake and 300 inhabitants living in the forest. Alternatively, using a change of variable to solve the system, define new sequences u t, v t by setting x t = P u t, where u t = (u t, v t ) T. Then x t = Ax t P u t = AP u t u t = P AP u t = Du t. The system ( ut v t ) = ( ) ut v t has solutions ( ut v t ) = u0 v 0 (0.2) t. Since u 0 = P x 0, you have u 0 = (00, 300) T, so that 5 00 x t = P u t = 3 300(0.2) t to obtain the same solution as above. Question 4 Let T denote the linear transformation T : R 4 R 3 given by x x 2 T (x) = T = x 3 x 4 x + 2x 2 + 3x 4 5x + 9x 2 x 3 + 5x 4 3x + x 2 + 7x 3 4x 4 (a) (i) Write down a matrix A T such that T (x) = A T x. (ii) Find all vectors x R 4 such that T (x) = b. (iii) Write down a basis for the null space N(T ). (iv) Find the range R(T ).. Let b =

14 MT73 Algebra (b) Let S be a linear transformation, S : R 3 R 2 such that R(S) = R 2 and N(S) = {tb t R}, where b is the vector given above. Write down a matrix A S such that the linear transformation S(x) = A S x has these properties. (c) Consider the composition ST of the linear transformations S and T. (i) Show how you are able to deduce from the ranges of T and S that the range of ST is R 2. (ii) Use the rank-nullity (dimension) theorem for linear transformations to determine the dimension of the null space of ST. (iii) Find a basis for N(ST ). (Note that it is possible to answer part (c) without using the matrix in part (b).) Reading for this question Linear transformations are the subject of Chapter 9 of the subject guide. This question tests the material in Sections You also need a facility with material from the previous chapters, including Gaussian elimination and knowledge of subspaces, basis and dimension. Approaching the question (a) (i) If T is the given linear transformation, then T (x) = A T x, where A T = (ii) To find all vectors x R 4 such that T (x) = b, solve the system of equations A T x = b using Gaussian elimination. Put the augmented matrix into reduced row echelon form, (A T b) = Set the non-leading variable equal to an arbitrary parameter t. Then all vectors x such that T (x) = b are of the form x = x y z w = + 2t 0 t = t (iii) A basis of the null space of T, N(T ), is t = p + tv, t R.

15 Examiners commentaries 204 (iv) Since the reduced row echelon form of A T has a leading one in every row, you can conclude that R(T ) = CS(A T ) = R 3. That is, since rank(a T ) = 3, you know that R(T ) = R(A T ) is a 3-dimensional subspace of R 3, so R(T ) = R 3. (b) Let S be a linear transformation, S : R 3 R 2 such that: R(S), the range of S, is R 2 the null space of S is the subspace, N(S) = {x x = tb, t R}, where b is the vector given above. A S will be a 2 3 matrix. Since R(S) = R 2, you can take the first two columns of A S to be the standard basis vectors in R 2. Then the third column can be obtained using the basis of the null space, which is the vector b. You have A S 2 0 = 0 2c 0c 2 + c 3 = 0, so that c 3 = 2e + 0e 2. Then a matrix A S with the required properties is given by 0 2 A S =. 0 0 Many other solutions for A S are also possible. (c) Consider the composition ST of the linear transformations S and T. transformation from R 4 to R 2. Then ST is a linear (i) You showed in part (a)(iv) that the range of T is R 3. Since the range of T is the entire domain of S and since R(S) = R 2, you can conclude that the range of ST is R 2. (The essential fact you need to state here is that the range of T is equal to the domain of S, so that the range of ST is equal to the range of S.) (ii) The rank-nullity (dimension) theorem for linear transformations states that if T is a linear transformation T : V W, then rank(t ) + nullity(t ) = n, where n is the dimension of the domain V (and rank(t ) = dim(r(t )), nullity(t ) = dim(n(t ))). Applying this to ST : R 4 R 2, you have n = dim(r 4 ) = 4 and rank(st ) = dim(r 2 ) = 2. Then 2 + nullity(st ) = 4 = dim(n(st )) = 2. (iii) You know from part (ii) that a basis of N(ST ) consists of two vectors. One of these will be the basis vector for N(T ), since T maps this to 0 and S maps 0 0. But S also maps b 0, since b is a basis vector of N(S). From part (a), you know that one of the vectors which T maps to b is p = (, 0, 0, ) T. Therefore ST also maps p 0, and a basis of N(ST ) consists of the two linearly independent vectors 0 0 Alternatively, you could calculate the matrix A ST = 2, = ( and then row reduce it to find a basis of the null space. Begin by the row operation R 2 5R, to find A ST from which you can deduce the same basis as above. ), 5

16 MT73 Algebra Question 5 (a) Let A be an n n matrix. State precisely what is meant by the statement λ is an eigenvalue of A with corresponding eigenvector v. (b) Consider the matrix A and the vector v, where A = and v = (i) Using the definition, show that v is an eigenvector of A and find the corresponding eigenvalue. (ii) Show that λ = 3 is an eigenvalue of A and find a corresponding eigenvector. Verify that the definition is satisfied for this eigenvalue and eigenvector. (iii) The matrix A defines a linear transformation T : R 3 R 3 given by T (x) = Ax. It is known that T (x) = x for some non-zero vector x. Use this information to determine another eigenvalue and corresponding eigenvector of A. Verify your result. (iv) Write down a diagonal matrix D = (d ij ), with d > d 22 > d 33, and an invertible matrix P, such that P AP = D. (v) Find P. (vi) Find a matrix E such that E 2 = D. Hence find a matrix B such that B 2 = A. Reading for this question Eigenvalues and eigenvectors, and the diagonalisation of a matrix are covered in Chapter 0 of the subject guide. The diagonalisation of a 3 3 matrix is one of the main skills you should have acquired in this course. Throughout the question you are being tested on whether you know and can apply the precise definition of an eigenvalue and an eigenvector of a matrix. You should read the question carefully and answer each part as it appears in the question. In this way you should be able to find all the eigenvalues without having to solve the characteristic equation. You are also tested in part (b)(v) on your ability to find an inverse matrix. Part (iv) tests your ability to use the diagonalisation to solve a problem you may not have seen before. Approaching the question (a) If A is an n n matrix and λ is a real number such that Av = λv for some v 0, v R n, then λ is an eigenvalue of A and v is a corresponding eigenvector. 6 (b) (i) Calculating Av for the given matrix A and the vector v, Av = = 2 2 = 2v So v is an eigenvector and the corresponding eigenvalue is λ = 2.

17 Examiners commentaries 204 (ii) To show that λ = 3 is an eigenvalue of A, solve (A 3I)v = 0 to find a corresponding eigenvector. Use Gaussian elimination, putting the matrix (A 3I) into reduced row echelon form. (The fact that the reduced row echelon form has a row of zeros confirms that this system has a non-trivial solution, and therefore shows that λ = 3 is an eigenvalue of A.) (A 3I) = = x = 2t t = t t 2, t R. So you can take v 2 = (2,, ) T as a corresponding eigenvector. To verify that the definition is satisfied, Av 2 = = = 3v 2. (iii) The matrix A defines a linear transformation T : R 3 R 3 by T (x) = Ax. It is known that for some non-zero vector x, T (x) = x. This is just another way of saying that T (x) = Ax = x, so that λ 3 = is an eigenvalue of A. Solving (A I)x = 0, (A I) = = Let v 3 = To check, Av 3 = = = v (iv) If D = and P = 2, then P AP = D, and d > d 22 > d 33. (v) You can find P using cofactors, or using row operations. Using cofactors, first calculate that P =. Then P = = You should immediately check your answer by calculating that P P = I. (vi) You should be familiar with the properties of diagonal matrices, and since D is a diagonal matrix with positive entries, you should be able to deduce that a matrix E such that E 2 = D is E =

18 MT73 Algebra To find a matrix B such that B 2 = A, use the diagonalisation in part (iv) to write A = P DP. Then A = P DP = P E 2 P = (P EP )(P EP ) = B 2, where B = P EP. Calculating B, B = P EP = B = B = (To find B, you cannot treat matrices as real numbers and take the square root of both sides of a matrix equation such as B 2 = A, and it does not follow from Section. that you can merely substitute the number 2 for the n in the statement An = P D n P. In Section. n is a positive integer, and the proof of the result requires n to be a positive integer. The expression A n is defined here only for positive integers n.) 8

19 Examiners commentaries 204 Examiners commentaries 204 MT73 Algebra Important note This commentary reflects the examination and assessment arrangements for this course in the academic year The format and structure of the examination may change in future years, and any such changes will be publicised on the virtual learning environment (VLE). Information about the subject guide Unless otherwise stated, all cross-references will be to the latest version of the subject guide (203). You should always attempt to use the most recent edition of any Essential reading textbook, even if the commentary and/or online reading list and/or subject guide refers to an earlier edition. If different editions of Essential reading are listed, please check the VLE for reading supplements if none are available, please use the contents list and index of the new edition to find the relevant section. Comments on specific questions Zone B Question Consider the following system of equations x + 7x 3 + x 4 + 3x 5 = 5 x + 2x 2 + x 3 3x 4 + 5x 5 = 3 2x + 3x 2 + 5x 3 + 5x 4 + 9x 5 = 7. (a) Write down a matrix A and a vector b for which this system of linear equations can be expressed in matrix form as Ax = b. Find the reduced row echelon form of the augmented matrix (A b). Hence find the general solution of the system of linear equations Ax = b (expressing your solution in vector form). (b) Let c, c 2,..., c 5 be the column vectors of the matrix A in part (a). (i) Can c 5 be expressed as a linear combination of c and c 2? Justify your answer. Write down a linear combination if one exists. (ii) With reference to the reduced row echelon form of A, explain why the set B = {c, c 2, c 4 } is a basis for R 3. Write down [b] B, the coordinates of the vector b in the basis B. (iii) Define what it means to say that a set of vectors {v, v 2,..., v n } is linearly dependent. Show that the set of column vectors {c, c 2, c 3 } is linearly dependent. (iv) Explain why {c, c 2 } is a basis of the subspace S = Lin{c, c 2, c 3, c 5 } of R 3. Deduce that the subspace S is a plane in R 3, and find the Cartesian equation of this plane. 9

20 MT73 Algebra Reading for this question This question examines basic skills from several parts of the subject guide. Part (a) tests your ability to solve systems of linear equations using Gaussian elimination, which is covered in Chapter 3. Then part (b) tests your ability to interpret information from the reduced row echelon form of the coefficient matrix. To answer the questions concerning basis and linear independence requires your knowledge of the basic material of Chapters 7 and 8, in particular Sections Approaching the question (a) In matrix form the system of equations is Ax = b, where A = x x , x = x x 4 x 5 b = To find the general solution of the system of linear equations Ax = b, put the augmented matrix into reduced row echelon form. (A b) = R2 R R R R2 3 R R R3 R2 R R3 R 2+2R Set the non-leading variables equal to arbitrary parameters; say, x 3 = s and x 5 = t. Then the general solution is: x 4 7s 3t x s t 2 3 x = x 3 = s = 0 + s + t 0 = p + sv + tv 2, s, t R. x x 5 t 0 0 You can easily check your solution by calculating Ap = b, Av = 0, Av 2 = 0.. (b) (i) There are several ways to do this question. You can just solve, = a 2 + b Equating the components, the first line immediately gives a = 3, and then the second line gives b =. Then you need to check that the third equation is also satisfied, 9 = 3(2) + 3(). So c 5 = 3c + c 2. Alternatively, the quickest way is to notice the relationship between c, c 2, c 5 as shown by looking at the corresponding columns of the RREF of A. If you denote these as ĉ, ĉ 2, ĉ 5, then it is clear that ĉ 5 = 3ĉ + ĉ 2, and the same relationship applies to the columns of A. 20

21 Examiners commentaries 204 Alternatively, you can see this relationship by using the null space vector v 2 = ( 3,, 0, 0, ) T and the basic fact that if A is an m n matrix and x = (x, x 2,..., x n ) T, then Ax = x c + x 2 c x n c n, where c, c 2,..., c n are the columns of A. This yields the equality Av 2 = 3c c 2 + 0c 3 + 0c 4 + c 5 = 0 from which you can again conclude that c 5 = 3c + c 2. (ii) The column vectors c, c 2, c 4 correspond to the columns with leading ones in the RREF of A, which indicates that they are linearly independent. Since R 3 has dimension three, three linearly independent vectors in R 3 are a basis. Using the dimension of R 3 to conclude that the three linearly independent vectors are a basis is an important part of this argument. Alternatively, you could deduce from the RREF of A that the reduced row echelon form of a matrix consisting of the three column vectors, c, c 2, c 4 would be the identity matrix, and therefore {c, c 2, c 4 } is a basis of R 3. From the solution, since Ap = b, you can write b = 4c 2c 2 + c 4, using the same basic fact as above. Therefore the coordinate matrix of b in this basis is [b] B = 4 2 Think about what it is you needed to know about a basis of R 3 in order to answer this part of the question. The material is all contained in Chapter 8, in particular Section 8.3. (iii) The set {v, v 2,..., v n } is linearly dependent if and only if the vector equation a v + a 2 v a n v n = 0 has a non-trivial solution; that is, a solution with not all coefficients a i = 0. While it is correct to say that the vectors are linearly dependent if and only if some vector in the set is a linear combination of the others, this is a consequence of the definition. Looking at the solution of Ax = b, the vector v = ( 7, 3,, 0, 0) T is a solution of the homogeneous system. Since Av = 0, you can deduce that B 7c + 3c 2 + c 3 = 0. This is a non-trivial linear combination showing that the vectors are linearly dependent. Alternatively, you could show that the determinant of the matrix whose columns are c, c 2, c 3 is equal to zero, and conclude that the equation a c + a 2 c 2 + a 3 c 3 = 0 has a non-trivial solution. (iv) The subspace Lin{c, c 2 } contains the vector c 5 by (i) since c 5 = 3c + c 2 is a linear combination of c, c 2. By (iii), c 3 = 7c 3c 2, so c 3 is also a linear combination of c, c 2. Therefore, S = Lin{c, c 2, c 3, c 4 } = Lin{c, c 2 }. Since the vectors c, c 2 are linearly independent (neither is a scalar multiple of the other, or this follows from (ii)), the set {c, c 2 } is a basis of S. You do need to state why the set {c, c 2 } spans S and why it is linearly independent. Then you can conclude it is a basis of S. The subspace S has a basis consisting of two vectors in R 3, and a two-dimensional subspace of R 3 is a plane through the origin. To find a Cartesian equation, you can evaluate the determinant (which is most easily expanded using column three). 0 x 2 y 2 3 z. = x 3y + 2z = 0. 2

22 MT73 Algebra You should understand why and how this works, as shown in Section 8.3 of the subject guide. Alternatively, you can use the parametric equation x = sc + tc 2, s, t R and eliminate s and t from the three resulting component equations to obtain a Cartesian equation in x, y, z. The plane has equation x + 3y 2z = 0. Check that this is correct; the vectors c, c 2, c 3, c 5 should all satisfy this equation. Question 2 (a) Let A be an m n matrix. Define what is meant by the null space of A, N(A). Prove that N(A) is a subspace of R n. (b) Let c, c 2, c 3 denote the columns of the matrix (i) Find rank(a). (ii) Find a basis for N(A). A = (iii) Find a basis for the range of A, R(A) (iv) Let d = c + c 2 + c 3. Before attempting to solve the system, explain why you know that Ax = d is consistent.. Write down a general solution of the system of equations Ax = d. (c) Let A be an m n matrix and let B be an n p matrix. Show that N(B) N(AB). State the rank-nullity (dimension) theorem for matrices, carefully defining each term. Use it to deduce that rank(ab) rank(b). Reading for this question This question, which is mainly about the null space of A is testing material from Chapters 3, 5, 7 and 8 of the subject guide. The null space, N(A), is first introduced in Section In Section 7.3. you will find the proof that N(A) is a subspace of R n. For the remainder of the question you need to know about basis and dimension, Chapter 8. In particular, the rank nullity theorem is Section 8.5. Part (b)(iv) is testing that you know that a system of equations Ax = b is consistent if and only if b is a linear combination of the columns of A, as shown in Section 5.4 and later in Section where it is stated that the range of A, R(A), is equivalent to the column space of A, CS(A). It then tests the basic fact from Section 5.3 that the general solution of a system of linear equations is composed of a particular solution plus the general solution of the homogeneous system. Part (c) requires knowledge of how to show a subspace is contained in another subspace. An example of this is shown in Section

23 Examiners commentaries 204 Approaching the question (a) Let A be an m n matrix. The null space of A, N(A) is the set N(A) = {x Ax = 0}. Since A is m n, x R n, so this is a subset of R n. To show it is a subspace, first note that it is non-empty. Since A0 = 0, 0 N(A) so it is non-empty. Next you need to show that it is closed under addition and scalar multiplication. For addition, let x, y N(A). Then Ax = 0 and Ay = 0, so that A(x + y) = Ax + Ay = = 0 = x + y N(A). For scalar multiplication, let x N(A) and α R. Then This proves that N(A) is a subspace of R n. A(αx) = α(ax) = α0 = 0 = αx N(A). (b) Let c, c 2, c 3 denote the columns of the matrix A. To answer the question, put the matrix A into reduced row echelon form A = (i) rank(a) = 2. (ii) To find a basis of N(A), find the solutions of Ax = 0. If x = (x, y, z) T, set the non-leading variable z = t, where t R is an arbitrary parameter. Then a general solution is x = x y = 9t 2t = t 9 2 = tv, t R. z t Therefore a basis of N(A) is {v}, where v = (9, 2, ) T. (iii) A basis of the range of A, R(A), is given by the columns of A which correspond to the leading ones in the reduced row echelon form of A, so a basis of R(A) is, (iv) If d = c + c 2 + c 3, then Ax = d is consistent because d is a linear combination of the columns of A; that is, d is in the column space of A, CS(A), and CS(A) = R(A). The range of A consists of all vectors b for which Ax = b is consistent. To write down a general solution of Ax = d, use the fact that a general solution of a system of linear equations is the sum of a particular solution and the general solution of Ax = 0, which you have already found in part (ii). You can find a particular solution using the fact that for any m n matrix A and vector x = (x, x 2,..., x n ) T, Ax = x c + x 2 c x n c n, where c, c 2,..., c n are the columns of A. Since d = c + c 2 + c 3, you can deduce that a particular solution p such that Ap = d is p = (,, ) T. Then a general solution of the system of equations Ax = d is x = p + tv = + t 9 2, t R. 23

24 MT73 Algebra Of course, you could also find a solution by finding d = (, 8, 3, 3) T and solving the system Ax = d by Gaussian elimination, which would take considerably more time. (c) Let A be an m n matrix and let B be an n p matrix. Then AB is an m p matrix. To show that N(B) N(AB), you need to show that any x N(B) is also contained in N(AB). (Both null spaces are contained in R p.) Let x N(B). Then Bx = 0, so that (AB)x = A(Bx) = A0 = 0 = x N(AB). Since any vector in N(B) is also in N(AB), this shows that N(B) N(AB). The rank-nullity (dimension) theorem for an m n matrix A, states that rank(a) + nullity(a) = n, where: n is the number of columns of A, rank(a) is the dimension of R(A), and nullity(a) is the dimension of N(A). Here notation is important. R(A) and N(A) denote the subspaces, not their dimensions. The rank-nullity theorem is about the dimensions of the subspaces. So you could also state the theorem as dim(r(a)) + dim(n(a)) = n. To deduce that rank(ab) rank(b), you need to use the fact that both B and AB are matrices with p columns, so that rank(ab) + nullity(ab) = p = rank(b) + nullity(b). Since N(B) N(AB), you can conclude that dim(n(b)) dim(n(ab)). Therefore, rank(ab) + nullity(ab) = rank(b) + nullity(b) rank(b) + nullity(ab), so that rank(ab) rank(b). Question 3 (a) A student takes out a loan of $5000. The monthly interest rate on this loan is 0.5% and interest is added monthly. The student also has to repay the loan by paying $P per month. If the loan is reduced monthly with her repayments and y t is the loan balance outstanding at the end of month t, explain why y t satisfies the difference equation y t+ =.005y t P. Solve the difference equation for y t in terms of P. Use your answer to find an expression for the amount P that the student needs to repay each month if the loan must be repaid at the end of five years. (b) In any year t, the fixed total population of 200 inhabitants of Island is divided into those living by the lake, and those living in the forest. Initially 600 inhabitants live by the lake. If x t denotes the number of inhabitants living by the lake in year t, and y t denotes the number living in the forest in year t, then the yearly population movements are given by the following system of difference equations x t = 0.5 x t y t y t = 0.5 x t y t. (i) Each year, what percentage of the population living by the lake moves to the forest? 24