Vector spacesMJ038 The linear transformation T : 4 4 is represented by the matrix A, where

A = 1 1 2 32 1 1 113 2 3 144 3 5 17

.Find the rank of A and a basis for the null space of T. [7]

The vector

1211 is denoted by e. Show that there is a solution of the equation Ax = Ae of the form

x = pq11

, where p and q are to be found. [4]At the outset, almost all responses showed the accurate reduction of A to the echelon form and the correctdetermination of its rank. Beyond this initial stage, responses divided into two categories in the first of which

the echelon form was used as a means of determining a basis for the null space. Thus the better

candidates, the minority, worked directly from equations x y 2z + 3t = 0, y + 3z + 5t = 0 and so

established expeditiously a possible form for . The less inspired, and much more error prone, method

which was adopted by the majority, was to start with Ax = 0 with the given form of A without reference to theechelon form previously obtained. This led to two independent linear equations in both of which there wereno zero coefficients and many candidates were then unable to work accurately from this situation to a correct

possible form of .

For the final part, less than a quarter of the candidature worked from

1

0

5

8

0

1

3

1

1

1

2

1

1

1

q

p

,

though those who did so usually obtained the required result.

In contrast, the majority ignored the version of obtained and so loaded themselves with a much more

formidable task. They first evaluated Ae as (2, 6, 4, 2)T and then went on to solve, in some way, the

system Ax = (2, 6, 4, 2)T with x

T preset to (p, q, 1, 1). The success rate for this strategy was less than

50%.

Answers: Rank of A = 2,

A basis for the null space is

1

0

5

8

,

0

1

3

1

,

p = 17, q = 18.

Question 8

3 Three n 1 column vectors are denoted by x1, x2, x3, and M is an n n matrix. Show that if x1, x2, x3are linearly dependent then the vectors Mx1, Mx2, Mx3 are also linearly dependent. [2]The vectors y1, y2, y3 and the matrix P are dened as follows:

y1 = (157) , y2 = (

234) , y3 = (

55155) ,

P = ( 1 4 30 2 50 0 7) .

(i) Show that y1, y2, y3 are linearly dependent. [2]

(ii) Find a basis for the linear space spanned by the vectors Py1, Py

2, Py

3. [2]

ON03

Question 3 In general, this question was not well answered. Some candidates appeared not to know the difference between linear dependence and linear independence. Moreover, many failed to see the relevance of the first result to what was required in part (ii). To begin with, there were a number of different strategies in evidence. The first was an attempt to produce

an argument such as == 0Mx0x iiii , so that xi linearly dependent not all i are zero Mxi are linearly dependent. However, many candidate responses along these lines were defective in that there was no mention of the

requirement that not all i are zero.

Other candidates argued that as xi linearly dependent det(x1 x2 x3) = 0, then det(Mx1 Mx2 Mx3) =

det[(M(x1 x2 x3)] = det(M) det(x1 x2 x3) = det M.0 = 0, so that Mx1, Mx2, Mx3 are linearly dependent. A further strategy, which was attempted by a few, was to argue that x1, x2, x3 linearly dependent e 0

such that (x1 x2 x3) e = 0 M(x1 x2 x3)e = M.0 = 0 (Mx1 Mx2 Mx3)e = 0 Mx1, Mx2, Mx3 are linearly

dependent. Such arguments, however, were seldom complete so that as a general conclusion it is clear that most candidates had only a limited understanding of the syllabus material appertaining to this question. (i) The majority of valid responses showed a linear dependence such as 9y1 2y2 y3 = 0. An

alternative strategy is to show that det(y1 y2 y3) = 0 and most working was both accurate and complete in this respect. Likewise the correct reduction of the matrix (y1 y2 y3) to the echelon form so as to establish its rank, followed by a properly argued conclusion, featured in some responses.

(ii) By this stage of the question, it should have been clear to candidates that the basis of the specified

linear space must consist of exactly two linearly independent vectors and could therefore be given immediately as {Py1, Py2} in numerical form. However, many responses showed the evaluation of {Py1, Py2, Py3}.

Answer: (ii)

14

7

13

,

49

45

2

.

10 The linear transformation T : 4 4 is defined by

T :

x

yt

Ax

yt

,where

A = 3 1 3 25 0 7 76 2 6 + 29 3 9

.

(i) Show that when 6, the dimension of the null space K of T is 1, and that when = 6, thedimension of K is 2. [4]

(ii) For the case 6, determine a basis vector e1 for K of the form x1y110

, where x1, y1, 1 areintegers. [2]

(iii) For the case = 6, determine a vector e2 of the form x2y20t2

, where x2, y2, t2 are integers, suchthat {e1, e2} is a basis of K. [3]

(iv) Given that = 6, b = 55

1015

, e0 = 1111

, show that x = e0 + k1e1 + k2e2 is a solution of theequation Ax = b for all real values of k1 and k2. [3]

MJ04

Question 10

This question highlighted much erroneous thinking with regard to the fundamentals of linear spaces and systems of linear equations.

(i) Many responses lacked clarity and frequently the impression was given that the candidate did not understand what was required. In fact, a, simple way to proceed is to obtain an echelon form, such

as

+

0000

6000

11650

5412

. It is then a simple matter to deal with the cases 6 and 6= . In

this respect what was required were arguments such as ( ) 134r4)(dim6 === AK and ( ) 224r4)(dim6 ==== AK . However, few candidates supplied such detail.

(ii) For the most part, candidates working was accurate.

(iii) A standard procedure here is to express x, y, z, t

x = 7 + 7, y = 6 11, z = 5, t = 5.

This not only provides a check of the result for e1 obtained in part (i), but also enables a result for e2 to be written down.

(iv) In general terms, it was expected that the candidate would verify clearly that bA =

=

15

10

5

5

1

1

1

1

, and

then set out working such as, Ax1 = Aeo + k1 Ae1 + k2Ae2 = b + k10 + k20 = b, for all k1, k2.

Some responses appeared to proceed along these lines though, it must be said, many such arguments were incomplete in some important way.

Answers: (ii) e1 =

0

5

6

7

; (iii) e2 =

5

0

11

7

.

1 The linear transformation T : 4 4 is represented by the matrix

1 5 2 62 0 1 73 1 2 104 10 13 29

.Find the dimension of the null space of T. [4]

ON04

Question 1 Most candidates produced a complete and correct response to this question. The most popular strategy was to reduce the given matrix, A to the row-echelon form and then to use the dimension theorem to argue that if K is the null space of T, then dim(K) = 4 r(A) = 1. However, some candidates, in defiance of the rubric, obtained the echelon form of A directly from a graphic calculator and so did not gain full credit. Yet others did not obtain an echelon form, but instead stopped the reduction process immediately a row of zeros had been obtained. This too led to loss of marks.

A popular alternative strategy was to show from the echelon form of A that

1

1

0

4

spans K, and hence that

dim(K) = 1. Actually, this vector can easily be obtained by working with linear equations based on the original form of A. However, among the few candidates who employed this strategy, there were some who did not complete their working by showing that within a multipicative constant no other non-zero vector satisfies Ax = 0. Answer: The dimension of the null space of T is 1.

11 The matrix A is defined by

A = ( 1 3 21 1 12 2

) .Find the rank of A, distinguishing between the cases 1 and = 1. [4]Consider the system S of equations:

x + 3y + 2 = 1,x y = 0,

2x + 2y + = 3 + 2.(i) Show that if 1 then S has a unique solution. Find this solution in the case = 0. [3]

(ii) Show that if = 1 and = 0 then S has an infinite number of solutions. [3](iii) Show that if = 1 and 0 then S has no solution. [2]

Question 11 Not all candidates attempted to answer the four parts of this question in a systematic way and so good responses were very much in a minority. In fact, there was a lot of poorly presented work in evidence together with a number of elementary errors. Because of the sequential nature of the reduction of a matrix to the echelon form, it is essential that the working be checked at each stage, especially as conclusions appertaining to the system of equations depend in an obvious way on the final form of the reduction.

About half of all candidates reduced the matrix A to

100

340

231

and so could determine r(A), the rank of

A, immediately, both for 1 and for the special case 1= . The rest arrived at a correct conclusion for at

least one of the above cases, even though there were errors in the working. For the first part of part (i), it is only necessary to point out that the system has an unique solution if r(A) = 3

which was proved previously to be the case for any 1 . However, most candidates ignored their previous

working and started again, or began with the unnecessary evaluation of det A.

Reduction of the augmented matrix to the form (R):

+

33100

1340

1231

together with setting 0=

will lead immediately to the required solution. Nevertheless, only a minority of candidates systematised their working in this way. Both parts (ii) and (iii) can be answered expeditiously using R, but again most candidates ignored their

earlier working, even if relevant to the problems here. A common error was to argue that [det(A) 0

system has a unique solution] [det A = 0 system has an infinite number of solutions.]

Answers: rank of A = 3 when 1 , rank of A = 2 when 1= ; (i) x = 1, y = 2, z = 3.

MJ05

11 Find the rank of the matrix A, where

A = 1 1 2 34 3 5 166 6 13 13

14 12 23 45

. [3]

Find vectors x0 and e such that any solution of the equation

Ax = 0213

(*)can be expressed in the form x0 + e, where . [5]Hence show that there is no vector which satisfies (*) and has all its elements positive. [3]

Question 11

Most candidates began by reducing A to an echelon form such as

=

0000

5100

4310

3211

E from which they

deduced correctly the rank of A. The better candidates then deduced that the dimension of the null space is 1 and hence, using an equation of the form Ex = 0, easily obtained a result for the vector e. They also obtained a correct result for x0 in a systematic and clearly intelligible way from the given equation (*). Candidates who did less well with this question generally had more difficulty in obtaining e than x0. Their method did not involve a separate consideration of Ex = 0 but instead, they generally worked with the 4 linear equations represented by (*) in a haphazard way. The final part of this question showed up a general deficiency in the understanding of inequality arguments, as was the case in Question 5 (iii). The majority of candidates started from the general solution

+

=

51

111

21

x , or, at least, something like it. Many subsequent arguments were either incomplete or

erroneous. Thus some considered only 0> and 0 , 051 >+ but still were unable to complete a convincing argument. In fact,

these 2 inequalities taken together imply that 11

1 which is clearly impossible.

Answers: 3;

=

0

1

1

1

0x ,

=

1

5

11

2

e

ON05

4 The linear transformation T : 4 4 is represented by the matrix

A = 1 1 2 32 3 4 55 6 10 144 5 8 11

.Show that the dimension of the range space of T is 2. [3]

Let M be a given 4 4 matrix and let S be the vector space consisting of vectors of the form MAx,where x 4. Show that if M is non-singular then the dimension of S is 2. [4]

Question 4 Responses to the first part of this question were generally complete and correct, but in contrast very few candidates made any significant progress with the remainder. Candidates generally established the echelon form

1 1 2 30 1 0 10 0 0 00 0 0 0

for the matrix A and hence obtained the dimension of RT, the range space of T. Some, however, unnecessarily worked with the transpose of A and so increased the risk of introducing errors into the working. For the last part of this question, suppose that b1, b2 are basis vectors of RT and consider the linear form L = Mb1 + Mb2. Since M is given to be non-singular, then M1L exists and equals b1 + b2 which, since dim(RT) = 2, cannot be the zero vector unless and are both zero. Thus Mb1 and Mb2 are linearly independent and so dim(S) = 2.

MJ06

5 Show that if a 3 then the system of equations2x + 3y + 4 = 5,4x + 5y = 5a + 15,6x + 8y + a = b 2a + 21,

has a unique solution. [3]

Given that a = 3, find the value of b for which the equations are consistent. [3] Question 5 Most candidates had a clear idea of what was expected of them. The most popular strategy was to attempt to reduce the matrix

+

+ a

a b a

2 3 4 54 5 1 5 156 8 2 21

to an echelon form such as

+ a

a b a

2 3 4 50 1 9 5 25 .0 0 3 7 11

From here both parts of the question can be answered immediately. However, there were many arithmetic errors in the working so that complete success was achieved only by a minority. Those who worked with equations did less well mainly because their algebra was badly organised. The syllabus does not demand a knowledge of the concept of the echelon form but nevertheless, it is clear that its application to problems of this type is more likely to lead to success than the undisciplined implementation of Guassian elimination. Answer: 10.

ON06

OR

The linear transformation T : 4 4 is represented by the matrix

M = 1 2 2 42 4 5 93 6 8 145 10 12 22

.

(i) Find the rank of M. [3]

(ii) Obtain a basis for the null space, K, of T. [3]

(iii) Evaluate

M1

234

,and hence show that any solution of

Mx = 5

111727

()has the form

1

234

+ e1 + e2,where and are constants and {e1, e2} is a basis for K. [3]

(iv) Find the solution x1 of () such that the first component of x1 is A, and the sum of all thecomponents of x1 is B. [5]

MJ07

Question 12 OR Most candidates attempting this alternative were able to reduce the matrix to echelon form and deduce that its rank was 2. Then, using a system of equations from the reduced matrix, they obtained, mostly successfully, the basis for the null space. Almost every candidate getting this far was able to evaluate the matrix product correctly. The next step, however, caused problems, because many candidates did not appreciate the distinction between showing that the given form was a solution of the equation and what they were asked to show, which was that any solution of the equation had the given form. There were very few good attempts at the last part of the question. This required setting up two equations, one for A and one for B, both in terms of and . The equations then had to be solved for and . Thus the solution could be given in terms of A and B.

Answers : (i) 2; (ii)

2 20 1

,1 01 0

; (iii)

2717115

; (iv)

+ +

11 32 21 32 2

AB A

AB

A B

.

10 The vectors b1, b2, b3, b4 are defined as follows:

b1 = 1000

, b2 = 1100

, b3 = 1110

, b4 = 1111

.The linear space spanned by b1, b2, b3 is denoted by V1 and the linear space spanned by b1, b2, b4 isdenoted by V2.

(i) Give a reason why V1 V2 is not a linear space. [1](ii) State the dimension of the linear space V1 V2 and write down a basis. [2]

Consider now the set V3 of all vectors of the form qb2 + rb3 + sb4, where q, r, s are real numbers.Show that V3 is a linear space, and show also that it has dimension 3. [3]

Determine whether each of the vectors

4425

and 5425

belongs to V3 and justify your conclusions. [4]

Question 10 This question was the least well-done question on the paper. A comment, with some evidence, to the effect that 21 VV was not closed under addition was required for part (i). In part (ii) identifying (b1, b2) as a basis gave the dimension of 21 VV as 2. Showing V3 is closed under addition was sufficient for the next mark. Identifying (b2, b3, b4) as a basis and showing the dimension of V3 to be 3 gained the next two marks. The final part could be tackled as follows.

3

5244

1111

5

0111

3

0011

2

5244

V

+

=

+

=

5244

5245

0001

and 3

0001

V

3

5245

V

or yxV

tzyx

=

3 not true for

5245

.

Answers: (i) Not closed under addition; (ii) 2, (b2,b3).

ON07

OR

The linear transformation T : 4 4 is represented by the matrix

1 2 1 11 3 1 01 0 3 10 3 4 1

.The range space of T is denoted by V .

(i) Determine the dimension of V . [3]

(ii) Show that the vectors

1110

, 2303

, 11

34 are linearly independent. [4]

(iii) Write down a basis of V . [1]

The set of elements of 4 which do not belong to V is denoted by W.(iv) State, with a reason, whether W is a vector space. [1]

(v) Show that if the vector x

yt

belongs to W then y t 0. [5]Question 12 OR Candidates choosing this alternative generally performed less well than those candidates who chose the other alternative. In many cases little was done beyond the first two parts. In part (i) candidates were well used to reducing the matrix to echelon form, but some made arithmetical errors or stopped well short and lost a mark in doing so. The mark for the dimension of V could be obtained nevertheless. In part (ii), the most popular approach was to show that if a linear combination of the vectors gave the zero vector then the only solution was the trivial one. A good number, however, used row reduction methods with matrices. A number of candidates did not appreciate the demands of the phrase show that and thought that a statement of what linear dependence meant was sufficient. Part (iii) following on from part (ii) should have been obvious, but some ignored part (ii). In part (iv), W was not a vector space and the absence of the zero vector or non-closure under addition were sufficient reasons. In part (v), again, show that was misinterpreted by some of those candidates who made an attempt at this part. The best approach was probably to use row reduction

on an augmented matrix to show that 0=

tzyV

tzyx

hence 0

tzyW

tzyx

. Other

approaches, via a system of equations, were equally acceptable.

Answers: (i) 3; (iii)

4311

,

3032

,

0111

; (iv) No, no zero vector or other valid reason.

MJ08

12 OR

(i) Transform given matrix to an echelon form, e.g.,

0000

1100

1010

1121

.

M1A1

3)( =Vdim OR BY EQUIVALENT METHOD If written down with no working 1/3 B1

(ii) 0,

4

3

1

1

3

0

3

2

0

1

1

1

321321====

+

+

0 shown linear independence M2A2

(iii)

4

3

1

1

,

3

0

3

2

,

0

1

1

1

B1

(iv) W not a vector space since W does not contain the zero vector, or equivalent

or not closed wrt addition B1

(v) Reduces

++

++

tzy

zyx

xy

x

t

z

y

x

000

23400

010

121

430

301

131

121

M1A1

0 iff =

tzyV

t

z

y

x

b or equivalent method M1A1

0 tzyWb A1

Alternative: suppose

+

+

=

4

3

1

1

3

0

3

2

0

1

1

1

t

z

y

x

(i.e. in V)

0...

...

...

...

...

==

=

=

=

=

tzy

t

z

y

x

Hence 0

tzyW

t

z

y

x

6 The matrix A is defined by

A = 1 1 2 32 1 7 23 3 6 7 6 17 17

.

(i) Show that if = 9 then the rank of A is 2, and find a basis for the null space of A in this case.[5]

(ii) Find the rank of A when 9. [2] Question 6 The vast majority of candidates knew that it was necessary to reduce the matrix A to echelon form and invariably this was done correctly. Those who worked with rather than 9 were able to use their reduced matrix in the final part of the question. Finding the basis for the null space of A caused problems for some. The usual method was to find a two-parameter solution for the two equations and separate the terms. There were six essentially different acceptable results, each of which could have opposite signs, or constant multipliers.

Answers: (i)

1041

,

0135

or

0135

340

17

, or

1041

,

340

17

or

51

170

,

0135

or

51

170

,

1041

or

51

170

,

340

17

;

(ii) 3.

6 (i) Reduction of A to echelon form, e.g.,

0000

9000

4310

3211

M1A1

= 9 last 2 rows consist entirely of zeros ( ) 2= Ar A1

A basis for the null space of A is ,

1

0

4

1

,

0

1

3

5

or equivalent M1A1

(ii) 09 M1

( ) 3=Ar A1

ON08

ORThe linear transformations T1 : >4 >4 and T2 : >4 >4 are represented by the matrices M1 andM2, respectively, where

M1 =

1 1 1 21 4 7 81 7 11 131 2 5 5

, M2 = 2 0 1 15 1 3 33 1 1 1

13 1 6 6 .

(i) Find a basis for R1, the range space of T1. [4]

(ii) Find a basis for K2, the null space of T2, and hence show that K2 is a subspace of R1. [5]

The set of vectors which belong to R1 but do not belong to K2 is denoted by W .

(iii) State whether W is a vector space, justifying your answer. [1]

The linear transformation T3 : >4 >4 is the result of applying T1 and then T2, in that order.(iv) Find the dimension of the null space of T3. [3]Question 12 ORThis was much the more popular of the two alternatives, but those candidates trying it found only limitedsuccess. For part (i), most of those attempting the question used row or column operations to reduce thematrix to echelon form. Some stopped short of this stage and lost one or both marks. Those who got acorrect echelon form mostly found a correct basis for R1. Similarly in part (ii), row operations were used onthe matrix, or its transpose, to obtain an echelon form. From this the basis of K2 was deduced, usually froma set of equations. Those who operated on the transpose frequently wrote results of the operations on the

vector

dcba

beside the matrix. This resulted in

++dc

cbaba

2.

The basis could then be written down from the 3rd and 4th elements, which corresponded to the zero rows inthe matrix. Few candidates obtained the fifth mark by showing that each basis vector was a linearcombination of the basis vectors of R1. In part (iii), the most straightforward justification for W not being avector space was to say that it did not contain the zero vector. Only a small number said this. It was notuncommon for statements to appear which could not be substantiated. Only a handful of candidates got

anywhere with part (iv). One way of tackling the problem was to find M2M1 =

909045020201003636180141470

, hence

nullity = 4 r(M2M1) = 4 1 = 3. Alternatively, for any vector x, M2M1x = M2(b1 + b2 + b3), where b1, b2,b3 are any 3 linearly independent basis vectors of R1, 2 of which must be basis vectors of K2. Hence if b1and b2 are basis vectors of K2, then the dimension of the null space of T3 = 4 1 = 3.

Answers: (i) e.g.

1100

,

1630

,

1111

or

1-100

,

7030

,

1003

; (ii) e.g.

0211

,

2011

or

1-100

,

1111

; (iv) 3.

MJ09

12 OR

(i) Reduces M1 to col echelon form by elementary col operations

or T1

M to row echelon form by elementary row operations

0000

1100

7030

1003

1

TM hence

1

1

0

0

,

7

0

3

0

,

1

0

0

3

M1

Any correct echelon form A1

Selects 3 li cols from reduced M1 (or equiv from reduced T

1M ) M1

(Allow M1A1M1 for any other valid method)

Any correct basis of R1, e.g.

1

1

0

0

,

1

6

3

0

,

1

1

1

1

or

1

1

0

0

,

7

0

3

0

,

1

0

0

3

A1

or for (i)

Reduces M1 to row echelon form by elementary row operations

0000

1200

2210

2111

M1

Any correct row echelon form A1

No working for echelon matrix here, or in (ii) gets B1

rr(M1) = 3 any 3 li columns form a basis of M1 May be implied by answer M1

Any correct basis of R1, e.g.,

5

11

7

1

,

2

7

4

1

,

1

1

1

1

A1

or

detM1 = 0 (calculator) M1

31 Mr A1

First 3 columns of M1 are clearly linearly independent M1

Hence basis A1

(ii) Reduces M2 to echelon form by elementary row operations

0000

0000

1120

1102

M1

Any correct row echelon form A1

Valid method to find basis of K2 M1

(Allow M1A1M1 for any other valid method)

Any correct basis of K2, e.g.,

0

2

1

1

,

2

0

1

1

or

1

1

0

0

,

1

1

1

1

A1

Shows each basis element of K2 is in R1 (AG) B1

First 4 marks

dc

2cba

b

a

0000

0000

1110

13352

d

c

b

a

6131

6131

1110

13352

2

++

=TM M1A1

Basis is

1

1

0

0

,

0

2

1

1

M1A1

(iii) Any valid argument, e.g., W does not contain zero vector so W not a vector space B1

(iv) For any vector x, M2M1x = M2( b1 + b2 + b3), where b1, b2, b3 are any 3 l.i. basis

vectors of R1, 2 of which must be basis vectors of K2 M1

Hence if b1 and b2 are basis vectors of K2, then M2M1x = M2b3 A1

Hence as dim(range of T3) = 1, then the dimension of the null space of T3 = 4 1 = 3 A1

or M2M1 =

9090450

2020100

3636180

141470

B1

Nullity = 4 r(M2M1) = 3 M1A1