math1231/1241 algebra s2 2007 test 1 v1b...math1231/1241 algebra s2 2007 test 1 v1b full solutions...

Draft

MATH1231/1241 Algebra S2 2007 Test 1

v1BFull Solutions

September 4, 2018

These solutions were written by Treves Li, typed up by Brendan Trinh and edited by Henderson

Koh and Aaron Hassan. Please be ethical with this resource. It is for the use of MathSoc

members, so do not repost it on other forums or groups without asking for permission. If you

appreciate this resource, please consider supporting us by coming to our events and buying our

T-shirts! Also, happy studying !

We cannot guarantee that our working is error-free, or that it would obtain full marks – please

notify us of any errors or typos at [email protected], or on our Facebook page. There

are sometimes multiple methods of solving the same question. Remember that in the real class

test, you will be expected to explain your steps and working out.

Please note quiz papers that are NOT in your course pack will not necessarily reflect the style

or di�culty of questions in your quiz.

1. (i) Rewrite the set as S =�x 2 R4|x1 � 5x3 � 2x4 = 0

.

• Clearly, 0 2 S, since 0� 5 (0)� 2 (0) = 0.

• Now, suppose that x,y 2 S, where x = (x1, x2, x3, x4)T and�!y = (y1, y2, y3, y4)

T .

Then we have the two conditions,

x1 � 5x3 � 2x4 = 0 and y1 � 5y3 � 2y4 = 0. (?)

1

Draft

Now, for x+ y to lie in set S, the following condition must be true:

(x1 + y1)� 5 (x3 + y3)� 2 (x4 + y4) = 0.

Considering the LHS,

(x1 + y1)� 5 (x3 + y3)� 2 (x4 + y4) = (x1 � 5x3 � 2x4) + (y1 � 5y3 � 2y4)

= 0 + 0 (using the conditions (?))

= 0.

Hence, x+ y 2 S, and so set S is closed under addition.

• Suppose that � 2 R. Now, for �x to lie in set S, the following condition must

be true,

(�x1)� 5 (�x3)� 2 (�x4) = 0. (†)

Considering the LHS,

(�x1)� 5 (�x3)� 2 (�x4) = � (x1 � 5x3 � 2x4)

= � (0) (†) (using the condition)

= 0.

Hence, �x 2 S, and so set S is closed under scalar multiplication.

Thus, by the Subspace Theorem, S is a subspace of R4.

(THOUGHT PROCESS: To prove that S is a subspace of a vector space V , we need

to show three things: non-empty (or equivalently, whether 0 is an element), closure

under addition, closure under scalar multiplication. Review the Subspace Theorem

on your course notes if you are not familiar with it. As always, make sure you name

the theorem you use in a test or an exam to get full marks.

For more information on this content, please refer to course notes pp.13-16.

For more practice on this type of question, go to course notes p.66 Problems 6.3

(Q12b, 14, 16, 18, 22, 24a in particular).)

(ii) There are an infinite possible number of choices. For example, we can pick a simple

one by choosing x2 = 0, x3 = 0, x4 = 1 such that x1 = 2. Thus, one non-zero element

in S is

0

BBBB@

2

0

0

1

1

CCCCA.

2. For these polynomials to span P2, every polynomial in P2 in the form b0+ b1t+ b2t2 where

b0, b1, b2 2 R must be expressible as a linear combination of p1(t), p2(t) and p3(t). Let

2

Draft

↵,�, �, � 2 R such that

↵p1(t) + �p2(t) + �p3(t) + �p4(t) = b0 + b1t+ b2t2.

Substituting in the polynomials and grouping the terms, we obtain

(↵+ 2� � �) + (�↵� � + 5� + �) t+ (2↵+ 3� � 5� � 2�) t2 = b0 + b1t+ b2t2.

By equating the coe�cients, we can obtain 4 simultaneous equations which can be written

in an augmented matrix, 0

B@1 2 �1 0 b0

�1 �1 5 1 b1

2 3 �5 �2 b2

1

CA .

Perform the row operations R2 R2 +R1 and R3 R3 � 2R1,

0

B@1 2 �1 0 b0

0 1 4 1 b0 + b1

0 �1 �3 �2 �2b0 + b2

1

CA .

Next, R3 R3 +R2,

0

B@1 2 �1 0 b0

0 1 4 1 b0 + b1

0 0 1 �1 �b0 + b1 + b2

1

CA .

If we let � = c where c can take any value from R, back-substitution can be performed to

find expressions for ↵,� and � in terms of c, b0, b1, b2. We observe that a solution always

exists for the system, and hence the polynomials form a spanning set for P2.

Alternatively, for these polynomials to span P2, exactly three of these polynomials must

be linearly independent. Use this fact to construct a matrix to show that three of them

are linearly independent.

3. Let ↵,�, � 2 R. For these vectors to be linearly independent, the following must only be

true for ↵ = � = � = 0,

↵

0

B@1

4

�2

1

CA+ �

0

B@2

5

5

1

CA+ �

0

B@1

3

1

1

CA =

0

B@0

0

0

1

CA .

Rewriting this in matrix form, 0

B@1 2 1 0

4 5 3 0

�2 5 1 0

1

CA .

3

Draft

Row-reducing, we first perform the row operations R2 R2 � 4R1 and R3 R3 + 2R1,

0

B@1 2 1 0

0 �3 �1 0

0 9 3 0

1

CA .

Next, R3 R3 + 3R2, 0

B@1 2 1 0

0 �3 �1 0

0 0 0 0

1

CA .

We observe that there are infinitely many solutions for ↵,� and � as not all columns are

leading, and hence the vectors are linearly dependent.

(THOUGHT PROCESS: The essential point of the definition of linear independence is

that the only way �1�!v1 +�2

�!v2 + · · ·+�n

�!vn =

�!0 is satisfied is that all the scalars are zero.

For more information and some detailed, worked examples on this content, please refer to

course notes pp.25-27.

For more practice on this type of question, go to course notes pp.69-70 Problems 6.5.)

4

MATH1231/1241 Algebra Test 1 2008 S0

v1AFull Solutions

August 23, 2018

These solutions were written by Treves Li and typed up by Brendan Trinh. Please be ethical

with this resource. It is for the use of MathSOC members, so do not repost it on other forums or

groups without asking for permission. If you appreciate this resource, please consider supporting

us by coming to our events and buying our T-shirts! Also, happy studying :)

We cannot guarantee that our working is correct, or that it would obtain full marks - please





& di�culty of questions in your quiz.

1. To show that it is a subspace, we need to show 3 things,

• Clearly, the 2⇥ 2 zero matrix lies in S, since 0 + 0 = 3 (0).

• Suppose A,B 2 S where A =

a11 a12

a21 a22

!and B =

b11 b12

b21 b22

!such that

a11 + a12 = 3a22, (1)

b11 + b12 = 3b22. (2)

1

Adding together equations (1) and (2), we obtain

(a11 + b11) + (a12 + b12) = 3 (a22 + b22)

which implies that A+B =

a11 + b11 a12 + b12

a21 + b21 a22 + b22

!2 S.

Thus, S is closed under addition.

• Suppose � 2 R. Now consider �⇥ (1),

� (a11 + a12) = � (3a22) .

This is the same as

(�a11) + (�a12) = 3 (�a22) .

This implies that �A =

�a11 �a12

�a21 �a22

!2 S.

Thus, S is closed under scalar multiplication.

Hence, by the Subspace Theorem, S is a subspace of M2,2 (R).

(THOUGHT PROCESS: To prove that S is a subspace of a vector space V , we need to

show three things: non-empty (or equivalently, whether 0 is an element), closure under ad-

dition, closure under scalar multiplication. Review the Subspace Theorem on your course

notes if you are not familiar with it. As always, make sure you name the theorem you use

in a test or an exam to get full marks.


For more practice on this type of question, go to course notes p.66 Problems 6.3 (Q12b,

14, 16, 18, 22, 24a in particular).)

2. (a) The vectors �!v1 ,�!v2 , . . . ,�!vn are linearly dependent in a vector space V if and only if

we can find scalars �1,�2, . . . ,�n not all zero such that

�1�!v1 + �2

�!v2 + · · ·+ �n�!vn =

�!0 .

(b) Let ↵,�, � 2 R. We consider

↵A1 + �A2 + �A3 = 0.

Substituting in the matrices, we find

↵+ 2� 2↵+ � + �

2↵+ � + 2� ↵+ �

!=

0 0

0 0

!.

2

Equating the elements, we obtain 4 linear equations which can be expressed in matrix

form, 0

BBBB@

1 0 2 0

2 1 1 0

2 1 2 0

1 0 1 0

1

CCCCA.

Performing the row operations R2 R2 � 2R1, R3 R3 � 2R1 and R4 R4 �R1,

0

BBBB@

1 0 2 0

0 1 �3 0

0 1 �2 0

0 0 �1 0

1

CCCCA.

Next, R3 R3 �R2, 0

BBBB@

1 0 2 0

0 1 �3 0

0 0 1 0

0 0 �1 0

1

CCCCA.

Next, R4 R4 +R3, 0

BBBB@

1 0 2 0

0 1 �3 0

0 0 1 0

0 0 0 0

1

CCCCA.

The solution is ↵ = � = � = 0, and hence the matrices are linearly independent.



�!v2 + · · ·+�n�!vn =





3. (i) A basis for W consists of a set of linearly independent vectors that span to form W .

To determine this, we construct a matrix with each vector as a column,

0

B@1 2 �1 1

3 7 0 1

1 1 �4 3

1

CA .

3

Performing the row-operations R2 R2 � 3R1 and R3 R3 �R1, we obtain

0

B@1 2 �1 1

0 1 3 �20 �1 �3 2

1

CA .

Next, R3 R3 +R2, 0

B@1 2 �1 1

0 1 3 �20 0 0 0

1

CA .

We observe that removing �!v3 and �!v4 (the columns that are non-leading) leaves us

with the linearly independent vectors �!v1 and �!v2 . Hence, �!v1 and �!v2 form a basis for

W .

(ii) The vector in question is �!v4 . We are simply trying to find ↵,� 2 R such that

↵�!v1 + ��!v2 = �!v4 .

Constructing an augmented matrix,

⇣�!v1 �!v2 �!v4

⌘=)

0

B@1 2 1

3 7 1

1 1 3

1

CA .

Now, row-reducing, we perform the row operationsR2 R2�3R1 andR3 R3�R1,

0

B@1 2 1

0 1 �20 �1 2

1

CA .

Next, R3 R3 +R2, 0

B@1 2 1

0 1 �20 0 0

1

CA .

Hence, we have � = �2 and ↵ = 5. That is,

5�!v1 � 2�!v2 = �!v4 .

(THOUGHT PROCESS: Definition of bases for a vector space is on p.35 of course notes.

For more information and worked examples on this content, please refer to course notes

pp.36-38.


4

MATH1231/1241 Algebra Test 1 2008 S2v3b

Answers/Hints

August 23, 2018

These answers were written by Treves Li and typed up by Brendan Trinh. Please be ethical

with this resource. It is for the use of MathSOC members, so do not repost it on other forums or


us by coming to our events and buying our T-shirts! Also, happy studying :)

We cannot guarantee that our answers are correct - please notify us of any errors or typos at

[email protected], or on our Facebook page. There are sometimes multiple methods of

solving the same question. Remember that in the real class test, you will be expected to explain

your steps and working out.



1. Very similar to other questions - you must prove that the set is closed under addition,

closed under scalar multiplication, and contains a zero vector. This set satisfies these

conditions and thus is a subspace by the Subspace Theorem.

(THOUGHT PROCESS: To prove that S is NOT a subspace of a vector space V , you

show a counterexample. To prove that S is a subspace, you show three things: non-empty

(or equivalently, whether 0 is an element), closure under addition, closure under scalar

multiplication. Review the Subspace Theorem on your course notes if you are not familiar

with it. As always, make sure you name the theorem you use to get full marks.


For more practice on this type of question, go to course notes p.66 Problems 6.3 (Q12b,

14, 16, 18, 22, 24a in particular).)

1

2. (i) Suppose that ~w is an element of R3. By considering the augmented matrix with ~w

as the subject and using Gaussian elimination, we observe that the system of linear

equations do not hold for all values of ~w (due to the bottom row of zeros equaling

12a1 � 5a2 + a3). Thus the set does not span R3. Vector ~w only spans the set of

vectors i↵ the condition 12a1 � 5a2 + a3 = 0 holds.

(ii) Vector ~u lies in the span of vectors if the above condition holds. Substituting u into

the equation, we obtain LHS = 7 6= 0 and thus ~u is not in the span of vectors.

(For more information on this content, please refer to course notes pp.20-24.


3. Consider the linear combination of the vectors and construct an augmented matrix of

these vectors with a column of zeros. If the solutions to these system of equations are

when the coe�cients of the vectors are zero, then the vectors are linearly independent

from each other. In this scenario, we can observe that this matrix has infinitely many

solutions for the system, and thus the vectors are linearly dependent.



�!v2 + · · ·+�n�!vn =





2

Draft


v2BFull Solutions September 4, 2018

These answers were written and typed up by Brendan Trinh and Aaron Hassan and edited by

Henderson Koh and Aaron Hassan. Please be ethical with this resource. It is for the use of

MathSoc members, so do not repost it on other forums or groups without asking for permission.

If you appreciate this resource, please consider supporting us by coming to our events and buying

our T-shirts! Also, happy studying !







1. Note that S does not contain the zero vector

0

0

!(because 3 ⇥ 0 � 2 ⇥ 0 = 0 6= 1).

Therefore, S cannot be a subspace of R2.

THOUGHT PROCESS:

It takes just a single violation of the vector space properties. See §6.3 in the course notes

for many similar examples.

2. We first need to establish whether the set containing these polynomials is linearly inde-

pendent or not. Noticing that we are working with quadratic polynomials, if it is linearly

independent, there can only be at most 3 polynomials in the set. But since we are working

with 4 polynomials, the set containing these polynomials is thus linearly dependent .

Next, to find which one of these polynomials can be written as a linear combination of

1

Draft

the others, we let ↵,�, �, � 2 R such that

↵p1(x) + �p2(x) + �p3(x) + �p4(x) = 0.

Substituting in the polynomials, we obtain

↵

�1� 3x+ 2x2

�+ �

�3� 8x+ 5x2

�+ �

�1 + x� 2x2

�+ �

��x+ 8x2

�= 0.

Grouping together like terms,

(↵+ 3� + �) + (�3↵� 8� + � � �)x+ (2↵+ 5� � 2� + 8�)x2 = 0.

By equating the coe�cients, we see that

↵+ 3� + � = 0

�3↵� 8� + � � � = 0

2↵+ 5� � 2� + 8� = 0.

We set up an augmented matrix,

0

B@1 3 1 0 0

�3 �8 1 �1 0

2 5 �2 8 0

1

CA . (?)

Row-reducing, we perform the operations R2 R2 + 3R1 and R3 R3 � 2R1,

0

B@1 3 1 0 0

0 1 4 �1 0

0 �1 �4 8 0

1

CA .

Next, R3 R3 +R2, 0

B@1 3 1 0 0

0 1 4 �1 0

0 0 0 7 0

1

CA .

As the first, second and fourth columns are leading, p1, p2, p4 are linearly independent

polynomials. However, the third column is non-leading. This implies that we can express

p3 as a linear combination of p1, p2 and p4.

To find this linear combination, set � to be a free parameter (as it belongs to column 3,

which is non-leading). From row 3, we see � = 0. Now, row 2 implies that � + 4� � � =

0) � = �4� (as � = 0. Now row 1 gives us ↵ = �3�� = 12�� = 11�, since � = �4�.

Hence substituting these into the original equation ↵p1(x)+�p2(x)+�p3(x)+�p4(x) = 0,

2

Draft

we have

11�p1(x)� 4�p2(x) + �p3(x) + 0p4(x) = 0, 8� 2 R

) �p3(x) = �11�p1(x) + 4�p2(x)

) p3(x) = �11p1(x) + 4p2(x) (setting � = 1).

Remarks/Tips. In your working out, you can just go straight to the augmented matrix

in (?). We have just provided detailed explanations before that step for your own benefit

(so that you may understand why we set up that augmented matrix). Also, near the end,

we set � = 1. You could also just set � = 1 at the start of the back-substitution (rather

than using an arbitrary free parameter �).

3. As we know, the vectors will form a basis for R3 if and only if the matrix whose columns

are v1,v2,v3 has every row and column leading in the row echelon form. Set this up as a

matrix: 0

B@1 3 4

1 5 3

�2 4 �7

1

CA .

Row-reducing by performing the row operations R2 R2 �R1 and R3 R3 + 2R1,

0

B@1 3 4

0 2 �1

0 10 1

1

CA .

Next, R3 R3 � 5R2, 0

B@1 3 4

0 2 �1

0 0 6

1

CA .

As we can see, every row and column in the row echelon form is leading. Hence v1,v2,v3

form a basis for R3.

THOUGHT PROCESS:

For the vectors to form a basis, they must (i) span the space, and (ii) be linearly indepen-

dent. We show both here by placing v1,v2,v3 into a matrix and row-reducing. See the

examples in §6.6.1 of the course notes.

4. (i) False. If they are linearly dependent (e.g. because one of them is the zero vector),

then it cannot be a basis.

(ii) True. A basis contains the smallest number of linearly independent vectors to span

the set, that is, there can only be 5 vectors in a basis for R5.

(iii) False. For example, the set {e1, 2e1, 3e1, . . . , 7e1} (where e1 is the first standard

3

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basis vector of R5) will not span R5 (its span is just span {e1}, so for example e2 is

not in this span, so the set is not spanning).

THOUGHT PROCESS:

See §6.5 and §6.6 of the course notes, on spans, linear independence and bases.

4


v1AFull Solutions

August 11, 2018

These solutions were written and typed up by Brendan Trinh. Please be ethical with this

resource. It is for the use of MathSOC members, so do not repost it on other forums or groups

without asking for permission. If you appreciate this resource, please consider supporting us by

liking our Facebook page and coming to our events. Also, happy studying :)

We cannot guarantee that our working is correct, or that it would obtain full marks - please






1. To show that S is a subspace of R3, we must show three things: that S is non-empty,

closed under addition and closed under scalar multiplication.

• Clearly 0 = (0, 0, 0)T 2 S as 7 (0)+ 3 (0) = 0 and (0)� 4 (0) = 0. So S is non-empty.

• To show that it is closed under addition, let x,y 2 S where x = (x1, x2, x3)T and

1

y = (y1, y2, y3)T . That is,

7x1 + 3x2 = 0 (1)

x2 � 4x3 = 0 (2)

7y1 + 3y2 = 0 (3)

y2 � 4y3 = 0. (4)

Adding (1) and (2) together, and (3) and (4) together, we obtain

(1) + (3) : 7x1 + 3x2 + 7y1 + 3y2 = 0

=) 7 (x1 + y1) + 3 (x2 + y2) = 0.

(1) + (4) : x2 � 4x3 + y2 � 4y3 = 0

(x2 + y2)� 4 (x3 + y3) = 0.

Hence, x+ y 2 S and S is closed under addition.

• Next, to show that it is closed under scalar multiplication, let � 2 R and consider

the following,

�⇥ (1) : � (7x1 + 3x2) = �⇥ 0

=) 7 (�x1) + 3 (�x2) = 0

�⇥ (2) : � (x2 � 4x3) = �⇥ 0

=) (�x2)� 4 (�x3) = 0.

Hence, �x 2 S and S is closed under scalar multiplication.

As S is non-empty, is closed under addition and scalar multiplication, by the Subspace

Theorem, S is a subspace of R3.

THOUGHT PROCESS: The solution is fairly straight forward and contains the necessary

explanation. The first property requires you to show that 0 satisfies the equation. Remem-

ber to revise the conditions, as they may provide more complicated equations.

For more practice on this type of question, please refer to page 14 of the course pack.

2. For b to be an element of span (v1,v2,v3), then there must exist ↵,�, � 2 R such that

↵v1 + �v2 + �v3 = b.

2

Substituting in the vectors writing this as an augmented matrix,

0

BBBB@

1 2 �2 b1

1 3 1 b2

2 4 1 b3

�1 1 6 b4

1

CCCCA.

For b to be in the span, we must be able to row-reduce to find solutions for ↵,� and �

which will be in terms of b1, b2, b3 and b4.

Row-reducing, R2 R2 �R1, R3 R3 � 2R1 and R4 R4 +R1,

0

BBBB@

1 2 �2 b1

0 1 3 �b1 + b2

0 0 5 �2b1 + b3

0 3 4 b1 + b4

1

CCCCA.

Next, R4 R4 � 3R2, 0

BBBB@

1 2 �2 b1

0 1 3 �b1 + b2

0 0 5 �2b1 + b3

0 0 �5 4b1 � 3b2 + b4

1

CCCCA.

Finally, R4 R4 +R3,

0

BBBB@

1 2 �2 b1

0 1 3 �b1 + b2

0 0 5 �2b1 + b3

0 0 0 2b1 � 3b2 + b3 + b4

1

CCCCA.

Hence, from R4, a condition for b to be an element of span(v1,v2,v3,v4) is that

2b1 � 3b2 + b3 + b4 = 0.

This is true since once this is true, we can perform back substitution for the rest of the

values of ↵,� and � without any problems (thus we can find a solution).

THOUGHT PROCESS: The first step follows from the fact that a vector can only be an

element of a span if it can be written as a linear combination of the other elements. Hence

there must be solutions for the three parameters.


3. (i) False. A basis for P4 requires 5 linearly independent polynomials in P4. Suppose the

set of 5 polynomials were linearly dependent. Then it would be untrue.

(ii) True. We require 5 linearly independent polynomials in P4 for a basis in P4.

3

THOUGHT PROCESS: The solution is fairly straight forward and contains the nec-

essary explanation.


4. To show that T is not a linear transformation, we simply need to show that it does

not satisfy a property of them. Knowing that linear transformations satisfy the scalar

multiplication condition, i.e.

T (�x) = �T (x) ,

(and noticing that the expression is not going to satisfy this), we try apply this to the

given transformation.

Let � 2 R and consider T (�x),

T (�x) = (�x) cos (�x)

= �x cos (�x) 6= �x cos (x) (as cos (�x) 6= cos (x) for all � 2 R.)

Thus, T is not a linear transformation as it does not satisfy the multiplicative condition.

THOUGHT PROCESS: The solution is fairly straight forward and contains the necessary

explanation. Save time in the exam by choosing the property which the proposed linear

transformation clearly does not satisfy.


4


v1BAnswers/Hints

September 4, 2018

These answers were written and typed up by Brendan Trinh. Please be ethical with this resource.

It is for the use of MathSOC members, so do not repost it on other forums or groups without

asking for permission. If you appreciate this resource, please consider supporting us by coming

to our events and buying our T-shirts! Also, happy studying :)

We cannot guarantee that our answers are correct - please notify us of any errors or typos at

[email protected], or on our Facebook page. There are sometimes multiple methods of

solving the same question. Remember that in the real class test, you will be expected to explain

your steps and working out.



1. (i) Many, many choices for your answer. Choose a value for x1 or x2 and solve for the

other. An example answer is x =� 2�1

�.

(ii) You can do this generally by using x =�x1x2

�where x 2 S or by using the x possibility

you found in part (i). Let � 2 R and consider �x and go from here to show that it

is not closed under scalar multiplication.

THOUGHT PROCESS: Show that if x satisfies the equation, then �x does not for

all real values of �.


2. (i) For q 2 span (p1, p2, p3), there exists ↵,�, � 2 R such that ↵p1+�p2+�p3 = q. Form

an augmented matrix and row-reduce to eventually find that q does lie in the span!

1

Draft

(ii) Recall to have a spanning set for P2, we require there to be 3 linearly independent

polynomials in P2. You need to try and show this. In the process, you will find that

only 2 are linearly independent, and thus the set {p1, p2, p3} does not span P2.

THOUGHT PROCESS: You can show this using a matrix and row reducing. You

will find that there is a non leading column, hence one of the vectors is not linearly

independent.


3. (a) To evaluate this, recall the property of linear transformations that for constants

↵,� 2 R, we have

T

✓1

0

◆= T

✓↵

✓2

3

◆+ �

✓1

2

◆◆= ↵T

✓2

3

◆+ �T

✓1

2

◆.

That is, you need to find ↵ and � such that ↵

�23

�+ �

�12

�=�10

�. You will find that

↵ = 2 and � = �3, and eventually that T�10

�=� 3�2

�.

THOUGHT PROCESS: The solution is fairly straight forward and contains the nec-

essary explanation.


(b) We are basically trying to find a matrix A that will multiply with the vector x you

plug in to get the result of the linear transformation. To do this, we can plug in

the standard basis vectors of the domain R2 to obtain the columns of the matrix A

one-by-one.

To prove this fact for you, notice that

a b

c d

!✓1

0

◆=

✓a

c

◆.

You will find that A =

3 1

�2 4

!.

THOUGHT PROCESS: The previous question required you to find T

�10

�, which is

equal to

�a

c

�. Use a similar method with T

�01

�to find

�b

d

�.


2

Draft


v2AAnswers/Hints & Worked Sample Solutions

September 4, 2018

These answers/hints were originally written and typed up by Brendan Trinh, with edits and

worked sample solutions provided by Henderson Koh and Aaron Hassan. Please be ethical with

this resource. It is for the use of MathSoc members, so do not repost it on other forums or


us by coming to our events and buying our T-shirts! Also, happy studying !







1. Hints

To show that S is a subspace of R2, we need to show that S is non-empty, is closed under

addition and closed under scalar multiplication. From there, we can use the Subspace

Theorem.

Recall matrix properties such as Ax+Ay = A (x+ y) to help answer the question. Here

is a sample answer.

Worked Sample Solutions

Note that S is a subset of R2, which is a vector space. Now, clearly S is nonempty, as the

zero vector of R2, 0R2 =

0

0

!, indeed satisfies A0R2 = 0R3 , where 0R3 is the zero vector

1

Draft

of R3, i.e.

0

B@0

0

0

1

CA (since any matrix times the relevant zero vector results in a zero vector).

Now, suppose x,y are vectors in S. Then x+ y 2 R2 and

A (x+ y) = Ax+Ay (Distributive Law of matrix multiplication)

= 0R3 + 0R3 (as x,y 2 S)

= 0R3

) x+ y 2 S.

Hence S is closed under addition.

Finally, suppose x is a vector in S, and let ↵ be a scalar. Then ↵x is a vector in R2

satisfying

A (↵x) = ↵ (Ax)

= ↵0R3 (as x 2 S)

= 0R3

) ↵x 2 S.

Hence S is closed under scalar multiplication. It follows from the Subspace Theorem that

S is a subspace of R2.

2. Hints

The column space of A is the span of the columns, i.e. if v1 =

0

BBBB@

1

�1

0

2

1

CCCCAand v2 =

0

BBBB@

4

�3

2

7

1

CCCCA,

then we want to see if b 2 span (v1,v2).

To do this, we need to check if there exists ↵,� 2 R such that ↵v1 + �v2 = b.

You will find that no solutions exist for ↵ and � such that ↵v1 + �v2 = b, and so b is

not in the column space of A. Here is a sample answer.


Note that b is in the column space of A if and only if there exist scalars ↵,� such that

↵v1 + �v2 = b, where v1 =

0

BBBB@

1

�1

0

2

1

CCCCAand v2 =

0

BBBB@

4

�3

2

7

1

CCCCA.

2

Draft

So we try and solve ↵

0

BBBB@

1

�1

0

2

1

CCCCA+ �

0

BBBB@

4

�3

2

7

1

CCCCA=

0

BBBB@

�1

5

5

�6

1

CCCCA.

Comparing the third entry of each vector (0↵+ 2� = 5), it is evident that � must be 2.5.

So comparing fourth entries now, we have

2↵+ 7� = �6

2↵+ 7⇥ 2.5 = �6

) 2↵ = �23.5

) ↵ = �11.75.

But now the first entries tell us that

↵+ 4� = �1

) �11.75 + 4⇥ 2.5 = �1

) �11.75 + 10 = �1

) �1.75 = �1,

which is a contradiction. So there is no solution for ↵,�, so b is not in the column space

of A.

Alternatively, you can create the corresponding augmented matrix:

0

BBBB@

1 4 �1

�1 �3 5

0 2 5

2 7 �6

1

CCCCA

and after row-reducing, realise that no solutions exist for ↵ and �.

3. (i) Sample Answer

If S is a basis for V , then it is a set of linearly independent vectors, and so m n.

But also, span (S) = V , and so n m. This implies that m = n.

Alternative answer: Recall that the definition of the dimension of a vector space is

the number of vectors in any basis for it (provided this is finite). Hence by definition,

m and n are equal.

(ii) Sample Answer

If S is linearly dependent, no relationship can be inferred between m and n. S can

be linearly dependent no matter what value m takes, so long as another one of the

vectors in S can be written as a linear combination of the others.

3

Draft

Alternative answer: Note that for any positive integer m � 2, the following set

is linearly dependent: {0,v2, . . . ,vm}, where v2, . . . ,vm are any vectors from V .

(Recall that any set with the zero vector is linearly dependent.)

For the case when S only has one vector, S can still be linearly dependent if its only

vector is the zero vector, 0.

This shows that there is no relation between m and n, since for any m, there exist

linearly dependent sets.

4. Hints

Very similar to MATH1231/1241 Algebra 2011 S2 Test 1 v1B Q3.

To check if T is a linear transformation, it would be easiest to check if the following

property holds: that T (↵x+ �y) = ↵T (x) + �T (y) where ↵,� 2 R. A sample answer is

provided below.


We will answer this question with a proof by contradiction.

Suppose that T is linear. Let e1 =

1

0

!, e2 =

0

1

!and w =

5

�1

!. Note that

w = 5e1 � e2. So we have

T (w) = T (5e1 � e2)

= 5T (e1)� T (e2) (linearity)

= 5

0

B@1

1

�1

1

CA�

0

B@2

3

2

1

CA (using given values)

=

0

B@3

2

�7

1

CA .

But we were told that T (w) =

0

B@4

�1

3

1

CA, which is not equal to

0

B@3

2

�7

1

CA. So we have a

contradiction. Hence T is not linear.

4

Draft


v1AAnswers/Hints & Worked Sample Solutions

September 4, 2018

These hints were originally written and typed up by Hub Wa, with edits and worked sample

solutions added by Aaron Hassan and Henderson Koh. Please be ethical with this resource.

It is for the use of MathSoc members, so do not repost it on other forums or groups without

asking for permission. If you appreciate this resource, please consider supporting us by coming

to our events and buying our T-shirts! Also, happy studying !







1. Hints

To show that S is a subspace of R3, you need to show that it is non-empty, is closed

under addition and under scalar multiplication. A core step you’ll be required to make is

x = (x1, x2, x3)T 2 R3 such that x1 � 2x2 = 0 and 5x2 + x3 = 0.

Worked Sample Solution

Note that clearly S is a subset of R3, which is a vector space. Since 0 � 2 ⇥ 0 = 0 and

5⇥ 0 + 0 = 0, the zero vector from R3, namely

0

B@0

0

0

1

CA, is in S, so S is non-empty.

1

Draft

Let x =

0

B@x1

x2

x3

1

CA and y =

0

B@y1

y2

y3

1

CA be in S. Then x + y =

0

B@x1 + y1

x2 + y2

x3 + y3

1

CA is an element of R3

since

(x1 + y1)� 2 (x2 + y2) = (x1 � 2x2) + (y1 � 2y2) (rearranging terms)

= 0 + 0 (since x,y 2 S)

= 0,

and

5 (x2 + y2) + (x3 + y3) = (5x2 + x3) + (5y2 + y3) (rearranging terms)

= 0 + 0 (since x,y 2 S)

= 0.

Thus S is closed under addition.

Now, suppose x =

0

B@x1

x2

x3

1

CA 2 S and ↵ 2 R, then ↵x =

0

B@↵x1

↵x2

↵x3

1

CA is a vector in R3. Moreover,

↵x1 � 2 (↵x2) = ↵ (x1 � 2x2)

= ↵⇥ 0 (as x 2 S)

= 0,

and

5 (↵x2) + ↵x3 = ↵ (5x2 + x3)

= ↵⇥ 0 (as x 2 S)

= 0.

Hence S is closed under scalar multiplication.

It follows by the Subspace Theorem that S is a subspace of R3.

2. Hints

Set up an augmented matrix and row-reduce. Your conditions on b to lie in the span(v1,v2,v3)

should look like

15b1 + 6b2 � b3 + b4 = 0.


As usual we set up the relevant augmented matrix, namely⇣v1 v2 v3 b

⌘, and row-

reduce. Note that this matrix corresponds to the equation ↵1v1 +↵2v2 +↵3v3 = b. This

2

Draft

is because (by definition of span) b 2 span(v1,v2,v3) if and only if this equation has a

solution for ↵1,↵2 and ↵3.

We can represent b in the augmented matrix either as

0

BBBB@

b1

b2

b3

b4

1

CCCCAor by creating a column each

for b1, b2, b3, b4. Here, we do the latter. And so, we have

2

66664

1 3 �2 1 0 0 0

�2 �5 5 0 1 0 0

0 7 1 0 0 1 0

�3 �8 1 0 0 0 1

3

77775R2 R2+2R1��!R4 R4+3R1

2

66664

1 3 �2 1 0 0 0

0 1 1 2 1 0 0

0 7 1 0 0 1 0

0 1 �5 3 0 0 1

3

77775

R3 R3�7R2��!R4 R4�R2

2

66664

1 3 �2 1 0 0 0

0 1 1 2 1 0 0

0 0 �6 �14 �7 1 0

0 0 �6 1 �1 0 1

3

77775

R4 R4�R3��!

2

66664

1 3 �2 1 0 0 0

0 1 1 2 1 0 0

0 0 �6 �14 �7 1 0

0 0 0 15 6 �1 1

3

77775.

Recall that b will be in the span if and only if the columns on the right-hand side of the

augmented matrix in row-echelon form are all non-leading. We see that from row 4 (the

only row with a zero row in the left-hand part), the necessary and su�cient condition on

b1, b2, b3, b4 for b to be in the span of {v1,v2,v3} is 15b1 + 6b2 � b3 + b4 = 0.

3. Hints

Remember that the dimension of P5 is 6 (not 5). (In general, the dimension of Pn is n+1,

for any n 2 N.) Also make sure to recall the relationships between dimension, basis, and

the number of elements in a set.


(i) False. A counterexample will be a set of six polynomials that are all constant poly-

nomials. Then, this set will not span P5, and so will not be a basis.

(ii) True, as the dimension of P5 is 6 and so six linearly independent polynomials

in P5 will form a basis for P5. For example, the standard basis for P5�the set

�1, t, t2, t3, t4, t5

�is a set of six polynomials in P5 that form a basis for P5.

3. Hints

We need to check which properties of linear transformations do not hold. Remember,

to show that a map T is not linear, we just need to produce a single example that

3

Draft

demonstrates that T does not satisfy a condition of linearity.


Note that T

�⇡2

�=�⇡2

�2sin

�⇡2

�=�⇡2

�2. Also, T

�2�⇡2

��= T (⇡) = ⇡

2 sin⇡ = 0. Hence

T

�2�⇡2

��6= 2T

�⇡2

�, and so T does not satisfy the scalar multiplication condition, whence

T is not linear.

4

Draft


v1BAnswers/Hints & Worked Sample Solutions

September 4, 2018

These answers were written and typed up by Hub Wa and edited by Henderson Koh and Aaron

Hassan. Please be ethical with this resource. It is for the use of MathSoc members, so do

not repost it on other forums or groups without asking for permission. If you appreciate this

resource, please consider supporting us by coming to our events and buying our T-shirts! Also,

happy studying !







1. (i) There are many, many di↵erent choices. But one easy example you can try is

1

0

!,

since 1� 5⇥ 0 < 2.

(ii) Hints

To show that a set is not closed under scalar multiplication, we need to find a vector

in the set, such that when we multiply this vector by a scalar, the resulting vector

will not be in S.


Using the vector in part (i)

1

0

!, we can multiply it by 2, so that the resulting vector

1

Draft

is

2

0

!. Clearly, 2 � 5(0) = 2. So the resulting vector is not in S. Hence, S is not

closed under scalar multiplication.

2. (i) Hints

To check if a particular polynomial is an element of the span of the given 3 vectors, we

just need to see if we can express this particular polynomial as a linear combination

of these 3 vectors, i.e. if there exist ↵1,↵2 and ↵3 such that ↵1p1(x) + ↵2p2(x) +

↵3p3(x) = q(x).


We construct the corresponding augmented matrix to represent this equation.

0

B@1 3 �1 1

�1 �1 0 2

2 0 1 �7

1

CA .

Performing the row operations R2 R2 + R1 and R3 R3 � 2R1 gives the aug-

mented matrix 0

B@1 3 �1 1

0 2 �1 3

0 �6 3 �9

1

CA .

Performing R3 R3 + 3R2 gives us the row-echelon form

0

B@1 3 �1 1

0 2 �1 3

0 0 0 0

1

CA .

The right-hand column is non-leading, so q is an element of span(p1, p2, p3).

(ii) Hints

Look at the row-echelon form of the matrix whose columns contain p1, p2, p3 (as

vectors), i.e. the left-hand part of the matrix in (i). Remember what the row-

echelon form has to look like in order for the set to be spanning.


The given set would span P2 if and only if the left-hand part of the augmented matrix

in part (i) (the matrix whose columns contain the coordinate vectors of p1, p2, p3 (with

respect to the standard basis of P2)) had no zero rows in its row-echelon form. But looking at

its row-echelon form that we found, we see that this is not the case (since the third

row is a zero row). Therefore, the given set does not span P2.

3. (i) Hints

To evaluate this, recall the property of linear transformations that for scalars ↵,� 2

2

Draft

R, we have

T

1

0

!= T

↵

5

3

!+ �

2

1

!!= ↵T

5

3

!+ �T

2

1

!(⇤).

We’re already given what T

5

3

!and T

2

1

!are. So, all we really need to do is find

↵ and � such that ↵

5

3

!+ �

2

1

!=

1

0

!. We can find using Gaussian Elimination

(or inspection) that the solution is ↵ = �1 and � = 3. From this, we find that

T

1

0

!=

1

�2

!, using Equation (⇤). Here is a worked sample solution.


We solve for ↵ and � in the equation

↵

5

3

!+ �

2

1

!=

1

0

!.

After row-reducing the augmented matrix

5 2 1

3 1 0

!R

2

R2

� 3

5

R1��! 5 2 1

0 �15 �3

5

!

we know that �15� = �3

5 and 5↵ + 2� = 1. Solving these simultaneously, we find

that ↵ = �1 and � = 3.

Thus

T

1

0

!= T

� 5

3

!+ 3

2

1

!!

= �T

5

3

!+ 3T

2

1

!(linearity of T )

= �

8

�1

!+ 3

3

�1

!(given data)

=

�8 + 9

1� 3

!

=

1

�2

!.

(ii) Hints

We want to find the matrix of T (with respect to standard bases). The standard

3

Draft

method to do this is to apply T to the standard basis vectors and place them in as

columns of the matrix. In other words, column 1 of A should be T

1

0

!and column

2 should be T

0

1

!. We have already calculated T

1

0

!=

1

�2

!earlier. We just

need to find the second column, which is simply T

0

1

!. To calculate T

0

1

!, do a

similar process to the first part of the question. A sample solution is below.


We have already calculated T

1

0

!=

1

�2

!earlier, so this is column 1 of A, so A

is of the form

1 ?

�2 ?

!.

Observe that

0

1

!= 2

2

3

!� 5

1

2

!(find this either by inspection or Gaussian

Elimination similar to the previous part). So

T

0

1

!= T

2

5

3

!� 5

2

1

!!

= 2T

5

3

!� 5T

2

1

!(linearity)

= 2

8

�1

!� 5

3

�1

!(given data)

=

1

3

!.

So this is the second column, and thus the required matrix is A =

1 1

�2 3

!.

4

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