MATH1251 ALGEBRA S2 2010 TEST 1 VERSION 1A Sample …

MATH1251 ALGEBRA S2 2010 TEST 1 VERSION 1A

Sample SolutionsSeptember 13, 2017

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Question 1

Firstly, we write −2 + 2i√

3 in polar form. Noting that the modulus −2 + 2i√

3 is 4 and its

argument is 2π/3, we have,

−2 + 2i√

3 = 4e2iπ3 .

Let the fourth roots of 4e2iπ3 be reiθ. This gives us the expression,

(reiθ)4 = 4e2iπ3 ,

Expanding the left side,

r4e4iθ = 4e2iπ3 ,

Equating the modulus and argument in the above expression,

r4 = 4, 4θ =2π

3.

Finally, we solve for r and θ, noting that r is a non-negative real value and θ lies between −πand π. This r =

√2 and we find that θ can be −5π6 , −π3 , π

6 or 2π3 . Hence, the roots are

√2e

π6 ,

1

√2e

2π3 ,

√2e−

π3 ,

√2e−

5π6 .

Question 2

We aim to prove that S = {p ∈ P3 | p(5) = 0} is a subspace of P3 using the subspace theorem.

Consider a polynomial p that performs the mapping x 7→ (x − 5)3. We note that this polyno-

mial is in P3 and p(5) = 0. Hence, p is in the set S, which implies that S is not empty. (An

alternative example you can use to show that S is non-empty is to use the zero polynomial,

since this also satisfies p(5) = 0, and belongs to P3.)

Let p and q be elements in S. Consider the polynomial p + q. Since P3 is a vector space, it is

closed under addition and hence p + q is in P3. By using the fact that p(5) = q(5) = 0, since

they are in S, we have (p + q)(5) = p(5) + q(5) = 0 + 0 = 0. Hence, p + q is also in S, which

implies that S is closed under addition.

Let λ be real and suppose p is in S. Since P3 is a vector space, it is closed under scalar

multiplication, so λp is in P3. By using the fact that p(5) = 0, since it is in S, we have

(λp)(5) = λp(5) = λ(0) = 0. Hence, λp is also in S, which implies that S is closed under scalar

multiplication.

Since S is a non-empty subset of P3 that is closed under addition and scalar multiplication, it

is a subspace of P3.

Question 3

(i) Let z = x+ iy and suppose that |z+ 1| = |z−1|. By squaring both sides of this expression

and substituting z, we have,

|(x+ 1) + iy|2 = |(x− 1) + iy|2.

Using the fact that |w|2 = ww,

((x+ 1) + iy)((x+ 1)− iy) = ((x− 1) + iy)(x− 1− iy).

Expanding,

(x+ 1)2 + y2 = (x− 1)2 + y2,

2

Further expanding and cancelling terms gives us,

(x2 + 2x+ 1) + y2 = (x2 − 2x+ 1) + y2,

⇒ 2x = −2x,

⇒ 4x = 0,

⇒ x = 0.

Therefore, x = 0 and z = iy. We have two cases to consider now, y ≥ 0 and y < 0.

Suppose y ≥ 0. Then Arg(z + 1) = Arg(1 + iy) = tan−1(y) and Arg(z − 1) = Arg(−1 + iy) =

tan−1(−y) + π. Thus,

Arg(z + 1) + Arg(z − 1) = tan−1(y) + tan−1(−y) + π,

= tan−1(y)− tan−1(y) + π,

= π.

Suppose y < 0. Then Arg(z + 1) = Arg(1 + iy) = tan−1(y) and Arg(z − 1) = Arg(−1 + iy) =

tan−1(−y)− π. Thus,

Arg(z + 1) + Arg(z − 1) = tan−1(y) + tan−1(−y)− π,

= tan−1(y)− tan−1(y)− π,

= −π.

Therefore, we have Arg(z + 1) + Arg(z − 1) = ±π.

(ii) Angles of isosceles triangles opposite to the sides of equal length are equal. To draw the

diagram for this question, draw two point on the y-axis (one above the x-axis and one below;

these correspond to the point z) and the triangle joining this point to the points (−1, 0) and

(1, 0), and label the base angles of this (isosceles) triangle as equal.

3

MATH1251 ALGEBRA S2 2010 TEST 1 VERSION 1B












Question 1

Let z = cos(θ) + i sin(θ). We can write z4 in two ways. Firstly, by using de Moivre’s formula,

we have

z4 = (cos(θ) + i sin(θ))4,

= cos(4θ) + i sin(4θ).

Secondly, by using a binomial expansion, we have

z4 = cos4(θ) + 4i cos3(θ) sin(θ)− 6 cos2(θ) sin2(θ)− 4i cos(θ) sin3(θ) + sin4(θ),

= cos4(θ)− 6 cos2(θ) sin2(θ) + sin4(θ) + i(4 cos3(θ) sin(θ))− 4 cos(θ) sin3(θ)),

Equating the real components of these two expressions gives,

cos(4θ) = cos4(θ)− 6 cos2(θ) sin2(θ) + sin4(θ).

4

Using the fact that sin2 θ = 1− cos2(θ),

cos(4θ) = cos4(θ)− 6 cos2(θ)(1− cos2(θ)) + (1− cos2(θ))2,

= cos4(θ)− 6 cos2(θ) + 6 cos4(θ) + (cos4(θ)− 2 cos2(θ) + 1),

= 8 cos4(θ)− 8 cos2(θ) + 1.

Question 2

We aim to prove that S = {x ∈ R3 | x1 − 2x2 = 0 and 4x1 + x3 = 0} is a subspace of R3 using

the subspace theorem. We start by verifying the necessary condition that S is a subset of R3.

This can be done by noting that all elements of S are required to be in R3, from the way S is

defined.

Consider the zero vector (0, 0, 0) ∈ R3. Since 0 − 2(0) = 0 and 4(0) + 0 = 0, we see that the

zero vector is in S, so S is non-empty.

Suppose u = (u1, u2, u3) ∈ R3 and v = (u1, u2, u3) ∈ R3 are in S. This means that the following

expressions are true:

u1 − 2u2 = 0,

4u1 + u3 = 0,

v1 − 2v2 = 0,

4v1 + v3 = 0.

Let w = u + v = (u1 + v1, u2 + v2, u3 + v3) ∈ R3. Observe that,

w1 + 2w2 = (u1 + v1) + 2(u2 + v2),

= (u1 + 2u2) + (v1 + 2v2),

= 0 + 0,

= 0,

and,

4w1 + w3 = 4(u1 + v1) + (u3 + v3),

= (4u1 + u3) + (4v1 + v3),

= 0 + 0,

= 0.

5

Hence, w is an element of S, which implies that S is closed under addition.

Let λ ∈ R and let y = λu = (λu1, λu2, λu3). Observe that,

y1 + 2y2 = λu1 + 2λu2,

= λ(u1 + 2u2),

= λ(0),

= 0.

and,

4y1 + y3 = 4λu1 + λu3,

= λ(4u1 + u3),

= λ(0),

= 0.

Hence, y is an element of S, which implies that S is closed under scalar multiplication.

We can therefore conclude that S is a subspace of R3.

Question 3

(i) Let α = e2πi/5. Raising α to the fifth power gives,

α5 =(e2πi/5

)5,

= e10πi/5,

= e2πi,

= cos 2π + i sin 2π,

= 1.

Hence, we have α5 = 1. Rearranging this gives us

α5 − 1 = 0,

which can be factorised to obtain

(α− 1)(1 + α+ α2 + α3 + α4

)= 0.

(You may expand this expression to confirm this factorisation.)

6

Since α is not equal to 1, α − 1 is not equal to zero. Therefore, we can divide both sides by

α− 1 to get

1 + α+ α2 + α3 + α4 = 0.

(ii) We can expand the required expression as follows

(z − α− α−1

) (z − α2 − α−2

)=(z2 − α2z − α−2z

)−(αz − α3 − α−1

)−(α−1z − α− α−3

),

= z2 −(α2 + α−2 + α+ α−1

)z +

(α+ α−1 + α3 + α−3

).

This can be simplified to

z2 + z − 1,

as is shown in the next part.

(iii) Consider putting α = e2πi/5 into the quadratic p(z) = (z − α− α−1)(z − α2 − α−2). By

noting that for any integer n,

αn + α−n = e2nπi/5 + e−2nπi/5,

= (cos(2nπ/5) + i sin(2nπ/5)) + (cos(−2nπ/5) + i sin(−2nπ/5)),

= (cos(2nπ/5) + i sin(2nπ/5)) + (cos(2nπ/5)− i sin(2nπ/5)),

= 2 cos(2nπ/5),

we can write the factorised form of p(z) as (z−2 cos(2π/5))(z−2 cos(4π/5)), which implies that

2 cos(2π/5) is a root of p(z). Using part (ii), we can write the expanded form of p(z) as

p(z) = z2 − (α2 + α−2 + α+ α−1)z + (α+ α−1 + α3 + α−3),

= z2 − α−5(α7 + α3 + α6 + α4)z + α−5(α6 + α4 + α8 + α2).

Using the fact that α5 = 1 and 1 +α+α2 +α3 +α4 = 0, we can simplify p(z) as follows,

p(z) = z2 − α−5(α5α2 + α3 + α5α1 + α4)z + α−5(α5α1 + α4 + α3α5 + α2),

= z2 − (α2 + α3 + α1 + α4)z + (α1 + α4 + α3 + α2),

= z2 − (−1)z + (−1),

= z2 + z − 1,

which is the quadratic polynomial we seek.

7

(iii) From the factorised form of p(z), which is (z − 2 cos(2π/5))(z − 2 cos(4π/5)), we can see

that 2 cos(4π/5) is the other root of p(z).

8













Question 1

Let z = cos(θ) + i sin(θ). We can write z5 in two ways. Firstly, by using de Moivre’s formula,

we have

z5 = (cos(θ) + i sin(θ))5,

= cos(5θ) + i sin(5θ).

Secondly, by using a binomial expansion, we have

z5 = cos5(θ) + 5i cos4(θ) sin(θ))− 10 cos3(θ) sin2(θ)− 10i cos2(θ) sin3(θ) + 5 cos(θ) sin4(θ) + i sin5(θ),

= (cos5(θ)− 10 cos3(θ) sin2(θ) + 5 cos(θ) sin4(θ)) + i(5 cos4(θ) sin(θ))− 10 cos2(θ) sin3(θ) + sin5(θ)).

Equating the imaginary components of these two expressions gives,

sin(5θ) = 5 cos4(θ) sin(θ))− 10 cos2(θ) sin3(θ) + sin5(θ)

Using the fact that cos2 θ = 1− sin2(θ),

sin(5θ) = 5(1− sin2(θ))2 sin(θ)− 10(1− sin2(θ)) sin3(θ) + sin5(θ),

9

= (5 sin5(θ)− 10 sin3(θ) + 5 sin(θ))− (10 sin3(θ)− 10 sin5(θ)) + sin5(θ),

= 16 sin5(θ)− 20 sin3(θ) + 5 sin(θ)

Question 2

We aim to prove that S = {x ∈ R3 | 5x1− 2x2 +x3 = 0} is a subspace of R3 using the subspace

theorem. Since all elements of S are in R3, S is a subset of R3.

Consider the zero vector (0, 0, 0). Since 5(0)− 2(0) + (0) = 0, we see that the zero vector is in

S, so S is non-empty.

Suppose u = (u1, u2, u3) and v = (u1, u2, u3) are in S. This means that the following expressions

are true,

5u1 − 2u2 + u3 = 0,

5v1 − 2v2 + v3 = 0.

Let w = u + v = (u1 + v1, u2 + v2, u3 + v3). Observe that,

5w1 − 2w2 + w3 = 5(u1 + v1)− 2(u2 + v2) + (u3 + v3),

= (5u1 − 2u2 + u3) + (5v1 − 2v2 + v3),

= 0 + 0,

= 0.

Hence, w is an element of S, which implies that S is closed under addition.

Let λ ∈ R and let y = λu = (λu1, λu2, λu3). Observe that,

5y1 − 2y2 + y3 = 5(λu1)− 2(λu2) + (λu3),

= λ(5u1 − 2u2 + u3),

= λ(0),

= 0.

Hence, y is an element of S, which implies that S is closed under scalar multiplication.

We can therefore conclude that S is a subspace of R3.

10

Question 3

(i) We can simplify the required expression as follows:

eiθ − 1

eiθ + 1=eiθ/2

eiθ/2eiθ/2 − e−iθ/2

eiθ/2 + e−iθ/2,

=eiθ/2 − e−iθ/2

eiθ/2 + e−iθ/2,

=2i sin(θ/2)

2 cos(θ/2),

= i tan

(θ

2

).

(ii) Since z is on the unit circle, we can write z as eiθ for some real θ. Conversely, since the

modulus of eiθ is 1, we know that eiθ represents a point on the unit circle for any real θ. From

part (i), we have

eiθ − 1 = i tan

(θ

2

)(eiθ + 1).

Substituting z = eiθ, where z is a point on the unit circle, and noting that i = eiπ/2,

z − 1 = eiπ/2 tan

(θ

2

)(z + 1)

We have two cases, tan (θ/2) ≥ 0 and tan (θ/2) < 0.

If tan (θ/2) ≥ 0, then tan (θ/2) is a non-negative real that is scaling the complex number z+ 1,

while eiπ/2 rotates the number by π/2. Therefore, we can see that z − 1 is a scaling of z + 1 by

tan(θ/2), followed by a rotation by π/2, which means that Arg(z − 1) = Arg(z + 1) + π2 .

If tan (θ/2) < 0, then tan (−θ/2) > 0 and we can write

z − 1 = −eiπ/2 tan

(−θ

2

)(z + 1).

By noting that −eiπ/2 = −i = e−iπ/2, we have

z − 1 = e−iπ/2 tan

(−θ

2

)(z + 1),

which we can interpret in a similar way as before. We see that z − 1 is a scaling of z + 1 by

tan(−θ/2), followed by a rotation by −π/2, which means that Arg(z − 1) = Arg(z + 1)− π2 .

11

Putting these two cases together allows us to deduce that

Arg(z − 1)−Arg(z + 1) = ±π2

(iii) Angles inscribed in a semicircle are always right angles.

12



These answers were written and typed up by 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 !







Question 1

(i) Place p1, p2, p3, q (as vectors) into an augmented matrix and row-reduce. We have 1 1 −1 3

3 5 −4 7

−1 3 −1 −7

R3 R3+R1−−−−−−−−→R2 R2−3R1

1 1 −1 3

0 2 −1 −2

0 4 −2 −4

R3 R3−2R2−−−−−−−−→

1 1 −1 3

0 2 −1 −2

0 0 0 0

.

This is now in row-echelon form, and the right-hand column is non-leading. Hence, q is an

element of span (p1, p2, p3).

(ii) Since the left-hand part of the row-echelon form in the matrix above contains a zero row,

the set {p1, p2, p3} is not spanning.

Remark. Make sure you know why the claims made above are true! For (i), the polynomial q

13

is in span (p1, p2, p3) if and only if there exist scalars α1, α2, α3 such that

α1p1+α2p2+α3p3 = q ⇐⇒ α1

(1 + 3x− x2

)+α2

(1 + 5x+ 3x2

)+α3

(−1− 4x− x2

)≡ 3+7x−7x2.

If we then equate coefficients of like powers of x above, we will end up with a system of linear

equations that is exactly described by the augmented matrix formed by placing the polynomials

p1, p2, p3, q (as vectors) into an augmented matrix. This is the reason why we consider such a

matrix. Also, such αi will exist if and only if this linear system has a solution, and we know

from first semester that the linear system will have a solution if and only if the right-hand

column is non-leading in the row-echelon form. This explains what we did, and is the general

explanation for why we do what we do for these types of questions.

For (ii), to check if the polynomials span P2, it is equivalent to the fact that their coordinate

vectors span R3 or C3 (why?), and this is the case if and only if the matrix formed by placing

them as columns has no zero rows in the row echelon form. This is because a set of vectors is

spanning if and only if any vector in the ambient space can be written as a linear combination

of them, i.e. for any right-hand vector in the augmented matrix, there should exist a solution

to the corresponding linear system Ax = b. There exists a solution to this for every b if and

only if the row-echelon form of A has no zero rows, as we know from first semester.

Question 2

(i) The domain and codomain of T are vector spaces. We must show that T preserves addition

and preserves scalar multiplication. Let x =

x1

x2

x3

x4

∈ R4 and α ∈ R. Then αx =

αx1

αx2

αx3

αx4

∈ R4,

so

T (αx) =

(αx1 + 2 (αx2)− (αx3)

3 (αx3) + (αx4)

)

=

(α (x1 + 2x2 − x3)α (3x3 + x4)

)

= α

(x1 + 2x2 − x3

3x3 + x4

)= αT (x) .

14

Thus T preserves scalar multiplication. Now, let x ∈ R4 as before and let y =

y1

y2

y3

y4

∈ R4.

Then x + y =

x1 + y1

x2 + y2

x3 + y3

x4 + y4

∈ R4, so

T (x + y) =

((x1 + y1) + 2 (x2 + y2)− (x3 + y3)

3 (x3 + y3) + (x4 + y4)

)

=

((x1 + 2x2 − x3) + (y1 + 2y2 − y3)

(3x3 + x4) + (3y3 + y4)

)

=

(x1 + 2x2 − x3

3x3 + x4

)+

(y1 + 2y2 − y3

3y3 + y4

)= T (x) + T (y) .

Thus T preserves addition. As T preserves scalar multiplication and addition, it is a linear

map.

(ii) By inspection of the formula for T

x1

x2

x3

x4

, the A is A =

(1 2 −1 0

0 0 3 1

).

Question 3

(i) Anticipating we would need to augment A with b in question (ii), we shall do this now, so

that we would not need to do the whole row-reduction again. We have

[A | b] =

−1 2 1 0 4

1 −1 0 −1 1

3 −2 1 −4 8

R2 R2+R1−−−−−−−−→R3 R3+3R1

−1 2 1 0 4

0 1 1 −1 5

0 4 4 −4 20

R3 R3−4R2−−−−−−−−→

−1 2 1 0 4

0 1 1 −1 5

0 0 0 0 0

.

The leading columns in the row echelon form of A are thus columns 1 and 2. Therefore, the

columns 1 and 2 of the original matrix A form a basis for the image of A. So a basis for im(A)

15

is −1

1

3

,

2

−1

−2

.

(ii) The right-hand column is non-leading in the row-echelon form we found in part (i). There-

fore, b is in the image of A.

16

MATH1251 ALGEBRA S2 2010 TEST 2 VERSION 1B












Question 1

Let S = {A1, A2, A3}. This set is linearly independent if and only if the set S̃ := {v1,v2,v3} is

linearly independent, where for each j ∈ {1, 2, 3}, vj is the coordinate vector of Aj with respect

to the standard basis of the space of 2×2 matrices (in other words, vj is Aj written as a vector).

Therefore, we place the vectors vj into a matrix and row-reduce. We get

(v1 v2 v3) =

1 3 −1

−1 −2 3

3 9 5

−1 1 9

R2 R2+R1−−−−−−−−−−−−−−−−→R3 R3−3R1, R4 R4+R1

1 3 −1

0 1 2

0 0 8

0 4 8

R4 R4−2R2−−−−−−−−→

1 3 −1

0 1 2

0 0 8

0 0 0

.

Every column in the row-echelon form is leading. Therefore, S̃ is a linearly independent set.

Thus S is a linearly independent set, so the answer to the question is “yes”.

17

Question 2

By inspection, we see that(9

1

)=

(1

−3

)+ 4

(2

1

), i.e. b = v1 + 4v2.

This means that the coordinate vector of b with respect to the basis B := {v1,v2} is

(1

4

).

Remark. If you couldn’t spot the required linear combination by inspection like was done

above, you could always just let the desired coordinate vector be

(α

β

), which gives us (by

definition of coordinate vector)

αv1 + βv2 = b, i.e. α

(1

−3

)+ β

(2

1

)=

(9

1

)⇐⇒

(1 2

−3 1

)(α

β

)=

(9

1

).

You could then solve this linear system (e.g. using Gaussian Elimination, inverse matrix, high

school methods, or whatever method you like) to obtain the required values α and β.

Question 3

Note that

(1

0

)∈ R2, 2 ∈ R, and 2

(1

0

)=

(2

0

)∈ R2. Therefore,

T

(2

(1

0

))= T

(2

0

)=

(22

02

)

=

(4

0

),

but

2T

(1

0

)= 2

(12

0

)

=

(2

0

)

6= T

(2

(1

0

))=

(4

0

).

Hence T does not preserve scalar multiplication and so cannot be linear.

Remember, to show that a map is not linear, it suffices to come up with a single numerical

example demonstrating that one of the properties of linear maps (preservation of addition or

scalar multiplication) is not obeyed.

18

Question 4

(i) We row-reduce A, and augment b as this will be useful for part (ii). We have

[A | b] =

−1 2 1 0 1 4

1 −1 0 −1 0 1

3 −2 1 −4 1 −3

R2 R2+R1−−−−−−−−→R3 R3+3R1

−1 2 1 0 1 4

0 1 1 −1 1 5

0 4 4 −4 4 9

R3 R3−4R2−−−−−−−−→

−1 2 1 0 1 4

0 1 1 −1 1 5

0 0 0 0 0 −11

.

As we can see, there are precisely two leading columns in the row-echelon form of A. Therefore,

we have rank(A) = 2. The number of non-leading columns in the row-echelon form of A is 3,

so nullity(A) = 3.

(ii) No, because the right-hand column in the row-echelon form in part (i) is leading.

19













Question 1

(i) We place (the coordinate vectors of) the matrices A1, A2, A3, A4 and B into an augmented

matrices (as columns), and row-reduce. We get1 1 3 1 2

2 1 7 0 1

1 0 4 −1 −1

−1 1 −5 3 4

R2 R2−2R1−−−−−−−−−−−−−−−→R3 R3−R1R4 R4+R1

1 1 3 1 2

0 −1 1 −2 −3

0 −1 1 −2 −3

0 2 −2 4 6

R3 R3−R2−−−−−−−−→R4 R4+2R2

1 1 3 1 2

0 −1 1 −2 −3

0 0 0 0 0

0 0 0 0 0

.

As the right-hand column in the row-echelon form is non-leading, B is an element of span {A1, A2, A3, A4}.

(ii) No, because not every row is leading in the row-echelon form of the left-hand part of the

augmented matrix above (i.e. there exists a zero row in the left-hand part).

20

Question 2

Method 1 – Inspection

By inspection, we see that(7

8

)= 5

(1

4

)−2

(−1

6

), i.e. b = 5v1−2v2.

This means that the coordinate vector of v with respect to the basis {v1,v2} is

(5

−2

).

Method 2 – Usual procedure

Let B = {v1,v2} and let the desired coordinate vector be [b]B =

(α

β

). Then by definition, we

have

αv1 + βv2 = b, i.e. α

(1

4

)+ β

(−1

6

)=

(7

8

).

Defining B to the matrix with columns v1 and v2, i.e. B = (v1 v2) =

(1 −1

4 6

), the above

equation can be equivalently expressed in matrix form as

B

(α

β

)= b ⇐⇒

(1 −1

4 6

)(α

β

)=

(7

8

)

This is a linear system that we can solve for

(α

β

)using a number of methods (e.g. Gaussian

Elimination, high school methods, inverse matrix, etc.). For example, using the inverse matrix

method, we obtain (α

β

)=

(1 −1

4 6

)−1(7

8

)

=1

6× 1− 1× (−4)

(6 1

−4 1

)(7

8

)

=1

10

(50

−20

)

=

(5

−2

).

21

Thus the answer is [b]B =

(α

β

)=

(5

−2

).

Remember, if we have a 2× 2 real or complex matrix A =

(a b

c d

), then its inverse is given by

A−1 = 1det(A)

(d −b−c a

), provided that det(A) 6= 0. In other words, switch the diagonal entries,

negate the others, and remember to divide by the determinant.

Question 3

Since

T (0) = 0 + 2× 0 + 3× 0 + 4 = 4 6= 0,

T is not a linear transformation.

Tip. Remember, a linear map must map the zero vector of its domain to the zero vector of its

codomain. So if you can show that a map does not map the zero vector of its domain to the

zero vector of its codomain, it cannot be linear. Of course, if a map does map the zero vector

of its domain to the zero vector of its codomain, we cannot conclude just from this whether or

not the map is linear.

Question 4

To find the kernel, we solve Ax = 0. To do this, we just row-reduce A, augmented with 0 (but

we don’t need to write the augmented part all the time since the zero vector does not change

when applying elementary row operations to it. So we will only write it at the last step.). We

have (keeping in mind that column j corresponds to xj , for j = 1, 2, 3, 4, 5)

A =

1 −1 3 2 0

2 −1 5 4 −1

3 −2 8 7 2

R2 R2−2R1−−−−−−−−→R3 R3−3R1

1 −1 3 2 0

0 1 −1 0 −1

0 1 −1 1 2

R3 R3−R2−−−−−−−→

1 −1 3 2 0 0

0 1 −1 0 −1 0

0 0 0 1 3 0

.

We set x3 = α and x5 = β (free parameters) as columns 3 and 5 in the row-echelon form are

non-leading. Then from row 3, we have

x4 = −3x5 ⇒ x4 = −3β .

Now, from row 2, we have

x2 = x3 + x5 ⇒ x2 = α+ β .

22

Now, from row 1, we have

x1 = x2 − 3x3 − 2x4 = (α+ β)− 3α− 2 (−3β)⇒ x1 = −2α+ 7β .

Therefore, we have Ax = 0 if and only if

x =

x1

x2

x3

x4

x5

=

−2α+ 7β

α+ β

α

−3β

β

for some scalars α, β

= α

−2

1

1

0

0

+ β

7

1

0

−3

1

.

Therefore, a basis for ker(A) is

−2

1

1

0

0

,

7

1

0

−3

1

.

Tip. If you have time, you should check your answer by checking that A multiplied by each of

the vectors you found for the basis of the kernel results in the zero vector, i.e.

0

0

0

. If this

does not happen, you must have made a mistake somewhere in your working, and should go

back and find it.

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MATH1251 ALGEBRA S2 2010 TEST 1 VERSION 1A Sample …