Ellis Math1131 ALG

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University of New South Wales

MATH 1131

Mathematics 1A Algebra

Semester 1 2016

Dr Bill Ellis ∗

[email protected]

February 23, 2016

∗Original notes by Mr Peter Brown & modified by DrBill Ellis

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Chapter 1 - Introduction to Vectors

1.1 Vector Quantities

You may have already met the notion of a vec-

tor in physics. There you will have thought of

it as an arrow, that was used to represent aforce. Adding forces corresponded to ‘adding

arrows’. In this topic we are going to look at

vectors from a geometric view point, although

we will include some examples based on simple

ideas from physics.

One of the most powerful developments in Math-

ematics came from the simple idea of the co-

ordinate plane. Indeed 2-dimensional co-ordinate

geometry was crucial in the development of

the Calculus. The obvious question arises as

to how we can generalise this to higher dimen-

sions. Vectors give us a way of generalising

co-ordinate geometry into higher dimensions

in a very straight forward manner.

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Definition: A vector is a directed line seg-

ment which represents a displacement fromone point P to another point Q.

The word vector comes from the Latin veho

(cf. vehicle ), meaning to carry .

We represent a vector either using the notation−→P Q or by using v. In the Algebra Course Notes (yellow book) (and in these notes), vectors are

represented using bold letters, v. You, when

hand writing, should represent vectors by using

a tilde sign under the letter, viz v

or v. This is

important, because you will need to carefullydistinguish between vectors, scalars (and later

matrices).

v

P

R

S

−→P Q

−→SR

Q

w

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A vector has both direction and length (or

magnitude). Two vectors are equal if they

have the same direction and the same magni-

tude. Hence in the diagram v = w.

We will denote the length (magnitude) of the

vector v by |v|. (In textbooks you will also seev.)

Two vectors are parallel if they have the same

direction. (Not necessarily the same length.)

We choose a fixed point O, in whatever dimen-

sional space we happen to be and call this the

origin. The position vector of a point in any

number of dimensions will be represented by a

vector from the origin to that point.

O

P

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Hence the vector−→OP in the diagram is called

the position vector of the point P . A positionvector gives the position of a point in space,whereas a direction vector is simply a vectorhaving direction and magnitude (length).

Addition of vectors:

To geometrically add two vectors there aretwo different methods (each important). If wethink of a force vector, then, the obvious wayto add two vectors is to put them tip to tailand join the tail of the first to the tip of the

second, as in the diagram.

vw

v + w

w

To add the vectors v and w, we move w andthen complete the triangle. This method of addition is known as the triangle law of addi-tion.

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You can see from this that one could obtain the

same vector by forming a parallelogram fromthe two vectors and taking the diagonal (often

called the resultant) as the sum of the two

vectors.

v

w

v + w

This method is known as the parallelogramlaw.

Subtraction of vectors is performed in a similar

way:

w

v

w − v

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To check this makes sense, add the vectors

that are tip to tail, v + (w − v) = w as ex-pected. Observe that the vector labelled w − vis not a position vector.

Thus if P and Q have position vectors v and

w respectively, then −→P Q = w − v. In general,−→P Q =

−→OQ − −→OP .

w

v

w − v

O

P

Q

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The Triangle Inequality:

Let us restrict ourselves, for the moment, to

the plane.

Since the sum of any two sides of a triangle

must exceed the third side, we can write

|u + v| ≤ |u| + |v|for any vectors u and v.

u

v

u + v

Example: When do we have equality in theTriangle Inequality?

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Scalar Multiplication:

We can multiply a vector by a scalar λ (gener-

ally just a real number).

This has the geometric effect of stretching the

vector if λ > 1, stretching and reversing itsdirection if λ


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Commutative and Associative Laws:

The commutative law of vector addition states

that a + b = b + a.

Geometrically this is obvious:

a

b

a + b = b + aba

The associative law of vector addition states

that a+(b+c) = (a+b)+c. Again the following

geometric proof will suffice.

a

b

b + ca + b

c(a + b) + c =

a + (b + c)

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Example: Construct a vector equation to de-

termine the midpoint of the line joining the

points P and Q.

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1.2 Vector Quantities and Rn

Thus far, much of what we have done works in

any number of dimensions. We are now going

to define n dimensional space and introduce a

co-ordinate system in which to place our vec-

tors.

We take an n-tuple

a1a2

...an

of real numbers and

think of each ai as lying on an axis xi. In 2 and3 dimensions, we identify these axes as the XY

and XY Z axes respectively, which are mutually

orthogonal.

The set of all such n-tuples will be called Rn.

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We say that the vector−→P Q in Rn has co-ordinates

a1a2

...an

, if we must move a1 units along the x1axis, a2 units along the x2 axis, and so on,

when moving from P to Q.

Hence an n-tuple

a1a2

...an

in Rn will interpretedas the position vector of a point P in Rn.

For example, in R3, the point P in the diagram

has position vector

231

.

P 2

31

O =

00

0

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We can then define the addition of two vectors

(algebraically) in Rn by

a1a2

...an

+

b1b2

...bn

=

a1 + b1a2 + b2

...an + bn

and multiplication by a scalar λ to be

λ

a1a2

...an

=

λa1λa2

...λan

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Multiplying a vector by a scalar λ merely stretches

the vector (if λ > 1 ) or shrinks it if 0 ≤ λ


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Note that we can now prove such rules as the

commutative law algebraically, viz:

a + b =

a1a2

...an

+

b1b2

...bn

=

a1 + b1a2 + b2

...an + bn

=

b1 + a1b2 + a2

...bn + an

=

b1b2

...bn

+

a1a2

...an

= b + a.

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Two vectors are defined to be parallel if one is

a non-zero multiple of the other. That is, v is

parallel to w if v = λw for some scalar λ = 0.

Example:

123

is parallel to

−2−4

−6

.

Example: Find the vectors−→P Q, and

−→QP if

P =

7−1

3

and Q =

21

−3

.

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Example: Suppose that A = 0

0 B = 1

4 ,C =

35

, D =

21

are the position vec-

tors for four points A,B,C,D. Prove that the

quadrilateral ABCD is a parallelogram.

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(cont.)

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1.3 Rn and Analytic Geometry

Basis Vectors:

The standard basis vectors in R2 are the vec-

tors 1

0 and 0

1 which are often denotedby i and j. Observe that every vector in R2 can

be written in terms of these basis vectors. For

example

1−2

can be written as i − 2 j.

In 3-dimensions, the basis vectors, i, j, k are 100

, 01

0

, 00

1

.

Note that every vector in R3 can be expressed

uniquely in terms of these basis vectors, viz: a1a2a3

can be expressed as a1i + a2 j + a3k.

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In higher dimensions, we label the basis vectors

as e1, e2, e3,... and so on.

Thus, in R4, we have

e1

=

10

00

, e2 =

01

00

, e3 =

00

10

, e4 =

00

01

.

Once again, we can represent any vector in Rn

uniquely in terms of the standard basis vectors

in Rn.

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Distances and Lengths:

Given a vector x =

a

b

in R2, we can use

Pythagoras’ Theorem to compute the length

of this vector as

a2 + b2. We use the notation

|x| = a2 + b2.In R3, given a vector x =

abc

, we can seefrom the diagram that

|OP

| = a2 + b2 and

then in ∆OAP we have |OA| = |x| = a2 + b2 + c2.

X

O

b

a

c

A

P

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In higher dimensions, we can define the length

of a vector by generalising this formula, i.e.

Definition: A vector x =

a1a2

...an

in Rn has

length |x| given by |x| = a21 + a22 + · · · + a2n.Example: Find the lengths of vectors a =

−13

2

and b =

12

34

.

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The distance between two points A and B in

Rn will be defined as the length of the vector−→AB, in other words, if A has position vector a =

a1a2

...an

and B has position vector b =

b1b2

...bn

,

then the length of −→AB is

| −→AB | = |b − a|=

(b1 − a1)2 + · · · + (bn − an)2.

Note: | −→AB | = | −→BA |

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Example: Find the distance between 1

−2 and

3−4

and between

−125

and 36

−1

.

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The ‘length function’

| · |, (sometimes called a

norm) has the following properties:

(i) |a| ≥ 0.

(ii) |a| = 0 if and only if a = 0.

(iii) |λa| = |λ||a|, for λ ∈ R.

A vector which has unit length is called a unitvector. Any vector can be made into a unit

vector by dividing by its length. The vector v̂

is the unit vector of v.

Example: Find a unit vector parallel to the

vector

2−31

.

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Example: Suppose A and B are points with

position vectors a and b respectively. Find a

vector (in terms of a and b) which bisects the

angle AOB, where O is the origin.

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Example: Are the points A(1, 2, 3), B(3, 1, 1)

and C (7, −1, −3) collinear?

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Example: If A(

−1, 3, 4), B(4, 6, 3), C (

−1, 2, 1)

and D are the vertices of a parallelogram, de-

termine all the possible co-ordinates for the

point D.

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(cont.)

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1.4 Lines

We seek to find the equation of a line in vector

form. The vector equation of a line is a formula

which gives the position vector x of every

point on that line. This equation is sometimes

referred to at the parametric vector form of the line. I will generally just say ‘vector equa-

tion’ of the line.

Suppose we have a line passing through the

origin which contains a vector u in R2

. Everypoint on that line will have a position vector

which is a multiple of u. Conversely, every mul-

tiple of u will correspond to the position vector

of a point on that line. Hence the equation of

the line can be written as x = λu where λ is

any real number.

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u

x = λu

0

For example, if u were the vector 23 thenthe equation of the line through u passing through

the origin would be x = λ

23

.

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Another way of denoting the set of all real mul-

tiples of a given vector is to call it the span

of the vector. Thus we could write λu as

span(u). This idea of span is extremely im-

portant. It will be developed more formally in

MATH 1231.

Example: In R2 what is the span of

10

?

What is span

11

?

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The advantage of this definition of the equa-

tion of a line is that it easily generalises to any

number of dimensions. For example, the equa-

tion of the line in R3 which passes through the

origin and is parallel to the vector

−1

3

−6

is

simply x = λ

−13−6

.If the line does not pass through the origin,

then we proceed as follows: To find the vectorequation of a line we need to know two things:

(i) The position vector a of a point on the line

(ii) The direction of the line, i.e., a vector b

parallel to the line.

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a

x = a + λb

O

bλb

A

X

x

Thus, the span of b will give a line through the

origin parallel to b and adding a will shift the

line to its proper position. Thus we can find

the position vector x of any point X on the

line by going from the origin along the vector

a and then moving along the line, by addingsome multiple of b until we reach X . Thus,

the vector−→OX is given by

−→OX =

−→OA +

−→AX and−→

AX is some multiple of b. Hence the equation

of the line is x = a + λb.

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Example: Find the vector equation of the line

passing through the point P with position vec-

tor

2−35

and parallel to the vector 16

−4

.

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passing through the two points P, Q with posi-

tion vectors p =

−126

and q = 4−2

3

.

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in 2-dimensions with Cartesian equation

y = 2x + 1.

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Example: Does the point (

−1, 2, 3) lie on the

line x = 34

11

+ λ 214

?

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1.4.2 Lines in R3

In R3 two lines can

• meet at a point

• be parallel

• neither meet nor be be parallel.

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Cartesian Equations of the Line: Given the

vector form of a line in 3-dimensions, we canwrite down the Cartesian equations (note the

plural) as follows. For example, suppose the

vector equation is x =

−1

23

+λ

3

−21

. Re-

call that x is simply an abbreviation for

x1x2x3

.Hence, equating co-ordinates, we can write

x1 = −1 + 3λ, x2 = 2 − 2λ, x3 = 3 + λ. Elimi-nating λ from these equations we have

x1 + 1

3 =

x2 − 2−2 =

x3 − 31

(= λ) .

These are called the Cartesian equations of the

line. Clearly this can be done for any such line

and so the general form is

x1 − av1

= x2 − b

v2=

x3 − cv3

(= λ),

where (a,b,c) is a point on the line and (v1, v2, v3)

is a direction vector of the line, provided that

none of the numbers v1, v2, v3 is zero.

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Hence in vector form this would be

x = ab

c

+ λ v1v2v3

. The following exampletells us what to do when one of the compo-

nents of the direction vector is zero.

Example: Convert x = 21

−5 + µ 20

−1

into Cartesian form.

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Observe that two lines will be parallel if their

direction vectors are parallel. Two direction

vectors are parallel if and only if one is a nonzero

multiple of the other. For example the lines

x =

156

+λ

32

−1

and x =

−2

47

+λ

64

−2

are parallel since their direction vectors are

32−1

and

64

−2

respectively and the second vec-

tor is simply a multiple of the first. Observe

also that the equations x =

156

+ λ 32

−1

and x =

15

6

+ λ

64

−2

represent the same

lines, since they are parallel and pass through

the same point.

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passing through (1, −2, 5) and parallel tox + 1

3 =

y − 1−2 =

z + 6

−4 .

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Example: Find the intersection (if possible)

of the lines x = −10

3

+ λ 12−1

andx =

35

−2

+ µ

21

−3

.

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(cont.)

Two non-parallel lines in space that never meet

are called skew lines.

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1.5 Planes

In 3-dimensions and higher, we can construct

planes. Suppose we seek the vector equation

of the plane passing through the origin parallel

to two given non-parallel vectors a and b.

a λaO

b

µb

X

P

Q

x

To reach any point X with position vector x

on the plane we need to ‘stretch’ the vector a

to P and ‘stretch’ the vector b to Q in such a

way that −→OX =−→OP + −→OQ. Thus, x = λa + µb.Conversely, if we take the vector which results

from adding a multiple of a and a multiple of b

then this will be the position vector of a point

on the plane.

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Example: Find the vector equation of the

plane passing through the origin parallel to the

vectors

15−3

and −24

7

.

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The plane generated by two such vectors is

called the span of the two vectors. So for ex-

ample, the span of

15−3

and −24

7

is sim-ply the set

λ

15

−3

+ µ

−24

7

: λ, µ ∈ R

.

This set is also referred to as the set of all lin-

ear combinations of the two vectors. Thus,

given any two vectors a, b, span{a, b} = {λa +µb : λ, µ ∈ R} and we say that x is a linearcombination of a and b if x = λa + µb for

some particular λ and µ.

Example: Describe the span of

123

,

−135

.

Repeat for span 123

, 246

.

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As with the equation of a line, to get the vector

equation of a plane not through the origin, we

simply shift the plane by adding any position

vector of a point which lies on the plane. Thus

to obtain the vector equation of a plane we

need:

(i) The position vector of a point on the plane.

(ii) Two non-parallel vectors which are parallel

to (or lie on) the plane.

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plane passing through the point P with posi-

tion vector

2−35

and parallel to the vectors

16

−4

and

2

−5

1

.

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plane passing through the three points P, Q, R

with position vectors P =

−1

262

, Q =

4−2

3−1

and R = 1

7−2

5

.

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Configurations

In R3, two (distinct) planes can be

• parallel

• meet in a line

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In R3, three (distinct) planes can be arranged

so that

• the three planes are parallel

• two planes are parallel and the third planeis parallel to neither the first two.

• they meet at a single point

• they meet in a line

• none are parallel, but no point lies on allthree planes.

In the next chapter we will learn how to analyse

these scenarios algebraically.

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Regions

Example: The set

S =

x = λ

120

+ µ

−124

, 0 ≤ λ ≤ 1, 0 ≤ µ ≤ 1

represents the parallelogram with vertices O, −124

, 12

0

, 04

4

in R3.Example: The set

S =x = λ

120

+ µ −124

, 0 ≤ λ ≤ 1, 0 ≤ µ ≤ λrepresents the triangle with vertices O,

120

,

044

in R3.

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1.5.3 Cartesian Form of a Plane in R3

As with lines, we can find the Cartesian equa-

tion of a plane by eliminating the two param-

eters λ and µ. This is generally quite fiddly

to do algebraically. Later in this course, you

will see a much better method, but for themoment, we will do it by algebra.

Example: Find the Cartesian equation of the

plane x =

12

3

+ λ

24

0

+ µ

−10

3

.

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(cont.)

The procedure can be reversed to find the vec-

tor equation of a plane from the Cartesian

equation.

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Example: Find the vector equation form of

the plane 3x − 6y + 2z = 12.

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Further Examples:

Example: Find the intersection of the planes

2x + y − z = 10 and 3x + 4y + 2z = 29.

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Example: Show that the line x = λ11

−5−3 isparallel to the plane

x =

256

+ µ 42

6

+ ν −13

5

.

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Example: Find the intersection of

x = 12

3

+ λ −14−2

and 2x + 3y − z = 29.

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Chapter 2 - VECTOR GEOMETRY

In Chapter 1, see looked at the rudiments of

vector geometry. We now have enough ma-

chinery to study this subject in greater depth.

2.1 Lengths

In higher dimensions, we can define the length

of a vector as follows:

Definition: A vector x =

a1a2...

an

in Rn haslength |x| given by

|x

| = a

21 + a

22 +

· · ·+ a2n.

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The distance between two points A and B in

Rn will be defined as the length of the vector−→AB, in other words, if A has position vector a =

a1a2

...an

and B has position vector b =

b1b2

...bn

,

then the length of −→AB is

| −→AB | = |b − a|=

(b1 − a1)2 + · · · + (bn − an)2.

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The ‘length function’

| · |, (sometimes called a

norm) has the following properties:

1. |a| ≥ 0.

2. |a| = 0 if and only if a = 0.

3. |λa| = |λ||a|, for λ ∈ R.

A vector which has unit length is called a unitvector. Any vector can be made into a unit

vector by dividing by its length. The vector v̂

is the unit vector of v.

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2.2 Scalar (Dot) Product

Definition: The scalar or dot product of two

vectors a and b in Rn is given by

a · b = a1b1 + a2b2 + · · · + anbn =n

k=1akbk .

The scalar product has a geometric interpre-

tation in R2 and R3.

Indeed for non-parallel vectors a and b in R3,

we consider the following triangle.

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c

b

a

We can write c = a − b and so taking lengths,we have

|c|2 = (a1 − b1)2 + (a2 − b2)2 + (a3 − b3)2= |a|2 + |b|2 − 2(a1b1 + a2b2 + a3b3)= |a|2 + |b|2 − 2 a · b.

Also, if we write down the cosine rule for this

triangle, we have |c|2 = |a|2 +|b|2−2|a||b| cos θ,where θ is the angle between the vectors a and

b. Comparing the two expressions we see that

a1b1 + a2b2 + a3b3 = a · b = |a||b| cos θ.This formula is fundamental and should be

committed to memory.

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Notice that if the angle between a and b is π

2(≡ 90◦) we have

a · b = a1b1 + a2b2 + a3b3 = 0.

Example: Find the scalar product and hence

the acute angle between the vectors a and b if

a =

1−12

and b = 21

1

.

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(cont.)

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2.2.1 Arithmetic Properties of the Scalar

Product

The scalar product has the following proper-

ties:

1. a · a = |a|2 and so |a| = √ a · a.

2. a ·b is a scalar (which is why it is called thescalar product ).

3. a · b = b · a. (Commutative law).

4. a · (λb) = λ(a · b).

5. a · (b + c) = a · b + a · c. (Distributive law).

The proofs of these are straightforward.

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Example: Prove that the diagonals of a rhom-

bus are perpendicular.

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2.2.2 Geometric Interpretation of the Scalar

Product in Rn

In R2 and R3, the equality a · b = |a||b| cos θimmediately yields,

|a

·b| ≤ |

a||

b|,

since | cos θ| ≤ 1 and |a| ≥ 0, |b| ≥ 0.

This is known as the Cauchy-Schwartz inequal-

ity.

It generalises to vectors in Rn, but the proof

required there is different.

Theorem 3: (Cauchy-Schwartz inequality).

If a, b ∈ Rn then

|a · b| ≤ |a||b| .

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Proof: The result is clearly true if either vector

is the zero vector, so we may suppose thatneither vector is zero.

Define the function f (t) by

f (t) =

|a

−tb

|2.

Observe that f (t) ≥ 0 for all t and that f (t) issimply a quadratic. Using rule (1) above, we

may write f (t) as

f (t) = (a − tb) · (a − tb) = t2|b|2 − 2ta · b + |a|2

and so f (t) has a minimum when t = a · b|b|2 .

Substituting into f , we see that the minimum

value of f is |a|2 − (a · b)2

|b|2

Thus, as f (t) ≥ 0 for all t, we have |a|2 −(a · b)2

|b|2 ≥ 0 which implies that (a·b)2 ≤ |a|2|b|2.

Taking positive square roots gives the result.

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In higher dimensions we can define the angle

θ between two vectors a and b by

cos θ = a · b|a||b|.

By the Cauchy-Schwarz inequality, the right-

hand side lies in [−1, 1], so this definition makessense.

Now we can give a general proof for the trian-

gle inequality for vectors in Rn.

Theorem 4: (The Triangle Inequality.)

If a, b ∈ Rn then |a + b| ≤ |a| + |b|.

Proof:

Again, using rule (1),

|a + b|2 = (a + b) · (a + b) = |a|2 + 2a · b + |b|2≤ |a|2 + 2|a||b| + |b|2= (|a| + |b|)2.

Taking positive square roots gives the result.

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2.3 Applications: Orthogonality and Pro-

jection

2.3.1 Orthogonality of Vectors

Two vectors a and b are said to be orthogonal

if a · b = 0.

In two and three dimensions, this is the same as

saying that the two vectors are perpendicular.

Example: Show that 25−1

is orthogonal to 31

11

.

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Example: Write down a unit vector which is

perpendicular to both 2−6

−3 and 43

−1.

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Example: Prove that the altitudes of a trian-

gle are concurrent.

Suppose our triangle is ABC which vertices at

position vectors a, b, c. Let P be the intersec-

tion of the altitudes AD and BE , and have

position vector p. Hence

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A set of vectors which are mutually orthogonal

is called an orthogonal set.

For example the set of vectors

111

,

1

−10

,

11

−2

is an orthogonal set.

A set of vectors which each have length 1 and

are also orthogonal is called an orthonormal

set.

For example the 3 basis vectors in R3, 10

0

, 01

0

, 00

1

form an orthonormal set.

The set 1√ 2 11 , 1√ 2 1−1 is an orthonor-mal set in R2.

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Example: Suppose that

{a1, a2, a3

} is an or-

thonormal set in Rn, (with n ≥ 3) andb = c1a1 + c2a2 + c3a3.

Find the scalars c1, c2, c3.

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2.3.2 Projections

In the diagram, we are given two (non-zero)

vectors a and b.

a

bprojbaB

A

P O

θ

Drop a perpendicular AP to OB from A.

The vector −→OP is called the projection of aonto b. It is parallel to b.

Now for θ acute, |−→OP | = |a| cos θ = a · b|b| using

the properties of the scalar product. Also−→OP =

| −→OP | b|b| and so, in two and three-dimensionalspace,

projba =

a · b|b|2

b =

a · bb · b

b.

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In higher dimensions we will take this for our

definition of the projection of a onto b.

Note that

a100

is the projection of a1a2

a3

onto the unit vector e1 and so on.

Example: Find the projection of

2−31

onto

−2

3

6 .

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Example: Find the length of the projection of 2−3

onto 56.

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2.3.3 Distance between a Point and a Line

in R3

Example: Find the shortest distance from the

point B =

038

to the line x =

123

+

λ 1−1

4

.

O

A

B

P

ad

Let A =

123

and thus a = −→OA. Similarly b =−−→OB. Also let P be the foot of the perpendicular

from B to the line.

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Then −→AP is the projection of

−→AB onto the di-

rection vector d = 1−1

4

of the line. Wecould then find the length of

−−→P B using the

length of −→AB and Pythagoras.

An alternate method is as follows. Notice −→AB =b − a = −→AP + −−→P B. Thus

−−→P B =

−→AB − −→AP = −→AB − projd

−→AB

= b − a − projd(b − a)

= b − a − (b − a) · d|d|2 dHence

|−−→P B| =

|−→AB|2 − |−→AP |2.

For our example we have

|−→AB| =|−→AP | =

|−−→P B| =

|−→AB|2 − |−→AP |2 =83


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2.4 Vector (Cross) Product

There is another way in which we can define

vector multiplication. This product only makes

sense in R3 and is known as the vector or cross

product.

Given two (non-zero) vectors a =

a1a2a3

andb =

b1b2

b3

, we seek a third vector x =

x1x2

x3

which is perpendicular to both of these. From

the scalar (dot) product we know that such a

vector must satisfy each of the equations

a1x1 + a2x2 + a3x3 = 0

b1x1 + b2x2 + b3x3 = 0.We can solve these to obtain

x = λ

a2b3 − a3b2a3b1 − a1b3a1b2 − a2b1

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We put λ = 1 and define the vector (cross)

product of a and b by

a × b = a2b3 − a3b2a3b1 − a1b3

a1b2 − a2b1

.It can be quite difficult to remember, but can

be written mnemonically as a determinant (tobe studied in detail at the end of the course):

a × b =

i j k

a1 a2 a3b1 b2 b3

where, recall,

i =

100

, j =01

0

, k =00

1

.That is, we expand the determinant and write

a × b = i a2 a3b2 b3− j a1 a3b1 b3

+ k a1 a2b1 b2 .

Each subdeterminant is then evaluated to give

the three co-ordinates of the vector product.

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Example: Use the determinant formula to find

the cross product of 2−1

2

and 6−2−3

.

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2.4.1 Arithmetic Properties of Vector Prod-

ucts

Suppose a, b and c are vectors in R3. Then

(i) a × a = 0.

(ii) b × a = −a × b.

(Note then that the cross product is NOT

commutative. Changing the order changes the

sign.)

(iii) a × (λb) = λ(a × b) and (λa) × b = λ(a × b)for λ ∈ R.

(iv) From (i) and (iii) we have a

× (λa) = 0.

In other words, the cross product of parallel

vectors is the zero vector.

(v) a × (b + c) = a × b + a × c and (a + b) × c =a × c + b × c. (Distributive Law).

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(vi) For the three standard basis vectors in R3

we have

e1 × e2 = e3, e2 × e3 = e1, e3 × e1 = e2.

(vii) |a × b| = |a||b| sin θ, where θ is the anglebetween the two vectors.

All of these may be proven by writing out the

co-ordinates. Some of the proofs are given in

the Algebra Course Notes (yellow book).

Example: If a, b

= 0 and a

× b = 0 explain

why the vectors must be parallel.

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2.4.2 A Geometric Interpretation of the

Vector Product

Note in particular result (vii) in the previous

section. We can use this to find:

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2.4.3 Area of a Parallelogram

Given a parallelogram S in R3, defined by

S = {λa + µb : 0 ≤ λ ≤ 1, 0 ≤ µ ≤ 1},the area of S is given by |a||b| sin θ.

0 a

b Area = |a × b|θ

But by (vii) above, this is simply |a × b|.

Example: Find the area of the parallelogram

spanned by

23

−1

and

−1

37

.

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2.5 Scalar Triple Product and Volume

The scalar triple product of three vectors is

defined by

a · (b × c).

We will see shortly that it also has a geometricinterpretation.

Note that in the evaluation of the triple prod-

uct, you must perform the cross product first.

Example: Find the scalar triple product of 2−13

, −31

4

, 5−2

1

.

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Theorem: For a, b, c we have

(i) a · (b × c) = (a × b) · c

(ii) a · (b × c) = −a · (c × b)

(iii) a · (a × b) = (a × a) · b = 0

(iv) The scalar triple product can be written

as a determinant, viz:

a · (b × c) = a1 a2 a3

b1 b2 b3c1 c2 c3

.

There is also a vector triple product which

arises in Physics. The vector triple product

of three vectors a, b, c is given by a × (b × c).Note that cross product is NOT associative so

a × (b × c) = (a × b) × c.

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2.5.1 Volume of Parallelepipeds

n

b B

A

a

O

C

c Volume = |a · (b × c)|

A parallelepiped is the 3-dimensional analogue

of a parallelogram.

(The word comes literally from the Greek

παραλληλǫπίπǫδoν meaning that (the top) is

parallel to (ǫπί) the base (literally plane )).

Suppose we have three non-coplanar vectors

a, b and c, which form the edges of the par-

allelepiped. The parallelepiped is said to be

spanned by these three vectors.

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To find the volume, we must multiply the area

of the base by the perpendicular height.

Let n be a normal to the base given by n =

b × c. The perpendicular height then, is givenby the length of the projection of a onto n

which is simply

|projna| = |a · n|

|n| = |a · (b × c)|

|b × c| .

Now the denominator of this expression is sim-

ply the area of the base parallelogram, and sothe volume is given by the magnitude of the

scalar triple product, i.e., |a · (b × c)|.

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Example: Find the volume of the parallelepiped

spanned by 2−4

1

, 3−46

, −32−5

.

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Note that if the three vectors are coplanar then

the volume of the parallelepiped that they span

will be zero, and conversely, so this gives us:

Theorem: Three vectors in R3 are co-planar

if and only if their scalar triple product is zero.

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2.6 Planes in R3

2.6.1 Equations of Planes in R3

We can use the scalar (dot) product to give analternate formula for the equation of a plane.

Suppose we know a point A on the plane anda vector n which is perpendicular to the plane.

0

X

n

Axa

Take any general point X on the plane. ThenAX and n are perpendicular and so their scalarproduct is zero. Hence n

·−−→AX = 0. If we let

x be the position vector of X and a be theposition vector of A, then

n · (x − a) = 0.This is called the point normal form of theplane.

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Example: Find the point normal form and

hence the Cartesian equation of the plane pass-

ing through

2−13

with normal 1−2

1

.

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Observe that if n = n1

n2n3, then when we

use the point-normal form of the equation, the

co-efficients in the Cartesian equation are pre-

cisely the components of this vector. We can

exploit this idea to convert a plane from vector

form to Cartesian form.

Example: Find the Cartesian form of the plane

whose equation is

x = 2

−1

2 + λ−1

−2

4 + µ3

−4

2 .

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Example: Convert to point-normal form, the

plane with equation 2x − 3y + 4z = 12.

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Example: Convert to point-normal form, the

plane with vector equation

x =

25−3

+ λ 30

−2

+ µ 01

−2

.

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2.6.2 Distance between a point and a plane

in R3

Find the shortest distance from the point

B =

1

−23

to the plane

x =

200

+ λ 110

+ µ −102

.

Let A be the point 2

00 and P be the foot of the perpendicular from B to the plane. Then

BP is simply the length of the projection of

AB onto the normal n to the plane.

A

P

B

n

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(cont.)

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Chapter 3 - Complex Numbers

3.1 & 3.2 Review of Number Systems, In-

troduction to Complex Numbers

Let us begin by trying to solve various algebraic

equations. Suppose we only know about the

set of natural numbers (written as N). Thenwe can solve the equation x − 3 = 2 and ob-tain the solution x = 5. On the other hand,

if we try to solve the equation x + 3 = 2 then

there is no solution! To solve this equation we

need a larger set of numbers which includes thenegative whole numbers as well as the positive

ones. This set is called the set of integers and

is denoted by Z. Continuing this idea:

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In Z: x + 3 = 2 3x = 2x = −1 No solution

In Q: 3x = 2 x2 = 2

x = 23 No solution

In R: x2 = 2 x2 + 1 = 0

x = ±

√ 2 No solution

N

ZQ R

C

10−1

−2

23

√ 2

π

i

2 − i

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At each stage in the above we are able to solve

each new type of equation by extending the set

of numbers in which we are working. Hence,

to solve the equation x2 = −1 we introducea new symbol i (much as we introduced the

symbol√

2 to solve x2 = 2.) We define i to

be the (complex) number whose square is −1.i.e. i2 = −1. Using this new symbol, we cannow solve x2 = −1 to obtain solutions x = ±i.Furthermore, we can define the complex num-

bers by:

Definition: The set of all numbers of the form

a + bi where a, b are real numbers, i.e., a, b ∈ Rand i2 = −1 is called the set of all complexnumbers, C.

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As with the set of real numbers and the set of

rational numbers, if we add, subtract, multiply,

or divide (with the exception of division by 0)

any two complex numbers, we again obtain a

complex number. This property is called clo-

sure . The complex numbers are closed under

addition, subtraction, multiplication and divi-sion (not by 0). A set of objects which has

these (and a number of other properties) is

called a field in mathematics.

Definition: Let F be a non-empty set of el-

ements for which a rule of addition (+) and

a rule of multiplication (×) are defined. Thenthe system is a field if the following twelve

axioms (or fundamental number laws) are sat-

isfied.

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1. Closure under Addition. If x, y

∈ F then

x + y ∈ F.

2. Associative Law of Addition.

(x + y) + z = x + (y + z) for all x, y, z ∈ F.

3. Commutative Law of Addition.

x + y = y + x for all x, y ∈ F.

4. Existence of a Zero. There exists an ele-

ment of F (usually written as 0) such that0 + x = x + 0 = x for all x ∈ F.

5. Existence of a Negative. For each x ∈F, there exists an element w

∈ F (usually

written as −x) such that x+w = w+x = 0.

6. Closure under Multiplication. If x, y ∈ Fthen xy ∈ F.

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7. Associative Law of Multiplication.

x(yz) = (xy)z for all x,y,z ∈ F.

8. Commutative Law of Multiplication.

xy = yx for all x, y ∈ F.

9. Existence of a One. There exists a non-

zero element of F (usually written as 1)

such that x1 = 1x = x for all x ∈ F.

10. Existence of an Inverse for Multiplica-tion. For each non-zero x ∈ F, there existsan element w of F (usually written as 1/x

or x−1) such that xw = wx = 1.

11. Distributive Law. x(y + z) = xy + xz for

all x,y,z ∈ F.

12. Distributive Law. (x + y)z = xz + yz for

all x,y,z ∈ F.

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Example: Is the set

{−1, 1

} a field?

Example: Is the set {0, 1} a field?

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Example: Is the set of integers Z a field?

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Example: How do we show that the set of

complex numbers C is a field?

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3.3 The Rules of Arithmetic for Complex

Numbers

Rules: Let z = a + ib, w = c + id be com-

plex numbers such that a,b,c,d ∈ R. Then

(i) z ± w = (a ± c) + i(b ± d)

(ii) zw = (ac − bd) + i(ad + bc).

(iii) z

w=

a + ib

c + id× c − id

c

−id

= (ac + bd) + i(bc − ad)

c2 + d2

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Example: If z = 1 + 2i then calculate z2.

114


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Example: If z = 2 + 5i then calculate 1

z

in

Cartesian form, i.e., 1

z= a + ib.

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3.4 Real Parts, Imaginary Parts and Com-

plex Conjugates

Definition: The real part of z = a + bi (writ-

ten Re(z)), where a, b ∈ R, is given by

Re(z) = a.

Definition: The imaginary part of z = a + bi

(written Im(z)), where a, b ∈ R, is given by

Im(z) = b.

Note: The imaginary part of a complex num-

ber is a real number!

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Example: If z = 2 + 5i what is Im(z)? Re(z)?

Example: If z = 3+4i what is Im(z2)? Re(z2)?

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Example: What is Im1 + 2i

3 − 4i?

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Example: If z = a+bi what is Im(iz)? Re(iz)?

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Example: For any two complex numbers z1

and z2 is the following true

Im(z1z2) = Im(z1)Im(z2) ?

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We noted in the previous lecture to calculate

1z where z = a + bi it was useful to multiplytop and bottom of 1z by a − bi. We give thiscomplex number a special name.

Definition: If z = a + bi, where a, b ∈ R, thenthe complex conjugate of z is z = a

−bi.

Properties of the Complex Conjugates.

1. z = z.

2. Re(z) = 1

2(z + z) and Im(z) =

1

2i(z − z).

3. zz ∈ R and zz ≥ 0.

4. z + w = z + w and z − w = z − w.

5. zw = z w and

z

w

=

z

w.

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Example: Prove that zz

≥ 0.

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Example: Prove that zw = z w.

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Example: Let P (z) = anzn + an−

1zn−1 +

· · ·+

a1z + a0 be a polynomial with real coefficients

an, an−1, . . . , a1, a0. If z1 is a complex root of P (z), i.e., P (z1) = 0, show that z1 is also a

root.

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(cont.)

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3.5 The Argand Diagram

Complex numbers can be represented using

the Argand plane, which consists of Cartesian

axes similar to that which you used to repre-

sent points in the plane. The horizontal axis is

used to represent the real part and the verticalaxis, (sometimes called the imaginary axis ), is

used to represent the imaginary part. For ex-

ample, the following points have been plotted:

−3, 2i, −3 + 2i, 4 + i.

−3

2i−3 + 2i4 + i

Complex numbers then are 2-dimensional, in

that we require two axes to represent them.

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Observe that a complex number z and its con-

jugate z are simply reflections of each other in

the real axis.

We lose the notion of comparison in the com-

plex plane. That is, we cannot say whether

one complex number is greater or lesser thananother.

Plotting real and imaginary parts leads to a

simple geometric picture of addition and sub-

traction of complex numbers. Addition andsubtraction is done by adding or subtracting

x-coordinates and y-coordinates separately, as

shown in the figure below.

z1z2

x2

y2

z1 + z2

0

z1z2

x2y2

z1 − z20

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3.6 Polar form, Modulus and Argument

You have already seen that complex numbers

can be expressed in Cartesian Form, a+ib,a,b ∈R. We can also specify a complex number z by

specifing the distance of z from the origin and

the angle it makes with the positive real axis.

Real axis I m a g i n a r y a x i s

x

y

θ

0

rz = x + yi

This distance is called the modulus (also called

magnitude and absolute value ) of z and written

as |z| while the angle is called the argument

and written as Arg(z). We insist, to removeambiguity, that −π < Arg(z) ≤ π.

Pythagoras’ theorem gives:

If z = x + yi then r = |z| =

x2 + y2.

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Care must be taken to find the correct argu-

ment. It is easiest to find the related angle θ

such that tan θ =ba and then use this to find

the argument in the correct quadrant recalling

that we use negative angles in the third and

fourth quadrant.

Example: Find the modulus and argument of

z = −1 + i√ 3 and w = 1 − 2i.

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Properties of Modulus:

(i) |zw| = |z||w|

(ii)

z

w

= |z||w|, provided w = 0.

(iii) |zn| = |z|n

(iv) |z| = 0 ⇒ z = 0.

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Example: Which of the following points lie

inside the circle |z − 1| = 2?1

2 + i, 2 + 3i,

√ 2 + i(

√ 2 + 1 ), −1

2 + i

√ 3

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Example: If z1 and z2 are arbitrary complex

numbers prove the triangle inequality |z1+z2| ≤|z1| + |z2|.

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Polar Form:

r sin θ

r cos θ

z = r(cos θ + i sin θ)

θ

r

From the diagram, we can see that the com-

plex number z can be written in the form z =

r(cos θ + i sin θ), where r is the modulus of zand θ is the argument of z. This is the called

the polar form of z.

Example: Write 1 − i in polar form.

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You will need to be able to convert a complex

number from Cartesian form, (a + bi), into po-

lar form r(cos θ + i sin θ) and vice-versa.

Example: Write√

3(cosπ

3 + i sin

π

3) in Carte-

sian form.

From the diagram we can also write down for

z = x + yi

cos θ = x x2 + y2

= x

r and sin θ = y x2 + y2

= y

r .

Also

Re(z) = x = r cos θ and Im(z) = y = r sin θ .

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Example: Write −2

1 + √ 3iin polar form.

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Example: Write 1

(1 − i)2

in polar form.

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Example: Determine a value for Arg(zz) if

z = 0.

You may have seen the abbreviation cis θ to

represent cos θ + i sin θ. You should not use

that here, since your tutor may not know what

it is. This form is NOT generally used in books

beyond High School. Moreover, as we shall

see, this polar form, is really a stepping stone

to a much better form which involves e.

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Example: Students were asked to write 1

(1 − i)2

in polar form. Different answers were given.

1

(1 − i)2 = 1

(1 − i)2(1 + i)2

(1 + i)2

= 2i

4

= i

2

= i sinπ

6

= 12cos π

2 + i sin π

2

Which one is correct?

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3.7 Properties and Applications of the Po-

lar Form

One important fact about the polar form is a

remarkable result called:

De Moivre’s Theorem:

For any real number θ, and any integer n, we

have

(cos θ + i sin θ)n = cos nθ + i sin nθ.

The proof of this, looking at the various cases

of n is given in the Algebra Course Notes (yel-

low book). The method of proof by induction

is used.

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Example: Let z = 1

−i√

3. Find z12.

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Let us write de Moivre’s theorem as follows:

Let f (θ) = cos θ + i sin θ, then (f (θ))n = f (nθ).

Also f (0) = 1.

Euler did the following:

He supposed we can differentiate the function,

treating i just like a real number.

If we momentarily ignore the logical difficulties

involved then f ′(θ) = i(cos θ + i sin θ).

Comparing this with ddθ(eiθ) = ieiθ, we can see

then that this function seems to have prop-

erties that are very similar to the exponential

function eiθ. We will therefore define:

Definition: eiθ

= cos θ + i sin θ (and hencee−iθ = cos θ − i sin θ).

This formula is sometimes called Euler’s for-

mula. We shall take it to be a definition of

the complex exponential.

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Thus, any complex number z can be expressed

in the polar form

z = reiθ

where r = |z| is the modulus and θ the argu-ment of z. For example,

z = 1 − i =

and z = −1 =

This last formula is quite remarkable since it

links together the four fundamental constants

of mathematics. In a very important sense,

this is the best way to write complex numbers.

We have used the term polar form in two dif-

ferent senses. From now on, when I say polarform, I will (generally) mean this new exponen-

tial form. You ought to be able to convert a

complex number from Cartesian form to polar

form and vice-versa.

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3.7.1 The Arithmetic of Polar Forms

Note the following important facts:

(i) The conjugate of the complex number

z = eiθ is given by z = e−iθ.

(ii) eiθ = ei(θ+2kπ) where k is an integer.

We can write cos θ and sin θ in terms of the

complex exponential as follows:

cos θ = eiθ + e−iθ

2 and sin θ =

eiθ − e−iθ2i

.

From the polar form, we can deduce the prop-

erties of modulus and argument which we listed

earlier. Let z = r1eiθ1 and w = r2e

iθ2 then

zw = r1r2ei(θ1+θ2) from which it follows that

|zw| = |z||w| and Arg(zw) = Arg(z) + Arg(w) + 2kπ zw = |z||w| and Arg

z

w

= Arg(z) − Arg(w) + 2kπ

where k is chosen so that −π < Arg(zw) ≤ πor −π


144/288

Example: Evaluate zw in polar form if

z = 2ei5π6 , w = 3eiπ3 .

Example: Evaluate the product

(1 + i)(1− i√ 3) to show that cos π12

= 1 +

√ 3

2√

2.

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Example: Find the square roots of 21

−20i.

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(cont.)

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Example: Find the roots (in Cartesian form)

of z2 − 3z + (3 − i) = 0.

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(cont.)

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3.7.2 Powers of Complex Numbers

Example: Find (√

3 − i)6.

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3.7.3 Roots of Complex Numbers

Definition: A complex number z is an nth

root of a number z0 if z0 is the nth power of

z, that is, z is the nth root of z0 if zn = z0.

To find roots of complex numbers, we will usethe polar form. Note that to find the nth root

of a complex number z0, we are really solving

zn = z0 and so we will convert z0 into polar

form. Such an equation will have n solutions!

To get all of these solutions we express z0 inpolar form, using the general argument, not

the principal one. An example will make this

clear.

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Example: Find the 7th roots of

−1.

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(cont).

If we plot these complex numbers we see that

they lie on a circle radius 1 and are equallyspaced around that circle.

z1

z2z3

z4

z5

z6

z7

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Example: Find the 5th roots of 4(1

−i).

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3.8 Trigonometric Applications of Com-

plex Numbers

Using Euler’s formula, we note that

eiθ = cos θ + i sin θ,

e−iθ

= cos(−θ) + i sin (−θ) = cos θ − i sin θ.

On first adding and then subtracting these for-

mulae, we obtain the important formulae

cos θ = 1

2 eiθ + e−iθ , sin θ =

1

2i eiθ

−e−iθ .

Thus, sines and cosines can be replaced by

exponentials with imaginary exponents.

A large variety of trigonometric formulae can

be obtained by using the above results and theBinomial Theorem. This theorem is stated be-

low, and a proof is given in the course notes.

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Theorem: If a, b

∈ C and n

∈ N, then

(a + b)n = an + nan−1b + n(n − 1)2!

an−2b2

+n(n − 1)(n − 2)

3! an−3b3 + · · ·

· · · + nabn−1 + bn

=

nk=0

nkan−kbk,

where the numbersn

k

=

n!

k!(n − k)! are thebinomial coefficients.

For small values of n the binomial coefficientsmay be easily calculated using Pascal’s trian-

gle.

n BINOMIAL COEFFICIENTS

012345

11 1

1 2 11 3 3 1

1 4 6 4 11 5 10 10 5 1

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Example: For n = 3, the coefficients are 1, 3,

3, 1, and hence

(a + b)3 = a3 + 3a2b + 3ab2 + b3.

Example Show that cos2 θ + sin2 θ = 1 using

the polar form of the trigonometric functions.

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Example Find a formula for cos3 θ in terms of

cosines of multiples of θ.

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Example Find a formula for cos 3θ in terms of

powers of sin θ and cos θ.

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3.9 Geometric Applications of Complex Num-

bers

In this section we see how to represent regions

in the argand plane algebraically.

Example: Give a geometric interpretation of |z − w| and Arg(z − w).

Let z = x + iy, w = a + ib. Note that z − w =(x − a) + i(y − b). Now

|z − w| = (x − a)2 + (y − b)2which is the distance between the points rep-

resenting z and w.

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To understand the geometric interpretation of

Arg(z − w), we represent the complex numberz − w by the directed line segment (the arrow)from w to z. We plot z and w on the Arganddiagram for each of the four cases as shown inthe figure below. The angle α satisfies

sin α =

y

−b

|z − w| and cos α = x

−a

|z − w|,and hence α = Arg(z − w).

0 Real axis I m a g i n

a r y a x i s

w

z

α

0 Real axis I m a g i n

a r y a x i s

w

z

α

0 Real axis I m a g i n a r y a x i s

z

wα

0 Real axis I m a g i n a r y a x i s

z

w α

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Example: The set A =

{z

∈ C :

|z

| ≤ 2

} repre-

sents the set of points whose distance from the

origin is less or equal to 2. (N.B. |z − α| mea-sures the distance between z and α.) Hence

this set represents a disc radius 2 centre the

origin.

2

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Similarly, the set B =

{z

∈ C :

|z + 1

| < 2

}represents the open disc centre (−1, 0) radius2.

−1 1−3

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The set C =

{x

∈ C : 0

≤ Arg(z)

≤

π

3} repre-

sents a wedge vertex at the origin, and arms

separated by an angle of 60◦.

o

Note that the origin in NOT included since theargument of 0 is not defined.

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Similarly the set

D = {z ∈ C : 0 ≤ Arg(z − i) ≤ π6

} represents awedge centre the point i as shown.

1o

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Example: Sketch

{z ∈ C : |z − i + 1|


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Example: Sketch

{z ∈ C : |z − 3| 0}

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More generally, to rotate complex number an-

ticlockwise around 0 through an angle θ, we

multiply it by eiθ.

Example: Rotate 3 − i anticlockwise about 0through an angle of

π

4.

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3.10 Complex Polynomials

Theorem: (Fundamental Theorem of Algebra

(FTA))

Suppose p(z) = anzn +an

−1z

n−1 +· · ·+a1z +a0is a polynomial, whose co-efficients an, . . . , a1, a0are all real (or complex) numbers, then the

equation p(z) = 0 has at least one root in the

complex numbers.

Corollary: The equation p(z) = 0 has exactlyn (complex) solutions in the complex numbers

(counting multiplicity).

(The last proviso ‘counting multiplicity’ refers

to polynomials which may, for example, have

factors such as (z − 2)4 in which case the rootz = 2 is counted four times.)

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The above result tells us (among other things)

that we do not need to find any larger set of

numbers if we want to solve polynomial equa-

tions. The complex numbers contain all the

roots of every polynomial.

The fundamental theorem of algebra, men-tioned above, tells us that in the complex plane,

all polynomials have all their roots. This is a

very powerful theoretical tool, but it does not

explicitly tell us how to find these roots for a

given polynomial. Moreover, if we know theroots then we also know how to factor the

polynomial. You will need to recall a num-

ber of basic facts about polynomials from High

School, which are:

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Remainder Theorem: If p(z) is a polynomial

then the remainder r when p(z) is divided byz − α is given by r = p(α).

Factor Theorem: If p(α) = 0 then (z − α) isa factor of p(z).

From the fundamental theorem of algebra, it isclear that over the complex numbers, all poly-nomials completely factor (at least in theory)into linear factors.

Theorem: Every polynomial (with real or com-

plex co-efficients) of degree n ≥ 1 has a fac-torisation into linear factors of the form: p(z) = a(z − α1)(z − α2) · · · (z − αn)

where α1, α2, · · · , αn are the (complex) roots of p(z).

This result, still does not tell us how to factor.

The key to factoring over the real numbers,is to firstly factor over the complex numberssince in the complex plane the polynomial ‘fallsto pieces’ into linear factors.

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Example: Factor z4 + 1 over the complex

numbers and hence over the real numbers.

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(cont.)

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Example: Factor z6 +8 over the complex and

real numbers.

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(cont.)

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Note that if the co-efficients of the polyno-

mial are real, then the roots occur in conjugate

pairs.

Theorem: Let p(z) = anzn + an−1zn−1 + · · · +a1z + a0 be a polynomial with real coefficients

an, an−1, . . . , a1, a0. If α is a complex root of p(z), i.e., p(α) = 0, show that α is also a root.

Proof:

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From this is follows that:

Theorem: A polynomial with real co-efficients

can be factored into a product of real linear

and/or real quadratic factors.

Proof: Factor p(z) over the complex numbersin the form

p(z) = a(z − b1) · · · (z − br) ×(z − α1)(z − α1) · · · (z − αs)(z − αs)

where the bi’s are real and the αi’s are complex(non-real). By the above theorem, these must

occur in conjugate pairs. Now each such pair

of factors containing the conjugate pairs, can

be expanded, viz:

(z − α)(z − α) = (z2 − (α + α)z + αα).Now α + α = 2Re(α) and so is REAL, and also

αα = |α|2 is also REAL. Hence the quadraticwe obtain has REAL co-efficients.

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Example: Show that z = i is a root of p(z) =

z4 − 2z3 + 6z2 − 2z + 5 and hence factor p overR and C.

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(cont.)

A very useful result, especially for some of the

tutorial problems, is given by

zn − 1 = (z − 1)(1+ z + z2 + · · · + zn−1), n ∈ N .179


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Chapter 4 - Linear Equations and Matrices

4.1 Introduction to Linear Equations

At School you learnt how to solve two-by-two

systems of linear equations such as

2x + 3y = 14x − 2y = 2

You did this by multiplying the equations by

various numbers and adding or subtracting so

as to eliminate one of the unknowns.

What does one do when faced with a system of

30 equations in 30 unknowns (such things turn

up routinely in the applications of mathematics

to the real world)? Ad hoc manipulation of the

equations is clearly out of the question.

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Example: Suppose A, B, C, D are arbitrary (but

fixed) real numbers. Find the (unique) poly-

nomial p(x) of degree 3, such that p(0) = A,

p′(0) = B, p(1) = C and p′(1) = D.

We need a SYSTEMATIC method for solving

systems of linear equations in many variables.

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This is provided by a method known as Gaus-

sian Elimination. It is one of the absolute fun-

damentals of the algebra course and will be

used constantly in MATH 1231 as well as in

this course. You will need to learn it carefully

and get lots of practice at it. Before we look

at this method, however, we wish to look care-fully at the geometric interpretation of solving

such systems.

In 2-dimensions, solving a system of 2 simulta-

neous equations corresponds to looking at theintersection of two lines. The two lines may be

parallel, in which case there is no intersection.

They may meet at one point, in which case

there will be one unique solution or they may

in fact be the same line in which case there

will be infinitely many solutions.

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In 3-dimensions, consider the following system

of linear equations:

x + y + z = 6x − y − z = 0

Geometrically, we are looking at the intersec-

tion of two planes. Now two planes could be

parallel, which would then mean that there are

no solutions to this system. The two planes

could be the same, in which case there are in-

finitely many solutions, all lying on that plane.

The third possibility is that the two planes

could meet in a line.

In our example,

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(cont.)

Observe that this is simply the equation of a

line! Hence the two planes in our example meet

in a line.

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Example: Consider the following set of simul-

taneous equations.

x + y + z = 7− 2y − z = 0

3z = 12

These represent three planes which intersect

in a single point.

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This technique is called back-substitution.

This last example was quite easy because at

each stage we could solve the last equation and

successively back substitute to solve the oth-

ers. We are going to learn a procedure which

will convert any set of equations into this or asimilar form.

Example: Solve

x + 2y + 3z = 6

2x + 4y + 6z = 11

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Example: Solve

x + 2y + 3z = 62x + 4y + 6z = 12

These represent two planes which are in fact

the same.

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The examples above were simply designed to

show you what kinds of things can happen

when we try to solve systems of equations and

how we can represent the solutions. What we

now need is a systematic method of reducing

the equations to a point where we can use the

above ideas to get all the solutions. This willbe achieved by the method known as Gaus-

sian Elimination. To explain this method we

need a new way of representing simultaneous

equations which we will discuss next lecture.

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Example: Solve

x1 + 2x2 + 3x3 = 52x1 + 5x2 + 8x3 = 12

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4.2 Systems of Linear Equations and Ma-

trix Notation

When we manipulate the equations, we are re-

ally only concerned with the co-efficients. The

symbols x, y and z or x1, x2, x3 are simply place

markers to separate off the co-efficients. Wewill therefore replace the equations with the

co-efficients properly aligned. For example the

system

x + y + z = 6

x − y − z = 0will be written as the block of numbers

1 1 1 61 −1 −1 0

.

This is called an augmented matrix. The list

consisting only of the co-efficients 1 1 11 −1 −1

is called a matrix.

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The word matrix (plural matrices ) is Latin for

womb . I suppose the idea was that a devel-

oping fetus is stored in the womb, just as a

matrix is used to store a block of numbers.

Example: The augmented matrix of

x + y + z = 7− 2y − z = 0

3z = 12

is

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An m

×n matrix then, is a block of numbers

of the form

a11 a12 a13 . . . a1na21 a22 a23 . . . a2na31 a32 a33 . . . a3n

... ... ... . . . ...

am1 am2 a33 . . . amn

Note the m×n matrix has m rows and n columns.At this stage, we will think of these numbers as

the co-efficients of the unknowns in some set

of equations. (In the next topic we will look

at matrices from a more general viewpoint.)

An alternate way to write a system of equa-

tions is the so-called vector form.

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Example: Write the following system of equa-

tions in augmented matrix and in vector form.

x1 + 3x2 − 6x3 + 7x4 = −2−2x1 + 5x3 − 4x4 = 3

7x1 − 5x4 = −10The augmented matrix form is

The vector form is

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If we write x for the vector x1

x2x3x4

in the aboveexample, A for the matrix

1 3 −6 7

−2 0 5 −47 0 0 −5

and b for the vector

−23−10

then the systemof equations will be written briefly as Ax = b

and the augmented matrix as (A|b). (Note

that we have as yet no defined matrix multi-

plication, so Ax is purely a formal symbol.)

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Thus, the system

a11x1 + a12x2 + · · · + a1nxn = b1

a21x1 + a22x2 + · · · + a2nxn = b2... ... . . . ... ...

am1x1 + am2x2 + · · · + amnxn = bmcan be written as Ax = b with augmented ma-

trix (A|b), written as

a11 a12 a13 . . . a1n b1

a21 a22 a23 . . . a2n b2a31 a32 a33 . . . a3n b3

... ... ... . . . ... ...am1 am2 a33 . . . amn bm

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4.3 Elementary Row Operations

Suppose we have our system of equations writ-

ten in augmented matrix form. There are a

number of operations we can perform on the

rows of the matrix without essentially altering

the underlying set of equations. These oper-ations will be called elementary row opera-

tions. We will describe these informally first,

using examples, and then write them down for-

mally later.

The systematic application of elementary row

operations is called Gaussian elimination.

The first such operation is to swap two rows.

Note that this is simply the same as writing

down the equations in a different order.

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Example: 2 −1 1 −11 1 1 43 2 −1 −2

R1↔R2−→ 1 1 1 42 −1 1 −1

3 2 −1 −2

Observe that this places a 1 in the top left-

hand corner. This is very desirable. Given any

augmented matrix, we swap the rows so asto have a non-zero number, preferably a 1 or

−1 in the top left-hand corner. A non-zeronumber in the top left hand corner is called a

pivot and the column is referred to as a pivot

column.

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The next operation involves subtracting a mul-

tiple of one row from another. This is precisely

the sort of thing you did back at school when

using the elimination method. In our example

I am going to subtract twice row 1 from row

2 to obtain

1 1 1 42 −1 1 −13 2 −1 −2

R2−2R1−→ 1 1 1 40 −3 −1 −93 2 −1 −2

Notice that this eliminates the x term from the

second equation. Now we subtract three times

row 1 from row 3 to obtain 1 1 1 40 −3 −1 −93 2 −1 −2

R3−3R1−→ 1 1 1 40 −3 −1 −9

0 −1 −4 −14

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Normally we would perform both of these op-

erations in one go and write 1 1 1 42 −1 1 −13 2 −1 −2

R2−2R1,R3−3R1−→

1 1 1 40 −3 −1 −90 −1 −4 −14

Observe now that the variable x (or x1) is ab-

sent from the second and third equations. This

process is called row reduction and is central

to the Gaussian elimination method. We seek

to reduce the co-efficients below the pivot to

0.

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We now move to the second line and observe

that the first non-zero number in that row is

−3. We could also call this a pivot element,but there is a −1 below it. So we swap R2 andR3 to obtain

1 1 1 40 −3 −1 −90 −1 −4 −14

R2↔R3−→ 1 1 1 00 −1 −4 −140 −3 −1 −9

This makes −1 our n

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