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University of New South Wales
MATH 1131
Mathematics 1A Algebra
Semester 1 2016
Dr Bill Ellis ∗
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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Example: Find the vector equation of the line
passing through the two points P, Q with posi-
tion vectors p =
−126
and q = 4−2
3
.
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Example: Find the vector equation of the line
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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Example: Find the vector equation of the line
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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Example: Find the vector equation of the
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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Example: Find the vector equation of the
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.
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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
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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