ES 153
LINEAR ALGEBRA AND TRIGONOMETRY
BSc
Compiled By
J. ACQUAH
A UGUST, 2020
LINEAR ALGEBRA AND
TRIGONOMETRY
ES 153
INSTRUCTOR: Dr JOE ACQUAH
Email: jacquah@umat.edu.gh
OFFICE: Mathematical Sciences Department / Phone: 031 27531
OBJECTIVES: The goal of the course is to develop a fundamental understanding in
mathematical describing and studying complex numbers, matrices, vectors and
trigonometry. We will look at theorems and how to apply such theorems for solving
problems.
PREREQUISITES: It is assumed that the student has some background in elective
mathematics and core mathematics.
GRADING CRITERIA and EVALUATION PROCEDURES: The grade for the
course will be based on homework, a class test and a final exam.
1.
Attendance: All students should make it a point to attend classes. Random
attendance will be taken to constitute 10% of the final grade.
2.
Homework: Homework is worth 5% of the final grade. Homework will be
assigned on Fridays and will be due at before 8 AM the following Monday.
3.
Class test: There shall be unannounced short class tests at any time
within the semester which will constitute 15% of the continuous assessment.
Also, one main test based on theoretical techniques worth 10% of the grade will
be given during class. The exam date will be announced one week in advance.
4.
Final exams: A final exam is worth 60% of the final grade.
Lecture Note prepared by Dr J Acquah
Page i
ASSESSMENT OF COURSE
➢ Assessment of students
The student’s assessment will be in two forms:
Continuous Assessment [40%] and
End of semester examination [60%]
➢ Assessment of Lecturer
At the end of the course each student will be required to evaluate the course and
the lecturer’s performance by answering a questionnaire specifically prepared to
obtain the views and opinions of the student about the course and lecturer. Please
be sincere and frank.
STUDENT RESPONSIBILITY: It is assumed that each student attends the lectures
and works all the assigned problems. The student is responsible for ALL material covered
in class and any assigned reading. Students may discuss homework problems but you are
responsible for writing your individual answers. If your name does not appear on the final
class roll, then you will not receive a grade for this course.
EXAM POLICY: No makeup for missed work will normally be given, unless
extenuating circumstances occur. Travel plans are not extenuating circumstances.
Acceptable medical excuses must state explicitly that the student should be excused
from class.
Lecture Note prepared by Dr J Acquah
Page ii
TABLE OF CONTENTS
TABLE OF CONTENTS
iii
CHAPTER 1 COMPLEX NUMBERS
1
1.0
Introduction
1
1.1
Complex numbers and the complex plane
1
1.2
Equality of Complex Numbers
2
1.2.1
Set of complex numbers
3
1.3
Addition and subtraction of complex numbers
4
1.4
Multiplication of Complex Numbers
5
1.5
Square Roots of a Complex Number
6
1.5.1
1.6
Multiplication by i
Conjugate of a complex number
7
8
1.6.1
Modulus of a complex number
9
1.6.2
Conjugate properties
9
1.7
Division of Complex Numbers
10
1.8
The modulus, argument and polar form of a complex number
12
1.8.1
1.9
Principal value of an argument
14
De Moivre's Theorem
17
CHAPTER 2 MATRIX ALGEBRA
27
2.0
Introduction
28
2.1
Important Concepts in Matrices
30
2.2
Special Matrices: Triangular, Diagonal
36
2.2.1
Addition of Matrices:
39
2.2.2
Subtraction of matrices
40
2.3
Multiplying a Scalar to a Matrix
41
2.4
Multiplication of Matrices
42
2.5
Determinants
46
2.5.1
Determinants of Matrices of Higher Order
Lecture Note prepared by Dr J Acquah
47
Page iii
2.6
Invertible Matrices
49
2.7
Linear Dependence and Independence of Matrices
55
2.8
Rank of a Matrix
56
2.9
Systems of Linear Equations: Gaussian Elimination
57
2.9.1
Matrix Representation of a Linear System
58
2.9.2
Gaussian Elimination.
59
2.9.3
Eigenvalues and Eigenvectors
63
CHAPTER 3 VECTORS
72
3.0
Introduction
72
3.1
Vector Elements or Components in a Coordinate Frame
3.2
Basic Properties
3.3
Vector Addition and Subtraction
3.4
Multiplication of a Vector by a Scalar: (Not the Scalar Product
3.5
Scalar, dot, or inner product
3.5.1
Geometrical interpretation of scalar product
3.6
Projection of one vector onto the other
3.7
Vector or cross product
3.7.1
3.8
Error! Bookmark not defined.
Geometrical interpretation of vector product
Scalar triple product
3.8.1
Geometrical interpretation of scalar triple product
CHAPTER 4 Trigonometry
REFERENCES
Lecture Note prepared by Dr J Acquah
105
Page iv
CHAPTER 1
COMPLEX NUMBERS
Objectives
At the end of this chapter you will be able to:
➢ Understand that i stand for
➢ Understand
that
all
1 and be able to reduce powers of i to
complex
numbers
are
in
i or
the
1.
form
i imaginary part .
real part
➢ Add, subtract and multiply complex numbers.
➢ Find the complex conjugate of a complex number.
➢ Divide complex numbers.
➢ Draw complex numbers.
➢ Convert a complex number from Cartesian to polar form and vice versa.
➢ Write a complex number in its exponential form.
1.0
Introduction
The set of complex numbers is an extension of the set of real numbers. An expression of
the form z
x
iy , where x and y are real numbers, is called a complex number, and
the symbol i is called the imaginary unit: i 2
1 . The numbers x and y are called,
respectively, the real part and imaginary part of z and denoted by
x
However, the set
ez
and
y
m z.
will not allow solutions to all quadratic equations. For example,
x2 + 1 = 0 . This implies that x 2 = −1 , which has no real solution since x 2 is always positive
for real numbers. This difficulty can be overcome by enlarging the set
to include a new
'number' denoted i with the property that i 2 = −1 .
1.1
Complex numbers and the complex plane
The introduction of this new number, i , overcome the problem of finding the square roots
of a negative number.
Example: Find the square roots of -25
Lecture Note prepared by Dr J Acquah
Page 1
Answer: −25 = 25  −1 = 25  i 2 = ( 5  i )
So this gives
2
−25 = 5i
With the existence of the square roots of a negative number, it is possible to find the
solutions of any quadratic equation of the form ax2 + bx + c = 0 using the quadratic
formula. If the discriminant ( b2 − 4ac )  0 , the equation has no real roots.
Example: Solve the quadratic equation x2 - 6x + 25 = 0
Answer: (here a = 1, b = -6, and c = 25)
−b2  b2 − 4ac
The quadratic formula x =
gives
2a
x=
6  36 − 100 6  −64
=
i.e. x = 3 + 4i and x = 3 − 4i
2
2
Self Check: Solve the following quadratics equation using the quadratic formula.
a.
x2 − 2x + 3 = 0
b.
x2 + 2x + 6 = 0
c.
x2 + x + 1 = 0
Notation
A typical complex number is usually denoted by z, where z is defined in terms of an
ordered pair
z
1.2
x, y
x, y of real number and the imaginary number i
x iy and component notation z
such that
x, y .
Equality of Complex Numbers
Two complex numbers x1 , y1 and x2 , y2
y1
1
are said to be equal if and only if x1
x2 and
y2 .
Examples
1. Given that z1 = 4 + i76 and z2 = 4 + i76 , then z1 = z2 since x1 = x2 = 4 y1 = y2 = 76 z1 = z2
Lecture Note prepared by Dr J Acquah
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2. z1 = 4 + i 6 , z2 = 4 + i76 , then z1  z2 since x1 = x2 = 4 and y1  y2 i.e. y1 = 6, y2 = 76
1.2.1
Set of complex numbers
The set of complex numbers is written as
= a + bi : a, b 

Note that it is customary to denote a complex number by the letter z .
If z = a + bi then:
•
•
The number a is called the real part of z.
The number b is called the imaginary part of z.
These are sometimes denoted e ( z ) for a and m ( z ) for b. That is, e ( a + bi ) = a and
m ( a + bi ) = b .
The real numbers can be represented on the number line.
Figure 1.1 Real Number Line
Is there a similar representation for the complex numbers?
The definition of a complex number involves two real numbers. Two real numbers give a
point on a plane. So complex numbers can be plotted in a plane by using the x-axis for
the real part and the y-axis for the imaginary part.
This plane is called The Complex Plane or Argand Diagram.
Lecture Note prepared by Dr J Acquah
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Figure 1.2 Argand Diagram
Example: Show the complex numbers z = -2 + 3i and w = 1 - 2i on an Argand diagram.
Answer: The complex numbers z and w are shown on the following diagram.
Figure 1.3 Complex Number on the Argand Diagram
a + bi = 0 if and only if a = b = 0
1.3
Addition and subtraction of complex numbers
The addition of two complex numbers, z1 and z2 , in general gives another complex
number. The real components and the imaginary components are added separately and
in a like manner to the familiar addition of real numbers:
➢ Rules for adding complex numbers
•
•
Add the real parts.
Add the imaginary parts.
i.e. ( a + bi ) + ( c + di ) = ( a + c ) + (b + d ) i
Example
1. Add 3 + 4i and 2 - i
Answer: The rule for adding these numbers is straightforward:
•
•
So
Add the real parts together to give the real part 5 ( i.e. 3 + 2 = 5)
Add the imaginary parts together to give the imaginary part 3 ( i.e. 4 + (-1) = 3)
(3 + 4i ) + ( 2 − i ) = (3 + 2) + ( 4 − 1) i = 5 + 3i
Lecture Note prepared by Dr J Acquah
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2. Given z1 = 6 + i9 and z2 = 12 + i7 , evaluate z1 + z2
Solution
z1 + z2 = (6 + 12) + i(9 + 7) = 18 + i16
➢ Rule for subtracting complex numbers
•
•
Subtract the real parts.
Subtract the imaginary parts.
i.e. ( a + bi ) − ( c + di ) = ( a − c ) + (b − d ) i
Example
1. Given that z1 = 7 + i and z2 = 3 − i evaluate z1 − z2
Solution:
z1 − z2 = ( 7 + i ) − ( 3 − i ) = ( 7 − 3) + (1 − ( −1) ) i = 4 + 2i
2. Evaluate z1 − z2 given that z1 = 16 − i 4, z2 = 12 − i 4
Solution:
z1 − z2 = (16 − i 4) − (12 − i 4) = 16 − 12 − i 4 + i 4 = 4
1.4
Multiplication of Complex Numbers
➢ Rule for multiplying complex numbers
Use the technique for multiplying out two brackets. ( a + bi )( c + di ) = ( ac − bd ) + ( bc + ad ) i
The technique for multiplying two complex numbers is similar to that used when
multiplying out two brackets. Now multiply the two complex numbers (3 + 2i) and (4 +
5i)
To multiply these together take the same approach.
Figure 1.0.4 Multiplication of Complex Numbers
Lecture Note prepared by Dr J Acquah
Page 5
This will give
( 3  4 ) + ( 3  5i ) + ( 2i  4 ) + ( 2i  5i )
= 12 + 15i + 8i + 10i 2
= 12 + 23i − 10 ( Remember that 10i 2 = 10 ( −1) = −10 )
= 2 + 23i
Self check
(i)
(ii)
1.5
Multiply (4 - 2i) and (2 + 3i)
Multiply (1 + 2i) and (2 - 3i)
Square Roots of a Complex Number
Earlier examples showed that the square roots of a negative real number could be found
in terms of i in the set of complex numbers. It is also possible to find the square roots of
any complex number.
Example: Find the square roots of the complex number 5 + 12i
Answer: Let the square root of 5 + 12i be the complex number
( a + bi ) so
( a + bi ) = 5 + 12i
2
Note: the trick is to equate the real parts and the imaginary parts to give
two equations, which can be solved simultaneously.
First of all multiply out the brackets. ( a + bi )( a + bi ) = a − b + 2abi
2
2
So ( a 2 − b2 ) + 2abi = 5 + 12i
Equate the real parts to give a2 − b2 = 5 (call this equation 1).
Equate the imaginary parts to give 2ab = 12 (call this equation 2).
Rearranging equation 2 to make b the subject gives
Substitute in equation 1 to gives
b=6
a
a4 − 5a2 − 36 = 0
Hence ( a2 − 9)( a2 + 4) = 0  a2 = −4 or 9
Lecture Note prepared by Dr J Acquah
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Since a2 + 4 = 0 then a2 = −4 is impossible. This leaves a 2 = 9 which gives a = 3
Substitute a = 3 in equation 2 to give b = 2 and then substitute a = −3 in equation 2 to
give b = −2
Hence the square roots are 3 + 2i and −3 − 2i
Self check: Find the square root of 3+4i
1.5.1 Multiplication by i
Multiplication by i has an interesting geometric interpretation. The following examples
should demonstrate what happens.
Example: Take the complex number 1 + 7i and multiply it by i
Answer: (1 + 7i)i = i + 7i2 = -7 + i
Figure 1.5 Multiplication of a Complex Number by i
Example: Take the complex number -6 - 2i and multiply it by i
Answer: (-6 - 2i)i = -6i - 2i2 = 2 - 6i
Lecture Note prepared by Dr J Acquah
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Figure 1.6 Complex Number Multiplication
Self check Take the complex number 8 - 4i and multiply it by i
In each of the examples it can be seen that the effect of multiplying a complex number by
i is a rotation of the point on the Argand diagram through 90° or
1.6
Conjugate of a complex number
The conjugate of the complex number z = a + bi is denoted as z and defined by z = a − bi
The conjugate is sometimes denoted as z * .
Geometrically the conjugate of a complex number can be shown in an Argand diagram as
the reflection of the point in the x-axis.
Figure 1.7 Conjuate of a Complex Number
Example Give the conjugate of the complex number z = −6 + 8i and show z and z on an
Argand diagram.
Answer: The conjugate is = -6 - 8i b
Lecture Note prepared by Dr J Acquah
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Figure 1.8 Example of Conjugate of a Complex Number
1.6.1
Modulus of a complex number
The modulus r of a complex number z = a + bi is written z and defined by z = a 2 + b 2
1.6.2 Conjugate properties
There are special identities which apply to complex numbers and their conjugates. These
are listed below.
Let the two complex numbers be z and w with conjugates z and w respectively.
1. z = z
2. z = z where z 
3. z + z = 2Re ( z )
4.
z = z
5. zz = z
2
( z + w) = z + w
7. ( zw) = zw
6.
8.
zw = z w
9.
z + w  z + w . This is commonly known as the triangle inequality.
Proofs of Special Identities
Proof (1):
z=z
Lecture Note prepared by Dr J Acquah
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Let z = a + bi for a and b  , By the definition of a conjugate z = a − bi = a + ( −b ) i .
Again, using the conjugate definition z = a + ( −b ) i = a − ( −b ) i = a + bi = z .
Proof (2):
If z 
z = z where z 
then z = a + 0i = a (since 0i = 0) for some a 
By the definition of a conjugate z = a − 0i , but 0i = 0, so z = a = z
Proof (3):
z + z = 2e ( z )
Let z = a + bi then z = a − bi , therefore, z + z = a + bi + a − bi = 2a = 2e ( z )
Proof (4):
z = z
2
2
2
Let z = a + bi where z = a 2 + b 2 and z = a − bi so z = a + ( −b ) = a + b = z
2
Proof (5): zz = z
2
Let z = a + bi and z = a − bi , Then zz = ( a + bi )( a − bi ) = a 2 + b 2
But z = a 2 + b 2 so z 2 = a 2 + b 2 , Thus zz = a 2 + b2 = z
2
Self check
1. Prove properties 6 to 9.
n
n
j =1
j =1
2. Prove the Triangle inequality for n complex numbers  z j   z j
1.7
Division of Complex Numbers
Consider, the two complex numbers z1
a bi and z2
c di . Writing the quotient in
component form we obtain
Lecture Note prepared by Dr J Acquah
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z1
z2
a bi
c di
➢ Rule for dividing complex numbers
•
•
•
Find the conjugate of the denominator.
Multiply the complex fraction, both top and bottom, by this conjugate to give an
integer on the denominator.
Express the answer in the form a + bi
a + bi ( a + bi )( c − di ) ( ac + bd ) + ( bc − ad ) i
=
=
c + di ( c + di )( c − di )
( c2 + d 2 )
( ac + bd ) + ( bc − ad ) i
=
(1.1)
(c + d ) (c + d )
2
2
Example 1
2
2
3 − 7i ( 3 − 7i )( 2 − 6i ) 6 − 14i − 18i − 42
=
=
2 + 6i ( 2 + 6i )( 2 − 6i )
( 22 + 62 )
=−
Example 2
36 32
− i
40 40
1000 + 1000i (1000 + 1000i )(10 + 10i ) 20000i
=
=
10 − 10i
200
(10 − 10i )(10 + 10i )
= 100i
Self check
i. Divide 5 – 2i by 4 + 3i
ii. If z1 = 2 – 3i and z2 = 3 – 4i, find
z1
z2
 3 − 2i 
 3 − 2i 
.
 and Im 
 1 + 4i 
 1 + 4i 
iii. Find Re 
Lecture Note prepared by Dr J Acquah
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1.8
The modulus, argument and polar form of a complex number
Remember the Argand diagram in which the point a, b corresponds to the complex
number z = a + bi
When the complex number is written as a + bi where a and b are real numbers, this is
known as the Cartesian form.
Figure 1.9 A pont in the Complex Plane
This point a, b can also be specified by giving the distance, r, of the point from the origin
and the angle, , between the line joining the point to the origin and the positive x – axis.
Figure 1.10 Modulus of a Complex Number
By some simple trigonometry it follows that a = r cos and b = r sin
Lecture Note prepared by Dr J Acquah
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Figure 1.11 Argument of a Complex Number
Thus, the complex number z can be written as r cos + i r sin . This is known as the polar
form of a complex number. r is called the modulus of z and
is the argument of z.
Argument of a complex number: The argument of a complex number is the angle
between the positive x-axis and the line representing the complex number on an Argand
diagram. It is denoted arg (z).
Polar form of a complex number: The polar form of a complex number is
z = r ( cos + i sin ) where r is the modulus and is the argument.
There will be times when conversion between these forms is necessary. Given a modulus
(r) and argument ( ) of a complex number it is easy to find the number in Cartesian
form.
Use the following steps to do this:
➢ Evaluate ' a ' = r cos and ' b ' = r sin
➢ Write down the number in the form a + bi
Example: If a complex number z has modulus of 2 and argument of − , express z in
6
the form a + bi and plot the point which represents the number in an Argand diagram.
Answer:
−1
 − 
3
 − 
= 3 and b = 2sin 
We have a = 2cos 
 = 2  = −1 , So a + bi =
 = 2
2
2
 6 
 6 
3 −i .
Lecture Note prepared by Dr J Acquah
Page 13
Figure 1.12 Fiinding the Argument of a Complex Number
Self check
Plot the complex number with modulus 2 and argument 11 in an Argand diagram.
6
1.8.1 Principal value of an argument
The principal value of an argument is the value which lies between − and  .
The values of the principal argument for a complex number in each quadrant are shown
on the following diagrams.
Figure 1.13 Principal Argument in the 1st Quadrant
arg z   where tan  
b
b
, i.e   tan −1  
a
a
Lecture Note prepared by Dr J Acquah
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Figure 1.14 Principal Argument in the 2nd Quadrant
arg z   where    − a and tan a 
b
b
, i.e a  tan −1  
a
a
In the previous two examples the complex number is represented by a point in either the
first or second quadrant and the principle argument is positive (lying between 0 and ).
In the next two examples when the complex number is represented in the third and
fourth quadrants the principal argument is negative (lying between 0 and - ).
Figure 1.15 Principal Argument in the 3rd Quadrant
arg z   where   − + a and tan a 
b
b
, i.e a  tan −1  
a
a
Figure 1.16 Principal Argument in the 4th Quadrant
arg z   where   −a and tan a 
−1
b
b
, i.e a  tan −1  
a
a
b
a
NB: Taking tan   without the modulus sign on a calculator may give a different value
than required.
Example 1
Lecture Note prepared by Dr J Acquah
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1.8.1 Multiplication and Division of two Complex numbers in Polar Form
1.8.2 Multiply of Complex Numbers in Polar Form
Consider the complex numbers z = r1 ( cos + i sin  ) and w = r2 ( cos + i sin  ) . Suppose we
seek for the product of the complex numbers,
Then zw = rr
1 2 ( cos cos − sin  sin  ) + i rr
1 2 ( sin  cos + sin  cos )


Using trig. formulae this simplifies to ZW = r1r2 cos ( +  ) + i sin ( +  )
So, the modulus of the product, r1 r2 , is the product of the moduli of z and w, namely
r1 and r2
The argument of the product,  +  , is the sum of the arguments of z and w, namely
 and 
Example: If z = 3( cos300 + i sin300 ) and w = 4 ( cos600 + i sin 600 ) , find zw
Answer:
(
)
ZW = 12 cos ( 30 + 60) − i sin ( 30 + 60) = 12 ( cos900 + i sin 900 ) = 12i
0
0
1.8.3 Division of Complex Numbers in Polar Form
Let z = r1 ( cos + i sin  ) and w = r2 ( cos + i sin  ) then
z r1 ( cos  + i sin  )
=
w r2 ( cos  + i sin  )
(1.2)
r ( cos  + i sin  )( cos  − i sin  )
= 1
r2 ( cos  + i sin  )( cos  − i sin  )
(1.3)
r cos  cos  + sin  sin  + i ( sin  cos  − cos  sin  )
= 1
r2
( cos2  + sin 2  )
(1.4)
Lecture Note prepared by Dr J Acquah
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r1
cos ( −  ) + i sin ( −  )
r2

(1.6)
z r1
= (cos ( −  ) + i sin( −  ))
w r2
z r1
z
and arg   = arg z − arg w
=
w r2
 w
Taking modulus of (1.6), we have
1.9
De Moivre's Theorem
Repeated use of multiplication shows how to compute powers of a complex number.
If
z
r cos
z2
r 2 cos2
z3
zz 2
(1.7)
i sin
then
(1.8)
i sin 2
and
r 3 cos3
(1.9)
i sin3
In general, we obtain the following result, which is named after the French mathematician
Abraham De Moivre.
Theorem:
zn
r n cos n
For
all
rational
values
of
n,
if
z
r cos
i sin
,
then
i sin n
This stipulate that to take the nth power of a complex number we take the nth power of
the modulus and multiply the argument by n.
Proof
The proof of De Moivre's theorem is in threefold.
Case I: n as a positive integer
Lecture Note prepared by Dr J Acquah
Page 17
The theorem states that if z = r ( cos + i sin ) then z = r ( cos n + i sin n ) for all n  .
n
n
This is a proof by induction.
For n = 1
r ( cos + i sin  ) = r ( cos + i sin  ) = r cos1 + i sin1  .
1
1
(1.10)
And so, it is true for n = 1
Now suppose that the result is true for n = k then
r ( cos + i sin  ) = r cos k + i sin k 
k
(1.11)
k
Consider n = k + 1 then
r ( cos + i sin  ) = r ( cos + i sin  )r ( cos + i sin  )
= r ( cos  + i sin  )r ( cos k + i sin k )
= r ( cos cos k − sin  sin k ) + i ( sin  cos k + cos sin k )
= r cos ( + k ) + i sin ( + k )
= r cos ( ( k + 1) ) + i sin ( ( k + 1) )
k +1
k
k
k +1
k +1
k +1
(1.12)
(1.13)
(1.14)
(1.15)
(1.16)
So the result is true for n = k + 1 if it is true for n = k . Since it is also true for n = 1 , then it
is true for all n  .
Case II: n as a negative integer
Let n
m where m is a positive integer.
r cos
1
n
i sin
r cos
1
r
m
cos m
= r m cos
r n cos n
i sin
r cos
rationalising gives
i sin m
m
i sin
1
n
m
i sin n
Lecture Note prepared by Dr J Acquah
i sin
m
(1.20)
(1.21)
(1.22)
(1.23)
Page 18
Case III: n as a fraction
Let n
p
were p and q are integers. By the result in CaseI and CaseII , We can write
q
p
cos
q
i sin
cos
q
1
q
i sin
such that cos
q
i sin
q
cos
i sin
p q
Again, by the result in CaseI and CaseII , we have
p
 
 p 
    p 
 cos + i sin   = cos    + i sin    
q
q    q 
 q 

(
Thus, for all rational values of n.
Example
Calculate 2  cos  + i sin   
5
 
5
5 
Answer: Using De Moivre's theorem
5
 

 
 
  
5
2  cos + i sin   = 2 cos  5   + i sin  5   
5
5 
5
5 

 
 
= 32 ( cos  + i sin  )
= −32
Example
1

 3
Calculate  cos + i sin 
4
4

Answer: By De Moivre's theorem for fractional powers
1
  3  1  
 
 1  
cos
+
i
sin

 = cos    + i sin    
4
4

 3 4 
 3 4 

 

=  cos + i sin 
12 
 12
Lecture Note prepared by Dr J Acquah
Page 19
Self check
Calculate
1
  
  3
27
cos
+
i
sin
1.  

2
2 
 
2
  
  3
2. 8  cos + i sin  
4
4 
 
1.9.1 Nth Roots of a Complex Number
Le z = r ( Cos  + iSin ) and let n be a positive integer. Then, z has the n distinct nth roots
1 
  + 2 k 
  + 2 k  
wk = r n Cos 
 + iSin 

n
n  




where k = 0,1, 2,
n−1
1
Notice that each of the nth roots of z has modulus wk = r n . Thus, all the nth roots of z lie on the
1
circle of radius r n in the complex plane. Also, since the argument of each successive nth root
exceeds the argument of the previous root by
2
, we see that the nth roots of z are equally spaced
n
on this circle.
EXAMPLE. Find the sixth roots of z = −8 and graph these roots in the complex plane.
Solution: In trigonometric form, z = 8 ( Cos  + iSin  ) .
1 
  + 2 k 
  + 2 k  
For n = 6 , we get wk = r 6 Cos 
 + iSin 

6
6





Lecture Note prepared by Dr J Acquah
Page 20
We get the six sixth roots of −8 by taking k = 0,1, 2,3, 4,5 in this formula:
 3 1 
1 


w0 = 8 6  Cos + iSin  = 2 
+ i 
6
6
2
2 


1 


w1 = 8 6  Cos + iSin  = 2i
2
2


1 
5
5 
3 1 
w2 = 8 6  Cos
+ iSin
+ i 
 = 2  −
6
6 
2
2 



1 
7
7 
3 1 
w3 = 8 6  Cos
+ iSin
− i 
 = 2  −
6
6 

 2 2 
1 
3
3 
w4 = 8 6  Cos
+ iSin  = − 2i
2
2 

 3 1 
1 
11
11 
w5 = 8 6  Cos
+ iSin
− i 
 = 2 
6
6 

 2 2 
All these points lie on the circle of radius
2.
th
1.9.2 n Roots of Unity
1
   + 2 k 
  + 2 k  
n
From Equation (12), we know that Wk = r n cos 
 + i sin 
  , where W = z is
n
n



 
the nth root of the complex number Z. For the special case where z = 1, that is unity,
z = r = 1 and arg z = 0. Suppose we represent nth root of unity by Z, then from Equation (12) it
can deduce that
 2 k 
 2 k 
Z k = n 1 = c os 
 + i sin 
,
 n 
 n 
where
n
(13)
1 is the nth root of unity and lie on a circle of radius one and center zero.
From Equation (13), the nth root of unity, 𝑍𝑘 can further be express as:
 2 k 
 2 k    2 k 
 2 k  
Z k = c os 
 + i sin 
 = c os 
 + i sin 

 n 
 n    n 
 n 
Lecture Note prepared by Dr J Acquah
k
(13a)
Page 21
 2 
 2 
 + i sin   , then Equation (13a) simplifies to
 n 
 n 
If we let w = c os 
Z k = wk , for k = 0,1, 2,...n
(14)

2

n −1
Therefore, the nth roots of unity is given by Zk = 1, w, w ,..., w
Example
Find the fourth root of unity.
Solution
For fourth root of unity, z
4
= 1.  n = 4 and k = 0, 1, 2, 3
Using the Identity zk = w , where w = cos 2 + i sin 2 , for n = 4 , One can compute w as:
k
n
w = cos
n
2
2


+ i sin
= cos + i sin
4
4
2
2
Substituting w into the identity zk = w , we obtain
k



Z k =  cos + i sin  , for k = 0,1, 2, 3
2
2

k


If k = 0 , Z 0 =  cos + i sin  = 1
2
2

0
If k = 1,



Z1 =  cos + i sin  = i = w
2
2

If k = 2,



Z 2 =  cos + i sin  = (i )2 = −1 = w2
2
2

If k = 3,



Z 3 =  cos + i sin  = (i )3 = −i = w3
2
2

1
2
3
Hence the fourth root of unity is Zk = Z0 , Z1 , Z2 , Z3  = 1, i, −1, −i
Activity One (Basic Operation of Complex Number)
Express the Complex Numbers complex numbers in standard form
Lecture Note prepared by Dr J Acquah
Page 22
1. 3 + −9 = 3 + 3i
( −75 ) = (5i 3 ) = 25i *3 = −75
2
3.)
2. − 3i 2 + i = −3(−1) + i = 3 + i
2
2
Perform the addition or subtract and write the result in standard form.
5. (11 − 2i ) + ( −3 + 6i ) = 8 + 4i
4. ( 4 + i ) + ( 7 − 2i ) = 11 − i
(
) (
) (
) (
)
6. 7 + −18 − 3 + 3i 2 = 7 + 3i 2 − 3 + 3i 2 = 4
7. 13i − (14 − 7i ) = −14 + 20i
8. 22 + ( −5 + 8i ) + 10i = 17 + 18i
 3 7   5 1  −3 7 5 1 −19 37
9. −  + i  −  − i  = − i − + =
− i
 4 5   6 6  4 5 6 6 12 30
Perform the multiplication and write the result in standard form.
10. −6 * −2 = 12 = 2 3
11. (1 + i )( 3 − 2i ) = 3 − 2i + 3i − 2(i )2 = 5 + i
12. ( 6 − 2i )( 2 − 3i ) = 6 ( 6 − 2i ) − 2i ( 2 − 3i ) = 6 − 22i
( 14 + i 10 )( 14 − i 10 ) = ( 14 ) − (i 10 ) = 14 − (−10) = 24
2
13.
2
Find the product of the number and its conjugate
14. ( 4 + 3i )( 4 − 3i ) = 16 − 9i 2 = 16 + 9 = 25
(
)(
)
15. −3 + i 2 −3 − i 2 = 9 − i 2 *2 = 11
Perform the division and write result in standard form
6 −i −6i
16. * = 2 = −6i
i −i −i
 4  ( 4 + 5i ) 16 + 20i 16 + 20i
17. 
=
=

41
 ( 4 − 5i )  ( 4 + 5i ) 16 − 25i
Lecture Note prepared by Dr J Acquah
Page 23
 8 − 7i  1 + 2i  22 + 9i
18. 

=
5
 1 − 2i  1 + 2i 
19.

  41 + 40i  41 + 40i 41 + 40i
1
=
=


=
2
( 4 − 5i )  ( 41 − 40i )   41 + 41i  412 + 402 3281
20.
( 2 − 3i )( 5i ) = (15 + 10i )( 2 − 3i ) = 60 − 25i = 60 − 25i
2 + 3i
13
( 2 + 3i )( 2 − 3i ) 22 + 32
1
Simplify the complex numbers and write it in the standard form
21. 4i 2 − 2i3 = 4(−1) − 2i(−1) = −4 + 2i
( −2 ) = (i 2 ) = 8
6
23.
22. − 5i5 = −5(−1)(−1)i = −5i
6
Activity 2 - Complex Numbers on the Complex Plane
Calculate the absolute value of each number and then graph each number on the complex
plane.
5.
7.
−2
6. 4 + 2i
−1 − i
8.
−1 − i =
( −1) + ( −1) = 2
2
2
Lecture Note prepared by Dr J Acquah
−1 + i
−1 + i =
( −1) + (1) = 2
2
2
Page 24
9. −2 + 2i
−2 + 2i =
10.
( −2) + ( 2) = 8 = 2 2
2
2
3 −i 3 =
11 . − 5 + i 5
− 5 +i 5 =
12.
( − 5 ) + ( 5 ) = 10
2
3 +i 3
2
Lecture Note prepared by Dr J Acquah
( 3) − ( 3) = 6
2
2
−3i
−3i = 02 + ( −3) = 3
2
Page 25
13.
14. −5 + 5i 5
2 − 2i 2
2 − 2i 2 =
( 2 ) + ( −2 2 ) = 10
2
2
−5 + 5i 5 =
( −5 ) + ( 5 5 ) = 150 = 5 6
2
2
Q15. Find each sum graphically ( 4 + i ) + ( −4 + 5i ) = 0 + 6i
16) ( 3 + 2i ) + ( −2 + 4i ) = 1 + 6i
Lecture Note prepared by Dr J Acquah
17) ( 6 − i ) + ( 3 − 2i ) = 9 − 3i
Page 26
Q. Use De Moivre’s Theorem to find the indicated power of the complex number. Express
the result in standard form:
1) (1 + i )
3
(2) ( −1 + i )
10
5
5 

6)  cos
+ i sin

4
4 

10
(3) 2
( 3 + i)
5
(
 (
)
(4) ( 3 − 2i ) (5) 5 cos 20 + i sin cos 20 
5
)
( 7) 3 cos150 + i sin150  (8) 4 ( cos 2.8 + i sin 2.8)


4
3
5
Q. Use the complex Roots Theorem to find the indicated roots of the complex number
and then represent each of the roots graphically, express the roots in standard form
4
4 

16  cos
+ i sin 
3
3 

1.
5
5 

32  cos
+ i sin

6
6 

2
Use the complex Roots Theorem to find all the solutions of the equation and represent
the solutions graphically.
1.
x4 − i = 0
2. x5 − 243 = 0
3. x3 + 64i = 0
CHAPTER 2
Lecture Note prepared by Dr J Acquah
Page 27
MATRIX ALGEBRA
Objectives
At the end of this chapter you will be able to:
➢ State the conditions for the equality of two matrices.
➢ Types of matrices.
➢ Addition, Subtraction and Multiplication of matrices.
➢ Multiplying a scalar to a matrix.
➢ Finding the determinant of a matrix.
➢ Solving systems of linear equations.
➢ Computing the eigenvalues and eigenvectors of a matrix.
2.0 Introduction
This chapter investigates matrices and algebraic operations defined on them. These
matrices may be viewed as rectangular arrays of elements where each entry depends on
two subscripts (as compared with vectors, where each entry depended on only one
subscript). Systems of linear equations and their solutions may be efficiently investigated
using the language of matrices.
DEFINITION
(Matrix) A rectangular array of numbers is called a matrix. We shall mostly be
concerned with matrices having real numbers as entries. The horizontal arrays of a matrix
are called its ROWS and the vertical arrays are called its COLUMNS. A matrix having
-rows and
A matrix
-columns is said to have the order m  n
of ORDER m  n can be represented in the following form:
 a11 a12
a
a22
A = aij =  21


 am1 am 2
a1n 
a2 n 
,


amn 
(2.1)
th
where aij is the entry at the intersection of the i th row and j column.
Lecture Note prepared by Dr J Acquah
Page 28
In a more concise manner, we also denote the matrix
by aij  by suppressing its order.
1 3 7 
Let A = 
 , then a11 = 1, a12 = 3, a13 = 7, a21 = 4, a22 = 5 and a23 = 6
4 5 6
A matrix having only one column is called a COLUMN VECTOR; and a matrix with only
one row is called a ROW VECTOR.
Example 1
Give the size of the matrices below
Solution


4
3 0
6 − 1 0


2 −4 −7 1 3 
0


1
−1 0 
 −6 1 15

2

Size 3*6
7
8

9
10
0
3
−1 
− 2 
0 
Size 3*3
In this matrix the number of rows is equal to the number of columns. Matrices that have the same
number of rows as columns are called Squared Matrices.
12 
 −4 


2 


 −17 
This matrix has a single column and often called Column Matrix.
3 −1 12 0 − 9
This matrix has a single Row and often called Row Matrix. Example
−2 . Often when dealing
with 1*1 matrix we drop the surrounding bracket and just write -2.
Lecture Note prepared by Dr J Acquah
Page 29
2.1
Important Concepts in Matrices
2.1.1
Principal Diagonal
The diagonal entries containing the elements a11 , a22 ,
, ann of a square matrix of order n,
is called the principal diagonal.
Example of the principal diagonal is given infigure 2.1
Figure 2.1 Diagonal of a Matrix
Example
 3 88 5 


A =  20 15 69  The principal diagonal of A is given by a11 = 3, a22 = 15, a33 = 23
 6 69 23 


2.1.2
Equality of two Matrices
Two matrices A = aij  and B = bij  having the same order m  n are equal if aij = bij for
each i = 1,2, , m and j = 1,2, , n .
In other words, two matrices are said to be equal if they have the same order and their
corresponding entries are equal.
Example:
(i)
10 8 5 
10 8 5 




A= 2 5 6  B= 2 5 6 
 6 2 23 
 6 2 23 




A = B since both have the same order 3  3 and their individual entries are equal i.e.
Lecture Note prepared by Dr J Acquah
Page 30
a11 = b11 = 10, a12 = b12 = 8, a13 = b13 = 5, a21 = b21 = 2, a22 = b22 = 5, a23 = b23 = 6
a31 = b31 = 6, a32 = b32 = 2, a33 = b33 = 23
(ii )
10 8 5 
10 8 5 




A =  2 15 6  B =  2 5 6 
 6 2 23 
 6 2 23 




In the above matrix, A  B even though both matrices have the same order. Here their
entries are not the same i.e. a22  b22 since a22 = 15 b22 = 5
2.1.3 Sub – Matrix
Any matrix obtained by omitting some row and columns from a given matrix A is called a
sub-matrix of A.
 a11 a12

 a21 a22
A   a31 a32

 a41 a42

 a51 a52
a13
a23
a33
a43
a53



A
A12 
   11

  A21 A22 



(2.2)
Example:
10 8 5 


Given that A =  2 15 6 
 6 2 23 


15 6 
then C = 
 is a sub-matrix of A
 2 23 
2.1.4 Zero-Matrix/ Null Matrix
A matrix in which each entry is zero is called a zero-matrix, denoted by 0. For example,
0 0
0 0 0
0 2 2 = 
and 023 = 


0 0
0 0 0
2.1.5 Nilpotent Matrix
A nilpotent matrix is a square matrix N such that N k = 0 , for some positive integer k.
The smallest k is sometimes called the degree of N.
Lecture Note prepared by Dr J Acquah
Page 31
Example:
0 1 
2
The matrix M = 
 , is nilpotent, since M = 0 . Thus the degree of matrix M = 2
0
0


2.1.5 Square Matrices
A square matrix is a matrix which has the same number of rows and columns. A n − by − n
matrix is known as a square matrix of order n.
10 8 5 


A =  2 15 6  is an example of a square matrix of order 3  3
 6 2 23 


Any two square matrices of the same order can be added and multiplied. A square matrix
A is called invertible or non-singular if there exists a matrix B such that
AB = I n
(2.3)
Where In is the identity element.
Example:
1
2
− 

1
1


3
3 then
Given that A = 

 has an inverse matrix B = 
1
 −1 2 
1


3
3
2
1
1
AB = I n = 
 3


 −1 2   1

3
Lecture Note prepared by Dr J Acquah
1
− 
1 0
3
=

1  0 1

3
1 0

0 1
where I = 
Page 32
2.1.6 Identity Matrix
The identity matrix or unit matrix of size n is the n − by − n square matrix with ones
on the main diagonal and zeros elsewhere. It is denoted by I n , or simply by I if the size is
immaterial or can be trivially determined by the context.
1 0 0 
1 0 
I1 = 1 , I 2 = 
, I 3 = 0 1 0  ,

0 1 
0 0 1 
1 0
0 1
, In = 


0 0
0
0 


1
The important property of matrix multiplication of identity matrix is that for m − by − n
matrix A.
I m A = AI n = A.
(2.4)
Example:
1 4 7


A =  2 6 1  , Show that I m A = AI n = A.
If
 3 2 4


Solution
Since A is a square matrix with order 3X 3 the expression becomes
I3 A = AI3 = A
Therefore
 1 0 0  1 4 7   1 4 7  1 0 0   1 4 7 


 

 

 0 1 0  2 6 1  =  2 6 1  0 1 0  =  2 6 1 
 0 0 1  3 2 4   3 2 4  0 0 1   3 2 4 


 

 

1+ 0 + 0 4 + 0 + 0 7 + 0 + 0 1+ 0 + 0 0 + 4 + 0 0 + 0 + 7 1 4 7

 
 

0 + 2 + 0 0 + 6 + 0 0 +1+ 0  =  2 + 0 + 0 0 + 6 + 0 0 + 0 +1 = 2 6 1
 0 + 0 + 3 0 + 0 + 2 0 + 0 + 4 3 + 0 + 0 0 + 0 + 2 0 + 0 + 4 3 2 4

 
 

Lecture Note prepared by Dr J Acquah
Page 33
1 4 7 1 4 7 1 4 7

 
 

2 6 1 = 2 6 1 = 2 6 1
 3 2 4  3 2 4  3 2 4

 
 

2.1.7 Transpose of a Matrix
The transposed matrix AT or A of a matrix A is defined to be the matrix which has rows
identical with the columns of A. Thus if A
ai j then AT
a j i . In short AT is obtained
by interchanging the rows and columns of A. Consider the matrix
a11 a12 a13
a21 a22 a23 then AT
a31 a32 a33
A
a11 a21 a31
a12 a22 a32
a13 a23 a33
(2.5)
Example 1:
1 4 7
1 2 3




T
Given that A =  2 6 1  then A =  4 6 2 
 3 2 4
7 1 4




Example 2:
 5 2 3
5 4 8
A =  4 7 1 AT =  2 7 5
8 5 9
 3 1 9
Theorem Given matrices A and B, then the following holds
i
A
B
ii
A
iii
kA
iv
AB
T T
T
AT
BT
A
T
kAT where k is a scalar.
T
BT AT
Lecture Note prepared by Dr J Acquah
Page 34
Observe in iv that the transpose of a product is the product of transposes, but in the
reverse order.
2.1.8 Conjugate Transpose
The conjugate transpose of an m − by − n matrix A with complex entries is the
n − by − m matrix A obtained from A by taking the transpose and then taking the complex
*
conjugate of each entry (i.e. negating their imaginary parts but not their real parts). The
conjugate transpose is formally defined by
(A ) = A
(2.6)
*
ji
ij
This definition can also be written as
( ) =A
A* = A
T
T
(2.7)
where AT denotes the transpose and A denotes the matrix with complex conjugated
entries.
Example 1
3 − i 2 + 2i 
 3 + i 5
If A = 
then A* = 

−i 
 5
 2 − 2i i 
2.1.9 Normal Matrix
A complex square matrix A is a normal matrix if
A* A = AA*
(2.8)
where A* is the conjugate transpose of A. That is, a matrix is normal if it commutes with
its conjugate transpose.
If A is a real matrix, then A = AT ; it is normal if AT A = AAT .
Lecture Note prepared by Dr J Acquah
Page 35
1.8.4
Symmetric Matrix
A symmetric matrix is a square matrix, A, that is equal to its transpose
A = AT
(2.9)
The entries of a symmetric matrix are symmetric with respect to the main diagonal (top
left to bottom right). So if the entries are written as A = ( aij ) , then aij = a ji for all indices i
and j. The following 3×3 matrix is symmetric:
1 2 3 
 2 4 −5


 3 −5 6 
2.1.11
Skew-Symmetric Matrix
A skew-symmetric (or antisymmetric or antimetric) matrix is a square matrix A
whose transpose is also its negative; that is, it satisfies the equation:
AT = − A
(2.10)
or in component form, if A = ( aij ) ; then aij = −a ji for all i and j. For example, the following
matrix is skew-symmetric:
 0 2 −1
 −2 0 −4


 1 4 0 
2.2
Special Matrices: Triangular, Diagonal
We have seen that a matrix is a block of entries or two dimensional data. The size of the
matrix is given by the number of rows and the number of columns. If the two numbers
are the same, we called such matrix a square matrix.
Lecture Note prepared by Dr J Acquah
Page 36
To square matrices we associate what we call the main diagonal (in short the diagonal).
Indeed, consider the matrix
a b 
A=

c d 
Its diagonal is given by the numbers a and d. For the matrix
a b c 


A = d e f 
g h k 


its diagonal consists of a, e, and k. In general, if A is a square matrix of order n and if aij
th
is the number in the i th - row and j - column, then the diagonal is given by the numbers
aii , for i = 1,
,n .
The diagonal of a square matrix helps define two type of matrices: upper – triangular
and lower – triangular. Indeed, the diagonal subdivides the matrix into two blocks:
one above the diagonal and the other one below it. If the lower-block consists of zeros, we
call such a matrix upper – triangular. If the upper-block consists of zeros, we call such
a matrix lower – triangular. For example, the matrices
a b e 
a b



 and  0 e f 
0 d 
0 0 k 


are upper-triangular, while the matrices
 a 0 0
a 0



 and  d e 0 
c d
g h k


are lower-triangular. Now consider the two matrices
 a 0 0
a d



A =  d e 0  and B =  0 e
g h k
0 0



g

h.
k 
The matrices A and B are triangular.
Lecture Note prepared by Dr J Acquah
Page 37
A diagonal matrix is a symmetric matrix with all of its entries equal to zero except may
be the ones on the diagonal. So a diagonal matrix has at most n different numbers. For
example, the matrices
 a 0 0
 a 0



 and  0 0 0 
0 b
0 0 b


are diagonal matrices. Identity matrices are examples of diagonal matrices.
Example 1 Consider the diagonal matrix
 a 0
A=

0 b
Define the power-matrices of A by
A0 = I 2 , A1 = A, A2 = AA, A3 = AAA etc.
Find the power matrices of A and then evaluate the matrices
I2 +
1
1
A + A2 +
1!
2!
+
1 n
A
n!
for n = 1, 2,
Answer. We have
 a 0  a 0   a 2
A2 = 

=
 0 b  0 b   0
0
.
b2 
and
 a2
A3 = A2 A = 
0
0  a

b2   0
0   a3
=
b  0
0
.
b3 
By induction, one may easily show that
 an
An = 
0
0
.
bn 
for every natural number n.
Lecture Note prepared by Dr J Acquah
Page 38
2.2.1 Addition of Matrices:
In order to add two matrices, we add the entries one by one.
Note: Matrices involved in the addition operation must have the same size.
So how do we add matrices? The answer is to add entries one by one. For example, we
have
a

d
b
e
c  
+
f  


   a +
=
   d + 
b+
e +
c + 
.
f +
Clearly, if you want to double a matrix, it is enough to add the matrix to itself. So we have
double of which implies
a

d
b
e
c a
+
f  d
b
e
c  a+a
=
f  d +d
b+b
e+e
c + c   2a
=
f + f   2d
2b
2e
2c 

2f 
Example 1
1 4 7
6 9 6




Given that A =  2 6 1  and B =  5 6 4  . Evaluate A + B
 3 2 4
 2 2 3




Solution
 1 4 7   6 9 6   7 13 13 

 
 

A + B =  2 6 1  +  5 6 4  =  7 12 5 
 3 2 4  2 2 3  5 4 7 

 
 

This suggests the following rule
Let A, B and C be matrices of order m  n , and let k , 
. Then
1.
A+ B = B+ A
(commutativity).
2.
( A + B) + C = A + ( B + C )
(associativity).
3. k ( A) = ( k ) A .
4.
( k + ) A = kA + A
5. A + ( − A) = ( − A) + A = 0
Lecture Note prepared by Dr J Acquah
Page 39
6. A + 0 = A
2.2.2 Subtraction of matrices
If M and N are two matrices, then we will/can write M − N = M + ( −1) N , we subtract the
entries one by one.
Note: Matrices involved in the subtraction operation must have the same size.
So how do we subtract matrices? The answer is to subtract entries one by one. For
example, we have
a

d
b
e
c  
−
f  


   a −
=
   d − 
b−
e −
c − 
.
f −
Clearly, if you want to get a zero matrix, it is enough to subtract the matrix from itself. So
we have
a

d
b
e
c a
−
f  d
b
e
c   a−a
=
f  d −d
b −b
e−e
c − c   0 0 0
=
.
f − f   0 0 0
Scalar Product. Consider the 3x1 matrices
a
 
 
 
X =  b  and Y =    .
c
 
 
 
The scalar product of X and Y is defined by
 
 
X Y = ( a b c )    = a + b + c
 
 
T
(2.11)
Example 1
For the following matrices perform the following operation if possible.
Lecture Note prepared by Dr J Acquah
Page 40
2 0 − 3 2 
A

 −1 8 10 − 5
2 0

0 − 4 − 7 2  C   −4 9
B
6 0
9 
12 3 7
2 
5 
− 6
(a) A + B
(b) B − C
(c) A + C
Solution
a. Both A and B are of the same size and so we know the addition can be done in this case.
Once we know addition can be done there isn’t much to do than to just add.
 2 − 4 − 10 4 
A+ B  

11 11 17 4 
b. Again since A and B are of the same size we can do the difference as the previous one,
there isn’t really much to do. All we need to be careful with is the order. Just like with
real number arithmetic B-C is different from A-B. So in this case we will subtract the
 −2 − 4 − 4 0 
B+ A 

13 − 5 − 3 14 
entries of A from entries of B
c. In this case because A and C are different sizes the addition cannot be done likewise, AC,C-A,B+C,C-B and B-C cannot be done for the same reason.
2.3
Multiplying a Scalar to a Matrix
Let A = aij  be an m  n matrix. Then for any element k 
1 4 5 
5
For example, if A = 
and k = 5 then 5 A = 

0 1 2
0
we define kA = kaij  .
20 25
.
5 10 
In order to multiply a matrix by a number, you multiply every entry by the given number.
Keep in mind that we always write numbers to the left and matrices to the right (in the
case of multiplication).
a
2
d
b
e
c   2a
=
f   2d
Lecture Note prepared by Dr J Acquah
2b
2e
2c 

2f 
Page 41
Example 1 Given that
8 1
2 3


B =  −7 0  C =  −2 5 
10 −6 
 4 −1
0 9
A =  2 −3
 −1 1 
Compute 3 A + 2 B − 1
C2
SOLUTION
So we are really being asked to compute a linear combination here. We will do that by first
computing the scalar multiplies and the perform the addition and subtraction. Note as
well that in the case of the third scalar multiple we are going to consider the scalar to be
a positive 1 and leave the miunus sign out in front of the mtrix. Here is the work for the
2
problem.
55 
 3  
2
 0 27   16 2   1 2  15
1 



3 A + 2 B − =  6 −9  +  −14 0  +  −1 5  = −7 −23 
2 
2

2
 −3 3   −8 3  5 −3   0 4 

 

2.4
Multiplication of Matrices
DEFINITION: (Matrix Multiplication / Product). Let A = aij  be an m  n
matrix and B = bij  be an n  r matrix. The product AB is a matrix C = cij  of
order m  r .
Observe that the product AB is defined if and only if number of columns of A is equal
to the number of rows of B.
1 2 1 
1 2 3 


For example, if A = 
 and B = 0 0 3  then
2
4
1


1 0 4 
 1+ 0 + 3
AB = 
 2 + 0 +1
2+0+0
4+0+0
1 + 6 + 12   4
=
2 + 12 + 4   3
Lecture Note prepared by Dr J Acquah
2
4
19 
18 
Page 42
In other words, we have
 a b  e


 c d  g
f   ae + bg
=
h   ce + dg
af + bh 

cf + dh 
In fact, we do not need to have two matrices of the same size to multiply them. Above, we
did multiply a ( 2 by 2) matrix with a ( 2 by 1) matrix (which gave a ( 2 by 1) matrix). In
fact, the general rule says that in order to perform the multiplication AB, where A is a
( m  n) matrix and B is a ( k 1) matrix, then we must have n = k . The result will be a ( m1)
matrix. For example, we have
a

d
 
c     a + b + c 
  =

f     d + e + f  
 
b
e
Remember that though we were able to perform the above multiplication, it is not possible
to perform the multiplication
 
  a
   d
 
 
b
e
c
.
f
Two square matrices A and B are said to commute if AB = BA .
1. Note that if A is a square matrix of order n then AI n = I n A . Also, a scalar matrix of
order n commutes with any square matrix of order n .
2. In general, the matrix product is not commutative. For example, consider the
1 1 
1 0 
following two matrices A = 
and B = 

 . Then check that the matrix
0 0
1 0 
product
 2 0  1 1
AB = 

 = BA .
 0 0  1 1
Lecture Note prepared by Dr J Acquah
Page 43
Properties involving Addition and Multiplication.
1. Let A, B and C be three matrices. If you can perform the appropriate products,
then we have
( A + B ) C = AC + BC
(2.12)
A( B + C ) = AB + AC
(2.13)
and
2. If  and  are numbers, A and B are matrices, then we have
 ( A + B) =  A +  B
(2.14)
( +  ) A =  A +  A
(2.15)
and
Example 1 Consider the matrices
 0 1
2
A=
 , B =   , and C = ( 0 1 5 )
 −1 0 
 −1
Evaluate ( AB ) C and A ( BC ) . Check that you get the same matrix.
Example 2
Compute AC and CA for the following two matrices. If possible
8 5 3
 −3 10 2 
1 −3 0 4 
A=
 C =  2 0 −4 
−
2
5
−
8
9




 −1 −7 5 
SOLUTION
Okay lets first do AC . Here are the sizes for A and C
A C= A C
2*4 4*3
2*3
So the inner numbers (4 and 4) are the same and so the multiplication can be done and
we can see that the new size of the matrix is 2*3. Now lets actually do the multiplication.
Lecture Note prepared by Dr J Acquah
Page 44
We will go through the first couple of entries in the product in detail and the remaining
entries a little quicker. To get the number in the first row and first column of AC we will
multiply the first row of A and the first column of C (1)(8)+(-3)(-3)+(0)(2)+(4)(-1)=13
If we want the entry in the first row and second column AC we will multiply the first row
of A by the second column of B as follows,
(1)(5)+(-3)(10)+(0)(0)+(4)(-7)=-53
Okay, at this point lets stop and insert these into the product so we can make sure that
we’ve got our bearings. Here is the product so far
8 5 3


1 −3 0 4   −3 10 2  13 −53
 −2 5 −8 9   2 0 −4  = 





 −1 −7 5 



As we can see we’ve got four entries left to compute. For these we will give row and column
multiplications but leave it to you to make sure we used the correct row/column and put
the result in the correct place her is the remaining work.
(1)( 3) + ( −3)( 2 ) + ( 0 )( −4 ) + ( 4 )( 5) = 17
( −2 )(8) + ( 5)( −3) + ( −8)( 2 ) + ( 9 )( −1) = −56
( −2 )( 5) + ( 5)(10 ) + ( −8) ()0 + ( 9 )( −7 ) = −23
( −2 )( 3) + ( 5)( 2 ) + ( −8)( −4 ) + ( 9 )( 5) = 81
Here is a complete product
8 5 3


1 −3 0 4   −3 10 2  13 −53 17 
 −2 5 −8 9   2 0 −4  =  −56 −23 81






 −1 −7 5 
Now lets do CA. here are the sizes of the product
C
4*3
A = CA
2*3 = N / A
Okay in this case the two inner numbers (3 and 2) are not the same and so this product
can’t be done.
Lecture Note prepared by Dr J Acquah
Page 45
Example 3 Compute BD and DB for the given matrices, if possible
 3 −1 7 
 −1 4 9 
B = 10 1 −8 D =  6 2 −1
 7 4 7 
 −5 2 4 
Solution
First notice that both of these matrices are 3*3 matrices and so both BD and DB are
defined.
Again its worth pointing out that this example differs from the previous example in that
both the products are defined in this example rather than only being defined as in the
previous example. Also know that in both case the product will be a new 3*3 matrix. In
this example we are going to leave the work of verifying the products to you it is good
practice so you should try and verify at least one of the following products.
 3 −1 7   −1 4 9   40 38 77 
BD = 10 1 −8  6 2 −1 =  −60 10 33 
 −5 2 4   7 4 7   45 0 −19 
 −1 4 9   3 −1 7   −8 23 −3
DB =  6 2 −1 10 1 −8 =  43 −6 22 
 7 4 7   −5 2 4   26 11 45 
2.5
Determinants
For any square matrix of order 2, we have found a necessary and sufficient condition for
invertibility. Indeed, consider the matrix
a b 
A=

c d 
a b
a b a b
determinant of 
= ad − bc
 = det 
=
c d
c d c d
The matrix A is invertible if and only if ad − bc  0 . We called this number the
determinant of A. It is clear from this that we would like to have a similar result for
bigger matrices (meaning higher orders). So is there a similar notion of determinant for
any square matrix, which determines whether a square matrix is invertible or not?
Lecture Note prepared by Dr J Acquah
Page 46
A matrix A is said to be a singular matrix if det ( A) = 0 It is called non-singular if
det ( A)  0
2.5.1 Determinants of Matrices of Higher Order
These occur in systems of three linear equations with three unknowns x1, x2 and x3 . The
determinant is defined by the equation
D
That is, D
a11 a12 a13
a21 a22 a23
a31 a32 a33
a11
a22 a23
a32 a33
a21
a12 a13
a32 a33
a31
a12 a13
a22 a23
(2.16)
a11M11 a21M 21 a31M 31 . The determinants M i j , obtained by deleting one row
and one column of D is called the minor of the elements that belong to the deleted row
and column.
Cofactors: The cofactors of the elements in D in the i-th row and j-th column are defined
as
1
i
j
Mi j
Ci j , i.e.
1
i
j
checkerboard pattern
times the minor of that element. The signs form a
. We may write D
a11C11 a21C21 a31C31, where C11 is
the cofactor of the element a11 in D.
Example 1 Evaluate
3 2 1
2 1 −3 .
4 0 1
We will use the general formula along the third row. We have
Lecture Note prepared by Dr J Acquah
Page 47
3 2 1
2 1
3 1
3 2
2 1 −3 = 4
−0
+1
= 4 ( −6 − 1) + 1( 3 − 4 ) = −29 .
1 −3
2 −3
2 1
4 0 1
Which technique to evaluate a determinant is easier ? The answer depends on the person
who is evaluating the determinant. Some like the elementary row operations and some
like the general formula. All that matters is to get the correct answer.
Properties of the Determinant
1. Any matrix A and its transpose have the same determinant, meaning
det A = det AT
2. The determinant of a triangular matrix is the product of the entries on the diagonal,
that is
a b a 0
=
= ad .
0 d b d
3. If we interchange two rows, the determinant of the new matrix is the opposite of
the old one, that is
a b
c d
=−
.
c d
a b
4. If we multiply one row with a constant, the determinant of the new matrix is the
determinant of the old one multiplied by the constant, that is
 a b
c
d
=
a b
a
b
=
.
c d c  d
In particular, if all the entries in one row are zero, then the determinant is zero.
5. If we add one row to another one multiplied by a constant, the determinant of the
new matrix is the same as the old one, that is
a + c b +  d a b
a
b
=
=
.
c
d
c d c +  a d + b
Note that whenever you want to replace a row by something (through elementary
operations), do not multiply the row itself by a constant. Otherwise, you will easily
make errors (due to Property 4).
Lecture Note prepared by Dr J Acquah
Page 48
6. We have
det ( AB ) = det ( A) det ( B ) .
In particular, if A is invertible (which happens if and only if det ( A)  0 ), then
det ( A−1 ) =
1
det ( A)
(2.17)
If A and B are similar, then det ( A) = det ( B ) .
2.6 Invertible Matrices
Invertible matrices are very important in many areas of science. For example, decrypting
a coded message uses invertible matrices
Definition. An n  n matrix A is called nonsingular or invertible iff there exists an
n  n matrix B such that
AB = BA = In
(2.18)
where In is the identity matrix. The matrix B is called the inverse matrix of A. The
inverse of an n by n matrix is another n by n matrix. If the first matrix is A, its inverse is
written A−1 (and pronounced "A inverse").
Example1 Let
 2 3
 −1 3 2 
A=
 and B = 

 2 2
 1 −1 
Solution
 2 3  −1 3 2   1 0 
AB = BA = 

=
 = I 2 . Hence A is invertible and B is its
 2 2  1 −1   0 1 
inverse.
−1
−1
Notation. A common notation for the inverse of a matrix A is A−1 . So AA = A A = I n .
Lecture Note prepared by Dr J Acquah
Page 49
Method 1
Example 2 Find the inverse of
 1 1
A=
.
 −1 2 
Solution:
a
Write A−1 = 
c
b

d
Since
b+d 
 a+c
AA−1 = 
 = I2 .
 −a + 2c −b + 2d 
we get
 a + c = 1
 −a + 2c = 0


 b + d = 0

 −b + 2d = 1
Therefore,
a=
2

Hence A−1 =  3
1

3
2
1
1
1
, b=− , c= , d = .
3
3
3
3
1
− 
3

1 

3 
The inverse matrix is unique when it exists. So if A is invertible, then A−1 is also invertible
and
(A ) = A.
−1 −1
The following basic property is very important:
If A and B are invertible matrices, then AB is also invertible and ( AB ) = B−1 A−1
−1
Lecture Note prepared by Dr J Acquah
Page 50
Method 2
If A  0 , the inverse of a 2  2 matrix A may be obtained from A as follows:
(i) Interchange the two elements on the diagonal.
(ii) Take the negatives of the other two elements.
(iii)
Multiply the resulting matrix by
1
or, equivalently, divide each element
A
by A .
In case A = 0 , the matrix A is not invertible.
Thus,
A−1 =
1
adj ( A)
A
(2.18)
Example 3 Given
a b 
A=

c d 
Solution:
d
adj ( A) = 
 −b
−c 
d
 =
a 
 −c
T
−b 
 and det A = ad − bc .
a 
which gives
A−1 =
1  d −b 


ad − bc  −c a 
(2.19)
 
Definition of cofactor :Let A = aij be n  n matrix. The cofactor of a ij is
defined as
Aij = (− 1)
i+ j
det(M ij )
Lecture Note prepared by Dr J Acquah
(2.20)
Page 51
where M ij is the (n − 1)  (n − 1) submatrix of A by deleting the i’th row of
j’th column o.f A
Example
Find the inverse of the matrix using the Inverse Method
0
3 
2
A = − 1 4 − 2
 1 − 3 5 
Solution:
Let
2
A = − 1
 1
0
4
−3
3 
− 2
5 
Then,
 4 − 2
 − 1 − 2
− 1 4 
M 11 = 
, M 12 = 
, M 13 = 


,
5
− 3 5 
1
 1 − 3
 0 3
2 3
2 0 
M 21 = 
, M 22 = 
, M 23 = 


,
− 3 5
 1 5
1 − 3
3
0 3 
2
 2 0
M 31 = 
, M 32 = 
, M 33 = 



 4 − 2
 − 1 − 2
 − 1 4
Thus,
Lecture Note prepared by Dr J Acquah
Page 52
A11 = (− 1)
det(M 11 ) = 1  4  5 − (−2)  (−3) = 14,
A12 = (− 1)
det(M 12 ) = (− 1)
1+1
(−1)  5 − (−2)  1 = 3
1+ 3
1+ 3
A13 = (− 1) det(M 13 ) = (− 1) (−1)  (−3) − 4  1 = −1
2 +1
2 +1
A21 = (− 1) det(M 21 ) = (− 1) 0  5 − (−3)  3 = −9
2+ 2
2+ 2
A22 = (− 1) det(M 22 ) = (− 1) 2  5 − 3  1 = 7
2+3
2+3
A23 = (− 1) det(M 23 ) = (− 1) 2  (−3) − 0  1 = 6
3+1
3+1
A31 = (− 1) det(M 31 ) = (− 1) 0  (−2) − 3  4 = −12
3+ 2
3+ 2
A32 = (− 1) det(M 32 ) = (− 1) 2  (−2) − 3  (−1) = 1
3+ 3
3+ 3
A33 = (− 1) det(M 33 ) = (− 1) 2  4 − 0  (−1) = 8
1+ 2
1+ 2
Important result:
  be an n  n matrix. Then,
Let A = aij
det( A) = ai1 Ai1 + ai 2 Ai 2 +  + ain Ain , i = 1,2,, n
= a1 j A1 j + a 2 j A2 j +  + a nj Anj , j = 1,2,, n
In addition,
ai1 Ak 1 + ai 2 Ak 2 +  + ain Akn = 0, i  k
a1 j A1k + a2 j A2 k +  + anj Ank = 0, j  k
A11 = 14, A12 = 3, A13 = −1, A21 = −9, A22 = 7, A23 = 6, A31 = −12, A32 = 1, A33 = 8
Thus,
det( A) = a11 A11 + a12 A12 + a13 A13 = 2  14 + 0  3 + 3  −1 = 25
= a 21 A21 + a 22 A22 + a 23 A23 = (−1)  (−9) + 4  7 + (−2)  6 = 25
= a31 A31 + a32 A32 + a33 A33 = 1  (−12) + (−3)  1 + 5  8 = 25
Also,
Lecture Note prepared by Dr J Acquah
Page 53
det( A) = a11 A11 + a21 A21 + a31 A31 = 2  14 + (−1)  (−9) + 1  (−12) = 25
= a12 A12 + a22 A22 + a32 A32 = 0  3 + 4  7 + (−3)  1 = 25
= a13 A13 + a23 A23 + a33 A33 = 3  (−1) + (−2)  6 + 5  8 = 25
In addition,
a11 A21 + a12 A22 + a13 A23 = 2  (−9) + 0  7 + 3  6 = 0
a11 A31 + a12 A32 + a13 A33 = 2  (−12) + 0  1 + 3  8 = 0
a 21 A11 + a 22 A12 + a 23 A13 = (−1)  14 + 4  3 + (−2)  (−1) = 0
a 21 A31 + a 22 A32 + a 23 A33 = (−1)  (−12) + 4  1 + (−2)  8 = 0
a31 A11 + a32 A12 + a33 A13 = 1  14 + (−3)  3 + 5  (−1) = 0
a31 A21 + a32 A22 + a33 A23 = 1  (−9) + (−3)  7 + 5  6 = 0
Similarly,
a1 j A1k + a 2 j A2 k + a3 j A3k = 0, j  k
Definition of adjoint:
The n  n matrix adj(A) , called the adjoint of A, is
 A11
A
adj ( A) =  12
 

 A1n
A21
A22

A2 n
 An1   A11
 An 2   A21
=
    
 
 Ann   An1
A12
A22

An 2
T
 A1n 
 A2 n 
   .

 Ann 
Important result:
A  adj( A) = adj( A)  A = det( A) I n
and
A −1 =
adj ( A)
det( A)
Lecture Note prepared by Dr J Acquah
(2.21)
Page 54
 A11
adj ( A) =  A12
 A13
A31  14 − 9
A32  =  3
7
A33  − 1 6
A21
A22
A23
− 12
1 
8 
and
A
2.7
−1
14 − 9
adj ( A)
1 
=
=
3
7
det( A) 25 
− 1 6
− 12
1 
8 
Linear Dependence and Independence of Matrices
Definition. A set of n, row or column vectors v1 , v2 ,
dependent if there exist n scalars
v
1 1
v
v
2 2
only true when
,
2
, , n , not all equal to zero, such that
0 . When no such scalars exist such that the above relationship is
n n
1
1
, vn are said to be linearly
2
n
0 , then N matrix is said to be linearly independent.
Test for Linear Dependence
Rows and columns of a square matrix A are said to be linear independent if A
Similarly, if A
0.
0 , then the rows and columns are of matrix A are said to be linearly
independent.
Example. Consider the matrix A
1
4 3
2 18 7 where A
4
6 1
0 . This implies linear
dependence exists between either rows or columns of A. Again the rows or columns are
linearly dependent because c2
2 c3 c1 where ci denotes column i. Consider another
Lecture Note prepared by Dr J Acquah
Page 55
example, B
1 1 0
3 2 1 where B
1 1 3
3 . So the rows or columns of B are linearly
independent.
2.8 Rank of a Matrix
There are two equivalent definitions of a rank of a matrix. Firstly, the rank of a matrix
may be defined in terms of linearly independent vectors, row or vector. Suppose that the
columns of an m
n matrix are interpreted as the components in a given basis of n (n-
component) vectors v1 , v2 ,
, vn , as follows:
A
v1 v2
vn
The rank of A, denoted by rank A or by R A , is defined as the number of linearly
independent vectors in the set v1 , v2 ,
, vn and equals the dimension of the vector space
spanned by those vectors.
Secondly, A second (equivalent) definition may be given of the rank of a matrix and uses
the concept of sub-matrices. A sub-matrix of A is any matrix that can be formed from the
elements of A by ignoring one, or more than one, row or column. It may be shown that
the rank of a general m
n matrix is equal to the size of the largest square sub-matrix of
A whose determinant is non-zero. Therefore, if a matrix A has an r
S
0 , but no r 1
r sub-matrix S with
r 1 sub-matrix with non-zero determinant then the rank of the
matrix is r. From either definition it is clear that the rank of A is less than or equal to the
smaller of m and n.
Example. Find the rank of each of the following matrices:
a
A
1 1 0
2 0 2
4 1 3
1
2
1
and
Lecture Note prepared by Dr J Acquah
b
B
1 3
8 9
2 1
1
4
2
Page 56
Solution
1 1 0
2 0 2
4 1 3
1 1
2 0
4 1
2
2
1
1 0
2 2
4 3
2
2
1
1 0
0 2
1 3
2
2
1
0.
This implies the rank of A is not 3. The next largest square sub-matrices of A are of
dimension 2 2 . Consider, for example, the 2 2 sub-matrix formed by ignoring the
third row and the third and fourth columns of A; this has determinant
1 1
2 0
1 0
2 1
2. Thus, A is of rank 2 and we need not consider any other
2 2 sub-matrices.
2.9
Systems of Linear Equations: Gaussian Elimination
Definition: The equation ax + by + cz + dw = h where a, b, c, d and h are known
numbers, while x, y, z and w are unknown numbers, is called a linear equation. If
h = 0 , the linear equation is said to be homogeneous. A linear system is a set of linear
equations and a homogeneous linear system is a set of homogeneous linear
equations.
For example,
= 1
x+ y−z

 x + 3 y + 3z = − 2
are linear systems, while
2 x − 3 y 2

x + y + z
= −1
= −2
is a nonlinear system (because of y2). The system is a non homogeneous linear system.
Lecture Note prepared by Dr J Acquah
Page 57
Remark. In more the general case in which m
n but det A
0 , the inverse does not
exist and so any procedure using A 1 would not work. In such circumstances, we consider
more carefully what solution means. Generally, when a solution vector x exists whose
elements simultaneously satisfy all the equations in the system – the equation will be said
to be consistent. Otherwise it is inconsistent
2.9.1 Matrix Representation of a Linear System
Matrices are helpful in rewriting a linear system in a very simple form. The algebraic
properties of matrices may then be used to solve systems. First, consider the linear system
 ax + by + cz + dw = e
 fx + gy + hz + iw = j

.

kx
+
ly
+
mz
+
nw
=
p


 qx + ry + sz + tw = u
Set the matrices
a

f
A=
k

q
b c d
e
x

 
 
g h i
j
y

, C=
, and X =  
 p
z
l m n

 
 
r s t
u
 w
Using matrix multiplications, we can rewrite the linear system above as the matrix
equation
AX = C
The matrix A is called the matrix coefficient of the linear system. The matrix C is called
the nonhomogeneous term. When C = 0 , the linear system is homogeneous. The
matrix X is the unknown matrix. Its entries are the unknowns of the linear system. The
augmented matrix associated with the system is the matrix  A | C  , where
a

f
 A | C  = 
k

q
Lecture Note prepared by Dr J Acquah
b
g
l
r
c d
h i
m n
s t
e

j
p

u 
(2.22)
Page 58
In general if the linear system has n equations with m unknowns, then the matrix
coefficient will be a n  m matrix and the augmented matrix an n  ( m + 1) matrix. Now we
turn our attention to the solutions of a system.
Definition: Two linear systems with n unknowns are said to be equivalent if and only
if they have the same set of solutions.
2.9.2 Gaussian Elimination.
Consider a linear system.
1. Construct the augmented matrix for the system;
2. Use elementary row operations to transform the augmented matrix into a
triangular one;
3. Write down the new linear system for which the triangular matrix is the associated
augmented matrix;
4. Solve the new system. You may need to assign some parametric values to some
unknowns, and then apply the method of back substitution to solve the new
system.
Elementary Row Operations
1. Interchange two rows.
2. Multiply a row with a nonzero number.
3. Add a row to another one multiplied by a number.
ie.
DEFINITION
(Forward/Gauss
Elimination
Method)
Gaussian
elimination is a method of solving a linear system Ax = b (consisting of m
equations in n unknowns) by bringing the augmented matrix
 a11 a12

a
a
 A | b =  21 22

 am1 am 2
Lecture Note prepared by Dr J Acquah
a1n
a2 n
amn
b1 

b2 


bm 
(2.23)
Page 59
to an upper triangular form
c11 c12

 0 c22


 0 0
c1n
c2 n
cmn
d1 

d2 


d m 
This elimination process is also called the forward elimination method.
For example, consider the matrix equation
 9 3 4  x1   7 

   
 4 3 4  x2  =  8 
 1 1 1  x   3 

 3   
In augmented form, this becomes
9 3 4 7   x1 

 
 4 3 4 8   x2 
1 1 1 3   x3 
Switching the first and third rows (without switching the elements in the right-hand
column vector) gives
1 1 1 3   x1 

 
 4 3 4 8   x2 
9 3 4 7   x3 
Subtracting 9 times the first row from the third row gives
1 1 1
3   x1 


8   x2 
4 3 4
 0 −6 −5 −20   x3 
Subtracting 4 times the first row from the second row gives
Lecture Note prepared by Dr J Acquah
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1 1
1
3   x1 


−4   x2 
0 −1 0
0 −6 −5 −20   x3 
Finally, adding -6 times the second row to the third row gives
1 1 1
3   x1 

 
0 −1 0 −4   x2 
0 0 −5 4   x3 
Restoring the transformed matrix equation gives
1 1 1   x1   3 
0 −1 0   x  =  −4 

 2  
0 0 −5   x3   4 
which can be solved immediately to give x3 = −4 , back-substituting to obtain x2 = 4
5
(which actually follows trivially in this example), and then again back-substituting to find
x1 = −
1
.
5
EXAMPLE: Solve the linear system by Gaussian elimination method.
y + z= 2
2x +
3z = 5
x + y + z= 3
Solution:
In this case, the augmented matrix is
0 1 1

2 0 3
1 1 1
Lecture Note prepared by Dr J Acquah
2

5
3 
Page 61
The method proceeds along the following steps.
1. Interchange 1st and 3rd equation.
x + y + z= 3
2x
+z =5
y + z =3
1 1 1

2 0 3
0 1 1
3

5
2 
2. Add -2 times the 1st equation to the 2nd equation.
x+ y+z = 3
− 2 y + z = −1
y+z = 2
1 1 1

0 −2 1
0 1 1
3

−1
2 
3. Add 2 times the 3rd equation to the 2nd equation.
x+ y+z = 3
− 2 y + z = −1
3z = 3
1 1 1

0 −2 1
0 0 3
3

−1
3 
4. Divide 2nd equation through by -2 and 3rd equation through by 3.
x+ y+z = 3
y − 12 z = 12
z = 1
1 1 1

1
0 1 − 2
0 0 1
3

2
1 
1
The last equation gives z = 1 the second equation now gives y = 1 . Finally the first
equation gives x = 1 . Hence the set of solutions is
( x, y, z ) = (1,1,1) , A UNIQUE
t
t
SOLUTION.
Definition. A linear system is called inconsistent or overdetermined if it does not
have a solution. In other words, the set of solutions is empty. Otherwise the linear system
is called consistent.
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2.9.3 Eigenvalues and Eigenvectors
The eigenvalue or characteristic value and its corresponding eigenvector or characteristic
vector of an N  N matrix A are defined as a scalar λ and a nonzero vector v satisfying
Av = v  ( A −  I n ) v = 0 ( v  0)
( )
where ( , v ) is called an eigenpair and there are N eigenpairs for the N  N matrix A.
How do we get them? Noting that
•
in order for the above equation to hold for any nonzero vector v, the matrix
 A −  In 
should be singular – that is, its determinant should be zero
( A −  I = 0) and
n
•
the determinant of the matrix is a polynomial of degree N in terms of λ,
we first must find the eigenvalue i ' s by solving the so-called characteristic equation
A −  I n =  N + aN −1 N −1 +
+ a1 + a0 = 0
( )
and then substitute the i ' s , one by one, into Eq. ( ) to solve it for the eigenvector vi ' s .
➢ Computation of Eigenvalues
For a square matrix A of order n, the number  is an eigenvalue if and only if there exists
a non-zero vector x such that
Ax =  x
Using the matrix multiplication properties, we obtain
( A −  In ) x = 0
This is a linear system for which the matrix coefficient is A −  I n . We also know that this
system has one solution if and only if the matrix coefficient is invertible, i.e.
det ( A −  I n )  0 . Since the zero-vector is a solution and x is not the zero vector, then we
must have det ( A −  I n ) = 0 .
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In general, for a square matrix A of order n, the equation det ( A −  I n ) = 0 will give the
eigenvalues of A. This equation is called the characteristic equation or
characteristic polynomial of A. It is a polynomial function in  of degree n. So we
know that this equation will not have more than n roots or solutions. So a square matrix
A of order n will not have more than n eigenvalues.
Example. Consider the matrix
 1 −2 
A=
.
 −2 0 
The equation det ( A −  I n ) = 0 translates into
1 −  −2
= (1 −  )( 0 −  ) − 4 = 0
−2 0 − 
which is equivalent to the quadratic equation
2 −  − 4 = 0
Solving this equation leads to
=
1 + 17
1 − 17
, and  =
2
2
In other words, the matrix A has only two eigenvalues.
Example. Find the eigenvalues/eigenvectors of the matrix
0 1 
A=

0 −1
First, we find its eigenvalues as
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 −
A − I = 
0
 (  + 1) = 0,
1 
= 2 +  = 0
−1 −  
1 = 0,
2 = −1
and then, get the corresponding eigenvectors as
0 1   v11   v21  0
A − 1I v1 = 
= 
  = 
0 −1 v21   −v21  0 
 (  + 1) = 0,
1 = 0,
2 = −1
where we have chosen v11, v12, and v22 so that the norms of the eigenvectors
become one.
Example. Consider the diagonal matrix
a

0
D=
0

0
0
b
0
0
0
0
c
0
0

0
0

d
Its characteristic polynomial is
a−
0
0
0
b−
0
det ( D −  I n ) =
0
0
c−
0
0
0
0
0
= ( a −  )( b −  )( c −  )( d −  ) = 0
0
d −
(2.24)
So the eigenvalues of D are a , b , c and d , i.e. the entries on the diagonal. .
We have some properties of the eigenvalues of a matrix.
Theorem. Let A be a square matrix of order n. If
is an eigenvalue of A, then:
1.  m is an eigenvalue of Am , for m = 1, 2,
2. If A is invertible, then 1 is an eigenvalue of A−1 .

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3. A is not invertible if and only if  = 0 is an eigenvalue of A.
4. If  is any number, then  +  is an eigenvalue of A +  I n .
5. If A and B are similar, then they have the same characteristic polynomial (which
implies they also have the same eigenvalues).
➢ Computation of Eigenvectors
Let A be a square matrix of order n and  one of its eigenvalues. Let x be an eigenvector
of A associated to  . We must have
Ax =  x or
( A −  In ) x = 0 .
(2.25)
This is a linear system for which the matrix coefficient is A −  I n . Since the zero-vector is
a solution, the system is consistent.
Remark. It is quite easy to notice that if x is a vector which satisfies Ax =  x , then the
vector y = cx (for any arbitrary number c) satisfies the same equation, i.e. Ay =  y . In
other words, if we know that x is an eigenvector, then cx is also an eigenvector associated
to the same eigenvalue.
Let us start with an example.
Example: Find the eigenvectors and eigenvalues of the linear transformation u = Ax
which has the component form
u1 = 3 x1 + 4 x2 ,
u2 = 5 x1 + 2 x2 .
Solution: Solving the characteristic equation
3−
5
4
=0
2−
or
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 2 − 5 − 14 = 0
we find the eigenvalues
1 = −2
2 = 7.
For 1 = −2 the eigenvector can be determined from the system
5 x1 + 4 x2 = 0,
5 x1 + 4 x2 = 0,
which implies
x1
4
=− .
x2
5
Similarly, for 2 = 7 we have
−4 x1 + 4 x2 = 0,
5 x1 − 5 x2 = 0,
which implies
x1
= 1.
x2
Thus the eigenvectors of the transformation A are
a1 = ( 4 , −5) ,
a2 = ( 1 , 1 ) ,
and all vectors collinear with a1 and a2 .
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Page 67
Remark. In general, the eigenvalues of a matrix are not all distinct from each other (they
can be equal root/repeated root or complex root). In the next two examples, we discuss
this problem.
 1 −4 
Example. Consider the matrix 
.
 4 −7 
The characteristic equation is given by
1− 
4
−4
2
= (  + 3) = 0
−7 − 
Hence the matrix A has one eigenvalue, i.e. -3. Let us find the associated eigenvectors.
These are given by the linear system
AX = ( −3) X
or
( A + 3I2 ) X = 0
which may be rewritten by



4x − 4 y = 0
4x − 4 y = 0
This system is equivalent to the one equation-system x − y = 0 .
So if we set x = c, then any eigenvector X of A associated to the eigenvalue -3 is given by
 x
 1
X =   = c   . y
 1
Let us summarize what we did in the above examples.
Summary: Let A be a square matrix. Assume  is an eigenvalue of A. In order to find
the associated eigenvectors, we do the following steps:
1. Write down the associated linear system
AX =  X
or
( A − In ) X = 0
2. Solve the system.
3. Rewrite the unknown vector X as a linear combination of known vectors.
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Page 68
The above examples assume that the eigenvalue  is real number. So one may wonder
whether any eigenvalue is always real. In general, this is not the case except for symmetric
matrices. The proof of this is very complicated. For square matrices of order 2, the proof
is quite easy. Let us give it here for the sake of being little complete.
Consider the symmetric square matrix
a b
A=

b c
Its characteristic equation is given by
det ( A −  I 2 ) =
a−
b
b
=  2 − ( a + c )  + ac − b 2 = 0
c−
This is a quadratic equation. The nature of its roots (which are the eigenvalues of A)
depends on the sign of the discriminant
 = ( a + c ) − 4 ( ac − b 2 ) .
2
Using algebraic manipulations, we get
 = ( a + c ) + 4b2
2
Therefore,
is a positive number which implies that the eigenvalues of A are real
numbers.
Remark. Note that the matrix A will have one eigenvalue, i.e. one double root, if and
only if  = 0 . But this is possible only if a = c and b = 0 . In other words, we have A = aI 2 .
➢ The Case of Complex Eigenvalues
First let us convince ourselves that there exist matrices with complex eigenvalues.
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Example. Consider the matrix
 3 −2 
A=

 4 −1 
The characteristic equation is given by
3−
−2
=  2 − 2 + 5 = 0 .
4
−1 − 
This quadratic equation has complex roots given by
=
2  i 16
= 1  2i
2
Therefore, the matrix A has only complex eigenvalues.
The trick is to treat the complex eigenvalue as a real one. Meaning we deal with it as a
number and do the normal calculations for the eigenvectors. Let us see how it works on
the above example.
We will do the calculations for  = 1 + 2i . The associated eigenvectors are given by the
linear system
AX = (1 + 2i ) X
which may be rewritten as



( 2 − 2i ) x − 2 y = 0
4 x − ( 2 + 2i ) y = 0
In fact, the two equations are identical since ( 2 + 2i )( 2 − 2i ) = 8 . So the system reduces to
one equation
(1 − i ) x − y = 0
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Set x = c , then y = (1 − i ) c . Therefore, we have
 1 
 x  c 
X = =
 = c

 y   c (1 − i ) 
 (1 − i ) 
where c is an arbitrary number.
Remark. It is clear that one should expect to have complex entries in the eigenvectors.
We have seen that (1 − 2i ) is also an eigenvalue of the above matrix. Since the entries of
the matrix A are real, then one may easily show that if  is a complex eigenvalue, then
its conjugate  is also an eigenvalue. Moreover, if X is an eigenvector of A associated to
 , then the vector X , obtained from X by taking the complex-conjugate of the entries of
X, is an eigenvector associated to  . So the eigenvectors of the above matrix A associated
to the eigenvalue (1 − 2i ) are given by
 1 
X = c

 (1 + i ) 
where c is an arbitrary number.
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Page 71
CHAPTER 3
VECTORS
Objectives
At the end of this chapter you will be able to:
➢ Vector elements or Components in a coordinate frame.
➢ Vector magnitude and unit vectors.
➢ Addition, Subtraction and Multiplication of vectors.
➢ Geometrical interpretation of scalar product.
➢ Projection of one vector onto the other.
➢ Geometrical interpretation of vector product.
➢ Geometrical interpretation of scalar triple product.
3.0
Introduction
A vector is a geometric entity characterized by a magnitude and a direction. A vector
possesses a definite initial point and terminal point. Such a vector is called a bound
vector. In other situations, when only the magnitude and direction of the vector matter,
then the particular initial point is of no importance, and the vector is called a free vector.
Thus two arrows AB and AB in space represent the same free vector if they have the
same magnitude and direction.
3.1
Definition of a vector
A vector may be defined as a quantity which has both magnitude and direction.
Geometrically, it may be represented by a directed line segment. If PQ is the directed
line segment, representing the vector, then P is called the initial point and Q the
terminal point.
Alternatively, the vector may also be represented by a small alphabet, a . The PQ
and a are notations for vectors.
The magnitude or length of the vector is then denoted by PQ or PQ .
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Examples of vectors
Some examples of vectors are velocity, force, momentum and acceleration. Each of these
quantities has both magnitude and a direction. Unlike vectors, quantities such as energy, distance
and speed have magnitude but no direction. These quantities are called scalar quantities. Others
are mass, length and temperature.
3.1
Types of Vectors
Certain terms are associated with vectors with special properties. Some of these terms are as
follows:
i.
The zero vector (null vector): This is the vector with magnitude zero and has infinitely
several directions. It is denoted by 0 . It could also be interpreted as its direction not
specified.
The unit vector: any vector that has a magnitude of 1 unit is called a unit vector. If a is
a
any vector with length a  0 , then aˆ =
is a unit vector.
a
ii.
iii.
Equal vectors: vectors are said to be equal if they have the same magnitude and are in the
same direction. This is regardless of whether they have the same starting point or not
Collinear vectors: two vectors are said to be collinear (they lie along the same line) if
ab = a b
iv.
v.
Concurrent vectors: vectors are said to be concurrent if they pass through the same point
or a common point.
Coplanar Vectors: two vectors are said to be coplanar, if they lie in the same plane.
vi.
3.2
Vector Algebra
The operations of addition, subtraction and multiplication familiar in the algebra of numbers are
with suitable definitions capable of extensions to algebra of vectors.
3.2.1 Laws of vector algebra
If a, b, c are vectors and m and n are scalars, then
i.
ii.
iii.
iv.
a+b = b+a
a + (b + c ) = ( a + b ) + c
m(na) = (mn)a = n(ma)
(m + n)a = ma + na
Lecture Note prepared by Dr J Acquah
Commutative law of addition
Associative law of addition
Associative law of multiplication
Distributive law
Page 73
v.
m(a + b) = ma + mb
1. Addition
Distributive law
Two vectors a and b may be added to produce a third vector c . That is, c = a + b . The vector
c is called the resultant of the two vectors a and b . This may be extended to find the resultant of
three or more vectors. For example, the resultant of the three vectors a, b, c is a + b + c . The
an is a1 + a2 + a3 +
resultant of n-vectors a1 , a2 , a3
+ an . The resultant of the vectors leads to
two important laws.
a. The Parallelogram Law of Vector Addition
The law states that if two vectors a and b have a common initial point and can be represented in
a magnitude and direction by the two sides OA and OB of a parallelogram OAPB (two adjacent
sides of a parallelogram) then their resultant a + b is represented in magnitude and direction by the
diagonal OP of the parallelogram.
B
O
P
A
Fig 3.1 Parallelogram Law of Vectors
b. The Triangle Law of Vector Addition
From the parallelogram law, we may deduce the triangle law which states that if two vectors a
and b are represented by the two sides of a triangle taken in order, then their resultant is
represented by the third side of the triangle taken in the reverse order.
Fig 3.2 Triangle Law of Vectors
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Note: the triangle law may be extended to give the polygon law of vectors which states that if (n1) vectors a1 , a2 , a3 an−1 are represented by the (n-1) sides of an n-sided polygon taken in order,
then their resultant is represented by the last side of the polygon taken in the reverse order.
Fig 3.3 Polygon Law of Vectors
2. Scalar Multiplication
A vector a may be multiplied by a scalar (or number)  to give another vector  a . For a situation
of this nature, the following are satisfied:
i.
ii.
iii.
The vector  a is in the same direction as a if   0
The vector  a is in the opposite direction as a if   0
The vector  a is a zero-vector if  = 0
If b vector, then the vector 5b is five times as long as the vector b and in the same direction.
−3b is three times as long as the vector b but in opposite direction.
Example: Suppose a car is traveling along a straight road at a constant velocity of v ms −1 . If it
starts at a time t = 0 , then after a time t sec, the distance traveled or displacement d is given by
d = vt .
d is the displacement and a vector, v is the velocity and a vector and t is the time and a scalar
quantity.
3. Subtraction of vectors:
Let a and b be two vectors. We may define the difference a − b as a − b = a + (−b) . Thus, a − b
is the resultant of the vectors a and −b .
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4. Position Vector:
Let O be a fixed point. The position of a point A may be specified with reference to the point O
(called the origin). The vector OA which reaches A from the origin O is called the position vector
of the point A relative to O or simply the position vector of the point A. We use the same letter to
represent a point and its position vector. Thus, OA = a and OB = b show the position vectors of
the points A and B relative to the origin.
A
B
O
Position Vectors of the Points A and B relative to O
Fig. 3.4
The position vector of the point B relative to A is AB and this can be expressed in terms of the
position vectors of the points A and B relative to the origin O .
B
A
O
Fig 3.5 The Position Vector of the Point B relative to the Point A
OA + AB = OB
AB = OB − OA
AB = b − a
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Thus, if a and b are respectively the position vectors of the points A and B relative to the origin
O , then the vector AB (the position vector of B relative to A) is given by
AB = b − a
Example: In the figure below, M is the midpoint of AB . If a and b are respectively the position
vectors of the points A and B find in terms of a and b the position vectors of M.
A
M
B
O
OM = OA + AM
= OA +
1
1
AB (since AM = AB )
2
2
m=a+
1
1
(b − a) = (a + b)
2
2
Exercise
The point E divides the line segment PQ internally in the ratio 1:2 and R is any point not on the
line PQ . If F is a point on QR such that QF : FR = 2 :1 , then show that EF is parallel to PR .
Components of a Vector:
If we take a point O as origin, we can refer to the position of another point P either by its position
vector r or by its coordinates relative to a set of mutually perpendicular axes OX, OY and OZ. It
is usual to take a right-hand set of axes, that is, axes for which a rotation from OX to OY threads a
right-handed screw in the direction OZ.
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Page 77
Z
Y
Z
Y
O
X
Y
X
O
Fig. 3.6
O
X
Z
If we let i , j and k be unit vectors along the OX, OY and OZ, we obtain Fig 3.7
P(x, y, z)
Fig. 3.7
The point P(x,y,z) can also be described as r = x i + yj + zk where x i , yj and zk are said to be
the component of r .
Example: if i , j are unit vectors in the direction East and North respectively. Find
a. The magnitude and direction of the vector a = 3 i − 4 j
b. The components along i and j of a displacement b which has magnitude 80 units on a bearing
of 320o .
Solution:
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From the figure above:
a = 32 + 42
and
tan  =
3
4
3
 
 = tan −1   = 37 o
4
= 9 + 16
= 25 = 5
80
Again, from the figure, b is seen to have components as below:
b = −80sin 40o i + 80cos 40o j = 51.4 i + 61.3 j
Space Vectors:
In space, we may consider the rectangular Cartesian coordinate system OXYZ with O as origin.
Let i , j , k be the unit vectors along the X, Y, Z-axes respectively. Then these unit vectors are
mutually perpendicular and form a right-handed set.
Z
Y
X
Fig 3.8 Mutually Perpendicular Unit Vectors
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Any vector a in three dimensions can be represented with initial point at the origin O of a
rectangular coordinates system (see Fig 3.9). Let (a1 , a2 , a3 ) be the rectangular coordinate of the
terminal point of vector a with initial point at O. The vectors a1 i , a2 j and a3 k are called the
rectangular components vectors or simply component vectors of a in the X, Y, and Z directions
respectively. a1 , a2 , a3 are called the rectangular components or simply components of A in the X,
Y, and Z directions respectively. The sum or resultant of a1 i , a2 j and a3 k is the vector a so that
we can write a as a = a1 i + a2 j + a3 k .
Z
Y
X
Fig. 3.9 Vector in Space
Equal Vectors:
Two vectors a and b are equal if and only if their corresponding components are equal.
Let a = a1 i + a2 j + a3 k and b = b1 i + b2 j + b3 k then a = b means that
a1 i + a2 j + a3 k = b1 i + b2 j + b3 k which means a1 = b1 , a2 = b2 , and a3 = b3
Thus, ( a1 − b1 ) i + (a2 − b2 ) j + (a3 − b3 ) = 0
Magnitude of a Vector
Let a = a1 i + a2 j + a3 k , then the magnitude of a denoted by a is defined by
a = a12 + a22 + a32
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Example: Given that a = 3 i − 4 j + k , b = 5 j + 2k and c = i − j − k . Find
i.
ii.
The resultant of a , b and c
iii.
a−b
2a + 3b − 4c
Solution:
i.
The resultant;
a + b + c = 3 i − 4 j + k + 5 j + 2k − 4 i − j − k
(
) (
) (
)
= 4 i + 2k
ii.
(
) (
) (
2a + 3b − 4 c = 2 3 i − 4 j + k + 3 5 j + 2k − 4 i − j − k
)
= 2 i + 11 j + 12k
iii.
(
) (
)
a − b = 3 i − 4 j + k − 5 j + 2k = 3 i − 9 j − k
a − b = 32 + (−9)2 + (−1)2 = 9 + 81 + 1 = 91
Example: If a = i + 2 j + 3k and b = 3i + j , find the unit vector in the direction of a + b .
3.3
Direction Angles and Direction cosines of Vectors
The direction angles of a non-zero vector a are the angles  ,  and 
in the interval 0,  
that the vector a makes with the positive x, y and z-axes. The cosines of these direction angles (
cos  , cos  and cos  ) are called the direction cosines of the vector a .
Consider a space vector a = a1 i + a2 j + a3 k and a = a . The scalars  ,  and  are such that
a1 =  a , a2 =  a and a3 =  a are called the direction cosines of a . Since a1  a , a2  a and
a3  a implies −1    1 , −1    1 , −1    1 for every non-zero vector a . When a is the zero
vector, then  =  =  = 0 . Let  ,  and  be the angles which a makes with the x, y and z-axes
respectively. Then the scalar components of a along these axes are:
Lecture Note prepared by Dr J Acquah
Page 81
a1 = a cos 
a2 = a cos 
a3 = a cos 
From the above equations:
a1 =  a
a1
a
=
a cos 
a
=
=
=
a3 =  a
a2 =  a
a2
a
=
a cos 
a
=
= cos 
= cos 
a3
a
a cos 
a
= cos 
Example: Find the direction cosines for the vector a = i − 2 j + 3k
Solution:
a = 12 + ( −2) + 32 = 1 + 4 + 9 = 14
2
From the given vector, a1 = 1 , a2 = −2 and a3 = 3
a1 =  a  1 =  14
 =
1
14
=
14
14
a2 =  a  −2 =  14
 =
−2 −2 14
=
14
14
Lecture Note prepared by Dr J Acquah
a3 =  a  3 =  14
 =
3
3 14
=
14
14
Page 82
 14 −2 14 3 14 
,
,
Thus, the direction cosines are 

14
14
14 

Now consider a vector a . Let â be the unit vector in the direction of a . Then,
a = a aˆ 
If a = a1 i + a2 j + a3 k , then, aˆ =
(
a
a
)
a
a
a
1
a1 i + a2 j + a3 k = 1 i + 2 j + 2 k
a
a
a
a
Hence, we have
â =  i +  j +  k
(3.1)
The Equation (3.1) implies that the direction cosines of a vector a are the components of the unit
vector in the direction of a .
Theorem: if  ,  and  are the direction cosines of a non-zero vector, then  2 +  2 +  2 = 1
Example: Given two vectors a and b where a = 2 i − j + k and b = 3 i + 2 j − 4k , if  ,  and
 are the direction cosines, show that  2 +  2 +  2 = 1 for each of the vectors.
Solution:
The direction cosines of a vector are the components of the unit vector in the direction of that
vector. Thus, if a = 2 i − j + k , then,
aˆ =
2i − j + k
2i − j + k
a
=
=
a
6
22 + (−1)2 + 12
aˆ =
2
1
1
i−
j+
k
6
6
6
 =
2
1
1
, =
,=
6
6
6
Lecture Note prepared by Dr J Acquah
Page 83
2
2
2
4 1 1 6
 2   −1   1 
 + + =
 +
 +
 = 6 + 6 + 6 = 6 =1
 6  6  6
2
2
2
For b = 3 i + 2 j − 4k , the unit vector is given by;
3 i + 2 j − 4k
3 i + 2 j − 4k
b
bˆ = =
=
b
29
32 + 22 + (−4)2
 bˆ =
3
2
−4
i+
j+
k
29
29
29
 =
3
2
−4
,=
,=
29
29
29
2
2
2
9
4 16 29
 3   2   −4 
 2 + 2 + 2 = 
 +
 +
 = 29 + 29 + 29 = 29 = 1
 29   29   29 
3.4
Multiplication of Vectors
Two types of vector multiplication will be considered. These are:
i.
ii.
The scalar (Dot) product
The vector (Cross) Product
3.4.1 The Scalar (Dot) Product
Consider two vectors a and b . Let  be the angle between the direction of
a and b then the
dot or scalar product of the two vectors a and b denoted by a b is defined as the product of the
magnitudes of a and b and the cosine of the angle between them.
Thus
a b = a b cos
0  
Lecture Note prepared by Dr J Acquah
Page 84
Note that the direction of the vectors a and b containing the angle should not be a vector of Head
to Tail since no direction is obtained if they are connected Head to Tail. Again, a b is a scalar not
a vector.
Properties of Scalar or Dot Product
i.
The dot product is commutative. That is,
ii.
a b=b a
The dot product is distributive over vector addition. That is,
a ( b + c ) = a b + a c and ( b + c ) a = b a + c a
iii.
 ( a b ) = ( a ) b = a ( b ) = ( a b ) 
iv.
(  a ) (  b ) =  ( a b )
v.
i i = j j = k k = 1 and i j = j k = k i = 0
vi.
If a b = 0 and a and b are not null vectors, then a and b are perpendicular.
a b = 0  a b cos = 0
 cos  = 0   =
vii.

2
If a and b are parallel, then  = 0 and
a b = a b cos 0 = a b
viii.
In this case, the dot product is ordinary multiplication of the magnitude of the two
vectors.
If a = a1 i + a2 j + a3 k and b = b1 i + b2 j + b3 k then,
(
) (b i + b j + b k ) = a b + a b + a b
a a = (a i + a j + a k ) (a i + a j + a k ) = a + a + a
b b = (b i + b j + b k ) (b i + b j + b k ) = b + b + b
a b = a1 i + a2 j + a3k
1
2
3
1
2
3
1
2
1
2
1
2
3
1 1
2 2
2
1
2
2
3
3
2
1
2
2
3 3
2
3
2
3
Applications of the Dot or Scalar Product
The dot or scalar product has many applications some of which are as follows:
i.
Finding the angle between two vectors:
Lecture Note prepared by Dr J Acquah
Page 85
Let a and b be two vectors expressed in component form as a = a1 i + a2 j + a3 k and
b = b1 i + b2 j + b3 k . Let  be the angle between them. Then, a b = a b cos .
(
a b = a1 i + a2 j + a3k
) (b i + b j + b k ) = a b + a b + a b
1
2
3
1 1
2 2
3 3
 a b cos = a1b1 + a2b2 + a3b3
 cos  =
a1b1 + a2b2 + a3b3
a b
a b + a b + a b 
  = cos −1  1 1 2 2 3 3 
a b


Example: Consider the two vectors q = 3 i + 2 j + k and r = − i + 4 j + 5k . Find the angle
between these two vectors
Solution: By definition, q r = q r cos
(
q r = 3i + 2 j + k
) ( − i + 4 j + 5k ) = 3(−1) + 2(4) +1(5) = 10
q = 32 + 22 + 12 = 14
 10 = 14
cos  =
r =
( −1) + 42 + 52 = 42
2
42 cos 
10
5
=
3
14 42 21
 5

3  = 65.645o
 21 
 = cos −1 
ii.
Finding the distance of a point from the origin O :
Let A be a point in space having position vector a . Then the distance of A form the origin O
is given by:
OA = a = a a
Lecture Note prepared by Dr J Acquah
Page 86
Z
A
O
Y
X
Example: A point P in space has coordinates ( 3, 2, −4) . How far is P from the origin
Solution: Let the position vector of the point P be r . Then
OP = r = r r =
( 3 i + 2 j − 4k ) ( 3 i + 2 j − 4k )
= 3(3) + 2(2) + (−4)(−4) = 9 + 4 + 16 = 29
iii.
Finding the work done by a constant force
Suppose that a force F pulls an object through a distance by the vector r along a line OA acting
at an angle  to it then, the components of the force F along OA is given by F cos and this is
the effective part of F pulling along the direction OA .
F
O
A
The components of F along a direction perpendicular to OA has no effect along OA . Since the
work done by a force is equal to the product of the product of the force and the displacement of
its point of application in the direction of the force, then
work done = F cos r = F r cos = F r
Lecture Note prepared by Dr J Acquah
Page 87
Example: A constant force of magnitude 5N in a direction i + 2 j − k , displaces a particle from
the point (1,0,1) to the point (3,4, -1). Calculate the amount of work done by the force.
Solution:
Let n̂ be a unit vector in the direction of the force F then,
nˆ =
i +2j −k
12 + 22 + (−1)2
F =5
(
(
=
1
i +2j −k
6
)
(
)
1
5
i +2j−k =
i +2j−k
6
6
)
(
)
The displacement vector r = 3 i + 4 j − k − ( i + k ) = 2 i + 4 j − 2k
Hence the work done by the force on the particle is W = F r
W =F=
=
5
6
(
5
i +2j −k
6
) ( 2 i + 4 j − 2k ) = 56 ( 2 + 8 + 2)
6
12 = 10 6 Joules
6
3.4.2 The Vector or Cross Product
The vector product of two vectors a and b denoted by a  b is defined as:
a  b = a b sin  nˆ
Where  is the angle between a and b and n̂ is the unit vector normal to plane containing a
and b . ( a  b is also called the cross product and pronounced as “ a cross b ”). This a  b is a
vector normal to the plane containing the vectors a and b .
Lecture Note prepared by Dr J Acquah
Page 88
Remarks:
i.
ii.
iii.
If the vectors a and b are parallel, then the unit vector n̂ is not defined. In this case
 = 0 and sin  = 0 , hence a  b = 0 .
If the vectors a and b are not zero vectors, then a  b = 0 only when sin  = 0 and
therefore a  b = 0 implies a is parallel to b .
If the vectors a and b are perpendicular, then  =
a  b = a b sin
iv.


and therefore
2
nˆ = a b nˆ
2
The vector product is a vector. It must be noted that the vector n̂ is a unit vector
ab
along a  b so that nˆ =
ab
Properties of Vector or Cross Product:
i.
From the definition, the cross product is not commutative that is
ab  ba
ii.
For the vectors a , b and c
a (b + c ) = a  b + a  c
That is, the cross product is distributive over vector addition and
(b + c ) a = b  a + c  a
(  a )  b =  (a  b ) = a  (  b )
a , b and scalars  and  . (  a )  (  b ) =  (a  b)
iii.
For the vectors a , b and a scalar  .
iv.
For the vectors
v.
For the unit vectors i , j and k which are mutually perpendicular:
a.
i  i = j j = k k = 0
b.
i  j = k ; j  k = i ; k  i = j and
Lecture Note prepared by Dr J Acquah
j  i = −k ; k  j = − i ; i  k = − j
Page 89
Vector Product in Component form:
Observe the cycle below:
The cross product of any two-unit vectors is the third unit vector, if the cycle is maintained but if
the cycle is reversed, then the result is the negative of the third vector. Considered two space
vectors a and b with components given as a = a1 i + a2 j + a3 k and b = b1 i + b2 j + b3 k . Then
(
) (
(
)
a  b = a1 i + a2 j + a3k  b1 i + b2 j + b3k
)
(
)
(
= a1 i  b1 i + b2 j + b3k + a2 j  b1 i + b2 j + b3k + a3k  b1 i + b2 j + b3k
(
= ( a1 i  b1 i + a1 i  b2 j + a1 i  b3 k ) + a2 j  b1 i + a2 j  b2 j + a2 j  b3 k
(
+ a3 k  b1 i + a3 k  b2 j + a3 k  b3 k
)
)
)
= a1b2 k − a1b3 j − a2b1k + a2b3 i + a3b1 j − a3b2 i
= (a2b3 − a3b2 ) i + (a3b1 − a1b3 ) j + ( a1b2 − a2b1 ) k
(3.2)
The Vector Product as Determinant
The components a  b could be given in the determinant form. Thus, in dealing with vector
products, it is necessary and sufficient for one to be conversant with evaluation of determinants.
Thus, the Equation (3.2) above may be expressed as:
i
j
k
a  b = a1 a2
b1 b2
a3
b3
=i
a2
b2
a3
a
−j 1
b3
b1
a3
a
+k 1
b3
b1
a2
b2
Lecture Note prepared by Dr J Acquah
Page 90
= (a2b3 − a3b2 ) i + (a3b1 − a1b3 ) j + ( a1b2 − a2b1 ) k
Example: Find the cross product for each of the following pairs of vectors:
i.
u = i + j and v = 2 i − j + 3k
ii.
x = 4 i − 5 j + k and y = − i + 2 j + 7k
Solution: Method I:
(
) (
) (
) (
)
= ( i  2 i ) + ( i  − j ) + ( i  3k ) + ( j  2 i ) + ( j  − j ) + ( j  3k )
u  v = i + j  2 i − j + 3k = i  2 i − j + 3k + j  2 i − j + 3k
i.
= −k − 3 j − 2k + 3 i = 3 i − 3 j − 3k
Method II
i
j
k
1 0
1 0
1 1
−j
+k
= 3 i − 3 j − 3k
uv = 1 1 0 = i
−1 3
2 3
2 −1
2 −1 3
Applications of the Cross/Vector Product in Geometry
We shall consider some geometrical applications of the vector product:
i.
The area of a parallelogram:
The quantity a  b is the area of a parallelogram with two adjacent sides represented by a
and b .
The area of a parallelogram = base length
perpendicular height
= a ( b sin  )
= a b sin 
= ab
Lecture Note prepared by Dr J Acquah
Page 91
Example: Determine the area of a parallelogram ABCD whose position vectors are
a = 2 i + j , b = i + 2 j − k , c = 3 i + 4 j − 2k , d = 4 i + 3 j − k
Solution:
(
) (
)
BA = a − b = 2 i + j − i + 2 j − k = i − j + k
(
) (
)
BC = c − b = 3 i + 4 j − 2k − i + 2 j − k = 2 i + 2 j − k
The area of a parallelogram = base length
perpendicular height
= a b sin  = a  b
i
j
k
= 2 2 −1
1 −1 1
= i
2 −1
2 −1
2 2
−j
+k
−1 1
1 1
1 −1
= i − 3 j − 4k = 12 + (−3)2 + (−4)2 = 26 square unit
ii.
The Area of a Triangle:
Every triangle may be regarded as one-half of a parallelogram.
D
A
B
Area of a triangle ABC=
C
1
1
(Area of parallelogram) = a  b
2
2
Lecture Note prepared by Dr J Acquah
Page 92
Example: Determine the area of triangle ABC whose position vectors are
a = 2 i + j , b = i + 2 j − k , c = 3 i + 4 j − 2k
Solution
(
) (
)
BA = a − b = 2 i + j − i + 2 j − k = i − j + k
(
) (
)
BC = c − b = 3 i + 4 j − 2k − i + 2 j − k = 2 i + 2 j − k
The Area of the triangle ABC =
1
1
a b sin  = a  b
2
2
i
j k
2 −1
2 2
1 2 −1
1
−j
+k
= 2 2 −1 = i
1 1
1 −1
2 −1 1
2
1 −1 1
=
1
1 2
1
i − 3 j − 4k =
1 + (−3) 2 + (−4) 2 =
26 square unit
2
2
2
iii.
The Angle Between Two Vectors
Let  be the angle between two vectors a and b and n̂ is a unit vector normal to a and b
then:
a  b = a b sin  nˆ
 ( a  b ) nˆ = a b sin 
But nˆ =
ab
ab
 a b sin  = ( a  b )
(a  b)
ab
ab
 a b sin  =
= ab
ab
2
Lecture Note prepared by Dr J Acquah
Page 93
 sin  =
ab
a b
(3.3)
Note that n̂ is a unit vector parallel to a  b and therefore ( a  b ) nˆ = a  b .
Example: Find the angle between the vectors u = i + 2 j − k and v = 2 i + 3 j + k
Solution: Let  be the angle between the two vectors. Then by the definition of cross
product:
(
) (
u  v = i + 2 j − k  2i + 3 j + k
i
j
)
k
= 1 2 −1 = ( 2 + 3) i − (1 + 2 ) j + ( 3 − 4 ) k = 5 i − 3 j − k
2 3 1
u = 12 + 22 + (−1)2 = 6
v = 22 + 32 + 12 = 14
a b 5i −3 j − k
52 + (−3) 2 + (−1) 2
=
=
Using sin  =
a b
2 21
6
4
( )( )
sin  =
35
2 3
3 1
=
15
3 6
1

  = sin −1  15  = 40.203
6

Lecture Note prepared by Dr J Acquah
Page 94
TRIGONOMETRY
4.1
Trigonometric Ratios
A right triangle is a triangle with one 90o . It is impossible to solve for the sides of a right triangle
if only the angles are known. However, if we have two sides or one acute angle and side, then it is
possible to solve for the remaining three quantities. This process is called solving the right triangle.
Trigonometric ratios are used to solve right triangles.
If two right triangles have the same acute angle θ, then the triangles are similar and ratios of
corresponding sides are equal.
Fig 4.1 Similar Right- Triangle
The 6 possible ratios of the sides of a right triangle depend only on the acute angle θ, not on the
size of the triangle. Theses 6 trigonometric ratios, are the sine, cosine, tangent, cotangent, secant
and cosecant of the angle θ. The values of these ratios at angle θ are denoted by sin 𝛉, cos 𝛉, tan
𝛉, cot 𝛉, sec 𝛉 and cosec 𝛉, respectively, as displayed below.
Fig 4.2 Trigonometric Ratios
Lecture Note prepared by Dr J Acquah
Page 95
Side b is often referred to as the side opposite angle θ, a as the side adjacent to angle θ, and c as
the hypotenuse. Using these designations for an arbitrary right triangle removed from a coordinate
system, we have the following:
4.2
Reciprocal Ratios
Each of the 6 trigonometric ratios is the reciprocal of another. For example,
cosec 𝜃 is the reciprocal of sin 𝜃. These reciprocal relationships make it easy to compute cosec 𝜃,
sec 𝜃, and cot 𝜃 from sin 𝜃, cos 𝜃 and tan 𝜃, respectively.
4.3
Evaluation of Trigonometric Ratios
o
4.3.1 Trigonometric Ratio for 45
Trigonometric ratios of certain angles can be evaluated using the Pythagorean theorem. The
diagonal of a square with sides of length 1 has length
2 .
Therefore,
Lecture Note prepared by Dr J Acquah
Page 96
4.3.2 Trigonometric Ratio for 60o
Similarly, by the Pythagorean theorem, an equilateral triangle with sides of length 2 has height
3 .
Therefore,
4.3.3 Trigonometric Ratio for 30o
Again, an equilateral triangle with sides of length two will have height of
Lecture Note prepared by Dr J Acquah
3 .
Page 97
4.3.4
2
=2
1
sin 30 =
1
2
cos 30 =
3
2
sec 30 =
2
2 3
=
3
3
tan 30 =
1
3
=
3
3
cot 30 =
3
= 3
1
cos ec30 =
Trigonometric Ratio for 0o
sin 0 = 0
cos ec0 = 
cos0 = 1
sec0 = 1
tan 0 = 0
cot 0 = 
4.3.5 Trigonometric Ratio for 90o
sin90 = 1
cos ec90 = 1
cos90 = 0
sec90 = 
tan 90 = 
cot 90 = 0
4.4
Trigonometric Identities
Identities enable us to simplify complicated expressions. They are basic tools of trigonometry used
in solving trigonometric equations, just as factoring, finding common denominators and using
special formulas are the basic tools of solving algebraic equations. Algebraic techniques are
constantly used to simplify trigonometric expressions. The fundamental trigonometric identities
we will consider are: the Pythagorean identities, the even-odd identities, the reciprocal identities
and the quotient identities.
Lecture Note prepared by Dr J Acquah
Page 98
4.4.1 The Pythagorean Identities
i.
sin 2  + cos2  = 1
ii.
1 + cot 2  = cos ec2
iii.
1 + tan 2  = sec2 
4.4.2 The Even-Odd Identities
i.
tan(− ) = − tan 
ii.
cot(− ) = − cot 
iii.
sin(− ) = − sin( )
iv.
cos ec(− ) = − cos ec
v.
cos(− ) = cos
vi.
sec(− ) = sec
Note that an odd function is one in which f (− x) = − f ( x) for all x in the domain.
4.4.3 Reciprocal Identities
4.4.4 Quotient Identities
4.5
Sum and Difference Identities
Continuing with the trigonometric identities, we will look at the sum and difference identities for
the cosine (cos), sine (sin) and tangent (tan).
Lecture Note prepared by Dr J Acquah
Page 99
4.5.1 Sum and Difference Identities for Cosine
Finding the exact value of the sine, cosine or tangent of an angle is often easier if we can rewrite
the given angle in terms of two angles that have known trigonometric values. The sum and
difference identities for the cosine are given below
Sum and Difference identities for Cosine
Sum identity for cosine
cos( A + B) = cos A cos B − sin A sin B
Difference identity for cosine
cos( A − B) = cos A cos B + sin A sin B
4.5.2 The Sum and Difference identities for Sine
The sum and difference identities for Sine is given below
Sum identity for sine
sin( A + B) = sin A cos B + cos A sin B
Difference identity for sine
sin( A − B) = sin A cos B − cos A sin B
4.5.3 The sum and difference identity for Tangent
The sum and difference identity for tangent is given below
Sum identity for tangent
tan( A + B) =
tan A + tan B
1 − tan A tan B
Difference identity for tangent
tan( A − B) =
tan A − tan B
1 + tan A tan B
4.6
Double Angle Identities
In the previous section, we used addition and subtraction identities for trigonometric functions.
Now we take another look at those same identities. The double angle identities are a special case
of the sum identities, where A=B.
4.6.1 Double Angle Identity for Sine
Deriving the double angle identity for sine begins with the sum identity.
sin( A + B) = sin A cos B + cos A sin B
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If we let A = B , then we have
sin( A + A) = sin A cos A + cos A sin A
sin(2 A) = 2sin A cos A
4.6.2 Double Angle Identity for Cosine
Deriving the double angle for cosine gives three options. First, starting from the sum identity,
cos( A + B) = cos A cos B − sin A sin B
If we let A = B , then we have
cos( A + A) = cos A cos A − sin A sin A
cos 2 A = cos2 A − sin 2 A
Using the Pythagorean properties, we can expand this double angle identity for cosine and get two
more identities. The first is
cos 2 A = cos2 A − sin 2 A
= (1 − sin 2 A) − sin 2 A
= 1 − 2sin 2 A
The second is
cos 2 A = cos2 A − sin 2 A
Lecture Note prepared by Dr J Acquah
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= cos2 A − (1 − cos2 A)
= 2cos2 A −1
4.6.3 Double Angle Identity for Tangent
Similarly, to derive the double angle identity for tangent, we let A = B in the sum and identity for
tangent. That is, given
tan( A + B) =
tan A + tan B
1 − tan A tan B
tan( A + A) =
tan A + tan A
1 − tan A tan A
Then,
tan(2 A) =
2 tan A
1 − tan 2 A
4.7
Half Angle Identities
The next set of identities is the set of half angle identities. The half angle identities for sine and
cosine are derived from two of the double angle identities for cosine.
4.7.1 Half Angle Identity for Cosine
We consider the double angle identity
cos 2 A = 2cos2 A − 1
Now let A =
D
,then we have
2
D
D
cos 2   = 2 cos 2   − 1
2
2
D
cos D = 2 cos 2   − 1
2
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D
1 + cos D = 2 cos 2  
2
1 + cos D
D
= cos 2  
2
2
1 + cos D
D
= cos  
2
2
4.7.2 Half Angle Identity for Sine
For the half angle identity for sine, we consider the double angle identity
cos 2 A = 1 − 2sin 2 A
We let A =
D
, then
2
D
D
cos 2   = 1 − 2sin 2  
2
2
D
cos D = 1 − 2sin 2  
2
D
2sin 2   = 1 − cos D
2
 D  1 − cos D
sin 2   =
2
2
1 − cos D
D
sin   =
2
2
4.8
The Factor Formula
4.8.1 The Factor Formula for Sine
Using the sum and difference identities for sine, the factor formula for sine is derived as follows;
Lecture Note prepared by Dr J Acquah
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sin( A + B) = Sin A cos B + cos A sin B
sin( A − B) = sin A cos B − cos A sin B
Adding the two equations gives
sin( A + B) + sin( A − B) = 2sin A cos B
Now let A + B = X and A − B = Y
Then,
X + Y = 2 A and therefore, A =
X +Y
2
Again,
X − Y = 2 B and therefore, B =
X −Y
2
Substituting the values of X and Y for A and B, we have
 X +Y 
 X −Y 
sin X + sin Y = 2sin 
 cos 

 2 
 2 
4.8.2 The Factor formula for Cosine
Similarly, the factor formula for cosine is given as
 X +Y 
 X −Y 
cos X + cos Y = 2 cos 
 cos 

 2 
 2 
Lecture Note prepared by Dr J Acquah
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REFERENCES
1. Stroud, K. A, and Dexter, B. (2013), Engineering Mathematics, 7th Edition, Macmillan
International, London
2. Bird, J. (2006), Higher Engineering Mathematics, 5th Edition, Elsevier Ltd.
3. Ayres, F. (1968), Matrices: Solved Problems. Schaum's Outline Series, McGraw-Hill,
USA
4. Anton, H. (1984), Elementary Linear Algebra, 4th edition. John Wiley, Princeton
5. Finkbeiner, D. T. (1978), Elements of Linear Algebra, 3rd edition. Freeman.
6. Kolman, B. (1986), Elementary Linear Algebra, 4th edition. Collier Macmillan, London
TRY EXERCISE 1
Question 1
Find z  such that
a) z = i(z − 1)
b. z 2 * z = z
c. z + 3i = 3 z
Question 2
Solve the equation
2z 2 − 2iz − 5 = 0 z 
Answer
3 1
+ i
2 2
z
Question
If w = − 9 + 3i Find the modulus and the argument of the complex number w.
1 − 2i
Answer
w =3 2
, arg w =
− 3
4
Question 3
Find the value of x and the value of y in the equation, Given further that x  , y  
Lecture Note prepared by Dr J Acquah
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(x + iy)(2 + i ) = 3 − i
Answer (x, y ) = (1,−1)
Question 4
z=
 + 4i
,   .
1 + i
Given that z is a real number, find the possible values of .
Answer  = 2
Question 5
Find the values of x and y in the equation
x(1 + i ) + y(2 − i ) = 3 + 10 x
2
2
x  , y  
Answer
x = 7, y = 1
Question 6
Find the values of x and y in the equation, , Given further that x  , y  
(x + iy)(3 + 4i ) = 3 − 4i
Answer
(x, y ) =  − 7 , − 24 
 25
25 
The complex number z satisfies the equation
1 − 18i
, where z denotes the conjugate of z .solve the equation, given the answer
2−i
in the form x + iy , where x and y are real numbers.
4 z − 3z =
Answer
z = 4−i
Lecture Note prepared by Dr J Acquah
Page 106
Question 7
z = −3 + 4i and zw = −14 + 2i
By showing clear workings, find…….
a) w in the form a + bi , where a and b are real numbers.
b) The modulus and the argument of w
Question 8
and z = 6 − 8i
z = 22 + 4i
w
By showing clear workings, find…….
c) w in the form a + bi , where a and b are real numbers.
d) The modulus and the argument of w
Answer
w= 5
w = 1+ 2i
arg w  1.11c
Question 9
z = (2 − i ) +
2
7 − 4i
−8
2+i
Express z in the form x + iy where x and ,are real numbers.
Answer= − 3 − 7i
Question 10
If complex conjugate of z is denoted by z.
Solve the equation
2 z − 3z =
− 27 + 23i
1+ i
Giving the answer in the form x + iy , where x and y real numbers.
Answer
z = 2 + 5i
Lecture Note prepared by Dr J Acquah
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QUESTION 11a
Solve the following equation
z 2 = 21 − 20i, z  C.
Give the answers in the form a + bi, where a   and b  
ANSWER
Z = (5 − 2i )
QUESTION 11b
The Cubic equation
2z 3 − 5z 2 + cz − 5 = 0
c   has a solution 0f z = 1− 2i
Find in any order
a) The other two solutions of the equation
b) The value of c
Answer
1
z2 = 1 + 2i, z3 = , c = 12
2
QUESTION 12
z − 8 = i(7 − 2 z ), z  C.
The complex conjugate of z is denoted by z.
Determine the value of z in the above equation, given your answer in the form x + iy
where x and y are real numbers.
Lecture Note prepared by Dr J Acquah
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TRY EXERCISE 2
1.
Compute the real and imaginary part of z = i − 4
2.
Compute the absolute value and the conjugate of
2i − 3
z = (1 + i )
w = i17
6
3.
Write in the “algebraic” form (a + ib) the following complex numbers
w = (3 + 3i )
8
z = i5 + i + 1
4.
Write in the “trigonometric” form ( (cos + i sin  )) the following complex numbers



c.  cos − i sin 
3
3

b.6i
a).8
7
5. Simplify
a) 1 + i − (1 + 2i )(2 + 2i ) + 3 − i
1− i
b. 2i(i − 1) +
1+ i
( 3 + i ) + (1 + i)(1 + i)
3
6. Compute the square root of z = −1 + i
7. Compute the cube root of z = −8
8. prove that there is no complex numbers such that z − z = i
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