Grade 9 IGCSE A1: Chapter 9 Matrices and Transformations
Materials Needed: Straightedge, Graph Paper
Exercise 1: Matrix Operations
1 minute Start Question:
Skill and Concepts:
Matrices are used in Linear Algebra to solve systems of linear equations. The next simple example takes a look at a
system of equations like
and changes its form to
3 −1 𝑥
8
3𝑥 − 𝑦 = 8
�
� �𝑦� = � �
2 3
6
2𝑥 + 3𝑦 = 6
Solution
Methods:
Solution Method:
a. Matrix Method
a. Elimination
b. Substitution
c. Graphical
Therefore, after becoming familiar with the Matrix Method, you will be able to solve a system of two linear equations
in four different ways.
So let’s understand what Matrices are and how to perform operations on them.
Definition: A matrix is a rectangular array of numbers arranged in rows and columns. Plural: Matrices
Symbolized by a Capital bold letter, example A (you will not need to write in bold, but be aware of the notational
difference  What does a non-bold upper case letter usually symbolize?)
1 2
1 2
An example of a 2X2 (two by two) matrix: 𝑨 = �
� or 𝑨 = �
�
3 4
3 4
Each matrix has a dimension/order: (number of rows) by (number of columns).
What is the dimension/order of the following matrix? (1 2 3)  ______________
A matrix with the same number of rows and columns is called a
?
matrix.
Each number in the array is called an element of the matrix. Each element (symbolized by an italicized lower case
letter) is designated by its position in the matrix ie. Row x, column y or say a23 means the 2nd row, 3rd column
element.
A column matrix (has only one column), a row matrix (has only one row) and zero matrix has only zeros as its
elements.
Matrix equality: Two matrices are equal iff they have the same order and corresponding elements.
1 2
1 2
1 2
−1 2
Example: �
�=�
� but �
�≠�
�
3 4
3 4
3 4
3 4
How might you express matrix equality using matrix notation? 𝑨 = 𝑩 ↔ _______________________
Matrix addition and subtraction: Add/subtract corresponding elements of only same order matrices.
1 2
1 2
2 4
Example: �
�+�
�=�
�
3 4
3 4
6 8
1 2
Example: �
� + (1 2 3) cannot be added together as the dimensions/orders are not the same.
3 4
1 2
1 2
0 0
Example: �
�−�
�=�
�=𝐎
3 4
3 4
0 0
In general: 𝑨𝑎𝑎𝑎 ±𝑩𝑎𝑎𝑎 = 𝑷𝑎𝑎𝑎
Notice that A + (-A) = O. Since adding A and its opposite –A produces the zero matrix which we called the additive
identity element, A and –A are called inverse elements of each other because of the following fact: A + (-A) = O
AND (-A) + A = O. This should remind you of the inverse property of functions.
Scalar multiplication: Multiply all elements by the scalar. There is no such thing as scalar division.
1 2
1 2
1 2
2 4
1 2
2 4
Example: 2 �
�=�
�+�
�=�
� therefore 2 �
�=�
�
3 4
3 4
3 4
6 8
3 4
6 8
1 2
2 4
In other words, if 𝑨 = �
�, then 2𝑨 = �
�
3 4
6 8
Do not write:
2
�
6
2
4
�
8
as this notation does not exist. How should you write it?
Matrix multiplication: Take a look at the textbook page 292 regarding 2x2 matrix multiplication. The orders of the
two matrices must satisfy the following property, otherwise matrix multiplication is not possible: 𝑨𝑚𝑚𝑚 𝑩𝑝𝑝𝑝 Also, if
matrix multiplication is possible, then the order of the resultant matrix will be 𝑨𝑚𝑚𝑚 𝑩𝑝𝑝𝑝 = 𝑹𝑚𝑚𝑚 .
Let’s practice some Matrix Multiplication
Notice that positive integer exponents to represent repeated matrix multiplication still holds true: 𝑨𝑨 = 𝑨𝟐
Is matrix multiplication commutative? In other words, is 𝑨𝑨 = 𝑩𝑩 always true? If you can find one example when
it is not true, then in mathematics, you cannot say that it is true (even though it may be true sometimes or even most of
the times).
As a result, if we multiply matrix A to B, we have to state whether we are multiplying A to the left or right of B as the
resultant matrix is different.
Left multiply A to B  AB
Right multiply A to B  BA
1 2
�
3 4
1 2 3)
There is no such thing as Matrix Division  (
�
HMWK:
Exercise 1:
C-type: 3 – 36 last column; 38 – 46 last column
A-type: 48, 49
Context/Concept:
is meaningless.
Exercise 2: Inverse of a Matrix
1 minute Start Question:
Skill and Concepts:
The notation for the inverse of a matrix is similar to that of a function. The inverse matrix of 𝑨 is written as 𝑨−1 .
What we want to do now is demonstrate
the matrix method to solve systems of
linear equations.
3𝑥 − 𝑦 = 8
2𝑥 + 3𝑦 = 6
Solution
Methods:
3
�
2
−1 𝑥
8
�� � = � �
3 𝑦
6
Solution Method:
a. Matrix Method
Remember though that we said that we
a. Elimination
cannot divide by a matrix. Therefore we
b. Substitution
have to introduce a matrix that is called the
c. Graphical
Identity Matrix represented by 𝐈 that is a
square matrix with only two different
elements, namely 0 and 1. All the 1’s are located on the diagonal that goes from the top left to the bottom right. All
the other elements are zero.
Examples: 𝐈𝟐𝟐𝟐 = �
1 0 0
1 0
1
� and 𝐈𝟑𝟑𝟑 = �0 1 0�. For the IGCSE course, we will only be using 𝐈𝟐𝟐𝟐 = �
0 1
0
0 0 1
0
�.
1
The important multiplicative property of 𝐈𝟐𝟐𝟐 is that if you multiply it to any matrix 𝐁𝟐𝟐𝟐 , it will not change the
identity of 𝐁𝟐𝟐𝟐 . Prove to yourself that this is true.
We need to investigate the inverse property.
Recall the inverse property of functions and inverse functions: 𝑓�𝑓 −1 (𝑥)� = 𝑥 = 𝑓 −1 (𝑓(𝑥))
Given that the multiplicative inverse of A, if it exists, it is written as 𝐀−1 and satisfies AA−1 = I = A−1 A .
3 −1 𝑥
8
So in order to solve �
� � � = � � we will write this in this form: 𝐀𝐀 = 𝐁 and solve for X.
2 3 𝑦
6
𝐀𝐀 = 𝐁 In order to solve for X, as there is no matrix division, we will left multiply each side by 𝐀−1 . Notice what
left multiply means in the next line.
𝐀−1 𝐀𝐀 = 𝐀−1 𝐁
𝐈𝐈 = 𝐀−1 𝐁
𝐗 = 𝐀−1 𝐁
So the question becomes how to solve for 𝐀−1 if you are given an A.
1
1
𝑎 𝑏
𝑑 −𝑏
𝑑
If 𝐀 = �
�, then 𝐀−1 =
�
�. We can also write it as 𝐀−1 = |𝐀| �
𝑎𝑎−𝑏𝑏
𝑐 𝑑
−𝑐 𝑎
−𝑐
called the determinant of A also written as detA.
−𝑏
� where |𝐀| = 𝑎𝑎 − 𝑏𝑏 is
𝑎
If |A| is not equal to zero, there is a unique solution to the equation AX=B. In this case, we say that the matrix A is
non-singular and A has an inverse matrix.
If |A| is equal to zero, there is no unique solution to the equation AX=B. In this case, we say that the matrix A is
singular and A does not have an inverse matrix.
Given two linear equations, when does the system not have a solution or an infinite number of solutions? Provide two
systems that represent this. What is evident from the matrix form of the solution?
HMWK:
Exercise 2:
C-type: 3, 9, 15, 17, 19, 21, 22, 25
A-type: 24, 26, 28
Context/Concept:
For all the Simple Transformations, use the links to see more examples and do activities on the web to
ensure that your understanding of the concepts is correct.
Exercise 3-4: Simple Transformations (Reflections)
1 minute Start Question:
Skill and Concepts:
A Graphical Transformation is a mathematical means by which to change a given object’s position, orientation or
shape in a consistent way.
Transformations are categorized as either Rigid or Non-Rigid Transformations.
Name of
Transformation
Reflection
Reflection in 2D
Wolfram Demo
Needed
Graphically
Algebraic
Line of Reflection
Must be able to draw lines of
reflections based upon
equations.
The line of reflection is
denoted with a dotted line and
labeled.
∆1 → ∆2
Triangle 1 has been
transformed to Triangle 2.
Coordinates are labelled in
this way:
A(a,b)  A’(c,d)
A is mapped onto A’
Notice the similar language
with the functional notation
𝑓: 𝑥 → 2𝑥 + 1
Primary Reflection Lines:
Horizontal lines: 𝑦 = 𝑎
Vertical Lines: 𝑥 = 𝑏
Diagonal Lines: 𝑦 = ±𝑥 + 𝑐
HMWK:
Exercise 3-4:
C-type: Exercise 3  3, 6, 9
Exercise 4  3
A-type: Exercise 4  4
Context/Concept:
The reflection is always
perpendicular to the mirror
line and on the opposite side
of the mirror.
Exercise 5-6: Simple Transformations (Rotations)
1 minute Start Question:
Skill and Concepts:
Name of
Transformation
Rotation
Rotations
Wolfram Demo
Needed
Graphically
Algebraic
Center of Rotation
Angle of Rotation
Direction of Rotation
Clockwise  Negative angle
measure
Counterclockwise  Positive
angle measure
Label Center as O
∆1 → ∆2 → ∆3
Triangle 1 has been
transformed to Triangle 2
which in turn has been
transformed to Triangle 3.
Coordinates are labelled in
this way;
A(a,b)  A’(c,d)A’’(e,f)
A is mapped onto A’ which is
mapped onto A’’
In order to find the center of Rotation given an original image and a
transformed image via rotation, if you construct the perpendicular bisectors
connecting corresponding coordinates, the intersection will be the center of
rotation.
How would you determine the angle of rotation?
HMWK:
Exercise 5-6:
C-type: Exercise 5  2, 4, 7
Exercise 6  3
A-type: Exercise 6  5
Context/Concept:
Exercise 7 - 9: Simple Transformations (Translation and Enlargement and all Transformations Combined)
1 minute Start Question:
Skill and Concepts:
Name of
Transformation
Translation
Enlargement
Needed
Graphically
Algebraic
Horizontal Translation
Vertical Translation
Combined into a Column
Vector
A translation is a slide of the
image without changing the
orientation or dimensions of
the image.
𝑥
�𝑦�
Center of Enlargement
Scale Factor = k
Center of Enlargement
designated with coordinate O.
Dotted lines emanating from
O to the respective vertices of
the shape.
Investigate different values of
k:
If 𝑘 > 1, then
If 𝑘 = 1, then
If 0 < 𝑘 < 1, then
If −1 < 𝑘 < 0, then
If 𝑘 < −1, then
Investigate different values of
k:
What happens to the image?
What happens to the image?
What happens to the image?
What happens to the image?
What happens to the image?
Enlargements
Wolfram Demo
In order to find center of an enlargement, connect corresponding
coordinates and the intersection of the lines will be the center of
enlargement.
How would you determine the scale factor?
HMWK:
Exercise 7 - 8:
C-type: Exercise 7  1
Exercise 8  2, 5
Exercise 9  1
A-type: Exercise 8  11, 13
Context/Concept:
Exercise 10-11: Combined and Inverse Transformations
1 minute Start Question:
Skill and Concepts:
Now that we’ve gone through all the Simple Transformations, how would you categorize each transformation
Rigid
Combined
Transformations
Use Matrices (and
simple column
vector) to represent
Transformations:
T  Translation
R  Rotation
F  Reflection
E  Enlargement
Non-Rigid
Needed
Graphically
Algebraic
A shape to transform
(usually a triangle)
All pertinent notation on the
graph. As there will be
combined transformations,
the order in which you
perform them is key.
We used this: ∆1 → ∆2 → ∆3
Coordinates are labelled in
this way;
A(a,b)  A’(c,d)A’’(e,f)
A is mapped onto A’ which is
mapped onto A’’
Now: 𝑇(∆1) = ∆2
All information required
to perform each
transformation
Now, we will designate what
transformation is being
applied to the shape.
So let’s say we do this:
𝑹𝑹(∆1) = 𝑹(∆2) = ∆3
What does the statement
mean?
𝑹𝟐 (∆1) = 𝑹𝑹(∆1)
Recall that the exponents can
only be positive integer
values.
Repeated
Transformations
Inverse
Transformations
Inverse Transformation Notation:
𝑻−1  Translation
𝑹−1  Rotation
𝑭−1  Reflection
𝑬−1  Enlargement
Will undo the
original
transformation.
𝑎
If 𝑻 = � �, then 𝑻−1 = � �
𝑏
If 𝑹 is given with Center at O and 90 degree rotation, what is 𝑹−1 ?
If 𝑭 is given with line of reflection 𝑦 = 𝑎, what is 𝑭−1?
If 𝑬 is given with Center at O and SF = k, what is 𝑬−1?
Notation: 𝑻−3 (∆1) = (𝑻−1 )3 (∆1) = 𝑻−1 𝑻−1 𝑻−1 (∆1)
HMWK:
Exercise 10-11:
C-type: Exercise 10  4  20 last column
A-type: Exercise 11  9, 10
Context/Concept:
Exercise 12, then 13: Transformation Matrices (Two Days)
1 minute Start Question:
Skill and Concepts:
Now let’s use Matrices to mathematically determine where the transformed coordinates will be located.
Graphical
Interpretation/Description
Algebraic Interpretation
𝑥1
IS: 𝐓 �𝑦
1
Reflect the point A(3, 5) and
B(-1, -2) about the line y = 0.
𝐓 represents either
The transformed point is
A’(3, -5) and B’(-1,2)
The Transformation F can be
written in the following matrix
form:
1 0
𝐹=�
�
0 −1
Transformation Matrices applied
to a shape Wolfram Demo
𝑥2
𝑥′1
𝑦2 � = �𝑦′1
𝑥1
�𝑦
1
𝑥′
� 1
𝑦′1
𝑥1
𝐓 �𝑦
1
𝑥′2
�
𝑦′2
𝐑  Rotation
𝐅  Reflection
𝐄  Enlargement
𝑥2
𝑦2 � represents two coordinate values of the original shape.
𝑥′2
�represents the corresponding two coordinate values of the
𝑦′2
transformed shape.
𝑥2
𝑦2 � represents the matrix multiplication of the two matrices.
Using the example to the left:
𝑥1 𝑥2
𝑥′ 𝑥′2
IS: 𝐓 �𝑦 𝑦 � = � 1
�
𝑦′1 𝑦′2
1
2
1 0
3 −1
3 −1
�
��
�=�
�
0 −1 5 −2
−5 2
Therefore the Transformed point is A’(3, -5)
Translation Matrix
Translations Matrices do not
involve matrix multiplication as
they merely translate using a
vector.
Translate the point A(3, 5) to the
right 1 unit and down 4 units.
The transformed point is
A’(4, 1)
The Transformation T is written
in the following vector form:
1
𝑇=� �
−4
𝑥1
𝑥′
IS: �𝑦 � + 𝐓 = � 1 �
𝑦′1
1
𝐓 represents the Translation Vector
𝑥1
�𝑦 � represents a coordinate value of the original shape.
1
𝑥′
� 1 � represents the corresponding coordinate value of the transformed shape.
𝑦′1
𝑥1
�𝑦 � + 𝐓 represents the addition of the two matrices
1
Using the example to the left:
𝑥1
𝑥′
IS: �𝑦 � + 𝐓 = � 1 �
𝑦′1
1
3
1
4
� �+� �=� �
5
−4
1
Therefore the Transformed point is A’(4, 1)
While you are doing these exercises, try to determine the link between the Matrix form of a Translation and the
Translation itself.
HMWK:
Exercise 12:
C-type: Exercise 12  2, 5
A-type: Exercise 12  9, 11cg
Next Day
Exercise 12-13:
C-type: Exercise 13  2, 4a, 5
A-type: Exercise 13  6, 7
Context/Concept:
Exercise 14: Determining the Matrix form of a transformation
1 minute Start Question:
Skill and Concepts:
Use base vectors to determine the transformation matrix 𝑇2𝑥2.
Description of Transformation
Graphical Interpretation
Algebraic Interpretation
Transformation that does not
change anything.
1
𝑰=�
0
Identity Matrix
Start with the Identity Matrix form,
then geometrically apply the
transformation to the basis vectors.
Then place the colored basis vectors
in their respective location in the
matrix.
0
�
1
0
𝑹90 𝑎𝑎𝑎𝑎𝑎 (0,0) = �
1
𝑹90 𝑎𝑎𝑎𝑎𝑎 (0,0)  Rotation of 90
degrees about the Origin (0, 0)
Follow the directions as the rotation
but this time apply a reflection.
−1
�
0
0 1
𝑭𝒚=𝒙 = �
�
1 0
𝑭𝒚=𝒙  Reflection about the
line 𝑦 = 𝑥
Follow the directions as the rotation
but this time apply an enlargement.
−2 0
𝑬𝑺𝑺=−𝟐 𝒂𝒂𝒂𝒂𝒂 (𝟎,𝟎) = �
�
0 −2
1 0
= −𝟐 �
�
0 1
= −𝟐𝐈
𝑬𝑺𝑺=−𝟐 𝒂𝒂𝒂𝒂𝒂 (𝟎,𝟎)  Enlargement
for Scale Factor -2 about the Origin
(0, 0)
Notice: If your transformations do not include the Origin, then the matrix forms of the transformations tend to be
more complicated. In order to still apply matrix transformations, it is matter of translating to center at (0, 0), apply the
desired transformations and then apply an inverse translation vector.
𝑥1 𝑥2
𝑥′ 𝑥′2
Also, if you know the original and transformed coordinates of the IS: 𝐓 �𝑦 𝑦 � = � 1
�, then you can apply
𝑦′1 𝑦′2
1
2
𝑥1 𝑥2 −1
a right multiply by �𝑦 𝑦 � to both sides of the equation to solve for T to determine the transformation matrix.
1
2
In other words,
𝑥1
𝐓 �𝑦
1
𝑥2 −1
𝑥 ′1
�
=
�
𝑦2
𝑦 ′1
𝑥′ 𝑥′2 𝑥1
𝐓𝐓 = � 1
��
𝑦′1 𝑦′2 𝑦1
𝑥′ 𝑥′2 𝑥1
𝐓=� 1
��
𝑦′1 𝑦′2 𝑦1
𝑥2 𝑥1
𝑦2 � �𝑦1
𝑥 ′ 2 𝑥1
��
𝑦 ′ 2 𝑦1
𝑥2 −1
𝑦2 �
𝑥2 −1
𝑦2 �
𝑥2 −1
𝑦2 �
HMWK:
Exercise 14:
C-type: 1 – 18 all
A-type: Find the Transformation Matrix equation that will perform the following transformation and then apply your
transformation matrix equation on A(4,5) to find the coordinate of A’.
a. A rotation of 90 degrees about O(1, 2)
b. An enlargement with SF = -2 about O(1, 2)
b. A reflection about 𝑦 = −𝑥 + 3
Context/Concept: