Chapter 4 Matrices 1
Learning Objectives
In this chapter, you will learn how to:
 carry out matrix operations;
 recognise the terms zero matrix, identity matrix, singular and
non-sigular
 evaluate determinants,find inverse matrices and know how to use
identities such as (��)−1 = �−1 �−1 ;
 apply geometric transformations, such as rotations and
enlargements, as well as being aware of the relation between
scale factor and determinants.
Introductions
Matrices are a way to present and change numbers and they help you
to use rules or find patterns within numbers. Matrices are used
extensively in scientific fields, as they can be used to manage large
amounts of data. They are widely used in computing processes, for
instance reflection and refraction in computing graphics or computer
modelling of probabilities for weather forecasting.
Many careers and industries use matrices, for example:
 If you were an accountant or bookkeeper, you might use matrices
to record, track and display sales and the movement of money or
goods.
1
 If you were a physicist you would use matrices in many fields,
from the study of quantum mechanics and electronics to optics.
 If you were responsible for determining the most efficient route
for a delivery driver, you could represent it by using matrices.
The ability to write the relationships between elements as
matrices means that you can then apply algorithms to a wide
range of inputs.
 If you were a video game programmer you would use matrices to
transform shapes as the field of view is moved by the player, and
to help illustrate the projection of 3D objects into 2D shapes This
needs to be done quickly and efficiently, so having a 'rule' in the
form of a matrix that the points of a shape follow allows the
game to move more quickly and smoothly.
 If you worked as a software encryption programmer you would
use matrices and inverse matrices to code or encrypt messages
sent over the internet-for example, bank details or business
documents.
So, what are the matrices?
Matrices(singular, matrix)are rectangular arrays of numbers,
variables or expressions arranged in rows and columns.
The size of a matrix is defined by its rows - m, and columns - n.We
refer to matrices by their order or size, m×n--we read ‘order m by n’.
2
For a general matrix of order m×n we have
amn inside represents an element of the matrix.
�11 �12 �13 ⋯
�21 �22 �23 ⋯
. Each
�31 �32 �33 ⋯
⋮ ⋮ ⋮ ⋱
For example, the column vector we have learned, is the simplest
form matrix.
1
2
3
is a matrix of order 3×1. A matrix that has the
same number of rows and columns is called a square matrix.
Example 1:
A retailer is recording which colours of T-shirt she sells, and
whether they are sold at her online shop or in her store in the high
street of a town. On Monday she records her sales in this grid.
The results can be written in a matrix, giving the matrix M shown
above.Working with matrices may help the retailer keep track of her
sales. We can use matrix M to represent number of sales on Monday.
4.1 Operations
M=
3 2 4
1 2 10
4.1.1 Matrix addition, subtraction
Rules: To add together two matrices you need to add the corresponding
elements. This is only possible if the two matrices are of the same order.
3
In the example above,the retailer wishes to add together number of sales
from Monday and Tuesday.
3 2
1 2
Monday: M =
4
10
Tuesday: T =
0 2 7
3 −1 8
Tips: A negative sign in this context would mean an item being returned.
3 2 4
0 2 7
+
1 2 10
3 −1 8
3+0 2+2 4+7
3
=
=
1 + 3 2 − 1 10 + 8
4
M+T =
4 11
1 18
A zero matrix, O, or null matrix is an m×n matrix consisting of all 0s.
Adding the zero matrix to another m×n matrix does not change it-in the
same way that adding 0 to a number does not increase or decrease it.
4.1.2 Scalar Multiplication of Matrix
To multiply a matrix by a scalar you must multiply each of the elements
within the matrix by the scalar. The scalar can be positive or negative,
and does not have to be an integer.
The retailer may notice that number of sales of all items on Wednesday
were double sales on Monday. Hence, W=2xM
W=2×
=
6
2
3
1
4
4
2 1
2×3
=
2 10
2×1
2
20
2×2 2×1
2 × 2 2 × 10
Some basic rules of these operations:
Commutative law; Distributive law; Associative law
4
4.1.3 Matrix Multiplication
Unlike addition and scalar multiplication, matrix multiplication does not
involve simply multiplying the equivalent elements.
For previous example, suppose the retailer's T-shirts have different prices,
as shown in this table.
Find the earnings of online and in-store on Monday.
4.5
You can write the prices as C = 3.5 .
4
4.5
3 2 1
3 × 4.5 + 2 × 3.5 + 1 × 4
× 3.5 =
So, M × C =
1 2 10
1 × 4.5 + 2 × 3.5 + 10 × 4
4
36.5
=
51.5
Example 2 Suppose Amy and Bob go to a baker’s shop to buy cakes,
doughnuts and eclairs. The numbers of each that they buy is
given by the array of numbers.
Cakes
Doughnuts
Eclairs
Amy
2
1
1
Bob
4
0
2
Suppose there are actually 2 shops, X and Y, that they could buy at, and
the prices are given in the table below, in pence.
Shop X
Shop Y
Cakes
40
45
Doughnuts
30
25
Eclairs
50
40
5
Now they want to compare how much they could spend in each shop.
The money Amy would spend in X is: 2 × 40 + 1 × 30 + 1 × 50 = 160.
The money Amy would spend in Y is: 2 × 45 + 1 × 25 + 1 × 40 = 155.
The money Bob would spend in X is: 4 × 40 + 0 × 30 + 2 × 50 = 260.
The money Bob would spend in Y is: 4 × 45 + 0 × 25 + 2 × 40 = 260.
40 45
2 1 1
The purchase matrix is P =
, the cost matrix is C = 30 25 . Therefore,
4 0 2
50 40
40 45
2 1 1
PC = P × C =
× 30 25
4 0 2
50 40
2 × 40 + 1 × 30 + 1 × 50 2 × 45 + 1 × 25 + 1 × 40
=
4 × 40 + 0 × 30 + 2 × 50 4 × 45 + 0 × 25 + 2 × 40
160 155
=
.
260 260
We could deduce the rule of matrix multiplication: the element in the first
row and the first column of PC is a11 calculated by the product of the first
row of P and the first column of C. More generally the element in the ith
row and jth column of PC is aij the product of the ith row of P and the jth
column of C.
Consider 2 matrices, A =
AB =
ae + bg af + bh
.
ce + dg cf + dh
a b
c d
and B =
e f
, the product of them
g h
The rules for matrix multiplication:
A(B+C) = AB+AC (B+C)A = BA+CA A(BC) = (AB)C
s(AB) = (sA)B = A(sB)
AO = OA = O
6
Questions: whether AB is equal to BA? Check by examples.
Generally, AB ≠ BA. The distributive law is not applied.
Investigate the multiplication of non-square matrices. How is the size of
the resulting matrix related to the matrices being multiplied together?
A matrix product may not exist.
��� ��� = ��
��
exits, the number of column of A should be equal
to the number of row of B. A and B are called conformable.
A special case, for square matrix A, AmAn = AnAm = Am+n.
4.1.3 Matrix Division - the inverse matrix
When we carry out the division of numbers, for example, ax = b ,
1
therefore we get x = a × b , the
1
�
is the inverse of a, or reciprocal. We
1
could check the answer ax = a × a × b = b.
Think about: If matrix equation AX = B or XA = B, can you find X using
similar method? Before solving the equation, we should find the inverse
of matrix A first. We will introduce a new definition.
The identity matrix, denoted by I, is a square form in which the
elements in the leading diagonal are all 1 and all other elements are 0,
I=
100⋯0
010⋯0
001⋯0 .
⋮ ⋮ ⋮ ⋱ ⋮
000⋯1
In general, we can say that ��� ��� = ��� ��� = ��� , provided
that the matrix multiplication is allowed. It is just like multiplying any
number by 1.
7
Question:
Give matrix A, could you find a matrix, such that AX = XA = I?
Answers: Sometimes, if A is squared.
If AC = CA = I , C is called the inverse of A, denoted by C = A−1 , A =
C−1 . The matrix A is said to be non-singular. If no such matrix exists, then
A is said to be singular.
A square matrix A has an inverse �−� , if and only if, A is non-singular.
Then:
A−1 AX = A−1 B
AX = B
X = A−1 B.
A−1 A X = A−1 B
IX = A−1 B
Then how to find an inverse? We will introduce 2 methods.
First we will look at the inverse of a 2 × 2 square matrix.
Example 3 Given A =
Method 1: Let A−1 =
2� + 2�
4� + 5�
2 2
, find A−1 .
4 5
w y
. We need
x z
w y
2 2
1 0
×
=
.
x
z
4 5
0 1
2� + 2�
1 0
=
, solve simultaneous equations of wx and yz, you
4� + 5�
0 1
5
will get w = 2 , x =− 2, y =− 1, z = 1.
Therefore A−1 =
5
2
−2
−1 = 1 5
2 −4
1
−2
.
2
� �
, determinant of A, denoted by det(A) is calculated by ad-bc,
� �
� �
i.e.,���(�) =
= �� − ��.
� �
In the previous example, the denominator of the coefficient, 2 is calculated by
2 2
det(A) =
= 2 × 5 − 2 × 4.
4 5
� �
So, the inverse of non-singular � × � matrix � =
is given as �−� =
� �
Actually, if � =
8
�
−�
be zero.
−�
. You could also notice that the det of a non-singular matrix cannot
�
�
��−��
Method 2: Row operation - augmented matrix � ⋮ � .
We use the method of row operation to find the inverse.
� ⋮ � ↔ (�−� � ⋮ �−� �) ↔ � ⋮ �−�
Row operations are used to change the elements of matrices. There are three types of
row operations:
 Row switching, where �� ↔ ��
 Row multiplication, where �� → ���
 Row addition, where �� → �� + ���
Therefore, the row operations carried out is generally the inverse operation, if A is
changed to I through these processes, then I will also be changed to �−1 through the
same operations.
22 ⋮10
45 ⋮01
Therefore, A−1 =
�2→�1 ×(−2)+�2
�1→�1÷(2)
5
2
−2
2 2 ⋮ 1 0 �1→�2×(−2)+�1 2 0 ⋮ 5 − 2
0 1 ⋮− 2 1
0 1 ⋮− 2 1
1 0 ⋮ 5/2 − 1
01⋮ −2 1
−1 = 1 5
2 −4
1
−2
.
2
Actually we usually use this method to calculate the inverse of higher order
non-singular square matrices.
1
Example 4 Given A = 0
2
We then use augmented matrix
1 2
0 −1
2 3
1⋮1 0 0
2⋮0 1 0
1⋮0 0 1
r3 →r2 ×(−1)+r3
r1→r2×2+r1
2 1
−1 2 , find A−1 .
3 1
1 2
0 −1
2 3
1 2
1⋮1
0 −1 2 ⋮ 0
0 −1 −1⋮−2
r3 →r1 ×(−2)+r3
1 2
1⋮1
0 −1 2 ⋮ 0
0 0 −3⋮−2
1 0
5⋮1
0 −1 2 ⋮ 0
0 0 −3⋮−2
0
1
−1
2
1
−1
1⋮1 0 0
2⋮0 1 0
1⋮0 0 1
0
0
1
0
0
1
r3 →r2 ×(−1)+r3
r1 →3r1 +5r3
9
to find the inverse.
0 0
1 0
0 1
1 2
1⋮1
0 0
0 −1 2 ⋮ 0
1 0
0 0 −3⋮−2 −1 1
3 0
0 ⋮−7
0 −1 2 ⋮ 0
0 0 −3⋮−2
1
1
−1
5
0
1
3 0
0 ⋮−7
0 −3 0 ⋮−4
0 0 −3⋮−2
r2 →3r2 +2r3
1
−1
−1
r1 → r1, r2 → r2 ,r3 → r3
3
3
3
−1
Therefore, �
=
1
1
−1
5
2
1
7 1
5
3 3
3
1 0 0⋮ 4
1
2
0 1 0⋮
−
−
3
3
0 0 1⋮ 3
2
1
1
−
3
3
3
7
−3
4
3
2
3
1
3
1
−
5
3
2
−3 −3
1
3
1
−3
1
=3
−7 1
5
4 −1 −2 .
2
1 −1
The left side of the augmented matrix is now in reduced row echelon form with ones
in the leading diagonal and zeros everywhere else.
After-class Question: What would happen if you performed column
operations instead of row operations? Would the same results be observed?
Can column operations work on augmented matrices? What is the
difference between the row and column operations?
Question:
How to find the inverse of the product matrix AB, denoterd by(��)−1 ?
(AB)C = C(AB) = I, so C is called the inverse of product matrix AB.
(AB)C = I
BC = A−1
A−1 (AB)C = A−1 I
B−1 (BC) = B−1 A−1
Therefore, (��)−1 = B−1 A−1 .
(A−1 A)BC = A−1
BC = A−1
(B−1 B)C = B−1 A−1
C = B−1 A−1
What about (���)−1 ,(���)−1 ?
Example 5
4 5
, show that �2 − 7� + 2� = 0. And use this to find �−1 .
2 3
26 35
�2 =
, �2 − 7� + 2� = 0.
14 19
M=
10
�2 − 7� + 2� = 0
�2 − 7� =− 2�
�(� − 7�) =− 2�
1
1
− �(� − 7�) = �. ∴ �−1 =− (� − 7�)
2
2
Tips: The notation of augmented matrix is �.
Therefore, the process of row operations is the multiplication by �−1 to
the left of matrix A, to change it to I.
11
4.1.4 Determinants
We have already know how to calculate the determinant of a 2 × 2 square
matrix. Now let us look at the 3 × 3 square matrix. First some definitions
2 1
should be given. Given a 3 × 3 square matrix M = 3 4
−1 0
a. Minor
−2
0 .
1
The minor of each element in the matrix, is the leaving part after crossing
out the corresponding row and column of the element, for example, the
minor of the 2 is
4
0
0
= 4 . If the element is denoted by a, then the
1
corresponding minor is denoted by A.
b. Co-factor
The co-factor of each element is the sign of the corresponding position
multiply the corresponding minor. The corresponding signs of each
element is determined by the position, for example, the sign of element aij
+
is (-1) . So the sign matrix of a 3 × 3 square matrix is −
+
3 0
The co-factor of 2 is 4, the co-factor of 1 is ( − 1)1+2
−1 1
i+j
c. The adjugate matrix of M is denoted by adjM.
a b c
If M = d e f , the co-factor matrix is
g h i
− +
+ − .
− +
=− 3.
A −B C
−D E −F , the
G −H I
A −D G
adjM = −B E −H , it is a transposed matrix of the co-factor
C −F
I
matrix, just swapping the rows and columns.
Then we can calculate the determinant of a 3 × 3 square matrix. The
12
overall determinant is solved by multiply each element of any row or
column by corresponding co-factors, i.e.,
detM = aA − bB + cC =− dD + eE − fF = gG − hH + iI(Row)
= aA − dD + gG =− bB + eE − hH = cC − fF + iI(Column)
In the given example,
detM = 2 × 4 + 1 × ( − 3) + ( − 2) ×
3 4
=− 3.
−1 0
We can easily find that some elements in the matrix are 0, then we could
use the row consisting of 0 to calculate determinants, which will save
some calculations. For example, using the last row,
detM = ( − 1) ×
1
4
−2
2 1
+1×
=− (0 + 8) + (8 − 3) =− 3.
0
3 4
We could also use adjM and detM to calculate the inverse of M.
4 −1 8
In the given example, adjM = −3 0 −6
4 −1 5
−3 0
0
1 0
M × adjM = 0 −3 0 = −3 0 1
0
0 −3
0 0
adjM
,
0
0 =− 3I = detM I,
1
M × detM = I, then according to the definition, �−1 =
Although AB ≠ BA, det(AB) = det(BA) = detA × detB.
1
����
× ����.
Question: Investigate the effects of 3 kinds of row operations on the
determinant of the matrix.
 Row switching, where �� ↔ �� ------------- multiply by (-1) for each switch
 Row multiplication, where �� → ��� ------ multiply by factor k
 Row addition, where �� → �� + ��� ------- do not change
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4.2 Transformations
Photo-editing software and video games need to be able to move points
and transform images. Matrices are useful in transformations - in
particular, transformation that keep origin fixed but move other points.
Let the transformation matrix be M. You can show the transformation of
x
x'
any point (x, y) to its image (x , y ) as: M y = ' . We are going to see
y
'
'
different kinds of matrices M, i.e., corresponding transformations.
We consider the rectangle A(1,1), B(1,3), C(5,3) and D(5,1),we can put
these coordinates into a matrix to get
1155
. This is formed by
1331
considering each vertex of the rectangle as a position vector. If we
multiply this matrix by a 2 × 2 matrix to produce a transformation, we
need to multiply in the correct order. For example, there is no change
when multiplying by the identity matrix.
1155
.
1331
1
0
0
1
1155
=
1331
4.2.1 Stretch
Consider the transformation
1 0
0 2
1155
1155
=
. Each y
1331
2662
value has doubled now. This is a stretch of scale factor 2 in the
y-direction: the distance relative to the x-axis has been doubled. If we
multiply
1144
1331
by the matrix
2 0
, we get
0 1
2 2 8 8
.
1 3 3 1
Similarly, this is a stretch of scale factor 2 in the x-direction: the distance
relative to the y-axis has been doubled. Combining the previous two
14
examples, we can see that matrix
2
0
0
2
has the effect of an
enlargement of the scale factor 2, measured from the origin, so we can
say the origin is the center of enlargement.
In each of these three examples we can see an invariant line or point. In
the first example, any point on the line y=0 will not change its y value. It
is, therefore,invariant. In the second example,any point on the line x=0
15
will not change its x value. It is also, therefore, invariant. In the third
example, both x=0 and y=0 are invariant lines. The point where they meet,
in this case the origin, is known as an invariant point.
� �
.
� �
� �
For a stretch in the y-direction of scale factor k,the matrix is represented by
.
� �
For an enlargement relative to the point (0,0) of scale factor k, the matrix is
� �
represented by
.
� �
For a stretch in the x-direction of scale factor k, the matrix is represented by
Any point that is unchanged by a transformation is known as an invariant point.
If that point is not the origin, then it must lie on an invariant line.
Now consider the matrix
1
2
0
0
1
2 2 8 8
and apply the transformation
2 6 6 2
. We have a scale factor again, but this time it is a reducing factor.
2
What is more interesting is if A =
1
2
0
0
1
2
0
1 2
0
, then A−1 = 4
2
0
0
=
2
. Applying an inverse matrix will revert any transformation to its
2
original state since �−1 � = �.
In all of the previous cases the area of the shape is increased by a factor
that is equal to the size of the determinant that is transforming the shape.
Take,for example, the matrix A =
3 0
. This stretches by factors 3 and
0 2
2, so the area of the new shape is six times as large since det(A)=6.
Question:
Apply the matrix
4 3
1 1
to the triangle represented by the points (1,1), (4,1) and
(4,5). What do you notice about the area of the image of the triangle? How does this
relate to the determinant of the matrix being applied?
16
As well as stretching and enlarging, we can also consider rotations and
reflections as transformations.
4.2.2 Reflections
Consider the triangle represented by the points(1,1),(1,4),(4,1). As
always,we write these coordinates in matrix form
1 1 4
.
1 4 1
Consider reflecting the triangle in the y-axis. That is, the effect of this
matrix is to change the sign of each x value but not to change the y values.
−1 0
, i.e.,
0 1
−1 0 1 1 4
−1 −1 −4
=
.
0 1 1 4 1
1
4
1
1 0
Similarly, the reflection matrix in the x-axis is
.
0 −1
So the reflection matrix in the y-axis is
4.2.3 Rotations
When we apply both of these transformations, in either order, to the
1 4
then the result is
4 1
1 0
−1 0 1 1 4
−1 0
=
0 −1
0 1 1 4 1
0 1
−1 −1 −4
.
−1 −4 −1
triangle
1
1
17
1
0
0
−1
1 1
1 4
4
=
1
This effect is a rotation about the origin by 180°. We multiply the two
reflection matrices, we get the rotation matrix of 180° is
This 180° can be considered to be clockwise or anticlockwise.
−1 0
.
0 −1
Now we will generalize the rotation matrix.
Example 6
Find the matrix which represents a rotation through an angle θ
anticlockwise about the origin.
The original points A(0,1) and C(1,0) are rotated through an angle θ anticlockwise
about the origin. The images are �' ( cos � , sin � ) and �' ( −sin � , cos � ) . So this
rotation matrix about the origin by θ in the anticlockwise direction is
cos � − sin �
.
sin � cos �
Example 7:
Find the matrix which represents a reflection in the line y = x.
Any point x, y reflects to point y, x in the line y = x.
b
, then we have the relation
d
1 3
2 4
0 1
a b
=
, we get
=
2 4
1 3
1 0
c d
Let the transformation matrix be
a b
c d
a
c
18
Question:
Find the matrix which represents a reflection in the line at an angle θ to
the x-axis.
Example 8
Describe the effect of the following matrices on the triangle given by
(1,1), (3,1), (1,3). Illustrate your findings.
a.
−2
0
Answers:
0
1
b.
a. Shape is reflected in the y-axis.
1 0
0 −2
c.
−2 0
0 −2
Shape is stretched by a scale factor of 2 in the x-direction.
Area is doubled since the magnitude of the determinant is 2.
b. Shape is reflected in the x-axis.
Shape is stretched by a scale factor of 2 in the y-direction.
Area is doubled since the magnitude of the determinant is 2.
c. Shape is rotated 180°clockwise or anticlockwise about the origin.
Shape is enlarged relative to the origin by a scale factor of 2.
Area is four times since the magnitude of the determinant is 4.
19
4.2.3 Shear
Consider the matrix
1 1
. If we apply it to the rectangle (1,1), (3,1),
0 1
(3,5), (1,5), then each point is translated parallel to the x-axis since the
new x-coordinates are based on x +y. This effect is called a shear. The
distance through which points are displaced depends on their distance
from the line x = 0.
Consider the rectangle (1,0), (3,0), (3,4), (1,4). If we apply the same
matrix to these points, we get a different shear. This time only the top
points move since the two bottom vertices lie on an invariant line,y = 0.
As we can see in this diagram, the top points have a y value that is
unchanged by the shearing factor. So a matrix of the form
1 �
0 1
will
produce a shear in the x-direction, and the distance each point is moved is
equal to ky0 , where is y0 is the original y value.
Similarly, a matrix of the form
1 0
2 1
should give a shearing effect in
the y-direction instead. Apply this matrix to the rectangle with vertices
20
(0,0), (0,3), (2,3) and (2,0). This gives
1
2
0
1
0 0 2 2
=
0 3 3 0
0 0 2 2
. We can see that the line x=0 is invariant and any point on
0 3 7 4
this line does not change.
So any matrix of the form
Question:
Apply the matrix
1 0
� 1
1 0
will produce a shear in the y-direction.
� 1
to sets of points and investigate with different
values of k. What happens in the cases where 0<k<1 or k<0?
4.2.4 Invariant points and invariant lines
Points that transform to themselves (i.e. are not affected by the
transformation) are called invariant points. In a reflection, the points on
the mirror line remain unchanged by the reflection. Enlargements, with
centre of enlargement (0, 0), only have one point that does not move,
namely the origin. Consider this product,
4 3
2 3
2
2
=
. It
−2
−2
means that the point (2,-2) is transformed onto itself, and so (2,-2) is an
invariant point for the matrix
4 3
.
2 3
To find the invariant points you can consider the general point x, y .
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4 3
2 3
x
x
x
4x + 3y
4x + 3y = x
=
→
=
.
So
→ y =− x.
y
y
y
2x + 3y
2x + 3y = y
So all the points on the line y=-x are invariant.
If you simplify both expressions and get different results it means that
there is only one invariant point: the origin.
Invariant lines are slightly less particular than invariant points. All points
on an invariant line will be transformed to a point on the same line.
KEY Information:
Invariant points transform to themselves.
Invariant lines have points that are transformed to another point on the
same line.
Invariant points may join up to form a line of invariant points.
For any reflection, invariant lines are all lines perpendicular to the mirror
line, together with the mirror line itself (which is called a line of invariant
points).The original points are labelled: D,E,F, etc. with their associated
images: D',E',F',etc.
A and B are on the mirror line, so
the image is not labelled.
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The diagram below illustrates that for an enlargement any line that passes
through the origin is an invariant line, even though the origin is the only
invariant point.
For most rotations there are no invariant lines, with one exception: a
rotation of 180. In this case, any line that passes through the origin is an
invariant line, even though the origin is the only invariant point.
So we understand that points on the invariant lines do not mean the points
are invariant. If we wanna find the invariant points, we could use the
x
x
previous method by solving equation A y = y . But for finding the
invariant lines, the method may be different.
Example 9
Find the invariant lines for the matrix
1 −2
.
−2 1
Suppose the invariant line is y = mx. The image of point x, y
invariant line is
on the
X, Y . And also there are relations y = mx, Y = mX .
Consider the matrix equation below:
1 −2 x
X
1 −2
=
→
y
−2 1
Y
−2 1
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x
X
=
mx
mX
→
x − 2mx = X
−2x + mx = mX
Divide the first equation by the second to get
1−2m
−2+m
=
1
m
. Solve the
equation to give �2 = 1, �� � =± 1. This means there are two invariant
lines, � =± � . They meet at the origin, which is always an invariant
point.
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