Basic Mathematics Form Four Notes – Matrices and Transformations
These Basic Mathematics Form Four notes cover Matrices and Transformations. The material is arranged in a clear mobile-friendly reading format while preserving the recognized definitions, explanations, examples, activities, calculations and revision material from the source notes.
Formula & Symbol Clarity
- For A = [[a,b],[c,d]], determinant: |A| = ad – bc
- When ad – bc ≠ 0: A⁻¹ = (1/(ad-bc)) [[d,-b],[-c,a]]
Operations on Matrices
The Concept of a Matrix
Explain the concept of a matrix
Definition: A matrix is an array or an Orderly arrangement of objects in rows and columns. Each object in the matrix is called an element (entity).
Consider the following table showing the number of students in each stream in each form.
Form ul Il Stream A 38 35 40 Stream B 36 40 34
Stream C 40 37 36
From the above table, if we enclose the numbers in brackets without changing their arrangement, then a matrix is farmed, this can be done by removing the headings and the bracket enclosing the
numbers (elements) and given a name (normally a capital letter).
Nowthe above information can be presented in a matrix form as
38 35 40 2 A=|36 40 34 39 40 37 36 35
Any matrix has rows and columns but sometimes you may find a matrix with only row without
Colum or only column without row.
In the matrix A above, the numbers 38, 36 an 40 form the first column and 38, 35, 40 and 28
form the first row. Matrix A above has three (3) rows and four (4) columns.
In the matrix A, 34 is the element (entity) in the second row and third column while 28 lies in the
first row and fourth column. The plural form of matrix is matrices.
Normallymatrices are named by capital letters and their elements by small letters which
represent real numbers.
_fa b). ' e.g. B-(e a| is a matrix. B is a matrix containing the elements a, b, c, and d.
C= f | is also a matrix which contains elements 1, 2, 3, and 4.
Order of a matrix (size of matrix)
The order of a matrix or size of a matrix is given by the number of its rows and the number of its
columns. So if A has m rows and n columns, then the order of matrix is m x n.
It is important to note that the order of any matrix is given by stating the number of its rows first
and then the number of its columns.
For example A -(}
3 2] is a 2 × 2 matrix or the order of matrix A is 2 × 2, and
def
la b Bt b j}isa 28 marx while C= dis a3 × 2 matrix.
e f
NB: mxn+#nxm since an m x nis a matrix with m rows and n columns while n x m is a matrix with n rows and m columns.
Types of matrices:
The following are the common types of matrices:-
(a) Zero matrixes; A zero is the matrix whose elements are all zeros. eg.
(og B=(o 0} c
(b) Square matrix: Is a matrix whose number of rows is equal to the number of columns.
For example
(c) Identity Matrix: Is the square matrix whose elements in the leading diagonal are ones and zeros elsewhere.
1 10
(d) Column matrix: Is the matrix with only one column.
+ of} oof |
(e) Row matrix: This is a matrix with only one row.
eg. D1 19 or E{3 -2 17)
Matrices of order up to 2 × 2 Add matrices of order up to 2 × 2
When adding or subtracting one matrix from another, the corresponding elements (entities) are
/added or subtracted respectively.
This being the case, we can only perform addition and subtraction of matrices with the same
orders.
Example 1
Given that
(> aan 2-[F A | find A+B
—_ _fa c¢ e f|_|ate ctf Solution: A+B =[' HE f) b+g dth
Matrices of order up to 2 × 2 Subtract matrices of order up to 2 × 2 Example 2
Given that
ae e ae jens oe
Example 3
Solve for x, y and z in the following matrix equation;
Now x-3=2, 2-z = 0 and 0-y=3 So x= 5, y=-3 and z=2
Exercise 1
Determine the order of each of the following matrices;
_(2 4 3 _|1 2 x= ( 1 ] (b) o-(} 2
Ff 23 4) @G=(9)
2. Given that
-fi 2 _(4 5 ' a=(5 | and B= (j 7] Find (a) A+B (b) B-A
3. Given that
(a) (A+B) +C (b) A-B+C (c) C-A-B
4. A house wife makes the following purchases during one week: Monday 2kg of meat and loaf of bread Wednesday, Ikg of meat and Saturday, kg of meat and one loaf of bread. The prices are 6000/= per kg of meat and 500/= per loaf of bread on each purchasing day
Write a 3 × 2 matrix of the quantities of items purchased over the three days .
Write a 2 × 1 column matrix of the unit prices of meat and bread.
5. Solve for x, y and z in the equation
646343
Additive identity matrix.
If M is any square matrix, that is a matrix with order mxm or nxn and Z is another matrix with
the same order as m such that
M+ Z= Z+M = M then Z is the additive identity matrix.
The 2 × 2 additive identity matrix is 2{°9 ; }
The additive inverse of a matrix.
If A and B are any matrices with the same order such that A+B = Z, then it means that either A is
an additive inverse of B or B is an additive inverse of A that is B=-A or A= -B
| B= fi '| and Z= (? °| then A+ B =Z implies B= Z-A
d 00
Find the additive inverse of A,
ea (2 3 if A f : |
Solution: The additive inverse of A is A= {2 3)
«The additive inverse of A is 5 =|
Example 5
Find the additive identity of B if B is a 3 × 3 matrix.
Solution: The additive identity of any nXn matrix is the nXn zero matrix.
So Z=| 0
A Matrix of Order 2 × 2 by a Scalar
Multiply a matrix of order 2 × 2 by a scalar
A matrix can be multiplied by a constant number (scalar) or by another matrix. Scalar multiplication of matrices:
Rule: If A is a matrix with elements say a, b, c and d, or
_Ja b : A-(2 d | and t is any real number, then
a-[4 b)_[ta tb
c d]} lte td
Example 6
Given that
_|7 8 Es) Find (a) 2A (b)-5A
Solution;
(a)A= (2 il. 2A=?
(b) -5A =?
– 7 8 -5SA=-5 [? 5]
-S§>x 7 -Sx 8 -5 x S§ -5 × 9
L
r
_|-35 —2S
\
r
—35 —25
.
Example 7
Given,
E '] _ Find B+B+B. 5 14
Solution;
B+B+B = 3B -2 / 2 4
=| 32 | 3 × 5 3 × 14
_[ 6 12 aB-{ § |
, _ 6 12 ~.B+B+B = (ss '|
Two Matrices of order up to 2 × 2
Multiply two matrices of order up to 2 × 2
Multiplication of Matrix by another matrix:
a Bb fe fF LetA=|— jjena8={5 ']
AB is the product of matrices A and B while BA is the product of matrix B and A.
_|a bile f sons-(2 'fe 1)
In AB, matrix A is called a pre-multiplier because it comes first while matrix B is called the post
multiplier because it comes after matrix A.
Rules of finding the product of matrices; L The pre —multiplier matrix is divided row wise, that is it is divided according to its rows.
2. The post multiplier is divided according to its columns.
3. Multiplication is done by taking an element from the row and multiplied by an element
from the column.
4. In rule (iii) above, the left most element of the row is multiplied by the top most element of the column and the right most element from the row is multiplied by the bottom most element
of the column and their sums are taken:
fp b _(e f Now tA =f Jana 8 =(5 h
Then AXB 4) ( f } c g\h
_| aet+bg af +bh ness ( cots of +48 |
Similarly BA = El . f
_| eat fe eb+ fd "| gat+ he gb+ hd
Therefore it can be concluded that matrix by matrix multiplication is only possible if the number
of columns in the pre-multiplier is equal to the number of rows in the post multiplier.
Example 8
Given That: A= : 7 Jand
(a) AB (b) BA Solution
-~{9 7 6 = wma (2 × 3 B Al 6|-1
8 6 . be 5
_(9 × 6+7x(-2) 9x(-1)+7 × 5 "(8 × 6+6x(-2) 8x(-1)+6 × 5
_ (54-14 -9435 48-12 -8+30
an (40 26
AB & 22
_ (6 -1 9|7 (b)BA= > (2 |
6 × 9+(-1)x8 oy re eel (-—2)x9+5 × 8 (—2)x7+5 × 6
_| 54-8 42-6 -18+40 -14+30
_{ 46 36 22 16
pa (46 36 Ba= (46 36]
From the above example it can be noted that AB#BA, therefore matrix by matrix multiplication
does not obey commutative property except when the multiplication involves and identity matrix
ie. AIFIA=A Example 9
Let,
_|4 2 _{5 " P-{t | and @={¢ | Find PQ
Solution: _(4 2 5 4xX5+2 × 6 Pe-fs i}(5] {rxstixd _| 20+12]_ |32 a { eet _[2) ppl 32 Po=(37
Example 10
Find CxD if
2 4 C=8 1 3 5
Solution:
(2 × 34+4 × 104+6 × 1 2 × 44+4xX6+6x (-2) 8 × 3+1xX104+0 × 1 8xX4+1 × 6+0x (-2) 3 × 34+5 × 10+7 × 1 3xX4+5 × 6+7x (-2) Nes.
(6+40+6 8+24-12 244+10+0 3246-0 [9+50+7 1243014
52 20 « CD=|34 38 66 28
Product of a matrix and an identity matrix:
If A is any square matrix and | is an identity matrix with the same order as A, then AI=IA=A
Example 11
Given;
_fi 3 _f1 0 A= é Jana = 4]
Find (a) Al (b) IA Solution:
Alt 3],{1 o]f1 × 1+3 × 0 1 × 0+3 × 1 4 5)"|0 1) |4 × 1+5 × 0 4 × 0+5 × 1 Ale (1+ 043) {1 3 4+0 0+5] 14 5 1 3
«Al= =
A F 5 A
_f1 0). fi 3 co) a= (5 kG '|
TS eeery 1 × 3 sl
Ox1+1 × 4 Ox34+1 × 5 aq (1+0 34+0)_f1 37)_ A ona al [3 rf A
Exercise 2
1. Given that A= (3 4) and
] find (a) AB (b) BA
3.Using the matrices
(a) 3A + 2(AC) (b) A2B_ (c) 3B-I where | is an identity matrix.
4.Find the values of x and y if x lly 1 -{6 4 3 2 y 4 17 14
Inverse of a Matrix The Determinant of a 2 × 2 Matrix
Calculate the determinant of a 2 × 2 matrix
Determinant of a matrix
d ding diagonal of matrix A while the elements (b and c) are in the main
If A= 7 "| where a, b, c and d are any real numbers, then the elements (a and d) are
in the diagonal.
f<l
a—+d = Leading diagonal b—+sc= Main diagonal
Now the determinant of matrix A is then defined as the difference of the product of elements in
the leading diagonal and the product of the elements in the main diagonal.
So if A= é ' then determinant of A is given by ad-bc. The determinant of A is denoted by |A[ or det (A)
SoA (i '| , then
|A| =ad-be
NB: Determinants exist for square matrices only
Example 12 Find
-|1 3 laina-[) ;
Solution: |A| = 1 × 4-2 × 3=4-6=-2 /|Al = -2
Example 13
Considering
_{1 0) ¢ ne-(3 | na |Al
Solution: |8|= 1 × 5-0 × 3=5-0=5 2 (B|=5
Example 14
Find the value of x
having determinant 46. Solution: (2 × 3 c-(7 ml Ic| = 2x.10 — (—2).3 = 46
20x + 6=46 20x=46-6
-.The value of x is 2.
Singular and non singular matrices: Definition:
Asingular matrix is a matrix whose determinant is zero, while non — singular matrix is the one
with a non zero determinant.
For example A= i 3 has determinant 1 × 8 -4 × 2 = 0,
So A is a singular matrix.
Also B = (s } is a non Singular matrix because its determinant is 6 × 5-4 × 1
=30-4=26 which is not equal to zero.
Example 15
Find the value of y
_{4 5). If o-(4, | is a Singular matrix.
Solution;
'4 3y
|B| =24-15y, but B is a singular matrix, then |B] = 0 So. |B| =24-15y=0
24-15y=0
24=15y
B= é|: |B|= 4 × 6-5x(3y)
The Inverse of a 2 × 2 Matrix
Find the inverse of a 2 × 2 matrix
Inverse of matrices
Definition: If A is a square matrix and B is another matrix with the same order as A, then B is
the inverse of A if AB=BA=I where | is the identity matrix.
Thus AB=BA=! means either A is the inverse of B or B is the inverse of A.
For 2 × 2 matrix, AB =BA=I= f A]
gf OC So meee |
The inverse of matrix A is denoted by A" How to find A*
(ac _(p r Let Ap a| and at |
Where B=A", that is B is the inverse of matrix A a c)\(P Yr)_f1 0 © 90369)
ap+cq ar+cs} |1 0 bp+dq br+ds\|o 1
From which we obtain
41-4-{1 0 antst=(5 {]
ap +Cq= 1 (i)
ar+cs =0 (ii) bp +dq =0 (iii) br+ds = 1 (iv)
Since we need the unknown matrix B, we can solve for p and q by using equations (i) and (iii)
and we solve for r and s using equations (ii) and (iv)
_ ap+cq =o} and
By elimination method
apt+cq =1… bp +dq=0.
abp + beq =b —jabp +adq=0
To get p proceed as follows
ap+cq =1 bp +dq=0
adp +dcq=d "chp + deq =0
(ad-bc) p=d __d p= ad—be
Alsoto get r and s, the same procedure must be followed:
bjar +cs = 0 ajbr+ds =1
ue
— alabr+bds=a (bc-ad)s = -a Or (ad-bc)s=a
= a ~ ad—be
djar+ces=0 c|jbr+ds=1
_ Jadr+cds =0 ber + cds =c
(ad-bc) r= -c
Or (bc-ad) r=c
c f= bce-ad
d _ b _ c ad-be' 7 be-ad' "be-ad
Therefore p= and s= 4
d-be
Remember B {? ; wich is the inverse of A where Af
So
d e ad—be be-ad
be—ad
But bc-ad = – (ad-bc
Hence A't=
But ad-bc = |A|
4-1} 4 l4l|—b
.At= tid -e "Kk "als |
Note that, if |A|= 0, Then
At= af, z) which is Undefined.
Example 16
Given that,
(a)A* (b) (A2y"
Solution:
|A? |= 22 × 7-15 × 10
det (A?) = 4
1 / 22 -15 4 – = (A*) ner "|
Which of the following matrices have inverses?
_{-1 3 _{3 6 (2.9 (a) B= E 4 (b) C= ( 3 (c) of 18 Solution:
_[-1 3 ial @e=[, 3}: IB| =-1 × 6-2 × 3 |B | =-12+0 «.B* exists
(b) C -f 6), Ic | = 3 × 8-4 × 6 =24-24=0
Ic | =0, so C* does not exist
(c)D '(i a ; |D| = 2 × 18-9 × 4= 36-36=0
:. D* does not exist.
Exercise 3
1. Find the determinant of each of the following matrices.
-|4 4 3 6 _|x 1 (a) m4 '] (b) of '| (c) o-[ )
2. Which of the following matrices are singular matrices?
1 2 _|3 2 12 3 mess '| (b) F-[2 | @o-(s 2
3. Findinverse of each of the following matrices.
= 6 _fe 2 (a) af sr] maf 7 where 946
2 × 2 Matrix to Solve Simultaneous Equations
Apply 2 × 2 matrix to solve simultaneous equations
Solving simultaneous equations by matrix method:
If A and B are two matrices Such that A f j and B -( |
ae x} _fax+ cy Tena 4 |: "ft 3)
Where a, b, c, d, x and y are any real numbers.
Let AB=C where C= | 4
_ ax+ cy)_|P AB=C means (it 3) #
Now by equating the corresponding elements, the following simultaneous equations are obtained.
QE CY BD on ce 00 0s con von one (1) Bx AY Gene ose coe 0s one v0e 200 0s (2)
Therefore for any system of simultaneous equations, a matrix method can be used if and only if ad-bc+0.
So, because ax+cy=p and bx+dy=q can be written, in a matrix equation form as f 4 x i = :] then we can Lta-{5 4. B= ; jana Cc F and the equation becomes AB=C If A" is multiplied on each side of the equation, then A* x AxB= A'xC (At A) x B= At xC But A* xA = | where | is an identity matrix. Also for any matrix, K, KI=IK = K Therefore (A-''xA) xB=IxB= A-'C
Then B= A'xC
And At = af4 7 where |A| = ad — be lall—b oa
By matrix method solve the following simultaneous equations:
5x+6y=11 7x+8y=15
Solution:
5x +by =11 (i)
The systems 2 5 gy = 15 (ii)
can be written in matrix form equation as
Multiplying A" an each side of the equation, gives,
—4xX11+3 × 15 -l7 × 2 +-sx=
"X=1 and y=1
Example 19
Solve
4x +2y = 40 x +3y=35
By matrix method
Solution: The above system is equivalent to the matrix equation
(t 3 b= &) Leta-f j
At= ——
4 × 3-1 × 2| -1
At=
3 × 40
10 5 "| 1 × 40 , 2 × 35
ag
«x =5 and y=10
Example 20 By using matrix method solve the following simultaneous equations:
3x-y=11 x+3y=-3
Solution: We can write the above equations
(i Gl)
A= i
~ 3 × 3-1x(-1)
"aau2) 3 1]. meals s}>
Multiplying A" on each side of the equation gives,
3 × 11 1 × 3
10 10 -1xi11 3 × 3
2 X=3and y=-2
Cramer's Rule
Cramer's rule is another method to solve the equations of the form hed –
By using the inverse of a matrix method we have seen that
xl 14d -—ce||P y| ad-bq—b al iq
So
x= aa iz i
and y = = : § Inx= | the numerator is obtained by replacing the first column by the column d al
[F , and iny = matrix after alec the second column by the column f While the denominator in
both cases is the determinant of [i a}
Example 21 Find
Sx +6y = 11
7x + By =15 by using Cramer's rule.
X and y from 1
Solution: The equation can be written as 5 6 [5] _ (11 7 8} 15
11 6 5 11
45_8! and y = "18 "& I;
_ 11 × 8-6 × 15 _ 88-90 ~ §x8=7 × 6 40-42
"X=1 and y=1
5 × 15-7 × 11 _75- 77 =
_=2 ~=2 = Tand y = 5 × 8-7 × 6 40-42 -2
Example 22
By using Cramer's rule
4x + 2y
, find x and y in xt3y =
ba 2| |* sal – and y = 4-35!
Ir al 1 3 1 3 = 40 × 3-35 × 2 _ 120-70 _ 50 5 214 y= 4 × 35-40 × 1 140-40 100 |
=== 10 4 × 3-1 × 2 12-2 10 4 × 3-1 × 2 12-2 10
=-x=5 and y=10
Example 23 Byusing Cramer's rule,
3x-y=11
Solve J 3y = -3
Solution:
The equations can be written as E
2 al ; =|
Sox "Foy and y =P ah 1 3 1 3
_ 11 × 3–3x(-1) _ 33-3 3 × 3-1x(-1) 941
+x=3 and y= -2
Exercise 4
1. Use the matrix method to solve the following systems of simultaneous equations.
3 3x(-—3)-1 × 11 -9-11 -20 = 30-3 andy = (3) = = =. 10 3 × 3-7 × 6 941 10
7x — 2y = 29 x+2y=12 3x-2y=7 oo () arse (©) {ers3y=40
4x -—6=3y (@) ft + Sy = —2x
Use Cramer's rule to solve the following simultaneous equation
3x+4y=8 2x -—3y=4 fartae=7 (2) 2x + 3y = 13 Seve © 4x-—Sy=2
3. Whythe system of simultaneous equations
3x+2y=5 fr +4ay=8 has no solution?
Definition: A transformation in a plane is a mapping which moves an object from one position to another within the plane. Figures on the plane can also be shifted from one position by a
transformation.
A new position after a transformation on is called the image.
Examples of transformations are (i) Reflection (ii) Rotation (iii) Enlargement (iv) Translation. Any Point P(X, Y) into P'(X',Y') by Pre-Multiplying (%) with a Transformation Matrix T
Transform any point P(X, Y) into P'(X',Y') by pre-multiplying (4) with a transformation matrix T
– Suppose a point P(x,y) in the x-y plane moves to a point P¢ (x¢,y¢) by a transformation T,
We say that P is mapped to P' by T and may be indicated as pt,p
¥
P(x, y) P(x. y) ~__ | CO
A transformation in which the size of the image is equal that of the object is called an ISOMETRIC MAPPING.
The Matrix to Reflect a Point P(X, Y ) in the X-Axis Apply the matrix to reflect a point P(X, Y ) in the x-axis Reflection:
When you look at yourself in a mirror you seem to see your body behind the mirror. Your body
is in front of the mirror as your image is behind it.
Mirror
> Horizontal surface
An object is reflected in the mirror to form an image which is;
The same size as the object
The same distance from the mirror as the object So reflection is an example of ISOMETRIC MAPPING. The mirror is the line of symmetry between the object and the image. Example 24
Find the image of the point A (2,3) after reflection in the x — axes.
Solution; Plot point A and its image A¢ such that AA¢ crosses the x — axis at B and also perpendicular to it.
For reflection AB should be the same as BA¢ i.e. AB = BA¢
+
-A' (2,-3)
From the figure, the coordinates of A ¢ are A¢ (2,-3). So the image of A (2,3) under reflection in the x-axis is A¢ (2,-3)
Normally the letter M is used to denote reflection and thus Mx means reflection in the x — axis.
So Mx(2,3) =- (2,-3).
Mx(x, y) =(x,-y) and My(x, y) = (-x, y)
Where Mx means reflection in the x — axis and My means reflection in the y-axis.
The Matrix to Reflect a Point P(X, Y) in the Y-Axis Apply the matrix to reflect a point P(X, Y) in the Y-Axis Example 25
Find the image of B(3,4) under reflection in the y- axis.
Solution:
From My (x.y)= (-x,y)
My (3 4 ) =( -3,4)
Therefore the image of B(3,4) is B'(-3,4) .
Reflection in the line y =x.
The line y=x makes an angle 45° with x and y axes. It is the line of symmetry for the angle YOX
formed by two axis. By using isosceles triangle properties, reflection of the point (1,0) in the line y=x will be ( 0,1) while the reflection of (0,2) in the line y=x will be ( 2, 0) it can be noticed that
the coordinates are exchanging positions. Hence the reflection of the point (x,y) in the line y=x is
(y,x).
+ t 1>X (1, 0) (2,0) (@,90) a
My=x(x, y) =(y, X)
Where My -xmeans reflection in the line y=x.
Example 26
Find the image of the point A(1,2) after reflection in the line y = x . Draw a sketch.
A' Q,1)
Therefore the image of A '(1, 2) is A' (2, 1). Reflection in the line y =-x The reflection of the point B(x,y) in the line y = -x is B'(-y,-x).
i.e.
My=-x (x y) = CY:-x)
Where by My =-x means reflection in the line y = -x.
Example 27
Find the image of B (3,4) after reflection in the line y=-x followed by another reflection in the line y=0.Draw a sketch.
Solution;
Reflection of B in the line y=-x is B'(-4,-3). The line y=0 is the x — axis. So reflection (-4,-3) in the x-axis is (-4,3)
Therefore the image of B (3,4) is B¢(-4,3).
The image of a point P(x,y) when reflected in the line making an angleawith positive x-axis
and passing through the origin.
If the line passes through the origin and makes an angle a with x — axis in the positive direction,
then its equation is y= xtana where tanais the slope of the line.
Consider the following diagram.
Y
(x,y)
From the figure above OF is inclined at f and the coordinates of P are (x, y)
_
OP' is the image of OF under reflection in the OF.
Now by physical laws of reflection PP' is perpendicular to the line of reflection, in this case the line of reflection is OS which is given by the equation y = xtana..
Alsop P'OS= POS= « – B and the line segment OP is equal to the line segment OP'.
Since the angle PR and PO are both perpendicular to the x — axis, then according to coordinate geometry the coordinates of Q are (x, 0) and that of R are (x', 0).
So OQ =x and OR =x' while QP = y and RP'= y'.
But OPQ is a right angled triangle. So x = OP Cosf and y = OPSinf .
Again OP¢R is a right angled triangle and the angle P¢QR = a -B + a- B+ B, this is due to the fact
that reflection is an isometric mapping.
Now the angle P¢OR = 2 a-f, then
x' = OP' [ Cos (a B — B)] and y' =OP' [Sin (2c-B)], on expanding x' = OP '[Cos 2aCos B+ Sin 2 a Sin]… And y' = OP'[ Sin 2 a CosB — SinBCos 2a] But OP = OP.'
Then x' = OP [ Cos2aCosf+ Sin2aSinf] And y'= OP [ Sin2a Cos — SinBCos2a ] Which implies that
X' (OPCosB) Cos2a + (OPSinB) Sin2a and y' = (OP Cosf) Sin2a – (OPSinB) Cos2a Remember x = OPCosf and y = OPSinB Therefore x' = xCos2a + ySin 2a
And y' = xSin2a – yCos2a
It follows therefore that if M is a reflection in the line inclined at a, then
MIx, y] = [x', y"] where x'=xCos2a+ySin2da and
y'= xSin2a – yCos2a
The above two linear equations are called transformation equations for reflection. These two equations can be written in matrix form as
x _|Cos2a = Sin2a | | * y' Sin2a —Cos2a) | ¥
!
Where is the position vector which is the image of y fonder reflection M,
Cos2a = Sin2a |. E& soz is the matrix of reflection.
= x NB aN (x, y) and f } &y)
Example 28
Find the image of the point A (1, 2) after a reflection in the line y = x.
Solution: The line y = x has a slope 1
So tana = 1, a= 45°
Cos2a Sin2a x) [x rom fo sae] ls ~ |
x'| _ |Cos90° Sin90° || * y= y'| > |sing0e —cos90° y | bute y) (1. 2)
Therefore (x', y') =(2, 1) which is the image of (1, 2).
Example 29
Find the image of B (3,4) after reflection in the line y = -x followed by another reflection in the
line y = 0.
Solution:
The line y = -x has slope -1 So tana =-1
a = -459 = 135°
Ma= Cos2a se] [ | Sin2a —Cos2) P
But (x, y) = (3. 4)
Mass°=[Sma7o —cos27oll 4 FL oll 4 FP3soH{ 3]
Now we need the image of (-4,-3) after being reflected in the line y = 0. But the line y = 0 has 0 slope because it is the x — axis, Therefore tan a = 0, a= 0°
Me=[Smor cools Fl SISOS SHS |
Nore that | J= [5 12, "L3 }
Also it should be noted that matrix multiplication does not obey communicative property, this is to say
MaiM o =Mo Ma ifand only ifai= on –
Example 30
Find the equation of the line y = 2x + 5 after being reflected in the line y = x,
Solution: The line y = x has a slope 1 So tan a= 1 which means a= 45°
To find the image of the line y = 2x + 5, we choose at least two points on it and find their
images, then we use the image points to find the equation of the image line.
The points (0,5) and (1,7) lie on the line
But Ma = [0520 al [3]
Sin2a —Cos2 So 2a=2 × 45° = 90°
Cos90° = Sin90° 0 Sin90° <coss0el Ls |
At ollS)-[otoH 1
Cos90° = Sin90° | [ | Sin90° —Cos90°L7
Pl ols )-[S30H 4]
So the image line is the line passing through (5,0) and (7,1) and it is obtained as follows;
For 0,5),[ | =|
Fort, 7).[*, |=
Therefore the equation of the line y = 2x + 5 after being reflected in the line y = x is 2y=x-5.
Exercise 5
Self Practice.
Find the image of the point D (4,2) under reflection in the x — axis Point Q (-4,3) is reflected in the y — axis. Find its image coordinates. Reflect the point (5,4) in the line y = x
Find the image of the point (1,2) after a reflection in the line y = x followed by another
reflection in the line y = -x.
3 Find the equation of the line y = 3x -1 after being reflected in the line x + y = 0.
A Matrix Operator to Rotate any Point P( X, Y ) Through 90° 180°, 270° and 360° about the Origin
Use a matrix operator to rotate any point P(X, Y ) through 90° 180°, 270° and 360° about the Origin
Rotation:
Definition; A rotation is a transformation which moves a point through a given angle about a
fixed point.
gy)
%, @ \. y)_»X
+ OF The point P(x, y) is rotated through angle 6 to form an image P'(x', y').
Rotation is an isometric mapping and it is usually denoted by R.
Therefore RO means rotation of an object through an angle9.
In the xy plane, when@ismeasured in the clockwise direction it is negative and when it is
measured in the anticlockwise direction it is positive.
Positive Rotation Negative Rotation
Example 31
Find the image of the point P(1,0) after a rotation through 90° about the origin in the anti
clockwise direction.
Solution:
P is on the x — axis, so after rotation through 90° about the origin it will be on the y — axis. Since P is lunit from O, P¢ is also 1 unit from O, the coordinates of P¢ (0,1) are P¢ (0,1). Therefore R 90°(1,0) = (0,1).
Example 32
Find the image of the point B (4,2) after a rotation through 90° about the origin in the anticlockwise direction.
Solution;
Consider the following figure,
From the figure, OB = OB' and QBO = POB'
APOB = AQBO So PB'=O0Q Hence y' = 4 and x' = -2
Therefore Roo (4, 2) = (-2, 4)
Exercise 6
Find the matrix of rotation through
a. 90° about the origin b. 45° about the origin c 270° about the origin
Find the image of the point (1,2) under rotation through 180° ant -clockwise about the origin.
Find the image of the point (-2,1) under rotation through 270° clockwise about the origin
Find the image of (1,2) after rotation of -90°.
Find the image of the line passing through points a (-2,3) and B(2,8) after rotation through
90° clockwise about the origin General formula for rotation
Consider the following sketch,
From the figure above, OP is inclined at angle B, also [OB is inclined angle B, also | OP | because they are radii of the circle whose center is the point (0, 0). Again OA = x, AP =y, OB =x' and BP'=y'
Now in the APOA, the angle POA=B
Sox=|OP|CosB and y =| OP | SinB
In the AP'OB, the angle P'OB = a+B
_loBl_ x So Cos (a+B) = jor = jor x' =| OP'| cos(a+f)
oe {PrBI F While sin(o+B) = | = op] —
y'= | OP! Sin (a+).
On expanding Cos (a+) and Sin (a+), we get x' = OP'(CosaCosf — SinaSinB) and
y' = OP'(SinaCos B + SinBCosa) but OP = OP' x' =OP(CosaCosB —SinaSinf ) and y' = OP(SinaCosf + SinBCosa) x' =| OP| CosaCof -| OP | SinaSinB and y'= | OP| SinaCosf +| OP | SinBCosa
but | OP| CosB = x and
|OP| SinB = y Then x' = xCosa — ySina
y'=xSina + yCosa
which can be written in matrix form as;
[ x _ [Coa | [ *]
y ina Cosaily.
Therefore
Ra= [Sine cose ll 9]
' Where Ra is the rotation of the point (x, y) through the angle Ol and [ | is the image of the
pi]
Example 33
Find the image of the point (1,2) under a rotation through 180° anticlockwise
Solution: Given (x,y) = (1,2) a= 180°
_ [Cosa —Sina][*] _) | Ra= Sina ol {*,
Sin180° Cos180° To <llz}-lo+%2)-[2
Therefore the image of (1, 2) after rotation through 180° anticlockwise is (-1,-2).
[% FI Co180° rae | |
Example 34
Find the image of the point (5,2) under rotation of 90° followed by another rotation of
180°anticlockwise.
Solution:
(x, y) = (5, 2), O1= 90° and a2 = 180°
From
» — [C0s90° —Sin90°)/ 5 -{*)] = Sin90° Cos90° 5] y'
"[t oll] Assol-[e] So R90°5,2)= (-2,5),
Also
R180° =
Plto Sls] 44"
Cos180° pace ied Sin90°—- Cos180° 5 yy
Therefore the image of (5,2) under rotation of 90° followed by another rotation of
180°anticlockwise is (2,-5) . Translation
Definition: A translation is a mapping of a point P (x, y) into P' (x', y') by the Vector (a, b) such that (x', y') = (x, y) + (a, b), translation is denoted by the letter T. So T maps a point (x, y) into x', y)
Where (x', y') = (x, y) + (a, b)
Or (x, y') = (x y) + (a, b)
Consider the triangle OPQ whose vertices are (0,0), (3,1) and (3,0) respectively which is mapped into triangle O¢P¢Q¢ by moving it 2 units in the positive x direction and 3 units in the positive y
direction
From the figure above, the image of AOPQ is AO'P'Q' whose vertices are (2.3), (5.4) and (5.3)
respectively.
Note that 0d= (2, 3) is called the vector of translation. Also the size of the image is equal to that of object.
Example 35 If T is a translation by the vector (4,3), find the image of (1, 2) under this translation. Solution;
From T(x, y) = (x'.y') where (x.y) =(%, y) + (a, b) (. y) = (1. 2) + 4.3) =G,5). Therefore the image of (1, 2) under the translation T is (5, 5).
Example 36
A translation T maps the point (-3, 2) into (4, 3). Find where (a) T maps the origin (b) T maps the point (7, 4).
Solution:
From T(x, y) = (x, y'). then
(x.y) =(& y)+@ b), where (a, b) is the translation vector,
4)]_[ -3 a 'on [5 J-l 2 }+[ 6 | Then 4=-3+a and 3 = 2+b, solving the two equations gives a= 7 and b= 1, so (a, b) = (7, 1)
@ T0.=[9]+[7] "1;
Therefore T maps the point (0,0) into (7,1)
(b) T (7,4) = (7,4)+07,1) = (14,5). Therefore T maps the point (7,4) into (14.5).
Example 37
Find the translation vector which maps the point (6,-6) into (7,16). Solution
Given that (x, y) = (6,-6) and (x¢, y¢) = (7,16), (a, b) =?
From T (x, y) = (x, y) + (a, b) = (x', y'),
then (7,16) = (6,-6)+(a,b) which means a=7-6 = 1 and b=16+6 = 22. Therefore translation vector (a,b) = (1,22).
The Enlargement Matrix E in Enlarging Figures
Use the enlargement matrix E in enlarging figures
Definition: Enlargement is the transformation which magnifies an object such that its image is proportionally increases on decreased in size by some factor k. The general matrix of
enlargement
| where k is a non-zero real number called the linear scale factor|
ito ells |
NB: Enlargement is not isometric.
Example 38 Find the image of the square with vertices O(0,0), A (1,0), B (1,1) and C (0,1) under the
enlargement matrix [ 7 : | Hence sketch both object and its image on the same axes. Solution:
— = = _ v3 k Oj]_[4 0 Given O= (0,0), A=(1,0), B= (1.1) and C=,1)and[ FP ]=[8 3]
From
Ito ells |
= [5 Moltol 4 =[6 alot otol=[ ol}:
Blo allillorsl-[ 4} eff alltll oral sh
So O' = (0,0), A'=(4,0), B' = (4,4) and C'= (0,4).
Sketch; A
Therefore the image of OABC is OA'B'C' as shown in the figure above.
Example 39
Find the image of (6, 9) under enlargement by the matrix
1 / 3 0 ["o ash Solution: From
Clto ills |
Plo sll $1]
o+3)=l 3}
Therefore the image of (6,9) is (2,3).
Example 40
Draw the image of a unit circle with center O (0,0) under
_[3 OQ) =[5 3. Solution:
Since the circle is centered at (0,0) and has a radius 1 unit, then it passes through the points , (0,1), (1,0) (0,-1), (-1,0) and other points.
Now the images of these points are (0,3), (3,0), (0,-3), (-3,0) and other points respectively, where
the centre remains (0,0) and the radius becomes 3 units.
Sketch; wv
I n the figure above, the circle with radius | unit and its image with radius 3 units C; and
C>respectively are shown. Linear Transformation: Definition:
For any transformation T, any two vectors U and V and any real number t, T is said to be a linear
transformation if and only if
T(t U) = tT(U) and T (U+V) = T(U) + TV)
Example 41
Show that the rotation by 90° about O(0,0) is a linear trans formation Solution
Let U=(U;,U2) and V =(V; , V2) be any two vectors in the plane and t be any real number To show that R90° is the linear transformation we must show that R90° (tU)= t R90° (U) and
R90° (U + V) = R90° (U) + R90° (V) R90° (U) = [0890° —Sin90°] (U1 Now R90" (U) =| sing0° Cos90° les
_fO —1)fU1)]_ [—U2
=[{ o llual= [ U1 R90 (U) =| "Y? FIM? | while #U = (0,102) because U= (Ui,U2) sors t)=[1 > ][tu2 =
—tU2
Therefore R90°(tU) = R90° (U) =| =
Or (-tU2, tUn)
nanan d= [BP] ant 150100 = [2]
sermeoemorco-[ 2]
_f —(U2 + v2) -[ (U1 + V1)
and U+V = (Ui,U2)+ (V1, V2) = (U1+V),(U2+V2) =((U1+ V1), U2 V2) )
(U1 + V1)
eanaieed BI —(U2 + V2) | |
R90? (U+V) = Li ri [ (U2 + V2)
Therefore, since R90° (U) + R90° (V) = R90° (U+V) and R90° (tU)= t R90° (U), then R90° is a
linear trans formation.
Example 42
Suppose that T is a linear transformation such that T(U) = (1,-2), T(V) = (-3,-1) for any vectors U and V, find
(a) T(U+ V) (b) T(8U) (c) T(3U -2V) Solution (a)Since T is a linear Transformation then
T( U+ V) =T(U) + T(V)
rere 4 J+ [A] 13 So T(U+ V)=(-2 .-3.)
(b) T (8U); T(SU) = 8 T(U) because T is a linear transformation.
So T(8U) = 8x (3 )=(6 )
Therefore T(8U) = (8,-16)
(c) T GU -2V); TGU-2V) = TU) – TV) =3T(U) -2T(V)
=3[_, ]-2[] = [2 ]+[3 ]-0. Therefore T(3U-2V) = (9,-4).
Exercise 7
Lf
'M= Cos2a Sin2a Sinda -Cos2
2. Is the matrix of reflection in a line inclined at angle a, U=(6,1) , V=(-1,4) and al13500, find (a) m(U+V) (b) m(2V)
If U =(2,-7) and V=(2,-3), find the matrix of linear transformation T such that T(2U)=(-4,14) and T(3V) = (6,9)
4. What is the image of (1,2) under the transformation
matrix | s] followed by (5 | ?
5. Given that I is the identify transformation such that I(U) =U for any Vector U, prove that I is a
linear transformation.
READ TOPIC 8: Linear Programming
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