ETHIO NATIONAL SCHOOL

2020 G.C 3RD QUARTER MATHEMATICS LESSON NOTE for GRADE 11

 Dear students:- Copy the note and do the given homework on your exercise book!

Subject: Mathematics Grade & section: 11(B –E)

Unit: 6&7
Main topic: Solving systems of linear equations using matrices & Set of complex
numbers
Sub topic/s:
Page range from: 256-279

Monday, April1 13, 2020

Objective of the lesson

At the end of this lesson you should be able to:
 Solve systems of equations by Crammer’s rule
CRAMER’S RULE

 Determinants can be used to solve systems of linear equations with equal number of
equations and unknowns.
 The method is practicable, when the number of variables is either 2 or 3.
 Consider the system

Let D= and = .Then, if ≠0

= .A similar calculation gives: =

The method is called Cramer’s rule for a system with two equations and two unknowns.
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Note:
Note:

 Dx and Dy are obtained by replacing the first and 2nd column by the constant column vector
respectively
 Under similar conditions, the rule holds for three

 Unknowns too.

The system of equations has exactly one solution, provided that the

Determinant of the coefficient matrix is non-zero. In this case the solution is:

Example 1 Use Cramer’s rule to find the solution set of

+ −2 =8
3 − =6
a) b) 2 + =0
+5 =2
3 +2 + =7
3 −1
Solution: a) = = 3 × 5 − 1 × −1 = 16 ≠ 0
1 5
∴ the system of equation has unique solution and the solution is

= = = 2 and = = = 0 ⟹ s.s={(2,0)}

1 1 −2
1 −2 1 −2
b) = 2 1 0 = (−2) + = −10 + 7 = −3
2 1 3 1
3 2 1

Then = =− , = = , = = − ∴s.s= − , ,−

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 2 −5
Class Work: a) If 2 1 3 = 5 and if determinant of the coefficient matrix is 5
−1 0 1 0
, then determine solution set of the system of equation.

2 −3 + =
b) For what value of a and b does − + 5 + 3 = 9
− − =1

i) have no solution ii) have exactly one solution iii) have infinite solution

Home work: Solve the following

3 + 2 − = 12 + + =1 − + ( − 1) + ( + 1) =
a) + =0 b) + + =1 c) ( + 1) − + ( − 1) =
−9 + 6 + 3 = 4 + + =1 ( − 1) + ( + 1) − =

Tuesday, April 07, 2020

At the end of this lesson you should be able to
Solving system of equations by matrix inversion
Consider the linear system (in matrix form), AX =B
If | | ≠ 0, then A is invertible and ( )=
⟹( ) =

⟹ =

⟹ =
Therefore, the system has a unique solution.

+ =9
Example: Solve the system of equation:
3 − =2

1 1 9
Solution: The system is equivalent to =
3 −1 2
1 1 1 1
The coefficient matrix is and = −1 × 1 − 3 = −4
3 −1 3 −1
1 1 9
⟹ =
3 −1 2

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 −1 −1 9
Hence the = ,i.e = and = solution is: =−
−3 1 2

−
+ + = 11 1 1 1 ⎛ ⎞
Class work:a) If − + 4 + = 12 and = −1 4 1 = ⎜− − ⎟,then
+ 15 + 3 = 10 1 15 3
−
⎝ ⎠
determine value of x,y and z

Wednesday, April 15, 2020

Lesson Objective
At the end of this lesson you should be able to;
 Define complex numbers
 Differentiate real and imaginary parts of complex numbers
Unit 7

The set of Complex Numbers

7.1 The concept of complex numbers

For any quadratic equation of the form + + = 0 to have solutions, you need a number
system in which√ −4 is defined for all numbers a, b and c.The number system which you
are going to define is called the complex number system.
To this end a new number which is called an “imaginary number” namely √−1 = (read as iota)
is introduced.
Example 1 using the notation introduced above, you have:
a)√−9 = √9 × −1 = √9√−1 = 3
b) √−36 = √36 × −1 = √36√−1 = 6
c) √−7 = √7 × −1 = √7√−1 = √7 = √7

Definition 7.1
A complex number z is an expression which is written in the form = + ,for some real
numbers x and y, where√−1 = ; the number x is called the real part of z and is denoted by Re(
)z and the number
y is called the imaginary part of z and is denoted by Im( )z .

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Notation:
The set of complex numbers denoted by ℂ is given by ℂ = {z|z = x + yi where x and y are real
numbers; and√−1 = }
Note that √−1 = ⟹ = −1
Example 2
a) For z = 2 - 5i, Re(z) = 2 and Im(z) = - 5
b )For z = 6 + 4i, Re(z) = 6 and Im(z) = 4
c) For z = 0 + 2i = 2i, Re(z) = 0 and Im(z) = 2
d) For z = 0 + 0i = 0, Re (z) = 0 and Im(z) = 0
e) For z = 4 + 0i = 4, Re(z) = 4 and Im(z) = 0

Example 3:Solve the following equations

a) −2 +2= 0 b) − +1=0

± ( ) ( ) ±√
Solution: a) = = = 1±

b) − + 1 = 0 ⇒ ( + 1)( − + 1) = 0 ⇒ ( + 1) = 0 &( − + 1) = 0

± ( ) ( ) ±√
⇒ = −1 and = =

Equality of complex numbers

Suppose = + and = + are two complex numbers; then we define the equality of
and , written as = , if and only if x = a and y = b.
Example 3 If 12 − 3 = 6 + 6 ⟹ 12 = 6 and −3 = 6 ⟹ = 2& = −2
Note The imaginary unit i can be raised to higher powers as shown below.

Notice
the

repeating pattern in each row of the table. The pattern allows you to express any power of i as
one of four possible values: i, –1, –i, or 1.

Example4 Simplify i63

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i63 = i  i62 …………………………….Rewrite as a product of i and an even power of i.

= i  (i2)31………………………………Rewrite i62 as a power of i2

= i  (–1)31 = i  –1 = –i ……………...Simplify.

Class Work:

1) Determine value of x and y if =2 −3 and = 9 − 8 are equal

2) Simplify ) b) c) d)

Home work: Exercise 7.1 page 268

Thursday, April 16, 2020

7.2.1Addition and Subtraction

Definition 7.2
Given two complex numbers = + and = + , we define the sum and difference of
complex numbers as follows:
i) + =( + )+( + )=( + )+( + )
ii) − =( + )−( + )=( − )+( − )

Example5

a) (3 + 4 ) + (7 + ) = (3 + 7) + (4 + 1) = 11 + 5
b) (8 + 6 ) − (2 + 3 ) = (8 − 2) + (6 − 3) = 6 + 3

The set of complex numbers is closed under

 Addition.
 Addition of complex numbers is commutative.
 Addition of complex numbers is associative
 0 is the additive identity element in C.
 For every z in C there is an additive inverse –z such that z + –z = 0 = –z +z.
7.2.2 Multiplication and Division of Complex Numbers Multiplication
Definition 7.3

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Given two complex numbers = + and = + , the product of and is Defined as
follows: =( + )( + )=( − )+( + )
.Example6 (2 + 3 )(4 − ) = (8 + 3) + (12 − 2) = 11 + 10

Note: The set of complex numbers is closed under

 Multiplication.
 Multiplication of complex numbers is commutative.
 Multiplication of complex numbers is associative.
 Multiplication is distributive over addition in C.
 1 is the multiplicative identity element in C

Class work: Compute

a)(9 − 7 ) − (7 − 8 ) b)(−12 − 10) + (4 + 3 ) c) (3 − ) + (2 − 5 )(2 )
Friday, April 10,2020
Division
You can think of division as the inverse process of multiplication, since for any two real Numbers
a and b with 0≠b the phrase “a is divided by b” can be symbolized as:

1
= : ≠0

Now division of complex numbers can be defined as follows:

Suppose = + and = + are given, then you have the following:

+ + − ( + ) ( − )
= = = +
+ + − + +

Example 6

( )
a) = = = 1−2 b) = ( )
= −

Class work: Write the following in + form

a) b) ( )( )
c)(1 − 4 ) −

Home work: Exercise 7.3 page 273

7.3 COMPLEX CONJUGATE AND MODULUS

Definition 7.5
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The complex conjugate (or conjugate) of a complex number = + denoted by ̅ is given by
̅= −
Example 7
a) = 2 + 3 ⇒ ̅ = 2 − 3 b) = −4 − ⇒ ̅ = −4 + c) = −4 ⇒ ̅ = 4

You can read the proof of the theorem on page 275 of your text
Example 8
Given = 1 − and = 2 + 3 ,then
i) = 1+ = 1− = ii) + = (1 − ) + (1 + ) = 2 = 2(1) = 2 ( )
iii) − = (1 − ) − (1 + ) = 2 = 2(− ) = 2 ( )
( )( ( )( )
iv) = )(
= = + and = =( )( )
=− +

∴ =

Example 9 = 3 + 4 ⇒ | | = √3 + 4 = 5

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Example 10

√
Find | | if = ( )
√

(√ )
√ √
Solution: | | = | |
= =( )(√
=
√ )
(√ ) ( )) √

Home work: Exercise 7.4 page 279

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