Lecture 2.1 Echelon method

Lecture 2.1: Echelon Method
Dr. Taoufik Ben Jabeur
Department of Mathematics, Physics and Statistics
Lecture Notes in Business Math1-Math-119
1

Learning Outcomes
By the end of this section In shaa Allah, you shall:
1. Manipulation between rows
2. Solving system of linear equations using Echelon Method for cases:
a) System 2x2, no solution
b) System 2x2, one solution
c) System 2x2, infinity of solutions
d) System 2x3, No solution
e) System 2x3, infinity of solutions (Line solution)
f) System 2x3, infinity of solutions (Surface solution)
2

Manipulation between rows
4000 people were watching a football match in a stadium. The admission price was
$10 for adults and $5 for children. The admission receipts were $35000, how many
adults and how many children attended?
System of linear equations
𝑥 + 𝑦 = 4000
10𝑥 + 5𝑦 = 35000
3

System of linear equations
𝑥 + 𝑦 = 4000
10𝑥 + 5𝑦 = 35000
10𝑅1 𝑥 + 𝑦 = 4000
x10 x10 x10
𝟏𝟎𝑹 𝟏: 𝟏𝟎𝒙 + 𝟏𝟎𝒚 = 𝟒𝟎𝟎𝟎𝟎
𝑅𝑜𝑤1 𝑅1:
𝑅𝑜𝑤2 𝑅2:
Multiplication a row by a number:
4

Possibility of exchange between rows:
To solve a system of linear equations, any row can be exchanged by a linear combination of many rows:
𝑎𝑅1 + 𝑏𝑅2 → 𝑅1, a and b are real numbers , a and b are not allowed to be 0 together
Case1: System with two rows
Case2: System with three rows
𝑎𝑅1 + 𝑏𝑅2 + 𝑐 𝑅3 → 𝑅2, a, b and c are real numbers, a, b and c are not allowed to be all 0 together
Warning: the following operations are not allowed: 𝑅1 𝑥𝑅2 → 𝑅2, 𝑅1 + 3𝑥 → 𝑅2 , 𝑅1 + 2 → 𝑅2
5

Example 1
Applying 10𝑅1 − 𝑅2 → 𝑅2 to the following system of linear equation
𝑥 + 𝑦 = 4000
10𝑥 + 5𝑦 = 35000
Solution 1 10 𝑅1
−
𝑅2:
_____
𝑅2:
10 𝑥 + 10𝑦 = 40000
10𝑥 + 5𝑦 = 35000
_____________________________________
0𝑥 + 5𝑦 = 5000
The system of linear equation becomes:
𝑥 + 𝑦 = 4000
5𝑦 = 5000
6

Example 2
Applying 3𝑅1 + 5𝑅2 → 𝑅2 to the following system of linear equation
−5𝑥 + 𝑦 = 6
3𝑥 + 2𝑦 = 3
Solution 1 3𝑅1
+
5𝑅1:
_____
𝑅2:
−15 𝑥 + 3𝑦 = 18
15𝑥 + 10𝑦 = 15
_____________________________________
0𝑥 + 13𝑦 = 31
−5𝑥 + 𝑦 = 6
13𝑦 = 31
7

Echelon Method:
Algorithm: 𝑅1: 𝑎1 𝑥 + 𝑎2 𝑦 + 𝑎3 𝑧 = 𝑑1
𝑅2: 𝑏1 𝑥 + 𝑏2 𝑦 + 𝑏3 𝑧 = 𝑑2
𝑅3: 𝑐1 𝑥 + 𝑐2 𝑦 + 𝑐3 𝑧 = 𝑑3
𝑅1: 𝑎1 𝑥 + 𝑎2 𝑦 + 𝑎3 𝑧 = 𝑑1
𝑅2: h𝑦 + 𝑘𝑧 = 𝑢
𝑅3: 𝑟𝑧 = 𝑣
From we obtain the value of z𝑅3
We substitute the value of z in the row to find y
𝑅1
We substitute the values of y and z in the row to find x
𝑅2
1
2
3
8

Echelon Method: case 2x2 : no solution
Example1 Solve the following system of linear equation
𝑥 + 𝑦 = 3
𝑥 + 𝑦 = 1
Solution 1 𝑅1
−
𝑅2:
_____
𝑅2:
𝑥 + 𝑦 = 3
𝑥 + 𝑦 = 1
_____________________________________
0𝑥 + 0𝑦 = 2
𝑥 + 𝑦 = 3
0 = 2 -3 -2 -1 0 1 2 3 4
-3
-2
-1
0
1
2
3
4
5
6
x+y=3
x+y=1
Impossible No solution
9

𝑥 + 𝑦 = 3
3𝑥 + 3𝑦 = 7
Solution 1 3𝑅1
−
𝑅2:
_____
𝑅2:
3𝑥 + 3𝑦 = 9
3𝑥 + 3𝑦 = 7
_____________________________________
0𝑥 + 0𝑦 = 2
𝑥 + 𝑦 = 3
0 = 2
10

𝑥 + 𝑦 = 3
𝑥 + 𝑦 = 1
Solution 1 𝑅1
−
𝑅2:
_____
𝑅2:
𝑥 + 𝑦 = 3
𝑥 + 𝑦 = 1
_____________________________________
0𝑥 + 0𝑦 = 2
𝑥 + 𝑦 = 3
0 = 2
11

Echelon Method: case 2x2 : one solution
𝑥 + 𝑦 = 3
−𝑥 + 𝑦 = 1
Solution 1 𝑅1
+
𝑅2:
_____
𝑅2:
𝑥 + 𝑦 = 3
−𝑥 + 𝑦 = 1
_____________________________________
0𝑥 + 2𝑦 = 4
𝑅1: 𝑥 + 𝑦 = 3
𝑅2: 2𝑦 = 4
2y=4; y=2 We substitute the value y=2 in 𝑅1, x+2=3, x=3-2=1
The solution is the order point (1,2)
12

Echelon Method: case 2x2 : one solution
2𝑥 + 𝑦 = 3
3𝑥 − 2𝑦 = 1
Solution 1 3𝑅1
−
2𝑅2:
_____
𝑅2:
6𝑥 + 3𝑦 = 9
6𝑥 − 4𝑦 = 2
_____________________________________
0𝑥 + 7𝑦 = 7
𝑅1: 2𝑥 + 𝑦 = 3
𝑅2: 7𝑦 = 4
7y=7; y=1 We substitute the value y=1 in 𝑅1, 2x+1=3, 2x=2; x=1
The solution is the order point (1,1)
13

Echelon Method: case 2x2 infinity of solutions
𝑥 + 𝑦 = 3
3𝑥 + 3𝑦 = 9
Solution 1 3𝑅1
−
𝑅2:
_____
𝑅2:
3𝑥 + 3𝑦 = 9
3𝑥 + 3𝑦 = 0
_____________________________________
0𝑥 + 0𝑥 = 0
𝑅1: 𝑥 + 𝑦 = 3
𝑅2: 0 = 0
Infinity of solution We use only one equation x+y=3; y=-x+3
The set of solution is the set {(x,-x+3), 𝑥 ∈ 𝑅}, 𝑡ℎ𝑎𝑡 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑠 𝑏𝑦 𝑎 𝑙𝑖𝑛𝑒
-3 -2 -1 0 1 2 3 4
-1
0
1
2
3
4
5
6
7
8
x+y=3
14

Echelon Method: case 2x2 infinity of solutions
3𝑥 − 2𝑦 = −3
−6𝑥 + 4𝑦 = 6
Solution 1 2𝑅1
+
𝑅2:
_____
𝑅2:
6𝑥 − 4𝑦 = −6
−6𝑥 + 4𝑦 = 6
_____________________________________
0𝑥 + 0𝑥 = 0
𝑅1: 3𝑥 − 2𝑦 = −3
𝑅2: 0 = 0
Infinity of solution We use only one equation 3x-2y=-3; 3x=2y-3; 𝑥 =
2𝑦−3
3
The set of solution is the set {(
2𝒚−3
3
,y), y∈ 𝑅}, 𝑡ℎ𝑎𝑡 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑠 𝑏𝑦 𝑎 𝑙𝑖𝑛𝑒
15

Echelon Method: case 2x3: No solution
𝑥 − 𝑦 + 𝑧 = −10
𝑥 − 𝑦 + 𝑧 = 5
Solution 1 𝑅1
−
𝑅2:
_____
𝑅2:
𝑥 − 𝑦 + 𝑧 = −10
𝑥 − 𝑦 + 𝑧 = 5
_____________________________________
0𝑥 + 0𝑥 = −15
𝑅1: 𝑥 − 𝑦 + 𝑧 = −10
𝑅2: 0 = −15
Impossible
No solution
-10
-5
0
5
10
-10
-5
0
5
10
-30
-20
-10
0
10
20
30
x-axisy-axis
z-axis
Surface 1: x-y+z=-10
Surface 2:x-y+z=5
16

Echelon Method: case 2x3: No solution
3𝑥 + 𝑦 − 2𝑧 = 10
−6𝑥 − 2𝑦 + 4𝑧 = 3
Solution 1 2𝑅1
+
𝑅2:
_____
𝑅2:
6𝑥 + 2𝑦 − 4𝑧 = 20
−6𝑥 − 2𝑦 + 4𝑧 = 3
_____________________________________
0𝑥 + 0𝑥 = 23
𝑅1: 3𝑥 + 𝑦 − 2𝑧 = 10
𝑅2: 0 = 23
Impossible
No solution
17

Echelon Method: case 2x3: Infinity of solutions (Line)
𝑥 − 𝑦 + 𝑧 = 4
3𝑥 − 2𝑦 + 𝑧 = 5
Solution 1 3𝑅1
−
𝑅2:
_____
𝑅2:
3𝑥 − 3𝑦 + 3𝑧 = 12
3𝑥 − 2𝑦 + 𝑧 = 5
_____________________________________
0𝑥 − 𝑦 + 2𝑧 = 7
The system of linear equation becomes: 𝑅1: 𝑥 − 𝑦 + 𝑧 = 4
𝑅2: 𝑦 + 2𝑧 = 7
Infiniy of solution;
We consider z as known, and we use the row 𝑅2 to get 𝑦, −𝑦 + 2𝑧 = 7 ↔ 𝑦 = 2𝑧 − 7;
We substitue the value of 𝑦 in the first row, 𝑥 − 2𝑧 − 7 + 𝑧 = 4, ↔ 𝑥 − 2𝑧 + 7 + 𝑧 = 4 ↔ 𝑥 − 𝑧 = 4 − 7 ↔
𝑥 = 𝑧 − 3
The solution is the set of the ordered points {(𝑧 − 3, 2𝑧 − 7, 𝑧), 𝑧 ∈ 𝑅}
-10
-5
0
5
10
-15
-10
-5
0
5
10
15
-60
-40
-20
0
20
40
60
x-axis
y-axis
z-axis
Surface : x-y+z=4
Surface: 3x-2y+z=5
Line : {(k-3,2k-7,k), k R}
1
2
18

Echelon Method: case 2x3: Infinity of solutions (Line)
2𝑥 − 𝑦 − 2𝑧 = 1
3𝑥 − 2𝑦 + 𝑧 = 5
Solution 1 3𝑅1
−
2𝑅2:
_____
𝑅2:
6𝑥 − 3𝑦 − 6𝑧 = 3
6𝑥 − 4𝑦 + 2𝑧 = 10
_____________________________________
0𝑥 + 𝑦 − 8𝑧 = −7
The system of linear equation becomes: 𝑅1: 2𝑥 − 𝑦 − 2𝑧 = 1
𝑅2: 𝑦 − 8𝑧 = −7
We consider z as known, and we use the row 𝑅2 to get 𝑦, 𝑦 − 8𝑧 = −7 ↔ 𝑦 = 8𝑧 − 7;
We substitue the value of 𝑦 in the first row, 2𝑥 − 8𝑧 − 7 − 2𝑧 = 1, ↔ 2𝑥 − 8𝑧 + 7 − 2𝑧 = 1 ↔ 2𝑥 − 10𝑧 = 1 − 7 ↔
2𝑥 = 10𝑧 − 6; 𝑥 = 5𝑧 − 3
The solution is the set of the ordered points {(5𝑧 − 3, 8𝑧 − 7, 𝑧), 𝑧 ∈ 𝑅}
1
2
19

Echelon Method: case 2x3: Infinity of solutions (Surface)
𝑥 − 𝑦 + 𝑧 = 5
3𝑥 − 3𝑦 + 3𝑧 = 15
Solution 1 3𝑅1
−
𝑅2:
_____
𝑅2:
3𝑥 − 3𝑦 + 3𝑧 = 15
3𝑥 − 3𝑦 + 3𝑧 = 15
_____________________________________
0𝑥 + 0𝑦 + 0𝑧 = 0
𝑅2: 0 = 0
We consider y and z as known, and we use the row 𝑅1 to get x, 𝑥 − 𝑦 + 𝑧 = 5 ↔ x = 𝑦 − 𝑧 + 5;
The solution is the set of the ordered points {(y-z+5, 𝑦, 𝑧); 𝑦, 𝑧 ∈ 𝑅}
1
-10
-5
0
5
10
-10
-5
0
5
10
-15
-10
-5
0
5
10
15
20
25
x-axisy-axis
z-axis
Surface: x-y+z=5
20

Echelon Method: case 2x3: Infinity of solutions (Surface)
2𝑥 − 𝑦 − 𝑧 = 1
4𝑥 − 2𝑦 − 2𝑧 = 2
Solution 1 2𝑅1
−
𝑅2:
_____
𝑅2:
4𝑥 − 2𝑦 − 2𝑧 = 2
4𝑥 − 2𝑦 − 2𝑧 = 2
_____________________________________
0𝑥 + 0𝑦 + 0𝑧 = 0
𝑅2: 0 = 0
We consider x and z as known, and we use the row 𝑅1 to get y, 2𝑥 − 𝑦 − 𝑧 = 1 ↔ 2𝑥 − 𝑧 = 𝑦 + 1; ↔ 2𝑥 − 𝑧 − 1 = 𝑦
The solution is the set of the ordered points {(x, 2𝑥 − 𝑧 − 1, 𝑧); 𝑥, 𝑧 ∈ 𝑅}
1
21

Lecture 2.1 Echelon method

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