1. Lecture 2.1: Echelon Method
Dr. Taoufik Ben Jabeur
Department of Mathematics, Physics and Statistics
Lecture Notes in Business Math1-Math-119
1
2. 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)
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3. 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
π₯ + π¦ = 4000
10π₯ + 5π¦ = 35000
3
4. Manipulation between rows
System of linear equations
π₯ + π¦ = 4000
10π₯ + 5π¦ = 35000
10π 1 π₯ + π¦ = 4000
x10 x10 x10
πππΉ π: πππ + πππ = πππππ
π ππ€1 π 1:
π ππ€2 π 2:
Multiplication a row by a number:
4
5. Manipulation between rows
Possibility of exchange between rows:
To solve a system of linear equations, any row can be exchanged by a linear combination of many rows:
ππ 1 + ππ 2 β π 1, 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
ππ 1 + ππ 2 + π π 3 β π 2, 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: π 1 π₯π 2 β π 2, π 1 + 3π₯ β π 2 , π 1 + 2 β π 2
5
6. Manipulation between rows
Example 1
Applying 10π 1 β π 2 β π 2 to the following system of linear equation
π₯ + π¦ = 4000
10π₯ + 5π¦ = 35000
Solution 1 10 π 1
β
π 2:
_____
π 2:
10 π₯ + 10π¦ = 40000
10π₯ + 5π¦ = 35000
_____________________________________
0π₯ + 5π¦ = 5000
The system of linear equation becomes:
π₯ + π¦ = 4000
5π¦ = 5000
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7. Manipulation between rows
Example 2
Applying 3π 1 + 5π 2 β π 2 to the following system of linear equation
β5π₯ + π¦ = 6
3π₯ + 2π¦ = 3
Solution 1 3π 1
+
5π 1:
_____
π 2:
β15 π₯ + 3π¦ = 18
15π₯ + 10π¦ = 15
_____________________________________
0π₯ + 13π¦ = 31
The system of linear equation becomes:
β5π₯ + π¦ = 6
13π¦ = 31
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8. Echelon Method:
Algorithm: π 1: π1 π₯ + π2 π¦ + π3 π§ = π1
π 2: π1 π₯ + π2 π¦ + π3 π§ = π2
π 3: π1 π₯ + π2 π¦ + π3 π§ = π3
π 1: π1 π₯ + π2 π¦ + π3 π§ = π1
π 2: hπ¦ + ππ§ = π’
π 3: ππ§ = π£
From we obtain the value of zπ 3
We substitute the value of z in the row to find y
π 1
We substitute the values of y and z in the row to find x
π 2
1
2
3
8
9. Echelon Method: case 2x2 : no solution
Example1 Solve the following system of linear equation
π₯ + π¦ = 3
π₯ + π¦ = 1
Solution 1 π 1
β
π 2:
_____
π 2:
π₯ + π¦ = 3
π₯ + π¦ = 1
_____________________________________
0π₯ + 0π¦ = 2
The system of linear equation becomes:
π₯ + π¦ = 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
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10. Echelon Method: case 2x2 : no solution
Example2 Solve the following system of linear equation
π₯ + π¦ = 3
3π₯ + 3π¦ = 7
Solution 1 3π 1
β
π 2:
_____
π 2:
3π₯ + 3π¦ = 9
3π₯ + 3π¦ = 7
_____________________________________
0π₯ + 0π¦ = 2
The system of linear equation becomes:
π₯ + π¦ = 3
0 = 2
Impossible No solution
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11. Echelon Method: case 2x2 : no solution
Example1 Solve the following system of linear equation
π₯ + π¦ = 3
π₯ + π¦ = 1
Solution 1 π 1
β
π 2:
_____
π 2:
π₯ + π¦ = 3
π₯ + π¦ = 1
_____________________________________
0π₯ + 0π¦ = 2
The system of linear equation becomes:
π₯ + π¦ = 3
0 = 2
Impossible No solution
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12. Echelon Method: case 2x2 : one solution
Example1 Solve the following system of linear equation
π₯ + π¦ = 3
βπ₯ + π¦ = 1
Solution 1 π 1
+
π 2:
_____
π 2:
π₯ + π¦ = 3
βπ₯ + π¦ = 1
_____________________________________
0π₯ + 2π¦ = 4
The system of linear equation becomes:
π 1: π₯ + π¦ = 3
π 2: 2π¦ = 4
2y=4; y=2 We substitute the value y=2 in π 1, x+2=3, x=3-2=1
The solution is the order point (1,2)
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13. Echelon Method: case 2x2 : one solution
Example2 Solve the following system of linear equation
2π₯ + π¦ = 3
3π₯ β 2π¦ = 1
Solution 1 3π 1
β
2π 2:
_____
π 2:
6π₯ + 3π¦ = 9
6π₯ β 4π¦ = 2
_____________________________________
0π₯ + 7π¦ = 7
The system of linear equation becomes:
π 1: 2π₯ + π¦ = 3
π 2: 7π¦ = 7
7y=7; y=1 We substitute the value y=1 in π 1, 2x+1=3, 2x=2; x=1
The solution is the order point (1,1)
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14. Echelon Method: case 2x2 infinity of solutions
Example1 Solve the following system of linear equation
π₯ + π¦ = 3
3π₯ + 3π¦ = 9
Solution 1 3π 1
β
π 2:
_____
π 2:
3π₯ + 3π¦ = 9
3π₯ + 3π¦ = 0
_____________________________________
0π₯ + 0π₯ = 0
The system of linear equation becomes:
π 1: π₯ + π¦ = 3
π 2: 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), π₯ β π }, π‘βππ‘ ππππππ πππ‘π ππ¦ π ππππ
-3 -2 -1 0 1 2 3 4
-1
0
1
2
3
4
5
6
7
8
x+y=3
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15. Echelon Method: case 2x2 infinity of solutions
Example2 Solve the following system of linear equation
3π₯ β 2π¦ = β3
β6π₯ + 4π¦ = 6
Solution 1 2π 1
+
π 2:
_____
π 2:
6π₯ β 4π¦ = β6
β6π₯ + 4π¦ = 6
_____________________________________
0π₯ + 0π₯ = 0
The system of linear equation becomes:
π 1: 3π₯ β 2π¦ = β3
π 2: 0 = 0
Infinity of solution We use only one equation 3x-2y=-3; 3x=2y-3; π₯ =
2π¦β3
3
The set of solution is the set {(
2πβ3
3
,y), yβ π }, π‘βππ‘ ππππππ πππ‘π ππ¦ π ππππ
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16. Echelon Method: case 2x3: No solution
Example1 Solve the following system of linear equation
π₯ β π¦ + π§ = β10
π₯ β π¦ + π§ = 5
Solution 1 π 1
β
π 2:
_____
π 2:
π₯ β π¦ + π§ = β10
π₯ β π¦ + π§ = 5
_____________________________________
0π₯ + 0π₯ = β15
The system of linear equation becomes:
π 1: π₯ β π¦ + π§ = β10
π 2: 0 = β15
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
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17. Echelon Method: case 2x3: No solution
Example2 Solve the following system of linear equation
3π₯ + π¦ β 2π§ = 10
β6π₯ β 2π¦ + 4π§ = 3
Solution 1 2π 1
+
π 2:
_____
π 2:
6π₯ + 2π¦ β 4π§ = 20
β6π₯ β 2π¦ + 4π§ = 3
_____________________________________
0π₯ + 0π₯ = 23
The system of linear equation becomes:
π 1: 3π₯ + π¦ β 2π§ = 10
π 2: 0 = 23
Impossible
No solution
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18. Echelon Method: case 2x3: Infinity of solutions (Line)
Example1 Solve the following system of linear equation
π₯ β π¦ + π§ = 4
3π₯ β 2π¦ + π§ = 5
Solution 1 3π 1
β
π 2:
_____
π 2:
3π₯ β 3π¦ + 3π§ = 12
3π₯ β 2π¦ + π§ = 5
_____________________________________
0π₯ β π¦ + 2π§ = 7
The system of linear equation becomes: π 1: π₯ β π¦ + π§ = 4
π 2: π¦ + 2π§ = 7
Infiniy of solution;
We consider z as known, and we use the row π 2 to get π¦, βπ¦ + 2π§ = 7 β π¦ = 2π§ β 7;
We substitue the value of π¦ in the first row, π₯ β 2π§ β 7 + π§ = 4, β π₯ β 2π§ + 7 + π§ = 4 β π₯ β π§ = 4 β 7 β
π₯ = π§ β 3
The solution is the set of the ordered points {(π§ β 3, 2π§ β 7, π§), π§ β π }
-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), kο R}
1
2
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19. Echelon Method: case 2x3: Infinity of solutions (Line)
Example2 Solve the following system of linear equation
2π₯ β π¦ β 2π§ = 1
3π₯ β 2π¦ + π§ = 5
Solution 1 3π 1
β
2π 2:
_____
π 2:
6π₯ β 3π¦ β 6π§ = 3
6π₯ β 4π¦ + 2π§ = 10
_____________________________________
0π₯ + π¦ β 8π§ = β7
The system of linear equation becomes: π 1: 2π₯ β π¦ β 2π§ = 1
π 2: π¦ β 8π§ = β7
Infiniy of solution;
We consider z as known, and we use the row π 2 to get π¦, π¦ β 8π§ = β7 β π¦ = 8π§ β 7;
We substitue the value of π¦ in the first row, 2π₯ β 8π§ β 7 β 2π§ = 1, β 2π₯ β 8π§ + 7 β 2π§ = 1 β 2π₯ β 10π§ = 1 β 7 β
2π₯ = 10π§ β 6; π₯ = 5π§ β 3
The solution is the set of the ordered points {(5π§ β 3, 8π§ β 7, π§), π§ β π }
1
2
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20. Echelon Method: case 2x3: Infinity of solutions (Surface)
Example1 Solve the following system of linear equation
π₯ β π¦ + π§ = 5
3π₯ β 3π¦ + 3π§ = 15
Solution 1 3π 1
β
π 2:
_____
π 2:
3π₯ β 3π¦ + 3π§ = 15
3π₯ β 3π¦ + 3π§ = 15
_____________________________________
0π₯ + 0π¦ + 0π§ = 0
The system of linear equation becomes: π 1: π₯ β π¦ + π§ = 4
π 2: 0 = 0
Infiniy of solution;
We consider y and z as known, and we use the row π 1 to get x, π₯ β π¦ + π§ = 5 β x = π¦ β π§ + 5;
The solution is the set of the ordered points {(y-z+5, π¦, π§); π¦, π§ β π }
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
21. Echelon Method: case 2x3: Infinity of solutions (Surface)
Example2 Solve the following system of linear equation
2π₯ β π¦ β π§ = 1
4π₯ β 2π¦ β 2π§ = 2
Solution 1 2π 1
β
π 2:
_____
π 2:
4π₯ β 2π¦ β 2π§ = 2
4π₯ β 2π¦ β 2π§ = 2
_____________________________________
0π₯ + 0π¦ + 0π§ = 0
The system of linear equation becomes: π 1: π₯ β π¦ + π§ = 4
π 2: 0 = 0
Infiniy of solution;
We consider x and z as known, and we use the row π 1 to get y, 2π₯ β π¦ β π§ = 1 β 2π₯ β π§ = π¦ + 1; β 2π₯ β π§ β 1 = π¦
The solution is the set of the ordered points {(x, 2π₯ β π§ β 1, π§); π₯, π§ β π }
1
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