1. System of Linear Equation (SLE)
Sub Chapter
Introduction
Solution of SLE using ERO
Solution of SLE using Inverse matrix and
Cramer’s Rule
Homogeneous SLE
Some Applications
Electrical system
Computer Network
Economy Model
etc.
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2. The problem can be expressed x + 2y = 5000 3x + y = 10000
Introduction
Linear Equation is mathematical expression where the variable does not contain exponential, trygonometri (as sin, cos, dll.), or multiplication with another variable or itself.
Example :
If Ady buy 1 Laptop (x) and 2 PC (y) then he have to pay $ 5000, if he buy 3 Laptop and 1 PC then he have to buy $
The problem can be expressed
x + 2y = 5000
3x + y = 10000
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3. Generally for of system of linear equation
In the multiplication of matrix :
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4. Let SLE Or AX = B where Example : A is said coeffisien matrix
X is said variable matrix
B is said constant matrix
Example :
Let SLE
x + 2y = 5000
3x + y = 10000
can be expressed in the form
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5. {x = 3000, y =1000 } is a solution of SLE
We have SLE :
x + 2y = 5000
3x + y = 10000
Then
{x = 3000, y =1000 } is a solution of SLE
{x = 1000, y =3000 } is not a solution of SLE
A SLE has three possibility about solution :
SLE has only one solution
SLE has infinitely many solution
SLE has no solution (inconstent)
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6. Ilustration of Solution in Cartesius
SLE 2x – y = x – y = 0
has one solution, i.e : x = 2, y = 2 x y
y = 2x - 2
(2, 2) is crossing point of each line
y = x 2
(2, 2) 1 2
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7. It can be expressed in cartesius
SLE
x – y = 0
2x – 2y = 2
It can be expressed in cartesius
There is no crossing point
It is mean that
SLE has no solution y
y = x
y = x – 1 x 1
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8. SLE
x – y = 0
2x – 2y = 0
2nd equation is twice of 1st equation
It can be expressed in cartesius
So,
SLE has infinitely many solution y
x – y = 0
2x – 2y = 0 x
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9. Get a solution of SLE using ERO
SLE can be expressed in the form augmented matrix
Using ERO to be reduced echelon row
Example :
Find a solution of SLE
3x – y = 5
x + 3y = 5
Answer :
Augmented Matrix of SLE
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10. Solution of SLE is x = 2 and y = 1 Example : a. a + c = 4 a – b = –1
SLE expresion is
Solution of SLE is x = 2 and y = 1
Example :
Determine a solution (if possible) for SLE :
a. a + c = 4
a – b = –1
2b + c = 7
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11. b. a + c = 4  a – b = –1 –a + b = 1 c. a + c = 4 –a + b = 2 Jawab :
Solution of SLE is
a = 1, b = 2, and c =3
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12.  b. or We have a + c = 4 dan b + c = 5.
Let c = t, where t is a parameter.
Solution of SLE is :
, where t is a parameter
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13. We can see on 3rd row c.  we have 0.a + 0.b + 0.c = 1.
It is not possible!
Then there is no a, b, and c
So, SLE has no solution.
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14. EXERCISES
Solve system of linear equations below
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15. Determine a such that SLE : a. Has one solution b. Has no solution
x + 2y – 3z = 4
3x – y + 5z = 2
4x + y + (a2 – 14) z = a+2
Determine a such that SLE :
a. Has one solution
b. Has no solution
c. Has infinitely many solution
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16. Augmented Matrix of SLE
Answer :
Augmented Matrix of SLE
a. SLE has one solution if:
a2 – 16  0 sehingga a   4
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17. SLE has no solution when a2 – 16 = 0 dan a– 4  0
b. Focus on 3rd row
0x + 0y + (a2 – 16a) z = a – 4
SLE has no solution when
a2 – 16 = 0 dan a– 4  0
such a =  4 and a  4.
so , a = – 4.
c. SLE has infinitely any solution
a2 – 16 = and a – 4 = 0
So , a = 4
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18. Get Solution SLE using inverse Matrixor
AX = B
If A is invertible then
A–1 A X = A–1 B
We have :
X = A–1 B
Remember that
A is invertible if and only if Det (A)  0.
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19. Determine solution of SLE : a + c = 4 a – b = –1 2b + c = 7 Answer :
Example :
Determine solution of SLE :
a + c = 4
a – b = –1
2b + c = 7
Answer :
We know
So, we have Inverse matrix
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20. X = A–1 B :
so, Solution of SLE
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21. HOMOGENEOUS LINEAR EQUATIONS SYSTEM
A system of linear equations is said to be homogeneous if the constant term are all zero; that is, the system has the form
Every homogeneous system is consistent
All system has x1=0,x2=0,…,xn=0 as a solution  trivial solution
Denoted by
If there are other solutions, they are called nontrivial solutions
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22. HOMOGENEOUS LINEAR EQUATIONS SYSTEM
There are two only possibilities for homogeneous linear system’s solutions
The system has only the trivial solution
The system has infinitely many solutions in addition to the trivial solution
Example
Solve the homogeneous system of linear equations
Augmented matrix
Reduced row-echelon
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23. HOMOGENEOUS LINEAR EQUATIONS SYSTEM
The solutions is
infinitely many solutions / nontrivial solutions
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24. EXERCISES 2p + q – 2r – 2s = 0 p – q + 2r – s = 0 –p + 2q – 4r + s = 0
Find solutions of homogeneous linear system Ax = 0 where
Find solutions of homogeneous linear system below
2p + q – 2r – 2s = 0
p – q + 2r – s = 0
–p + 2q – 4r + s = 0
3p – 3s = 0
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25. Get solution of SLE using Cramer’s Rule Let SLE is :
The following sequence step for Cramer’s Rule :
Find determinant of A (If det(A)=0, we cancel it)
Determine Ai  matrix from A where ith column is replaced by B.
Example:
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26. Find b of SLE : Calculate |Ai| Solution of SLE for variable xi is
Example :
Find b of SLE :
a + c = 4
a – b = –1
2b + c = 7
Answer :
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27. So Solution of b is 2
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28. Find a solution of a ?
Find a solution of c ?
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29. Example : Look the ilustration :  a b   c Show that :
a2 = b2 + c2 – 2bc cos
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30. Answer : we have : c cos + b cos = a c cos + a cos = b
b cos + a cos = c or
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31. Using Cramer’s rule :
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32. So, we have :
a2 = b2 + c2 – 2bc cos
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33. 1. Determine solution of SLE :
Exercise
1. Determine solution of SLE :
2. Determine solution of SLE :
2p – 2q – r + 3s = 4
p – q + 2s = 1
–2p +2q – 4s = –2
3. Determine solution of Homogeneous SLE :
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34. 5. Let 4. Let SLE AX = B Determine Solution of SLE using: ERO
Inverse matrix
Cramer’s rule
5. Let
Find
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35. 6. Let HSLE (with variable p, q, and r)
Determine k such that SLE has one solution
7. Let
Determine a not zero vector such that
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