Matakuliah
Tahun
: S0262-Analisis Numerik
: 2010
Linear Equation System
Pertemuan 4
Material Outline
• Linear equation system
– Gauss Elimination
– Gauss-Jordan Elimination
•
Linear Equations
The following equation is the general form of linear equation:
a1 x1  a2 x2    an xn  b
where: a1, a2,…, an, and b are constants and x1, x2,…, xn are
variables which also called as the unknowns.
•
Linear Systems
A linear system is a finite set of linear equations in the variables
x1, x2,…, xn .
A sequence s1, s2,… sn is a solution to linear system if x1= s1 ,
x2= s2 … xn =sn is a solution to every equation in the system.
28-Jun-16
DR. Paston Sidauruk
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•
System of Linear Equations (Linear system)
Note: Every linear system has either no
infinitely many solutions
Examples:
4 x1  x2  3 x3  1
solution, exactly one solution, or
A linear system with infinite many
solutions, i.e, x1=1, x2=2, and x3=-1
3 x1  x2  9 x3  4
x1  x2  3
A linear system with exactly one
x1  x2  1
x1  x2  4
x1  x2  3
solution that is: x1=2, and x2=1
A linear system with no solution
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•
A general form of a linear system
m equations in n unknowns
a11 x1  a12 x2    a1n xn  b1
a21 x1  a22 x2    a2 n x n  b2



am1 x1  am 2 x2    amn xn  bm
•
A linear system in Augmented Matrices
 a11
a
 21
 

am1
a12
a 22

am 2


a1n
a2 n

 amn
b1 
b2 


bm 
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•
Example
3 equations in 3 unknowns
x1  x2  2 x3  9
2 x1  4 x2  3 x 3  1
3 x1  6 x2  5 x3  0
Augmented matrix
1
2

3
1
2
4
6
3
5
9
1
0
The basic method of solving an LS is to replace the given system with a new system which is easier to
solve. The following steps are commonly used
1.
Multiply an equation by a non-zero constant
2.
Interchange two equations
3.
Add a multiple of one equation to another
Note: the rows in Augmented Matrix correspond to the equations in the system, the above steps can
be applied to AM in which equation replaced by row
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How to use row operations to find a solution of Linear System:
x1  x2  2 x3  9
2
9
1 1
2 4  3 1
2 x1  4 x2  3 x 3  1


3 x1  6 x2  5 x3  0

 3 6  5 0

1. Subtract 2 times 1st row to 2nd row and 3 times
1st row to the third row
x1  x2  2 x3  9
2
9 
1 1
0 2  7

2 x2  7 x 3  17

17


3x2  11x3  27

0 3  11  27

2. Multiply 2nd row by ½ etc., finally we obtain the following
1
x1
x2
2
x3  3
1
0


0
0
1
0
0
0
1
1
2

3

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•
Gauss Elimination(GE)
•
GE is a systematic procedure to solve LS by reducing the augmented matrix of given
LS to a simple augmented matrix that allow the solution of LS sought by inspection.
Procedure: Reduced the Augmented matrixrow-echelon form.
Row-echelon form:
1. The 1st non-zero number in the row is a 1 called a leading 1
2. All rows that consist entirely zero are put at the bottom of the matrix
3.
The leading 1 in the lower row occurs farther to the right of the leading 1 in
the higher row.
A matrix in reduced row-echelon form is a matrix in row-echelon form that each
column that contains a leading 1 has zero every where else.
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•
Matrices in row-echelon form
Examples:
1 4  3 7
0 1 6 2 ;


0 0 1 5
1 1 0
0 1 0 


0 0 0
•
Matrices in reduced row-echelon form
Examples:
1 0 0 4 
0 1 0 7 ;


0 0 1  1
1 0 0
0 1 0


0 0 1
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•
Steps in Gauss Elimination
1.
Interchange the top row with another row, if necessary, such that the top of the most left
column is non-zero entry
Suppose that the top of left most column entry is a non-zero entry then multiply the first row
to introduce a leading 1
Add suitable multiples of the top row to the rows below such that all entries in the column
below leading 1 are zero
Apply the steps 1, 2, and 3 to the remaining sub matrix to produce a matrix in row-echelon
form
2.
3.
4.
Note: Gauss Elimination will produce matrix in row-echelon form
Gauss Jordan elimination will produce matrix in reduced row-echelon form
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•
1.
2.
Steps in Gauss-Jordan Elimination
Steps 1, 2, 3, and 4 the same as steps 1, 2, 3, and 4 in Gauss
Elimination, respectively.
Beginning with the last nonzero row, add suitable multiples of each
row to the rows above to introduce all zeros above the leading 1
Note: Gauss Elimination will produce matrix in row-echelon form
Gauss Jordan elimination will produce matrix in reduced row-echelon form
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Note: 1. Gauss Elimination will produce matrix in row-echelon form
2. Gauss Jordan elimination will produce matrix in reduced rowechelon form.
3. The next step is after finding the row-echelon or reduced rowechelon form
matrix begin with the last row used the back
substitution to find the solution of the
given LS.
•
Example:
Find the solution of the following linear system using Gauss and Gauss-Jordan eliminations:
x  y  2z  9
2 x  4 y  3z  1
3x  6 y  5 z  0
x1  3 x2  2 x3  2 x5  0
2 x1  6 x2  5 x3  2 x4  4 x5  3 x6  1
5 x3  10 x4  15 x6  5
2 x1  6 x2  8 x4  4 x5  18 x6  6
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