B3001/UNIT9/1
Prepared by : Siti Sharmila Osmin Page 1
Unit
9
NUMERICAL METHOD
Understand and solve linear simultaneous
equations using numerical methods
At the end of this unit, students will be able to:
1. Solve linear simultaneous equations
using Gaussian Elimination
Specific Objectives
General Objectives
B3001/UNIT9/2
Prepared by : Siti Sharmila Osmin Page 2
9.0 INTRODUCTION
The objective of this unit is to introduce two more methods in order to solve simultaneous
equations by using Gaussian Elimination and LU Decomposition method.
9.1 GAUSSIAN ELIMINATION METHOD
Gaussian elimination is an algorithm that can be used to solve a system of linear equations.
Elementary row operations are used throughout the algorithm. In this unit we only discuss three-
equation linear systems.
The following is a set of 3 equations ;
a11x1 + a12x2 + a13x3 = b1
a21x1 + a22x2 + a23x3 = b2
a31x1 + a32x2 + a33x3 = b3
Then change the equation into matrix form
333231
232221
131211
aaa
aaa
aaa
3
2
1
x
x
x
=
3
2
1
b
b
b
INPUT
B3001/UNIT9/3
Prepared by : Siti Sharmila Osmin Page 3
Next, the Gaussian Elimination algorithm is applied to the augmented matrix of the system above
which, at the end of the first part of the algorithm looks like this:
''
33
'
23
'
21
131211
a00
aa0
aaa
3
2
1
x
x
x
''
3
'
2
1
b
b
b
The zero elements in this matrix are eliminated using Gaussian Elimination method. The process to
eliminate those elements is known as Transformation. There are two steps in Transformation. Only the
second and the third row of the matrix are involved in the first transformation. But, in the second
transformation process only the third row of the matrix is involved.
Transformation I:
0a
aaaa
11
211121
'
21
11
211222
'
22a
aaaa
11
211323
'
23a
aaaa
11
2112
'
2a
abbb
11
31
1131
'
31a
aaaa = 0
11
31
1232
'
32a
aaaa
11
31
1333
'
33a
aaaa
and
11
31
13
'
3a
abbb
B3001/UNIT9/4
Prepared by : Siti Sharmila Osmin Page 4
So, the results of the first are:
'
33
'
32
'
23
'
22
131211
0
0
aa
aa
aaa
3
2
1
x
x
x
'
3
'
2
1
b
b
b
Transformation II:
0a
aaaa
'
22
'
32'
22
'
32
''
32
'
22
'
32'
23
'
33
''
33a
aaaa
and
'
22
'
32'
2
'
3
''
3a
abbb
So, after the second transformation process the results are:
''
33
'
23
'
22
131211
a00
aa0
aaa
3
2
1
x
x
x
''
3
'
2
1
b
b
b
Then we will be able to get''33
''3
3a
bx . The value of 2x and 1x can be retrieved by substituting the
value of 3x and 2x in the linear equation that we get from the matrix equations.
''
33
''
3
3a
bx
'
22
3
'
23
'
2
2a
xabx
11
3132121
1a
xaxabx
B3001/UNIT9/5
Prepared by : Siti Sharmila Osmin Page 5
Example 9.1:
Solve this linear equation by using Gauss Elimination Method:
x + y + z = 8
3x + 2y + z = 49
5x – 3y + z = 0
Solution:
Step 1: Transform the linear equation form to matrix form
0
49
8
z
y
x
135
123
111
Step 2: Transformation I,
01
313
a
aaaa
11
211121
'
21
11
312
a
aaaa
11
211222
'
22
21
311
a
aaaa
11
211323
'
23
251
3849
a
abbb
11
2112
'
2
01
515
a
aaaa
11
31
1131
'
31
81
513
a
aaaa
11
31
1232
'
32
41
511
a
aaaa
11
31
1333
'
33
401
580
a
abbb
11
31
13
'
3
B3001/UNIT9/6
Prepared by : Siti Sharmila Osmin Page 6
Result for first Transformation I,
40
25
8
z
y
x
480
210
111
Step 3: Transformation II:
01
818
a
aaaa
'
22
'
32'
22
'
32
''
32
121
824
a
aaaa
'
22
'
32'
23
'
33
''
33
2401
82540
a
abbb
'
22
'
32'
2
'
3
''
3
Result for Transformation II,
240
25
8
z
y
x
1200
210
111
So, 240z12
2012
240z
25z2y
25202y
15y
8zyx
zy8x
20158x
13x
So, x = 13, y = 15 and z = -20
B3001/UNIT9/7
Prepared by : Siti Sharmila Osmin Page 7
Example 9.2:
Solve this linear equation using Gauss Elimination Method:
x + 3y + 3z = 4
2x –3y –2z= 2
3x + y + 2z = 5
Solution:
Step 1: Transform the linear equation to matrix form
z
y
x
=
5
2
4
Step 2: Transformation I,
01
212
a
aaaa
11
211121
'
21
91
233
a
aaaa
11
211222
'
22
81
232
a
aaaa
11
211323
'
23
61
242
a
abbb
11
2112
'
2
01
313
a
aaaa
11
31
1131
'
31
81
331
a
aaaa
11
31
1232
'
32
71
332
a
aaaa
11
31
1333
'
33
71
345
a
abbb
11
31
13
'
3
213
232
331
B3001/UNIT9/8
Prepared by : Siti Sharmila Osmin Page 8
Result for Transformation I,
7
6
4
z
y
x
780
890
331
Step 3: Transformation II:
09
898
a
aaaa
'
22
'
32'
22
'
32
''
32
9
1
9
887
a
aaaa
'
22
'
32'
23
'
33
''
33
9
15
9
8)6(7
a
abbb
'
22
'
32'
2
'
3
''
3
Result for Transformation II,
915
6
4
z
y
x
9100
890
331
So, 9
15z
9
1
1599
15z
6z8y9
z86y9
9
)15(86y
14y
4z3y3x
z3y34x
B3001/UNIT9/9
Prepared by : Siti Sharmila Osmin Page 9
153)14(34x
7x
So, x = 7, y = 14 and z = -15.
B3001/UNIT9/10
Prepared by : Siti Sharmila Osmin Page 10
ACTIVITY 9a
9a.1 Solve this linear equation using Gauss Elimination Method:
a) x + 2y – 3z = 3
2x – y – z = 11
3x + 2y + z = -5
b) x –4y – 2z = 21
2x + y + 2z = 3
3x + 2y – z = -2
B3001/UNIT9/11
Prepared by : Siti Sharmila Osmin Page 11
ANSWER 9a
9a.1 a) x = 2
y = -4
z = -3
b) x = 3
y = -5
z = 1
B3001/UNIT9/12
Prepared by : Siti Sharmila Osmin Page 12
SELF ASSESSMENT
9.1 Solve this linear equation by using Gauss Elimination Method:
a- 3x + 6y –2z = 8
2y –x + 4z = 21
x + y – 4z = 14
b- 2t + 3s – u = 3
t – 3s + 3u = 15
-2t – 5s + 3u = -2
B3001/UNIT9/13
Prepared by : Siti Sharmila Osmin Page 13
ANSWER
9.1 a) x =1.5
y = 4
z = -2
b) t = -1.6
s = 2.4
u = 1.8
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