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EE 216 Class Notes
Pages 1 of 21
Numerical Solutions of Linear Systems of Equations
Linear Dependence and Independence An equation in a set of equations is linearly independent if it cannot be generated by any linear combination of the other equations. If an equation in a set of equations can be generated by a linear combination of the other equations then it is called a dependent equation. In order to have a unique solution for a set of unknowns, the number of independent equations must be at least equal to the number of unknowns. Therefore, to solve for three unknowns in a unique way, we need at least three independent equations. Let us consider the following set of equations.
2x - 4y + 5z = 36 … … (1)
- 3x + 5y + 7z = 7 … … (2)
4x - 8y + 10z = 72 … … (3)
It can be easily noticed that Eqn. (3) can be generated by multiplying Eqn. (1) by 2. Therefore, Eqn. (3) is a linearly dependent equation. Let us consider the following set of equations.
2x - 4y + 5z = 36 … … (1)
- 3x + 5y + 7z = 7 … … (2)
- x + y + 12z = 43 … … (3)
In the above set, Eqn. (3) can be generated by adding Eqn. (1) to Eqn. (2). Therefore, Eqn. (3) is a dependent equation. Let us consider the following set of equations.
2x – 4y + 5z = 36 … … (1)
- 6x + 8y + 43z = 136 … … (2)
- 3x + 5y + 7z = 7 … … (3)
EE 216 Class Notes
Pages 2 of 21
In the above set of equations, Eqn. (2) can be generated by adding 3 times of Eqn. (1) to 4 times of Eqn. (3).
3 X Eqn. (1): 6x – 12y + 15z = 108
4 X Eqn. (3): -12x + 20y + 28z = 28
- 6x + 8y + 43z = 136 Therefore, Eqn. (2) is a dependent equation. Dependent equations do not add any new information to the solution process. Let us consider the following set of equations.
2x - 4y + 5z = 36 … … (1)
- 3x + 5y + 7z = 7 … … (2)
5x + 3y - 8z = - 31 … … (3) All three equations in the above set are linearly independent with respect to each other. These three equations should provide a unique solution for the three unknowns: x, y and z. The solution is:
x = 2 y = -3 z = 4
EE 216 Class Notes
Pages 3 of 21
GAUSS ELIMINATION The operations in the Gauss elimination are called elementary operations. Elementary operations for equations are:
(I) Interchange of two equations.
(II) Multiplication of an equation by a nonzero constant.
(III) Addition of a constant multiple of one equation to another equation.
Let us consider the set of linearly independent equations.
2x - 4y + 5z = 36 … … (1)
- 3x + 5y + 7z = 7 … … (2)
5x + 3y - 8z = - 31 … … (3) Augmented matrix for the set is:
−−−−−−−−−−−−
−−−−
31835775336542
Step 1
Eliminate x from the 2nd and 3rd equation.
212 RR23R ++++→→→→′′′′
313 RR25R ++++−−−−→→→→′′′′
EE 216 Class Notes
Pages 4 of 21
Step 2
Eliminate y from the 3rd equation.
323 RR13R ++++→→→→′′′′
From Row 3, 168z = 672 Therefore, z = 672/168 = 4
From Row 2, -y + 14.5z = 61
or, - y + 14.5 (4) = 61
or, y = - 3 From Row 1, 2x – 4y + 5z = 36
or, 2x – 4 (- 3) + 5 (4) = 36
or, x = 2
Let us consider another set of linearly independent equations.
- 4x + 3y + 2z = - 5 … … (1)
−−−−−−−−−−−−−−−−
1215.20130615.141036542
2x – 4y + 5 = 36 - y + 14.5z = 61 13y –20.5z = - 121
−−−−−−−−
67216800615.141036542
2x – 4y + 5 = 36 - y + 14.5z = 61
168z = 672
EE 216 Class Notes
Pages 5 of 21
5x + y - 3z = 32 … … (2)
3x - 4y + z = - 15 … … (3)
The augmented matrix for this set is:
−−−−−−−−−−−−
−−−−−−−−
1514332315
5234
Step 1 Eliminate x from the 2nd and 3rd equation.
212 R4R5R ++++→→→→′′′′ 313 R4R3R ++++→→→→′′′′
Step 2 Eliminate y from the 3rd equation.
323 R19R7R ++++→→→→′′′′
- 4x + 3y + 2z = - 5 19y - 2z = 103
- 7y + 10z = - 75
−−−−−−−−−−−−
−−−−−−−−
7510701032190
5234
−−−−−−−−
−−−−−−−−
704176001032190
5234
- 4x + 3y + 2z = - 5 19y - 2z = 103
176z = - 704
EE 216 Class Notes
Pages 6 of 21
From Row 3, 176z = -704 Therefore, z = - 704/176 = - 4 From Row 2, 19y – 2z = 103 or, 19y – 2 (- 4) = 103 or, y = 5 From Row 1, - 4x + 3y + 2z = -5 or, - 4x + 3 (5) + 2 (- 4) = -5 or, x = 3 Elementary operations for equations and matrices are:
(I) Interchange of two equations.
(II) Multiplication of an equation by a nonzero constant.
(III) Addition of a constant multiple of one equation to another equation.
When would you interchange two equations (rows)?
Let us consider the following set of equations.
3y + 2z = 16 … … (1)
4x + 2y - 3z = - 10 … … (2)
3x + 4y + z = 9 … … (3)
The corresponding augmented matrix is:
−−−−−−−−9143
1032416230
EE 216 Class Notes
Pages 7 of 21
Eqn. (1) (Row 1) cannot be used to eliminate x from Eqns. (2) and (3) (Rows 2 and 3).
Interchange Row 1 with Row 2.
The augmented matrix becomes:
−−−−−−−−
91431623010324
Now follow the steps mentioned earlier to solve for the unknowns.
The solution is:
x = - 3
y = 4
z = 2
Let us consider the following set of linear equations.
Gauss elimination suggests doing elementary row operations from top to bottom. A slightly modified form, known as Gauss-Jordan elimination suggests doing elementary row operations from bottom to top as well.
2x1 – 4x2 + 5x3 = 36 … … (1)
- 3x1 + 5x2 + 7x3 = 7 … … (2)
5x1 + 3x2 – 8x3 = - 31 … … (3)
EE 216 Class Notes
Pages 8 of 21
−−−−−−−−
−−−−
−−−−
31
7
36
835
753
542
313212 25
23 RRRRRR ++++−−−−→→→→′′′′++++→→→→′′′′
−−−−
−−−−
−−−−
121-
61
36
241130
22910
542
33 131 RR →→→→′′′′
−−−−
−−−−
−−−−
121/13-
61
36
264110
22910
542
323 RRR ++++→→→→′′′′
−−−−
−−−−
672/13
61
36
1316800
22910
542
33 16813 RR →→→→′′′′
−−−−
−−−−
4
61
36
100
22910
542
322311 229 5 RRRRRR −−−−→→→→′′′′−−−−→→→→′′′′
−−−−
−−−−
4
3
16
100
010
042
EE 216 Class Notes
Pages 9 of 21
211 4RRR −−−−→→→→′′′′
−−−−
4
3
4
100
010
002
)...()...()...(
34 2314 2
3
2
1
========−−−−====
xx
x
Definitions
Homogeneous system
3y + 2z = 0 … … (1)
4x + 2y - 3z = 0 … … (2)
3x + 4y + z = 0 … … (3)
Notice the values on the right hand side of each equation.
Non-homogeneous system
3y + 2z = 16 … … (1)
4x + 2y - 3z = 0 … … (2)
3x + 4y + z = 0 … … (3)
At least one equation with a nonzero element on the right hand side makes a system non-homogeneous. If a system is homogeneous, it has at least the trivial solution of x = 0, y = 0 and z = 0.
EE 216 Class Notes
Pages 10 of 21
SYSTEMS OF LINEAR EQUATIONS Solution by Cramer’s Rule
Consider the following set of linear equations.
1313212111 bxaxaxa ====++++++++ … … (1)
2323222121 bxaxaxa ====++++++++ … … (2)
3333232131 bxaxaxa ====++++++++ … … (3)
The system of equations above can be written in matrix form as:
====
3
2
1
3
2
1
333231
232221
131211
bbb
xxx
aaaaaaaaa
[A][x] = [B]
where
[[[[ ]]]]
====
333231
232221
131211
Aaaaaaaaaa
[[[[ ]]]] [[[[ ]]]]
====
====
3
2
1
3
2
1
Bbbb
andxxx
x
If 0≠≠≠≠D , then the system has a unique solution as shown below (Cramer’s Rule).
DDx
DDx
DDx 3
32
21
1 and, ============
EE 216 Class Notes
Pages 11 of 21
where
333231
232221
131211
aaaaaaaaa
D ====
33323
23222
13121
1
aabaabaab
D ====
33331
23221
13111
2
abaabaaba
D ==== 33231
22221
11211
3
baabaabaa
D ====
Let us consider the following set of linear equations.
2x1 – 4x2 + 5x3 = 36 … … (1)
- 3x1 + 5x2 + 7x3 = 7 … … (2)
5x1 + 3x2 – 8x3 = - 31 … … (3)
[A][x] = [B]
where
[[[[ ]]]]
−−−−−−−−
−−−−====
835753542
A
[[[[ ]]]] [[[[ ]]]]
−−−−====
====317
36B
3
2
1
andxxx
x
EE 216 Class Notes
Pages 12 of 21
336835753542
−−−−====−−−−
−−−−−−−−
====D
67283317575436
1 −−−−====−−−−−−−−
−−−−====D
100883157735362
2 ====−−−−−−−−
−−−−====D
13443135753
3642
3 −−−−====−−−−
−−−−−−−−
====D
23366721
1 ====−−−−−−−−========
DDx
3336
100822 −−−−====
−−−−========
DDx
4336
134433 ====
−−−−−−−−========
DDx
EE 216 Class Notes
Pages 13 of 21
Solve the following set of linear equations.
x1 + x2 + x3 = 2 … … (1)
2x1 + 5x2 + 3x3 = 11 … … (2)
- x1 + 2x2 + x3 = 3 … … (3)
Solution by Matrix Inversion
Consider the following set of linear equations.
1313212111 bxaxaxa ====++++++++ … … (1)
2323222121 bxaxaxa ====++++++++ … … (2)
3333232131 bxaxaxa ====++++++++ … … (3)
The system of equations above can be written in matrix form as:
====
3
2
1
3
2
1
333231
232221
131211
bbb
xxx
aaaaaaaaa
[A][x] = [B]
where
[[[[ ]]]]
====
333231
232221
131211
Aaaaaaaaaa
EE 216 Class Notes
Pages 14 of 21
[[[[ ]]]] [[[[ ]]]]
====
====
3
2
1
3
2
1
Bbbb
andxxx
x
If 0≠≠≠≠D , then the system has a unique solution.
[x] = [A]-1[B]
A Traffic Light Assembly
6 m
3 m
6 m
3 m
3 m
4 m
4 m
4 m4 m
x
y
h
z
A
B
C
D
Figure 1.1 A traffic light assembly
The mass of the traffic light assembly is 20 kg and h = 3.5 m. For equilibrium, the forces acting on the light assembly in all three directions (x, y and z) should equal to zero.
0=∑ xF , 0=∑ yF and 0=∑ zF
EE 216 Class Notes
Pages 15 of 21
Based on the above, a set of equations in terms of the three unknown forces can be written as:
07960.08381.05970.0 ====++++−−−− ADACAB FFF … … (1)
05970.04191.07960.0 ====−−−−−−−− ADACAB FFF … … (2)
2.1960995.03492.00995.0 ====++++++++ ADACAB FFF … … (3)
[A][F] = [B]
where
[[[[ ]]]]
−−−−−−−−−−−−
====0995.03492.00995.05970.04191.07960.07960.08381.05970.0
A
[[[[ ]]]] [[[[ ]]]]
====
====2.19600
BandFFF
F
AD
AC
AB
[[[[ ]]]] [[[[ ]]]]B]A[ 1−−−−====F
[[[[ ]]]]
−−−−−−−−−−−−====
−−−−
8867.06207.06799.01056.20421.02948.07737.17685.03547.0
A1
[[[[ ]]]]
−−−−−−−−−−−−====
2.19600
8867.06207.06799.01056.20421.02948.07737.17685.03547.0
F
EE 216 Class Notes
Pages 16 of 21
[[[[ ]]]]
====
====978.173126.413997.347
AD
AC
AB
FFF
F
Therefore, FAB = 347.997 N, FAC = 413.126 N and FAD = 173.978 N
An Electrical Circuit
Based on the principles of voltage drops (Kirchoff’s voltage law) and current flow
(Kirchoff’s current law), the following equations can be written in terms of the unknown current flow.
I1 I2
I3
20 ohms 15 ohms
12 ohms80 V
+
-120 V
+
-
Figure 2.1 An electrical circuit
801220 31 ====++++ II … … (2.1)
1201215 32 ====++++ II … … (2.2)
0321 ====−−−−++++ III … … (2.3)
The set of equations can be solved by substitution method in the following manner.
EE 216 Class Notes
Pages 17 of 21
[A][C] = [B]
where
[[[[ ]]]]
−−−−====
1111215012020
A
[[[[ ]]]] [[[[ ]]]]
====
====0
12080
B
3
2
1
andIII
C
[[[[ ]]]] [[[[ ]]]]B]A[ 1−−−−====C
[[[[ ]]]]
−−−−−−−−
−−−−====
−−−−
4167.00278.00208.03333.00444.00167.02500.00167.00375.0
A1
[[[[ ]]]]
−−−−−−−−
−−−−====
0120
80
4167.00278.00208.03333.00444.00167.02500.00167.00375.0
C
[[[[ ]]]]
====
====541
3
2
1
III
C
Therefore, I1 = 1 A, I2 = 4 A and I3 = 5 A.
EE 216 Class Notes
Pages 18 of 21
Flow Through a Piping Network
A
B
C
D
0.03 m3/s
0.06 m3/s
0.09 m3/sd = 0.5 mL = 150 m
d = 0.5 mL = 150 m
d = 0.5 mL = 300 m
d = 0.5 mL = 300 m
d = 0.5 mL = 300 m
d = 0.75 mL = 300 m
d = 0.75 mL = 300 m
d = 0.75 mL = 300 m
Figure 3.1 A piping network
A set of equations can be written based on the following principles:
1. At each junction, flow in = flow out.
2. Kinetic energy and potential energy of the fluid makes the fluid move through the piping network.
3. For a given fluid flow, the velocity increases and the pressure decreases when the pipe diameter is decreased.
4. Frictional loss is directly proportional to the length of the pipe and the viscosity of the fluid and inversely proportional to the diameter of the pipe.
A set of equations as functions of the five unknown flow rates can be written as:
03.0====++++ ACAB QQ … … (3.1)
06.0====++++++++−−−− CDCBAC QQQ … … (3.2)
0====−−−−++++ BDCBAB QQQ … … (3.3)
09.39.299.29 ====−−−−−−−− CBACAB QQQ … … (3.4)
09.199.79.3 ====++++−−−− BDCDCB QQQ … … (3.5)
EE 216 Class Notes
Pages 19 of 21
It should be noted that Eqns. (3.1) – (3.3) have been derived based on the principle of conservation of mass. Eqns. (3.4) and (3.5) have been derived based on the principle of pressure drop along pipe sections. The set of equations can be solved by substitution method in the following manner.
[A][Q] = [B]
where
[[[[ ]]]]
−−−−−−−−−−−−
−−−−−−−−
====
9.199.79.300009.39.299.29101010111000011
A
[[[[ ]]]] [[[[ ]]]]
====
====
000
06.003.0
Band
QQQQQ
Q
BD
CD
CB
AC
AB
[[[[ ]]]] [[[[ ]]]]B]A[ 1−−−−====Q
EE 216 Class Notes
Pages 20 of 21
[[[[ ]]]]
−−−−−−−−−−−−−−−−
−−−−−−−−−−−−−−−−−−−−−−−−
====−−−−
0318.00019.03675.02511.03093.00318.00019.06325.07489.06907.00298.00139.05938.02357.0179.00019.00158.00387.00154.05117.00019.00158.00387.00154.04883.0
A1
[[[[ ]]]]
−−−−−−−−−−−−−−−−
−−−−−−−−−−−−−−−−−−−−−−−−
====
000
06.003.0
0318.00019.03675.02511.03093.00318.00019.06325.07489.06907.00298.00139.05938.02357.0179.00019.00158.00387.00154.05117.00019.00158.00387.00154.04883.0
Q
[[[[ ]]]]
====
====
0243.00657.00087.00144.00156.0
BD
CD
CB
AC
AB
QQQQQ
Q
Therefore,
QAB = 0.0156 m3/s, QAC = 0.0144 m3/s
QCB = 0.0087 m3/s, QCD = 0.0657 m3/s
QBD = 0.0243 m3/s
EE 216 Class Notes
#21
Solve the following set of linear equations.
x1 + x2 + x3 = 2 … … (1)
2x1 + 5x2 + 3x3 = 11 … … (2)
- x1 + 2x2 + x3 = 3 … … (3)
Exercise Sets Solve the following sets of linear equations by Gaussian elimination, Gauss-Jordan elimination, Cramer’s rule and matrix inversion.
(1)
11423252
84
321
321
321
=++=++
=+−
xxxxxx
xxx
(2)
3133
23
21
321
321
=+−=+−
=+−
xxxxx
xxx
(3)
8221
851
41
31
961
51
41
321
321
321
=++
=++
=++
xxx
xxx
xxx
(4)
534
632
321
321
321
=−+−=−+
=+−
xxxxxx
xxx
(5)
1051262432
321
321
321
=+−=+−=−+
xxxxxxxxx
(6)
453303
12
321
321
321
=++=++
−=+−
xxxxxxxxx