![1.5. SOLUTION OF LINEAR SYSTEM OF EQUATIONS
CHAPTER 1. NUMERICAL SOLUTIONS FOR
LINEAR SYSTEM OF EQUATIONS
1.5.4 Iterative Techniques
Consider the linear system of equations,
a11x1 + a12x2 + a13x3 + ... + a1nxn = b1
a21x1 + a22x2 + a23x3 + ... + a2nxn = b2
a31x1 + a32x2 + a33x3 + ... + a3nxn = b3
.
.
.
an1x1 + an2x2 + an3x3 + ... + annxn = bn
Rewrite it as,
x1 =
1
a11
[b1 − a12x2 − a13x3 − ... − a1nxn]
x2 =
1
a22
[b2 − a21x1 − a23x3 − ... − a2nxn]
x3 =
1
a33
[b3 − a31x1 − a32x2 − ... − a3nxn]
.
.
.
xn =
1
ann
bn − an1x1 − an2x2 − ... − an(n−1)xn−1
In general,
xn =
1
ann
bn −
n
i=1
anixi ; ∀ i = n
To get a solution with this method we should satisfy this conditions:
1. ann = 0 ∀ n ; existence condition.
2. A matrix should be diagonally dominant, |aii| > n
j=1 |aij| ∀ j; j = i ; convergence condition.
Methods of solution:
1. Jacobi method
We start the solution by an initial vector,
x(0)
= x
(0)
1 x
(0)
2 x
(0)
3 ... x
(0)
n
T
Then get the fist iteration values,
x
(1)
1 =
1
a11
b1 − a12x
(0)
2 − a13x
(0)
3 − ... − a1nx(0)
n
x
(1)
2 =
1
a22
b2 − a21x
(0)
1 − a23x
(0)
3 − ... − a2nx(0)
n
x
(1)
3 =
1
a33
b3 − a31x
(0)
1 − a32x
(0)
2 − ... − a3nx(0)
n
.
.
.
x(1)
n =
1
ann
bn − an1x
(0)
1 − an2x
(0)
2 − ... − an(n−1)x
(0)
n−1
Mohamed Mohamed El-Sayed Atyya Page 7 of 13](https://image.slidesharecdn.com/numericalsolutionsforlinearsystemofequations-190107185809/85/Numerical-solutions-for-linear-system-of-equations-7-320.jpg)
![1.5. SOLUTION OF LINEAR SYSTEM OF EQUATIONS
CHAPTER 1. NUMERICAL SOLUTIONS FOR
LINEAR SYSTEM OF EQUATIONS
Iteration No. x y z
0 0 0 0
1 1.714286 1.678572 0.845238
2 0.753401 1.015731 1.025652
3 0.985929 0.983656 1.003379
4 1.006039 0.999820 0.999349
5 1.000263 1.000391 0.999927
6 0.999853 0.999999 1.000016
7 0.999996 0.999991 1.000001
8 1.000004 1.000001 0.999999
9 1.000000 1.000001 1.000000
| error| 0.000004 0.000000 0.000001
1.5.5 Choleski Decomposition (Square-Root) Method for Positive Definite System of
Matrices
Theorem: A symmetrical n × n matrix is positive definite if and only if det(Ak) > 0 ;
k = 1, 2, 3, ..., n. where Ak is the k × k matrix formed by the kth
first row and
column of matrix A.
Theorem: For a symmetric positive definite matrix A, |aij|2
≤ |aiiajj| and the maximum
element of A lies on the main diagonal.
Theorem: Let A be a symmetric positive definite matrix, there is a unique upper-triangular
matrix R with the diagonal elements, such that RT
R = A or LLT
= A. Where
L = lower-triangular matrix, L = RT
.
A = LLT
a11 a12 a13 ... a1n
a21 a22 a23 ... a2n
. . . .
. . . .
. . . .
an1 an2 an3 ... ann
=
l11
l21 l22
. . .
. . .
. . .
ln1 ln2 . . . lnn
l11 l21 . . . ln1
l22 . . . ln2
. .
. .
. .
lnn
l11 =
√
a11
lij =
1
lji
aij −
j−1
k=1
likljk ; j = 1, 2, 3, ..., i − 1
lii = aii −
i−1
k=1
l2
ik ; i = 1, 2, 3, ..., n
li−1, j = 0 ; j = i, i + 1, ..., n
Also,
A = RT
R
AX = B ⇒ RT
RX = B
RT
[RX − D] = AX − B ⇒ RT
D = B , RX = D
Mohamed Mohamed El-Sayed Atyya Page 10 of 13](https://image.slidesharecdn.com/numericalsolutionsforlinearsystemofequations-190107185809/85/Numerical-solutions-for-linear-system-of-equations-10-320.jpg)
![1.5. SOLUTION OF LINEAR SYSTEM OF EQUATIONS
CHAPTER 1. NUMERICAL SOLUTIONS FOR
LINEAR SYSTEM OF EQUATIONS
r13 =
a13
Pivot 1
r22 =
b11
Pivot 2
r23 =
b12
Pivot 2
r33 = Pivot 3
1.5.6 LU Decomposition Using Gauss Elimination
Consider a linear system of equations
AX = B
Which A can be expressed as
A = LU
Then we get
LUX = B
L [UX − D] = AX − B ⇒ LD = B , UX = D
Example
Consider the following linear system of equations
x + 6y + 5z = 28
2x + 4y + 9z = 37
3x + 7y + 8z = 41
A B Check Raw operations
1 6 5
2 4 9
3 7 8
28
37
41
−16
−22
−23
−2 × R1 + R2 → R2
−3 × R1 + R3 → R3
1 6 5
0 −8 −1
0 −11 −7
28
−19
−43
−16
10
25
−
11
8
× R2 + R3 → R3
1 6 5
0 −8 −1
0 0 −45/8
28
−19
−135/8
−16
10
45/4
U =
1 6 5
0 −8 −1
0 0 −45/8
L =
1 0 0
2 1 0
3 11/8 1
LD = B ⇒ D = 28 −19 −135/8
T
UX = D ⇒ X = 1 2 3
T
Note: As we see the raw operation in general is aij × Rj + Ri → Ri, this used to generate L matrix as lij = −aij
Mohamed Mohamed El-Sayed Atyya Page 12 of 13](https://image.slidesharecdn.com/numericalsolutionsforlinearsystemofequations-190107185809/85/Numerical-solutions-for-linear-system-of-equations-12-320.jpg)
Numerical solutions for linear system of equations
- 1. Contents 1 NUMERICAL SOLUTIONS FOR LINEAR SYSTEM OF EQUATIONS 2 1.1 Linear System of Equations in General Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.2 Linear System of Equations in Matrix Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.3 Homogeneous and Non-Homogeneous Linear System of Equations . . . . . . . . . . . . . . . . . . . . . . . 2 1.4 Consistency of Linear System of Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.5 Solution of Linear System of Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.5.1 Gauss Elimination Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.5.2 Gauss-Jordan Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.5.3 Inverse of A if Exists . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.5.4 Iterative Techniques . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.5.5 Choleski Decomposition (Square-Root) Method for Positive Definite System of Matrices . . . . . . 10 1.5.6 LU Decomposition Using Gauss Elimination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 1.6 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 1.7 Contacts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 1
- 2. Chapter 1 NUMERICAL SOLUTIONS FOR LINEAR SYSTEM OF EQUATIONS 1.1 Linear System of Equations in General Form a11x1 + a12x2 + a13x3 + ... + a1nxn = b1 a21x1 + a22x2 + a23x3 + ... + a2nxn = b2 a31x1 + a32x2 + a33x3 + ... + a3nxn = b3 . . . an1x1 + an2x2 + an3x3 + ... + annxn = bn 1.2 Linear System of Equations in Matrix Form AX = B A = a11 a12 a13 ... a1n a21 a22 a23 ... a2n a31 a32 a33 ... a3n . . . . . . . . . . . . an1 an2 an3 ... ann , X = x1 x2 x3 . . . xn , B = b1 b2 b3 . . . bn 1.3 Homogeneous and Non-Homogeneous Linear System of Equations • If B = 0 ⇒ Homogeneous linear system of equations. • If B = 0 ⇒ Non-homogeneous linear system of equations. 1.4 Consistency of Linear System of Equations • If AX = B has no solution ⇒ Inconsistent linear system of equations. • If AX = B has at least one solution ⇒ Consistent linear system of equations. – The linear system of equations has one solution if the number of unknowns is equal to the number of indepen- dent equations (A has inverse or |A| = 0). 2
- 3. 1.5. SOLUTION OF LINEAR SYSTEM OF EQUATIONS CHAPTER 1. NUMERICAL SOLUTIONS FOR LINEAR SYSTEM OF EQUATIONS – The linear system of equations has more than one solution if the number of unknowns is greater than the number of independent equations(A has not inverse or |A| = 0). NOTE: Any homogeneous linear system of equations is a consistent linear system of equations. 1.5 Solution of Linear System of Equations 1.5.1 Gauss Elimination Method 1. Backward substitution (Upper triangular system) × × × × × × × × A × × × × × × × × × × B × × i=n i=1 a1i − b1 i=n i=1 a2i − b2 Check i=n i=1 a3i − b3 i=n i=1 a4i − b4 Raws operations ↓ ↓ ↓ ... × × × × 0 × × × 0 0 × × 0 0 0 × × × × × Then apply backward substitution to get the unknowns. Example Consider the following linear system of equations x + 6y + 5z = 28 2x + 4y + 9z = 37 3x + 7y + 8z = 41 A B Check Raws operations 1 6 5 2 4 9 3 7 8 28 37 41 −16 −22 −23 −2 × R1 + R2 → R2 −3 × R1 + R3 → R3 1 6 5 0 −8 −1 0 −11 −7 28 −19 −43 −16 10 25 − 11 8 × R2 + R3 → R3 1 6 5 0 −8 −1 0 0 45/8 28 −19 −135/8 −16 10 45/4 z = 135 45 = 3 y = −19 + z −8 = 2 x = 8 − 5z − 6y = 2 Mohamed Mohamed El-Sayed Atyya Page 3 of 13
- 4. 1.5. SOLUTION OF LINEAR SYSTEM OF EQUATIONS CHAPTER 1. NUMERICAL SOLUTIONS FOR LINEAR SYSTEM OF EQUATIONS 2. Forward substitution (Lower triangular system) × × × × × × × × A × × × × × × × × × × B × × i=n i=1 a1i − b1 i=n i=1 a2i − b2 Check i=n i=1 a3i − b3 i=n i=1 a4i − b4 Raws operations ↓ ↓ ↓ ... × 0 0 0 × × 0 0 × × × 0 × × × × × × × × Then apply forward substitution to get the unknowns. Example Consider the following linear system of equations x + 6y + 5z = 28 2x + 4y + 9z = 37 3x + 7y + 8z = 41 A B Check Raws operations 1 6 5 2 4 9 3 7 8 28 37 41 −16 −22 −23 − −9 8 × R3 + R2 → R2 − −5 8 × R3 + R1 → R1 −7/8 13/8 0 −11/8 −31/8 0 3 −7 8 19/8 −73/8 41 −13/8 31/8 −23 1 8 × R1 → R1 1 8 × R2 → R2 −7 13 0 −11 −31 0 3 −7 8 19 −73 41 −13 31 −23 13 31 × R2 + R1 → R1 −360/31 0 0 −11 −31 0 3 7 8 −360/31 −73 41 0 31 −23 x = 1 y = −73 + 11x −31 = 2 z = 41 − 3x − 7y 8 = 3 Mohamed Mohamed El-Sayed Atyya Page 4 of 13
- 5. 1.5. SOLUTION OF LINEAR SYSTEM OF EQUATIONS CHAPTER 1. NUMERICAL SOLUTIONS FOR LINEAR SYSTEM OF EQUATIONS 1.5.2 Gauss-Jordan Method × × × × × × × × A × × × × × × × × × × B × × i=n i=1 a1i − b1 i=n i=1 a2i − b2 Check i=n i=1 a3i − b3 i=n i=1 a4i − b4 Raws operations ↓ ↓ ↓ ... I X Example Consider the following linear system of equations x + 6y + 5z = 28 2x + 4y + 9z = 37 3x + 7y + 8z = 41 A B Check Raws operations 1 6 5 2 4 9 3 7 8 28 37 41 −16 −22 −23 −2 × R1 + R2 → R2 −3 × R1 + R3 → R3 1 6 5 0 −8 −1 0 −11 −7 28 −19 −43 −16 10 25 − 1 8 × R2 → R2 1 6 5 0 1 1/8 0 0 45/8 28 −19/8 −135/8 −16 −5/4 45/4 −6 × R2 + R1 → R1 11 × R2 + R3 → R3 1 0 17/4 0 1 1/8 0 0 45/8 55/4 −19/8 −135/8 −17/2 −5/4 45/4 − 8 45 × R3 → R3 1 0 17/4 0 1 1/8 0 0 1 55/4 −19/8 3 −17/2 −5/4 −2 − 17 4 × R3 + R1 → R1 − 1 8 × R3 + R2 → R2 1 0 0 0 1 0 0 0 1 1 2 3 0 −1 −2 x = 1 y = 2 z = 3 Mohamed Mohamed El-Sayed Atyya Page 5 of 13
- 6. 1.5. SOLUTION OF LINEAR SYSTEM OF EQUATIONS CHAPTER 1. NUMERICAL SOLUTIONS FOR LINEAR SYSTEM OF EQUATIONS 1.5.3 Inverse of A if Exists A I Check Raws operations ↓ ↓ ↓ ... I A−1 X = A−1 B Example Consider the following linear system of equations x + 6y + 5z = 28 2x + 4y + 9z = 37 3x + 7y + 8z = 41 A I Check Raw operations 1 6 5 2 4 9 3 7 8 1 0 0 0 1 0 0 0 1 11 14 17 −2 × R1 + R2 → R2 −3 × R1 + R3 → R3 1 6 5 0 −8 −1 0 −11 −7 1 0 0 −2 1 0 −3 0 1 11 −8 −16 − 1 8 × R2 → R2 1 6 5 0 1 0.125 0 −11 −7 1 0 0 0.25 −0.125 0 −3 0 1 11 1 −16 −6 × R2 + R1 → R1 11 × R2 + R3 → R3 1 0 4.25 0 1 0.125 0 0 −5.625 −0.5 0.75 0 0.25 −0.125 0 −0.25 −1.375 1 5 1 −5 − 1 5.625 × R3 → R3 1 0 4.25 0 1 0.125 0 0 1 −0.5 0.75 0 0.25 −0.125 0 2/45 11/45 −8/45 5 1 8/9 −0.125 × R3 + R2 → R2 −4.25 × R3 + R1 → R1 1 0 0 0 1 0 0 0 1 −31/45 −13/45 34/45 11/45 −7/45 1/45 2/45 11/45 −8/45 11/9 8/9 8/9 ∴ A−1 = −31/45 −13/45 34/45 11/45 −7/45 1/45 2/45 11/45 −8/45 x y z = A−1 B = 1 2 3 Mohamed Mohamed El-Sayed Atyya Page 6 of 13
- 7. 1.5. SOLUTION OF LINEAR SYSTEM OF EQUATIONS CHAPTER 1. NUMERICAL SOLUTIONS FOR LINEAR SYSTEM OF EQUATIONS 1.5.4 Iterative Techniques Consider the linear system of equations, a11x1 + a12x2 + a13x3 + ... + a1nxn = b1 a21x1 + a22x2 + a23x3 + ... + a2nxn = b2 a31x1 + a32x2 + a33x3 + ... + a3nxn = b3 . . . an1x1 + an2x2 + an3x3 + ... + annxn = bn Rewrite it as, x1 = 1 a11 [b1 − a12x2 − a13x3 − ... − a1nxn] x2 = 1 a22 [b2 − a21x1 − a23x3 − ... − a2nxn] x3 = 1 a33 [b3 − a31x1 − a32x2 − ... − a3nxn] . . . xn = 1 ann bn − an1x1 − an2x2 − ... − an(n−1)xn−1 In general, xn = 1 ann bn − n i=1 anixi ; ∀ i = n To get a solution with this method we should satisfy this conditions: 1. ann = 0 ∀ n ; existence condition. 2. A matrix should be diagonally dominant, |aii| > n j=1 |aij| ∀ j; j = i ; convergence condition. Methods of solution: 1. Jacobi method We start the solution by an initial vector, x(0) = x (0) 1 x (0) 2 x (0) 3 ... x (0) n T Then get the fist iteration values, x (1) 1 = 1 a11 b1 − a12x (0) 2 − a13x (0) 3 − ... − a1nx(0) n x (1) 2 = 1 a22 b2 − a21x (0) 1 − a23x (0) 3 − ... − a2nx(0) n x (1) 3 = 1 a33 b3 − a31x (0) 1 − a32x (0) 2 − ... − a3nx(0) n . . . x(1) n = 1 ann bn − an1x (0) 1 − an2x (0) 2 − ... − an(n−1)x (0) n−1 Mohamed Mohamed El-Sayed Atyya Page 7 of 13
- 8. 1.5. SOLUTION OF LINEAR SYSTEM OF EQUATIONS CHAPTER 1. NUMERICAL SOLUTIONS FOR LINEAR SYSTEM OF EQUATIONS In general, x(k) n = 1 ann bn − n i=1 anix (k−1) i ; ∀ i = n Continue till get a solution with a certain percentage of error. Example Consider the following linear system of equations x + y + 9z = 11 7x + 3y + 2z = 12 −2x + 8y + 4z = 10 and initial values x(0) = 0 0 0 T with accuracy k = 5 ⇒ x(i+1) − x(i) < 0.5 × 10−5 A|B = 1 1 9 7 3 2 −2 8 4 11 12 10 not diagonally dominant ⇒ A|B = 7 3 2 −2 8 4 1 1 9 12 10 11 ∴ x(i+1) = 1 7 12 − 3y(i) − 2z(i) y(i+1) = 1 8 10 − 2x(i) − 4z(i) z(i+1) = 1 9 11 − x(i) − y(i) Iteration No. x y z 0 0 0 0 1 1.714286 1.250000 1.222222 2 0.829365 1.067461 0.892857 3 1.001700 1.010912 1.011463 4 0.992048 0.994694 0.998599 5 1.002674 0.998713 1.001473 6 1.000131 0.999932 0.999850 7 1.000072 1.000108 0.999993 8 0.999975 1.0000215 0.999980 9 0.999997 0.999999 1.000002 10 0.999999 0.999998 1.000000 | error| 0.000002 0.000001 0.000002 2. Gauss-Seidel method We start the solution by an initial vector, x(0) = x (0) 1 x (0) 2 x (0) 3 ... x (0) n T Mohamed Mohamed El-Sayed Atyya Page 8 of 13
- 9. 1.5. SOLUTION OF LINEAR SYSTEM OF EQUATIONS CHAPTER 1. NUMERICAL SOLUTIONS FOR LINEAR SYSTEM OF EQUATIONS Then get the fist iteration values, x (1) 1 = 1 a11 b1 − a12x (0) 2 − a13x (0) 3 − ... − a1nx(0) n x (1) 2 = 1 a22 b2 − a21x (1) 1 − a23x (0) 3 − ... − a2nx(0) n x (1) 3 = 1 a33 b3 − a31x (1) 1 − a32x (1) 2 − ... − a3nx(0) n . . . x(1) n = 1 ann bn − an1x (1) 1 − an2x (1) 2 − ... − an(n−1)x (0) n−1 In general, x(k) n = 1 ann bn − i<n i=1 anix (k) i + i=n i>n anix (k−1) i Continue till get a solution with a certain percentage of error. Example Consider the following linear system of equations x + y + 9z = 11 7x + 3y + 2z = 12 −2x + 8y + 4z = 10 and initial values x(0) = 0 0 0 T with accuracy k = 5 ⇒ x(i+1) − x(i) < 0.5 × 10−5 A|B = 1 1 9 7 3 2 −2 8 4 11 12 10 not diagonally dominant ⇒ A|B = 7 3 2 −2 8 4 1 1 9 12 10 11 ∴ x(i+1) = 1 7 12 − 3y(i) − 2z(i) y(i+1) = 1 8 10 − 2x(i+1) − 4z(i) z(i+1) = 1 9 11 − x(i+1) − y(i+1) Mohamed Mohamed El-Sayed Atyya Page 9 of 13
- 10. 1.5. SOLUTION OF LINEAR SYSTEM OF EQUATIONS CHAPTER 1. NUMERICAL SOLUTIONS FOR LINEAR SYSTEM OF EQUATIONS Iteration No. x y z 0 0 0 0 1 1.714286 1.678572 0.845238 2 0.753401 1.015731 1.025652 3 0.985929 0.983656 1.003379 4 1.006039 0.999820 0.999349 5 1.000263 1.000391 0.999927 6 0.999853 0.999999 1.000016 7 0.999996 0.999991 1.000001 8 1.000004 1.000001 0.999999 9 1.000000 1.000001 1.000000 | error| 0.000004 0.000000 0.000001 1.5.5 Choleski Decomposition (Square-Root) Method for Positive Definite System of Matrices Theorem: A symmetrical n × n matrix is positive definite if and only if det(Ak) > 0 ; k = 1, 2, 3, ..., n. where Ak is the k × k matrix formed by the kth first row and column of matrix A. Theorem: For a symmetric positive definite matrix A, |aij|2 ≤ |aiiajj| and the maximum element of A lies on the main diagonal. Theorem: Let A be a symmetric positive definite matrix, there is a unique upper-triangular matrix R with the diagonal elements, such that RT R = A or LLT = A. Where L = lower-triangular matrix, L = RT . A = LLT a11 a12 a13 ... a1n a21 a22 a23 ... a2n . . . . . . . . . . . . an1 an2 an3 ... ann = l11 l21 l22 . . . . . . . . . ln1 ln2 . . . lnn l11 l21 . . . ln1 l22 . . . ln2 . . . . . . lnn l11 = √ a11 lij = 1 lji aij − j−1 k=1 likljk ; j = 1, 2, 3, ..., i − 1 lii = aii − i−1 k=1 l2 ik ; i = 1, 2, 3, ..., n li−1, j = 0 ; j = i, i + 1, ..., n Also, A = RT R AX = B ⇒ RT RX = B RT [RX − D] = AX − B ⇒ RT D = B , RX = D Mohamed Mohamed El-Sayed Atyya Page 10 of 13
- 11. 1.5. SOLUTION OF LINEAR SYSTEM OF EQUATIONS CHAPTER 1. NUMERICAL SOLUTIONS FOR LINEAR SYSTEM OF EQUATIONS Example Consider the following linear system of equations 7x + y + 5z = 13 x + 8y + 2z = 11 5x + 2y + 9z = 16 7 1 5 1 8 2 5 2 9 ⇒ 55/7 9/7 9/7 38/7 ⇒ 287/55 Pivot 1 = √ 7 Pivot 2 = 55/7 Pivot 3 = 287/55 R = √ 7 1/ √ 7 5/ √ 7 0 55/7 9/ √ 385 0 0 287/55 RT = √ 7 0 0 1/ √ 7 55/7 0 5/ √ 7 9/ √ 385 287/55 RT D = B ⇒ D = 13/ √ 7 64/ √ 385 287/55 T RX = D ⇒ X = 1 1 1 T For clarification a11 a12 a13 a21 a22 a23 a31 a32 a33 ⇒ b11 b12 b21 b22 ⇒ c11 Pivot 1 = √ a11 Pivot 2 = √ b11 Pivot 3 = √ c11 b11 = a22 − a12 Pivot 1 × a21 Pivot 1 b12 = a23 − a13 Pivot 1 × a21 Pivot 1 b21 = a32 − a12 Pivot 1 × a31 Pivot 1 b22 = a33 − a13 Pivot 1 × a31 Pivot 1 c11 = b22 − b12 Pivot 2 × b21 Pivot 2 R = r11 r12 r13 0 r22 r23 0 0 r33 r11 = a11 Pivot 1 r12 = a12 Pivot 1 Mohamed Mohamed El-Sayed Atyya Page 11 of 13
- 12. 1.5. SOLUTION OF LINEAR SYSTEM OF EQUATIONS CHAPTER 1. NUMERICAL SOLUTIONS FOR LINEAR SYSTEM OF EQUATIONS r13 = a13 Pivot 1 r22 = b11 Pivot 2 r23 = b12 Pivot 2 r33 = Pivot 3 1.5.6 LU Decomposition Using Gauss Elimination Consider a linear system of equations AX = B Which A can be expressed as A = LU Then we get LUX = B L [UX − D] = AX − B ⇒ LD = B , UX = D Example Consider the following linear system of equations x + 6y + 5z = 28 2x + 4y + 9z = 37 3x + 7y + 8z = 41 A B Check Raw operations 1 6 5 2 4 9 3 7 8 28 37 41 −16 −22 −23 −2 × R1 + R2 → R2 −3 × R1 + R3 → R3 1 6 5 0 −8 −1 0 −11 −7 28 −19 −43 −16 10 25 − 11 8 × R2 + R3 → R3 1 6 5 0 −8 −1 0 0 −45/8 28 −19 −135/8 −16 10 45/4 U = 1 6 5 0 −8 −1 0 0 −45/8 L = 1 0 0 2 1 0 3 11/8 1 LD = B ⇒ D = 28 −19 −135/8 T UX = D ⇒ X = 1 2 3 T Note: As we see the raw operation in general is aij × Rj + Ri → Ri, this used to generate L matrix as lij = −aij Mohamed Mohamed El-Sayed Atyya Page 12 of 13
- 13. 1.6. REFERENCES CHAPTER 1. NUMERICAL SOLUTIONS FOR LINEAR SYSTEM OF EQUATIONS 1.6 References 1. A First Course in Numerical Analysis, 2 nd edition, Anthony Ralston and Philip Rabinowitz. 2. Elementary Numerical Analysis, An Algorithmic Approach, 3 rd edition, S. D. Conte and Carl de Boor. 3. Numerical Methods for Engineers, 6 th edition, Steven C. Chapra and Raymond P. Canale. 4. Numerical Methods and Algorithms, Milan Kubicek, Miroslava Dubcova, Drahoslava Janovska, VSCHT Praha 2005. 1.7 Contacts mohamed.atyya94@eng-st.cu.edu.eg Mohamed Mohamed El-Sayed Atyya Page 13 of 13