![Method
For most non-singular matrix [A] that one could
conduct Naïve Gauss Elimination forward elimination
steps, one can always write it as [A] = [L][U]
where
[L] = lower triangular matrix
[U] = upper triangular matrix
3](https://image.slidesharecdn.com/ludecomposition-100721061419-phpapp01/85/Lu-decomposition-3-320.jpg)
![Operation mode
If solving a set of linear equations [A][X] = [C]
If [A] = [L][U] then [L][U][X] = [C]
Multiply by [L]-1
Which gives [L]-1[L][U][X] = [L]-1[C]
Remember [L]-1[L] = [I] which leads to [I][U][X] = [L]-1[C]
Now, if [I][U] = [U] then [U][X] = [L]-1[C]
Now, let [L]-1[C]=[Z]
Which ends with [L][Z] = [C] (1)
and [U][X] = [Z] (2)
4](https://image.slidesharecdn.com/ludecomposition-100721061419-phpapp01/85/Lu-decomposition-4-320.jpg)
![Way to use
Given [A][X] = [C]
1. Decompose [A] into [L] and [U]
2. Solve [L][Z] = [C] for [Z]
3. Solve [U][X] = [Z] for [X]
5](https://image.slidesharecdn.com/ludecomposition-100721061419-phpapp01/85/Lu-decomposition-5-320.jpg)
![[A] to [L] and [U]
1 0 0 u11 u12 u13
A LU 21 1 0 0 u 22 u 23
31 32 1 0 0 u 33
[U] is the same as the coefficient matrix at the end of the forward
elimination step.
[L] is obtained using the multipliers that were used in the forward
elimination process
6](https://image.slidesharecdn.com/ludecomposition-100721061419-phpapp01/85/Lu-decomposition-6-320.jpg)
![LU Decomposition in linear
equations
Given the next linear
25 5 1 x1 106.8
equations system, solve
by using LU 64 8 1 x2 177.2
decomposition 144 12 1 x3 279.2
Using the procedure for finding the [L] and [U] matrices
1 0 0 25 5 1
A L U 2.56 1 0 0 4.8 1.56
5.76 3.5 1 0 0 0.7
7](https://image.slidesharecdn.com/ludecomposition-100721061419-phpapp01/85/Lu-decomposition-7-320.jpg)
![1 0 0 z1 106.8
Set [L][Z] = [C] 2.56 1 0 z2 177.2
5.76 3.5 1 z 3 279.2
z1 10
Solve for [Z] 2.56 z1 z2 177.2
5.76 z1 3.5 z 2 z3 279.2
8](https://image.slidesharecdn.com/ludecomposition-100721061419-phpapp01/85/Lu-decomposition-8-320.jpg)
![Completing the substitution to solve [Z]
z1 106.8
z1 106.8
z2 177.2 2.56 z1
177.2 2.56 106.8 Z z2 96.21
96.2 z3 0.735
z3 279.2 5.76 z1 3.5 z 2
279.2 5.76 106.8 3.5 96.21
0.735
9](https://image.slidesharecdn.com/ludecomposition-100721061419-phpapp01/85/Lu-decomposition-9-320.jpg)
![Set [U][X] = [Z]
25 5 1 x1 106.8
0 4.8 1.56 x2 96.21
0 0 0.7 x3 0.735
Solve for [X]
The 3 equations become 25a1 5a2 a3 106.8
4.8a2 1.56a3 96.21
0.7a3 0.735
10](https://image.slidesharecdn.com/ludecomposition-100721061419-phpapp01/85/Lu-decomposition-10-320.jpg)
Lu decomposition
- 1. 1
- 2. LU Decomposition LU Decomposition is another method to solve a set of simultaneous linear equations In linear algebra, the LU decomposition is a matrix decomposition which writes a matrix as the product of a lower triangular matrix and an upper triangular matrix. The product sometimes includes a permutation matrix as well. This decomposition is used in numerical analysis to solve systems of linear equations or calculate the determinat. 2
- 3. Method For most non-singular matrix [A] that one could conduct Naïve Gauss Elimination forward elimination steps, one can always write it as [A] = [L][U] where [L] = lower triangular matrix [U] = upper triangular matrix 3
- 4. Operation mode If solving a set of linear equations [A][X] = [C] If [A] = [L][U] then [L][U][X] = [C] Multiply by [L]-1 Which gives [L]-1[L][U][X] = [L]-1[C] Remember [L]-1[L] = [I] which leads to [I][U][X] = [L]-1[C] Now, if [I][U] = [U] then [U][X] = [L]-1[C] Now, let [L]-1[C]=[Z] Which ends with [L][Z] = [C] (1) and [U][X] = [Z] (2) 4
- 5. Way to use Given [A][X] = [C] 1. Decompose [A] into [L] and [U] 2. Solve [L][Z] = [C] for [Z] 3. Solve [U][X] = [Z] for [X] 5
- 6. [A] to [L] and [U] 1 0 0 u11 u12 u13 A LU 21 1 0 0 u 22 u 23 31 32 1 0 0 u 33 [U] is the same as the coefficient matrix at the end of the forward elimination step. [L] is obtained using the multipliers that were used in the forward elimination process 6
- 7. LU Decomposition in linear equations Given the next linear 25 5 1 x1 106.8 equations system, solve by using LU 64 8 1 x2 177.2 decomposition 144 12 1 x3 279.2 Using the procedure for finding the [L] and [U] matrices 1 0 0 25 5 1 A L U 2.56 1 0 0 4.8 1.56 5.76 3.5 1 0 0 0.7 7
- 8. 1 0 0 z1 106.8 Set [L][Z] = [C] 2.56 1 0 z2 177.2 5.76 3.5 1 z 3 279.2 z1 10 Solve for [Z] 2.56 z1 z2 177.2 5.76 z1 3.5 z 2 z3 279.2 8
- 9. Completing the substitution to solve [Z] z1 106.8 z1 106.8 z2 177.2 2.56 z1 177.2 2.56 106.8 Z z2 96.21 96.2 z3 0.735 z3 279.2 5.76 z1 3.5 z 2 279.2 5.76 106.8 3.5 96.21 0.735 9
- 10. Set [U][X] = [Z] 25 5 1 x1 106.8 0 4.8 1.56 x2 96.21 0 0 0.7 x3 0.735 Solve for [X] The 3 equations become 25a1 5a2 a3 106.8 4.8a2 1.56a3 96.21 0.7a3 0.735 10
- 11. From the 3rd equation Substituting in a3 and using the second equation 0.7 a3 0.735 4.8a2 1.56 a3 96 .21 0.735 96.21 1.56a3 a3 a2 0.7 4.8 a3 1.050 96.21 1.56 1.050 a2 4.8 a2 19.70 11
- 12. 25 a1 5a2 a3 106 .8 Substituting in a3 and a2 using 106.8 5a2 a3 a1 the first equation 25 106.8 5 19.70 1.050 25 0.2900 a1 0.2900 Hence the Solution Vector is: a2 19.70 a3 1.050 12