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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.
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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
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Operation mode
If solving a set of linear equations
If [A] = [L][U] then
Multiply by
Which gives
Remember [L]-1[L] = [I] which leads to
Now, if [I][U] = [U] then
Now, let
Which ends with
and
[A][X] = [C]
[L][U][X] = [C]
[L]-1
[L]-1[L][U][X] = [L]-1[C]
[I][U][X] = [L]-1[C]
[U][X] = [L]-1[C]
[L]-1[C]=[Z]
[L][Z] = [C] (1)
[U][X] = [Z] (2)
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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]
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[A] to [L] and [U]
 1 0 0 u11 u12 u13 
A  LU    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
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LU Decomposition in linear
equations
Given the next linear
equations system, solve
by using LU
decomposition
 25 5 1  x1  106.8 
 64 8 1  x   177.2 

  2 

144 12 1  x3  279.2
Using the procedure for finding the [L] and [U] matrices
0 0 25
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1 
 1
A  LU   2.56 1 0  0  4.8  1.56
5.76 3.5 1  0
0
0.7 
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Set [L][Z] = [C]
0 0  z1  106.8 
 1
2.56 1 0  z   177.2 

 2  

5.76 3.5 1  z 3  279.2
z1  10
Solve for [Z]
2.56 z1  z 2  177.2
5.76 z1  3.5 z 2  z 3  279.2
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Completing the substitution to solve [Z]
z1  106.8
z 2  177.2  2.56 z1
 177.2  2.56106.8
 96.2
z3  279.2  5.76 z1  3.5 z 2
 z1   106.8 




Z    z2    96.21
 z3   0.735 
 279.2  5.76106.8  3.5 96.21
 0.735
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Set [U][X] = [Z]
1   x1   106.8 
25 5
 0  4.8  1.56  x    96.21

  2 

 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
0.7 a3  0.735
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From the 3rd equation
Substituting in a3 and using the
second equation
0.7 a3  0.735
 4.8a2  1.56a3  96.21
0.735
a3 
0.7
a3  1.050
 96.21  1.56a3
a2 
 4.8
 96.21  1.561.050 
a2 
 4.8
a2  19.70
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25a1  5a2  a3  106.8
Substituting in a3 and a2 using
the first equation
Hence the Solution Vector is:
106.8  5a2  a3
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106.8  519.70   1.050

25
 0.2900
a1 
 a1  0.2900
a    19.70 
 2 

 a3   1.050 
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BIBLIOGRAPHY
•http://numericalmethods.eng.usf.edu/topics/lu_decomposi
tion.html
•http://en.wikipedia.org/wiki/LU_decomposition
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