LU Factorization
 A11

A  A21

 A31
 L11

L  L21

 L31
A12
A22
A32
0
L22
L32
A13 

A23

A33 
0

0

L33 
A  LU
U11 U12 U13 


U  0 U 22 U 23


 0
0 U 33 
 L11
L
 21
 L31
 A11

 A21

 A31
0
L22
L32
A12
A22
A32
0  U11 U12 U13 



0
0 U 22 U 23


L33   0
0 U 33 
A13 

A23

A33 
 L11U 11
L U
 21 11
 L31U 11
 A11

 A21

 A31
L11U 12
L21U 12  L22U 22
L31U 12  L32U 22
A12
A22
A32


L21U 13  L22U 23

L31U 13  L32U 23  L33U 33 
L11U 13
A13 

A23

A33 
Equating the elements of the First Row :-
L11U11  A11
L11U12  A12
L11U13  A13
Equating the elements of the 2nd Row :-
L21U11  A21
L21U12  L22U22  A22
L21U13  L22U23  A23
Equating the elements of the 3rd Row :-
L31U11  A31
L31U12  L32U22  A32
L31U13  L32U23  L33U33  A33
We have 12 unknowns but only 9 equations.
We need some sort of compromise.
Crout’s Method
Set
U11  U 22  U 33  1
Dolittle’s Method
Set
L11  L22  L33  1
Crout's Method
U
1 1
 1
U
2 2
A
L
1 1

L
2 1
2 3
1 2
U
2 1
2 2
1 1


L
2 1
L
2 2
A
1 2
U
1 3
L
2 2
L
1 3
U
2 1
U
2 2
U
 1
1 1

L
2 3
3 3
A
U
A
U
U
1 1

U
A
1 1
A
 1

1 2
L
1 3
1 1
A
L
3 1

3 3
3 1
L
3 2
U
1 1
A
L
A

3 3
L
U
3 1
1 3
U
3 3

3 2
L
U
2 2
L
U
3 2
U
3 1
2 3
1 2
Use of LU factors in solving systems
of linear equations
Ax  b
LUx  b
Ly  b
Solve for y, and then solve
for x.
Ux  y
2 1 3 
A   1 1 2 


 3 2 4 
 13 
B   7 
 
 5 
2 0 0 
L   1 0.5 0 


 3 3.5 25 
 1 0.5 1.5 
U   0 1 7 


0 0 1 
LUX = B
LY = B
0
0  Y1   13 
2
1 0.5
 Y    7 
0

 2   
3  3.5  25 Y3   5
B L
B
Y 
1
1
Y 
3
2
L
Y  L
Y
L
1 1
B L
3
2 1
2
Y 
3 1
1
L
3 3
Y
1
2 2
3 2
2
 6.5 
Y 1 


 0.84 
UX = Y
1 0.5 1.5  X 1   6.5 
0 1  7   X    1 

 2  

0 0
1   X 3  0.84
Y U
Y
X 
3
3
X 
2
U
3 3
Y U
X 
1
1
2
U
1 1
3
U
2 2
X  U
1 2
X
2 3
2
X
1 3
3
 1.8 
X   6.88 


 0.84 
Elementary Matrices and The LU Factorization
Definition: Any matrix obtained by performing
a single elementary row operation (ERO) on
the identity (unit) matrix is called an
elementary matrix.
There are three elementary operations:
Permute rows i and j
Multiply row i by a non-zero scalar k
Add k times row i to row j
Corresponding to the three ERO, we have then
three elementary matrices:
Type 1: - permute rows i and j in In.
Type 2: - multiply row i of In by a non-zero
scalar k
Type 3: - Add k times row i of In to row j
Permutation matrix:
Scaling matrix:
Row combination:
0 1 0 
P12  1 0 0
0 0 1
1 0 0
M 2 k   0 k 0
0 0 1
 1 0 0
A12 k   k 1 0
 0 0 1
Pre-multiplying a matrix A by an elementary matrix E has
the effect of performing the corresponding ERO on A.
2  3 1
A

4
7
5


Example: We can multiply the First row of the matrix A
by 3 (an elementary row operation). The resulting matrix
will become
6  9 3 
4 7 5 


We can achieve the same result by pre-multiplying A by the
corresponding elementary matrix.
 3 0   2  3 1  6  9 3 
M 1 3A  





 0 1   4 7 5  4 7 5 
An ERO can be performed on a matrix by premultiplying the matrix by a corresponding elementary
matrix. Therefore, we can show that any matrix A can be
reduced to a row echelon form (REF) by multiplication by
a sequence of elementary matrices.
E1 E 2 E k A  R
where R denotes an REF of A.
Since the unique reduced row echelon form
(RREF) of a matrix is the unit matrix
E1 E 2  E k A  I n
E1 E 2  E k A  I n
1
A A  In
A
A
1
 E1 E 2  E k
1
 E1 E 2  E k I n
A nonsingular matrix can be reduced to an upper
triangular matrix using elementary row operations
of Type 3 only. The elementary matrices
corresponding to Type 3 EROs are unit lower
triangular matrices. We can write
E1 E 2  E k A  U
E1 E 2  E k A  U
Since each elementary matrix is nonsingular (meaning
their inverse exist) we can write
1
k
A E E
1
k 1
1
2
1
1
E E U
We know that the product of two lower triangular
matrices is also a lower triangular matrix. Therefore
A  LU
1
k
L E E
1
k 1
1
2
E E
1
1
Inverses of the three elementary matrices are:
M i k   M i 1 k 
1
Pij
1
 Pij
Aij k   Aij ( k )
1
Determine the LU factorization of the matrix
2 5 3
A   3 1  2
 1 2 1 
First, let us do the EROs to reduce A into an upper
triangular matrix.
2 5 3 
 3 1  2  A  3 2, A 1 2 
12
13


  1 2 1 
5
3 
2
0  13 2  13 2


0
92
5 2 
 A23 9 13
5
3 
2


  0  13 2  13 2
 0
0
 2 
These EROs can be written in terms of their
equivalent elementary matrices as
5
3 
2
E 1 E 2 E 3 A  0  13 2  13 2
0
0
 2 
E1  A23 9 13, E 2  A13 1 2, E 3  A12  3 2
5
3 
2
U   0  13 2  13 2
 0
0
 2 
L  E 31 E 21 E 11
E1
1
 A23  9 13, E 2
1
 A13  1 2, E 3
1
 A12 3 2
0 0 1
0
0
 1 0 0  1
L   3 2 1 0  0
1 0 0
1
0
 0 0 1   1 2 0 1 0  9 13 1
0
0
 1
L   3 2
1
0
  1 2  9 13 1
0
0  2
5
3 
 1
A   3 2
1
0  0  13 2  13 2
  1 2  9 13 1  0
0
 2 
We can construct the lower triangular matrix L
without multiplying the elementary matrices if we
utilize the multipliers obtained while we converted
matrix A into an upper triangular matrix.
E1  A23 9 13, E 2  A13 1 2, E 3  A12  3 2
Definition: When using ERO of Type 3, the multiple
of a specific row i that is subtracted from row j to put
a zero in the ji position is called a multiplier, and is
denoted as m ji
m 21  3 2 , m 31   1 2 , m 32   9 13
m 21  3 2 , m 31   1 2 , m 32   9 13
0
0
 1


L 3 2
1
0
  1 2  9 13 1
If we notice the unit lower triangular matrix
L carefully, we see that the elements beneath
the leading diagonal are just the
corresponding multipliers. This relationship
holds in general. Therefore, we can do
elementary row operations of Type 3 to
reduce A to upper triangular form and then
utilize the corresponding multipliers to write
L directly.