Page 60 - 35Linear Algebra

60 Systems of Linear Equations

What if we stop at a different point in elimination? We could multiply
rows so that the entries in the diagonal are 1 next. Note that the EROs that
do this are diagonal. This gives a slightly different factorization.

Example 30 (LDU factorization building from previous example)

     3 1 
2 0 −3 1 2 0 −3 1 1 0 −
2 2
0 1 2 2 0 1 2 2 0 1 2 2
  E 3 E 2 E 1   E 4  
M =   ∼   ∼  
−4 0 9 2 0 0 3 4 0 0 3 4
     
0 −1 1 −1 0 0 0 −3 0 0 0 −3
1 0 − 1 0 −
 3 1   3 1 
2 2 2 2
0 1 2 2   0 1 2 2 
E 5 E 6

∼   ∼   =: U
 0 0 1 4   0 0 1 4 
3 3
0 0 0 −3 0 0 0 1
The corresponding elementary matrices are
1 0 0 0 1 0 0 0 1 0 0 0
     
2
0 1 0 0 0 1 0 0 0 1 0 0
     
E 4 =   , E 5 =  1  , E 6 =   ,
 0 0 1 0   0 0 0   0 0 1 0 
3
0 0 0 1 0 0 0 1 0 0 0 − 1
3
     
2 0 0 0 1 0 0 0 1 0 0 0
0 1 0 0  −1  0 1 0 0  −1  0 1 0 0 
−1

E =   , E =   , E =   .
4 5 6
 0 0 1 0   0 0 3 0   0 0 1 0 
0 0 0 1 0 0 0 1 0 0 0 −3
The equation U = E 6 E 5 E 4 E 3 E 2 E 1 M can be rearranged as
M = (E −1 E −1 E −1 )(E 4 −1 E 5 −1 E 6 −1 )U.
2
3
1
We calculated the product of the first three factors in the previous example; it was
named L there, and we will reuse that name here. The product of the next three
factors is diagonal and we wil name it D. The last factor we named U (the name means
something different in this example than the last example.) The LDU factorization
of our matrix is

       3 1 
2 0 −3 1 1 0 0 0 2 0 0 0 1 0 −
2 2
0 1 2 2 0 1 0 0 0 1 0 0 0 1 2 2
       
  =       .
−4 0 9 2 −2 0 1 0 0 0 3 0 0 0 1
       4 
3
0 −1 1 −1 0 −1 1 1 0 0 0 −3 0 0 0 1
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