Page 163 - 35Linear Algebra

7.7 LU Redux 163

so we would like to perform the row operations

1 2
R 2 → R 2 − R 1 and R 3 → R 3 − R 1 .
3 3
If we perform these row operations on M to produce

 
6 18 3
U 1 =  0 6 0  ,
0 3 1

we need to multiply this on the left by a lower triangular matrix L 1 so that
the product L 1 U 1 = M still. The above example shows how to do this: Set
L 1 to be the lower triangular matrix whose first column is filled with minus
the constants used to zero out the first column of M. Then

 
1 0 0
 1 1 0 .

3
L 1 = 
2 0 1
3
By construction L 1 U 1 = M, but you should compute this yourself as a double
check.
Now repeat the process by zeroing the second column of U 1 below the
1
diagonal using the second row of U 1 using the row operation R 3 → R 3 − R 2
2
to produce
 
6 18 3
U 2 =  0 6 0  .
0 0 1
The matrix that undoes this row operation is obtained in the same way we
found L 1 above and is:
 
1 0 0
0 1 0 .
 
0 1 1
2
Thus our answer for L 2 is the product of this matrix with L 1 , namely
   
1 0 0  1 0 0  1 0 0
 1   1 
 0 1 0  1 0 .
3 1 0 =  3
L 2 = 
2 0 1 0 1 1 2 1 1
3 2 3 2
Notice that it is lower triangular because

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