linear-guest

7.7 LU Redux 163
so we would like to perform the row operations
R 2 → R 2 − 1 3 R 1 and R 3 → R 3 − 2 3 R 1 .
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 ⎟
L 1 = ⎝ 1 0
3 ⎠ .
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
diagonal using the second row of U 1 using the row operation R 3 → R 3 − 1R 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 2
1
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 ⎟
L 2 = ⎝ 1 0
3 ⎠ ⎝0 1 0⎠ ⎜ 1 ⎟
= ⎝ 1 0
2
0 1 0 1 3 ⎠ .
1 2 1
3 2 1
3 2
Notice that it is lower triangular because
163