Chapter 3 Solution of Linear Systems - Math/CS

62 CHAPTER 3. SOLUTION OF LINEAR SYSTEMS
Here the superscripts on the entries a (k)
ij
are used to indicate entries of the matrix that are modified
during the kth elimination step. Recall that, in general, the multipliers are computed as
mij =
element to be eliminated
current pivot element
If the process does not break down (that is, all pivot elements, a11, a (1)
22
the matrix A can be factored as A = LU where
⎡
⎢
L = ⎢
⎣
1
m21
m31
.
0
1
m32
.
0
0
1
· · ·
· · ·
. ..
⎤
0
0 ⎥
0 ⎥
. ⎦
and
⎡
⎢
U = ⎢
⎣
mn1 mn2 mn3 · · · 1
The following example illustrates the process.
Example 3.3.1. Consider the matrix
⎡
A = ⎣
1 2 3
2 −3 2
3 1 −1
Using Gaussian elimination without row interchanges, we obtain
⎡
1
⎣ 2
2
−3
⎤
3
2 ⎦ −→
⎡
1
⎣ 0
2
−7
⎤
3
−4 ⎦ −→
3 1 −1
0 −5 −10
and thus
⎡
L = ⎣
1 0 0
m21 1 0
m31 m32 1
m21 = 2
m31 = 3
3 5
7
⎤
⎦ .
.
, a(2) 33 , . . ., are nonzero) then
upper triangular matrix
after elimination is
completed
⎡
⎣
m32 = 5
7
⎤ ⎡
1 0 0
⎤
⎡
⎦ = ⎣ 2 1 0
1
⎦ and U = ⎣
It is straightforward to verify2 that A = LU; that is,
⎡
1
⎣ 2
2
−3
⎤ ⎡
3 1
2 ⎦ = ⎣ 2
0
1
⎤ ⎡
0
0 ⎦ ⎣
3 1 −1
1
3 5
7
If we can compute the factorization A = LU, then
1 2 3
0 −7 −4
0 0 − 50
7
Ax = b ⇒ LUx = b ⇒ Ly = b, where Ux = y.
Therefore, the following steps can be used to solve Ax = b:
• Compute the factorization A = LU.
• Solve Ly = b using forward substitution.
• Solve Ux = y using backward substitution.
1 2 3
0 −7 −4
0 0 − 50
7
1 2 3
0 −7 −4
0 0 − 50
7
2 Recall from Section 1.2 that if A = LU, then the i, j entry of A can be obtained by multiplying the ith row of L
times the jth column of U.
⎤
⎦ .
⎤
⎦
⎤
⎦ .
⎤
⎥ .
⎥
⎦