Gaussian Elimination with Partial Pivoting Example
Apply Gaussian elimination with partial pivoting to A=
1 2 −4 3
2 5 −6 10
−2−7 3 −21
2 8 15 38
and solve Ax =b
for b=
0
9
−28
42
.
Solution:
Apply Gaussian elimination with partial pivoting to Ausing the compact storage mode where the
multipliers (= elements of L) are stored in Ain the locations of Athat are to be made zero.The elements
of Lare in red. Note that when one interchanges rows of the current A, one must also interchange rows
of the current L. For example, in the step at the third arrow, below, we switch the second and fourth
rows of A2and L2. No row interchanges are required for A3, so we have skipped pivoting for A3.
λ A1
1|1 2 −4 3
2|2 5 −6 10
3| −2−7 3 −21
4|2 8 −5 38
→
λ P1A1
2|2 5 −6 10
1|1 2 −4 3
3| −2−7 3 −21
4|2 8 −5 38
→
λL2and A2
2|2 5 −6 10
1|1/2−1/2−1−2
3|−1−2−3−11
4|13 1 28
→
λP2L2and P2A2
2|2 5 −6 10
4|13 1 28
3|−1−2−3−11
1|1/2−1/2−1−2
→
λL3and A3
2|2 5 −6 10
4|13 1 28
3|−1−2/3−7/3 23/3
1|1/2−1/6−5/6 8/3
→
λL4and A4
2|2 5 −6 10
4|13 1 28
3|−1−2/3−7/3 23/3
1|1/2−1/6 5/14 −1/14
So U=A4=
2 5 −6 10
0 3 1 28
0 0 −7/3 23/3
0 0 0 −1/14
and L=L4=
1 0 0 0
1 1 0 0
−1−2/3 1 0
1/2−1/6 5/14 1
.
P will have a one in columns 2, 4 , 3 and 1, in that order, (see λ) so P=
0100
0001
0010
1000
. The per-
mutation vector, p, describing the matrix P is p = [2 4 3 1]. P A =A(p, :) =
2 5 −6 10
2 8 15 38
−2−7 3 −21
1 2 −4 3
.
Here A(p, :) uses Matlab notation. LU =P A can be checked.
To solve Ax =bnote that P Ax =Pb =b
bor LUx =b
b. Therefore we can solve Ax =bin three steps:
(1) Let b
b=P b =
9
42
−28
0
.
(2) Solve Lw =b
busing forward substitution. We get w=
9
33
3
1/14
.
(3) Solve Ux =wusing back substitution. We get x=
3
1
2
1
.
1