矩阵LU分解的高斯消元法
Posted 木卜
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A=[1,-1,1,-4;5,-4,3,12;2,1,1,11;2,-1,7,-1]
L=eye(length(A))
%开始消元过程
for k=1:(length(A))
a=A(k,k)
for i=k+1:(length(A))
c=-A(i,k)
L(i,k)=-c./a
for j=1: (length(A))
A(i,j)=A(i,j)+c.*A(k,j)./ a
end
end
end
L
U=A
A =
1 -1 1 -4
5 -4 3 12
2 1 1 11
2 -1 7 -1
L =
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
a =
1
c =
-5
L =
1 0 0 0
5 1 0 0
0 0 1 0
0 0 0 1
A =
1 -1 1 -4
0 -4 3 12
2 1 1 11
2 -1 7 -1
A =
1 -1 1 -4
0 1 3 12
2 1 1 11
2 -1 7 -1
A =
1 -1 1 -4
0 1 -2 12
2 1 1 11
2 -1 7 -1
A =
1 -1 1 -4
0 1 -2 32
2 1 1 11
2 -1 7 -1
c =
-2
L =
1 0 0 0
5 1 0 0
2 0 1 0
0 0 0 1
A =
1 -1 1 -4
0 1 -2 32
0 1 1 11
2 -1 7 -1
A =
1 -1 1 -4
0 1 -2 32
0 3 1 11
2 -1 7 -1
A =
1 -1 1 -4
0 1 -2 32
0 3 -1 11
2 -1 7 -1
A =
1 -1 1 -4
0 1 -2 32
0 3 -1 19
2 -1 7 -1
c =
-2
L =
1 0 0 0
5 1 0 0
2 0 1 0
2 0 0 1
A =
1 -1 1 -4
0 1 -2 32
0 3 -1 19
0 -1 7 -1
A =
1 -1 1 -4
0 1 -2 32
0 3 -1 19
0 1 7 -1
A =
1 -1 1 -4
0 1 -2 32
0 3 -1 19
0 1 5 -1
A =
1 -1 1 -4
0 1 -2 32
0 3 -1 19
0 1 5 7
a =
1
c =
-3
L =
1 0 0 0
5 1 0 0
2 3 1 0
2 0 0 1
A =
1 -1 1 -4
0 1 -2 32
0 3 -1 19
0 1 5 7
A =
1 -1 1 -4
0 1 -2 32
0 0 -1 19
0 1 5 7
A =
1 -1 1 -4
0 1 -2 32
0 0 5 19
0 1 5 7
A =
1 -1 1 -4
0 1 -2 32
0 0 5 -77
0 1 5 7
c =
-1
L =
1 0 0 0
5 1 0 0
2 3 1 0
2 1 0 1
A =
1 -1 1 -4
0 1 -2 32
0 0 5 -77
0 1 5 7
A =
1 -1 1 -4
0 1 -2 32
0 0 5 -77
0 0 5 7
A =
1 -1 1 -4
0 1 -2 32
0 0 5 -77
0 0 7 7
A =
1 -1 1 -4
0 1 -2 32
0 0 5 -77
0 0 7 -25
a =
5
c =
-7
L =
1.0000 0 0 0
5.0000 1.0000 0 0
2.0000 3.0000 1.0000 0
2.0000 1.0000 1.4000 1.0000
A =
1 -1 1 -4
0 1 -2 32
0 0 5 -77
0 0 7 -25
A =
1 -1 1 -4
0 1 -2 32
0 0 5 -77
0 0 7 -25
A =
1 -1 1 -4
0 1 -2 32
0 0 5 -77
0 0 0 -25
A =
1.0000 -1.0000 1.0000 -4.0000
0 1.0000 -2.0000 32.0000
0 0 5.0000 -77.0000
0 0 0 82.8000
a =
82.8000
L =
1.0000 0 0 0
5.0000 1.0000 0 0
2.0000 3.0000 1.0000 0
2.0000 1.0000 1.4000 1.0000
U =
1.0000 -1.0000 1.0000 -4.0000
0 1.0000 -2.0000 32.0000
0 0 5.0000 -77.0000
0 0 0 82.8000
>> A=L*U
A =
1 -1 1 -4
5 -4 3 12
2 1 1 11
2 -1 7 -1
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