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第三讲
消去法求逆矩阵
一．并行变换法
AX=B
经
n
次初等变换
A
?
B
?
E
?
E
?
X
?
A
a
< br>a
(
k
)
kj
?
1
?
a
?
a
(<
/p>
k
?
1
)
kj
/
a
(
k
?
1
)
kk
(
k
)
ij
(
k
< br>?
1
)
ij
?
a
(
k
?
1
)
ik
.
a
(
k
?
1
)
kj
/
a
(
k
?
1
)
kk
(k)
表示第
k
步消元
k=1
……
..m
I=1,2,
…
.,k-1,k+1
……
.m
(i
≠
k)
J=1,2,
…
.,n,n+1,
……
..,2m+1
SUB INVERSE1(M as
Integer)
for K=1 To M
for
I=1 to M
IF I<>K then
c = A(I, K) / A(K, K)
For J=1 to
2*M+1
A(I,J)=A(I,J)-C* A(K,J)
Next j
Endif
Next I
Next k
For k=1 to M
C=a(K,K)
For J=1 to
2*M+1
A(K,J)=A(K,J)/C
Next j
Next k
END sub
二．
紧凑变换法
AX=B
经
n
次初等变换
A
?
B
?
A
?
1
?
< br>X
a
(
k
)
(
k
?
1
)
1
)
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)
?
1
)
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?
a
ij
?
a
(
k
?
ik
.
a
(
k
?
kj
/
a
(
k
kk
a
(
k
)
k
?
1<
/p>
)
kj
?
a
(
kj
/
a
(
k
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1
)
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J
a
(
k
)
(
k
?
1
)
ik
?
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< br>a
ik
/
a
(
k
?
1
)
kk
I
≠
K
≠
K
≠
K
I,J
a
(
k
)
kk
?
1
/
a
(
k
)
kk
(k)
表示第
k
步消元
k=1
……
..n
I=1,2,
…
.,n
J=1,2,
…
.,n,n+1
SUB INVERSE2(M as integer)
for K=1 to M
FOR I=1 TO M
IF I<>K THEN
FOR
J=1,M+1
IF J<>K THEN
A(I,J)=A(I,J)-A(I,K)*A(K,J)/A(K,K)
ENDIF
NEXT J
End
if
Next i
for I=1 to M
IF (I<>K) then
A(I,K)=-A(I,K)/A(K,K)
endif
Next i
for J=1 to M+1
IF (j<>K) then
A(K,J)=A(K,J)/A(K,K)
ENDIF
Next j
A(K,K)=1/A(K,K)
Next k
END
三．
列主元消去法求逆
Sub
InvZuQu(M As Integer)
Dim L1(5, 5)
As Double
Dim L2(5, 5) As Double
Dim TT As Double
For I = 1 To M
For J = 1 To
M
If (I = J) Then
L1(I, J) = 1#
Else
L1(I, J) = 0#
End If
Next J