VB逆序代码 求解！！！！！！！！

Private Sub Command1_Click()

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Label1.Caption = StrReverse(Text1.Text)

End Sub

vb中n阶方阵求逆

1、求逆的必要条件是N阶方阵A的行列式不等于0;

2、然后求N阶方阵A的行列式的值|A|;

3、求N阶方阵A的伴随矩阵A*;

4、N阶方阵A的逆矩阵=A*/|A|;

怎样用VB编程矩阵求逆

矩阵求逆的VB程序

Private Function MRinv(N As Integer, mtxA() As Double) As Boolean

'****************************************************************************************

' 功能： 实现矩阵求逆的全选主元高斯－约当法

' 参数： n - Integer型变量，矩阵的阶数

' mtxA - Double型二维数组，体积为n x n。存放原矩阵A；返回时存放其逆矩阵A-1。

' 返回值：Boolean型，失败为False，成功为True

'****************************************************************************************

ReDim nIs(N) As Integer, nJs(N) As Integer

Dim i As Integer, j As Integer, k As Integer

Dim D As Double, p As Double

' 全选主元，消元

For k = 1 To N

D = 0#

For i = k To N

For j = k To N

p = Abs(mtxA(i, j))

If (p D) Then

D = p

nIs(k) = i

nJs(k) = j

End If

Next j

Next i

' 求解失败

If (D + 1# = 1#) Then

MRinv = False

Exit Function

End If

If (nIs(k) k) Then

For j = 1 To N

p = mtxA(k, j)

mtxA(k, j) = mtxA(nIs(k), j)

mtxA(nIs(k), j) = p

Next j

End If

If (nJs(k) k) Then

For i = 1 To N

p = mtxA(i, k)

mtxA(i, k) = mtxA(i, nJs(k))

mtxA(i, nJs(k)) = p

Next i

End If

mtxA(k, k) = 1# / mtxA(k, k)

For j = 1 To N

If (j k) Then mtxA(k, j) = mtxA(k, j) * mtxA(k, k)

Next j

For i = 1 To N

If (i k) Then

For j = 1 To N

If (j k) Then mtxA(i, j) = mtxA(i, j) - mtxA(i, k) * mtxA(k, j)

Next j

End If

Next i

For i = 1 To N

If (i k) Then mtxA(i, k) = -mtxA(i, k) * mtxA(k, k)

Next i

Next k

' 调整恢复行列次序

For k = N To 1 Step -1

If (nJs(k) k) Then

For j = 1 To N

p = mtxA(k, j)

mtxA(k, j) = mtxA(nJs(k), j)

mtxA(nJs(k), j) = p

Next j

End If

If (nIs(k) k) Then

For i = 1 To N

p = mtxA(i, k)

mtxA(i, k) = mtxA(i, nIs(k))

mtxA(i, nIs(k)) = p

Next i

End If

Next k

' 求解成功

MRinv = True

End Function

来源:

矩阵求逆编程

算法的大致思想是通过行列式初等变换来求。

代码如下：

private double[,] ReverseMatrix( double[,] dMatrix, int Level )

{

double dMatrixValue = MatrixValue( dMatrix, Level );

if( dMatrixValue == 0 ) return null;

double[,] dReverseMatrix = new double[Level,2*Level];

double x, c;

// Init Reverse matrix

for( int i = 0; i Level; i++ )

{

for( int j = 0; j 2 * Level; j++ )

{

if( j Level )

dReverseMatrix[i,j] = dMatrix[i,j];

else

dReverseMatrix[i,j] = 0;

}

dReverseMatrix[i,Level + i ] = 1;

}

for( int i = 0, j = 0; i Level j Level; i++, j++ )

{

if( dReverseMatrix[i,j] == 0 )

{

int m = i;

for( ; dMatrix[m,j] == 0; m++ );

if( m == Level )

return null;

else

{

// Add i-row with m-row

for( int n = j; n 2 * Level; n++ )

dReverseMatrix[i,n] += dReverseMatrix[m,n];

}

}

// Format the i-row with "1" start

x = dReverseMatrix[i,j];

if( x != 1 )

{

for( int n = j; n 2 * Level; n++ )

if( dReverseMatrix[i,n] != 0 )

dReverseMatrix[i,n] /= x;

}

// Set 0 to the current column in the rows after current row

for( int s = Level - 1; s i;s-- )

{

x = dReverseMatrix[s,j];

for( int t = j; t 2 * Level; t++ )

dReverseMatrix[s,t] -= ( dReverseMatrix[i,t]* x );

}

}

// Format the first matrix into unit-matrix

for( int i = Level - 2; i = 0; i-- )

{

for( int j = i + 1 ; j Level; j++ )

if( dReverseMatrix[i,j] != 0 )

{

c = dReverseMatrix[i,j];

for( int n = j; n 2*Level; n++ )

dReverseMatrix[i,n] -= ( c * dReverseMatrix[j,n] );

}

}

double[,] dReturn = new double[Level, Level];

for( int i = 0; i Level; i++ )

for( int j = 0; j Level; j++ )

dReturn[i,j] = dReverseMatrix[i,j+Level];

return dReturn;

}

private double MatrixValue( double[,] MatrixList, int Level )

{

double[,] dMatrix = new double[Level, Level];

for( int i = 0; i Level; i++ )

for( int j = 0; j Level; j++ )

dMatrix[i,j] = MatrixList[i,j];

double c, x;

int k = 1;

for( int i = 0, j = 0; i Level j Level; i++, j++ )

{

if( dMatrix[i,j] == 0 )

{

int m = i;

for( ; dMatrix[m,j] == 0; m++ );

if( m == Level )

return 0;

else

{

// Row change between i-row and m-row

for( int n = j; n Level; n++ )

{

c = dMatrix[i,n];

dMatrix[i,n] = dMatrix[m,n];

dMatrix[m,n] = c;

}

// Change value pre-value

k *= (-1);

}

}

// Set 0 to the current column in the rows after current row

for( int s = Level - 1; s i;s-- )

{

x = dMatrix[s,j];

for( int t = j; t Level; t++ )

dMatrix[s,t] -= dMatrix[i,t]* ( x/dMatrix[i,j] );

}

}

double sn = 1;

for( int i = 0; i Level; i++ )

{

if( dMatrix[i,i] != 0 )

sn *= dMatrix[i,i];

else

return 0;

}

return k*sn;

}

调用如下：

double[,] dMatrix = new double[3,3]{{0,1,2},{1,0,1},{4,2,1}};

double[,] dReturn = ReverseMatrix( dMatrix, 3 );

if( dReturn != null )

{

for( int i=0; i 3; i++ )

Debug.WriteLine( string.Format( "{0} {1} {2}",

dReturn[i,0], dReturn[i,1],dReturn[i,2] ) );

}