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C#矩阵求逆

时间:2016-05-23 13:09:56      阅读:235      评论:0      收藏:0      [点我收藏+]

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来源:http://zhidao.baidu.com/link?url=DiqAbq9YUYn3z7QjxGGoF0PLZwN-Y9ecqKB7Gy38JWRD1riMIYukVKXKq88pxtWLwIl6-iOUf2p3liE51phEe_

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] ) );
    }

 

C#矩阵求逆

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原文地址:http://www.cnblogs.com/144823836yj/p/5519540.html

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