三元二次方程组如何解？谢谢
已知方程：
1 (x-a)*(x-a) + (y-b)*(y-b) + (z-c)*(z-c)=k
2 (x-d)*(x-d) + (y-e)*(y-e) + (z-f)*(z-f)=l
2 (x-h)*(x-h) + (y-i)*(y-i) + (z-j)*(z-j)=m
除 x, y, z以外所有参数都是常量
怎样快速算出x,y,z的解。谢谢各位。
没有多少分了。
[解决办法]
通过1-2, 1-3得到2个关于x，y，z的三元一次方程，可以写出x，y关于z的一次表达式x=f(z), y=g(z)
带入到任意一个式子中就得到只关于z的一元二次方程，求得z的值，x，y的值也容易得到了
[解决办法]
分两次发给你
多元一次方程组求解封装类：1
using System;
using System.Collections.Generic;
using System.Text;
namespace _1fangcheng
{
public class Matrix
{
public double[,] Data;
public Matrix(int size)
{
this.Data = new double[size, size];
}
public Matrix(int rows, int cols)
{
this.Data = new double[rows, cols];
}
public Matrix(double[,] data)
{
Data = data;
}
public static Matrix operator +(Matrix M1, Matrix M2)
{
int r1 = M1.Data.GetLength(0); int r2 = M2.Data.GetLength(0);
int c1 = M1.Data.GetLength(1); int c2 = M2.Data.GetLength(1);
if ((r1 != r2) || (c1 != c2))
{
throw new System.Exception("Matrix dimensions donot agree");
}
double[,] res = new double[r1, c1];
for (int i = 0; i < r1; i++)
{
for (int j = 0; j < c1; j++)
{
res[i, j] = M1.Data[i, j] + M2.Data[i, j];
}
}
return new Matrix(res);
}
public static Matrix operator -(Matrix M1, Matrix M2)
{
int r1 = M1.Data.GetLength(0); int r2 = M2.Data.GetLength(0);
int c1 = M1.Data.GetLength(1); int c2 = M2.Data.GetLength(1);
if ((r1 != r2) || (c1 != c2))
{
throw new System.Exception("Matrix dimensions donot agree");
}
double[,] res = new double[r1, c1];
for (int i = 0; i < r1; i++)
{
for (int j = 0; j < c1; j++)
{
res[i, j] = M1.Data[i, j] - M2.Data[i, j];
}
}
return new Matrix(res);
}
public static Matrix operator *(Matrix M1, Matrix M2)
{
int r1 = M1.Data.GetLength(0); int r2 = M2.Data.GetLength(0);
int c1 = M1.Data.GetLength(1); int c2 = M2.Data.GetLength(1);
if (c1 != r2)
{
throw new System.Exception("Matrix dimensions donot agree");
}
double[,] res = new double[r1, c2];
for (int i = 0; i < r1; i++)
{
for (int j = 0; j < c2; j++)
{
for (int k = 0; k < r2; k++)
{
res[i, j] = res[i, j] + (M1.Data[i, k] * M2.Data[k, j]);
}
}
}
return new Matrix(res);
}
public static Matrix operator /(double i, Matrix M)
{
return new Matrix(scalmul(i, INV(M.Data)));
}
public static bool operator ==(Matrix M1, Matrix M2)
{
bool B = true;
int r1 = M1.Data.GetLength(0); int r2 = M2.Data.GetLength(0);
int c1 = M1.Data.GetLength(1); int c2 = M2.Data.GetLength(1);
if ((r1 != r2) || (c1 != c2))
{
return false;
}
else
{
for (int i = 0; i < r1; i++)
{
for (int j = 0; j < c1; j++)
{
if (M1.Data[i, j] != M2.Data[i, j])
B = false;
}
}
}
return B;
}
public static bool operator !=(Matrix M1, Matrix M2)
{
return !(M1 == M2);
}
public override bool Equals(object obj)
{
if (!(obj is Matrix))
{
return false;
}
return this == (Matrix)obj;
}
public double this[int i, int j]
{
get
{
return this.Data[i, j];
}
set
{
this.Data[i, j] = value;
}
}
public void Display()
{
int r1 = this.Data.GetLength(0); int c1 = this.Data.GetLength(1);
for (int i = 0; i < r1; i++)
{
for (int j = 0; j < c1; j++)
{
Console.Write(this.Data[i, j].ToString("N2") + " ");
}
Console.WriteLine();
}
Console.WriteLine();
}
public void Display(string format)
{
int r1 = this.Data.GetLength(0); int c1 = this.Data.GetLength(1);
for (int i = 0; i < r1; i++)
{
for (int j = 0; j < c1; j++)
{
Console.Write(this.Data[i, j].ToString(format) + " ");
}
Console.WriteLine();
}
Console.WriteLine();
}
public Matrix Inverse()
{
if ((this.IsSquare) && (!this.IsSingular))
{
return new Matrix(INV(this.Data));
}
else
{
throw new System.Exception(@"Cannot find inverse for non square /singular matrix");
}
}
public Matrix Transpose()
{
double[,] D = transpose(this.Data);
return new Matrix(D);
}
public static Matrix Zeros(int size)
{
double[,] D = new double[size, size];
return new Matrix(D);
}
public static Matrix Zeros(int rows, int cols)
{
double[,] D = new double[rows, cols];
return new Matrix(D);
}
public Matrix LinSolve(Matrix COF, Matrix CON)
{
return COF.Inverse() * CON;
}
public double Det()
{
if (this.IsSquare)
{
return det(this.Data);
}
else
{
throw new System.Exception("Cannot Determine the DET for a non square matrix");
}
}
public bool IsSquare
{
get
{
return (this.Data.GetLength(0) == this.Data.GetLength(1));
}
}
public bool IsSingular
{
get
{
return ((int)this.Det() == 0);
}
}
public int rows { get { return this.Data.GetLength(0); } }
public int cols { get { return this.Data.GetLength(1); } }
static double[,] INV(double[,] a)
{
int ro = a.GetLength(0);
int co = a.GetLength(1);
try
{
if (ro != co) { throw new System.Exception(); }
}
catch { Console.WriteLine("Cannot find inverse for an non square matrix"); }
int q; double[,] b = new double[ro, co]; double[,] I = eyes(ro);
for (int p = 0; p < ro; p++) { for (q = 0; q < co; q++) { b[p, q] = a[p, q]; } }
int i; double det = 1;
if (a[0, 0] == 0)
{
i = 1;
while (i < ro)
{
if (a[i, 0] != 0)
{
Matrix.interrow(a, 0, i);
Matrix.interrow(I, 0, i);
det *= -1;
break;
}
i++;
}
}
det *= a[0, 0];
Matrix.rowdiv(I, 0, a[0, 0]);
Matrix.rowdiv(a, 0, a[0, 0]);
for (int p = 1; p < ro; p++)
{
q = 0;
while (q < p)
{
Matrix.rowsub(I, p, q, a[p, q]);
Matrix.rowsub(a, p, q, a[p, q]);
q++;
}
if (a[p, p] != 0)
{
det *= a[p, p];
Matrix.rowdiv(I, p, a[p, p]);
Matrix.rowdiv(a, p, a[p, p]);
}
if (a[p, p] == 0)
{
for (int j = p + 1; j < co; j++)
{
if (a[p, j] != 0)// Column pivotting not supported
{
for (int p1 = 0; p1 < ro; p1++)
{
for (q = 0; q < co; q++)
{
a[p1, q] = b[p1, q];
}
}
return inverse(b);
}
}
}
}
for (int p = ro - 1; p > 0; p--)
{
for (q = p - 1; q >= 0; q--)
{
Matrix.rowsub(I, q, p, a[q, p]);
Matrix.rowsub(a, q, p, a[q, p]);
}
}
for (int p = 0; p < ro; p++)
{
for (q = 0; q < co; q++)
{
a[p, q] = b[p, q];
}
}
return (I);
}
