假设有二维数组
int a[][]=new int[4][4];
我想找到矩阵的行列式，请帮助我知道如何找到数学，但我试图在programaticaly中找到它 我正在使用语言java和c＃但在这种情况下我认为c ++也会有所帮助
答案 0 :(得分：8)
如果你被固定为4x4，最简单的解决方案就是硬编码公式。
public double determinant(int[][] m) {
return
m[0][3] * m[1][2] * m[2][1] * m[3][0] - m[0][2] * m[1][3] * m[2][1] * m[3][0] -
m[0][3] * m[1][1] * m[2][2] * m[3][0] + m[0][1] * m[1][3] * m[2][2] * m[3][0] +
m[0][2] * m[1][1] * m[2][3] * m[3][0] - m[0][1] * m[1][2] * m[2][3] * m[3][0] -
m[0][3] * m[1][2] * m[2][0] * m[3][1] + m[0][2] * m[1][3] * m[2][0] * m[3][1] +
m[0][3] * m[1][0] * m[2][2] * m[3][1] - m[0][0] * m[1][3] * m[2][2] * m[3][1] -
m[0][2] * m[1][0] * m[2][3] * m[3][1] + m[0][0] * m[1][2] * m[2][3] * m[3][1] +
m[0][3] * m[1][1] * m[2][0] * m[3][2] - m[0][1] * m[1][3] * m[2][0] * m[3][2] -
m[0][3] * m[1][0] * m[2][1] * m[3][2] + m[0][0] * m[1][3] * m[2][1] * m[3][2] +
m[0][1] * m[1][0] * m[2][3] * m[3][2] - m[0][0] * m[1][1] * m[2][3] * m[3][2] -
m[0][2] * m[1][1] * m[2][0] * m[3][3] + m[0][1] * m[1][2] * m[2][0] * m[3][3] +
m[0][2] * m[1][0] * m[2][1] * m[3][3] - m[0][0] * m[1][2] * m[2][1] * m[3][3] -
m[0][1] * m[1][0] * m[2][2] * m[3][3] + m[0][0] * m[1][1] * m[2][2] * m[3][3];
}
对于一般的 NxN ，问题相当困难，各种算法的顺序为O(N!)，O(N^3)等。
答案 1 :(得分：3)
您可以查看以下链接：Determinant of matrix in Java
答案 2 :(得分：3)
static double DeterminantGaussElimination(double[,] matrix)
{
int n = int.Parse(System.Math.Sqrt(matrix.Length).ToString());
int nm1 = n - 1;
int kp1;
double p;
double det=1;
for (int k = 0; k < nm1; k++)
{
kp1 = k + 1;
for(int i=kp1;i<n;i++)
{
p = matrix[i, k] / matrix[k, k];
for (int j = kp1; j < n; j++)
matrix[i, j] = matrix[i, j] - p * matrix[k, j];
}
}
for (int i = 0; i < n; i++)
det = det * matrix[i, i];
return det;
}
这种方法不一定有效，因为你必须除以矩阵的一个成员，如果矩阵的一个成员是0，你可以得到结果det = NaN。
答案 3 :(得分：2)
如果您知道如何以数学方式进行，那么应用这些知识并编写与您必须手动计算行列式（在纸上）时完全相同的代码。正如伊格纳西奥在评论中告诉你的那样，请告诉我们你尝试了什么，也许你会得到更好的答案。我很乐意编辑我的答案并帮助你。
编辑：
因为看起来这里的问题不是公式本身，而是理解如何使用数组，我建议像这个教程（我假设你使用C＃）： how to: arrays in C#
答案 4 :(得分：1)
生成整数1..N的所有变换，并且对于每个这样的序列s_1..s_N，计算单元格M（i，s_i）的值的乘积乘以符号值p（s_1..s_i） ），如果i-s_1是偶数则为1，否则为-1 。总结所有这些产品。
<强>后记
正如polygene所说，算法效率低下，而且这个算法是O（N！），因为它会不断重新计算共享的子产品。但是如果懒洋洋地做的话，它的直观性和空间效率。
哦，上面的符号函数是错误的：P（s_1..s_i）是+1，如果s_i在序列1..N中有奇数索引，s_1..s_ {i-1}被删除，并且 - 1表示偶数索引。
答案 5 :(得分：1)
我可以确认以前的解决方案适用于3x3和4x4但不适用于5x5等。 遵循适用于任何维度（也是5x5或更多）的解决方案（非常简单）。
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace MatrixTest
{
public class Matrix
{
private int dimension; //number of rows & colums for matrix
private double[][] matrix; //holds values of matrix itself
/// <summary>
/// Create dim*dim matrix and fill it with data passed to this constructor.
/// </summary>
/// <param name="double_array"></param>
/// <param name="dim"></param>
public Matrix(double[][] double_array)
{
matrix = double_array;
dimension = matrix.Length;
// check square matrix:
for (int i = 0; i < dimension; i++)
if (matrix[i].Length != dimension)
throw new Exception("Matrix is not square");
}
/// <summary>
/// Get determinant of current matrix
/// </summary>
/// <returns></returns>
public double Determinant()
{
if (dimension == 1)
return matrix[0][0];
// else ricorsive call:
double det = 0;
for (int j = 0; j < dimension; j++)
{
if (j % 2 == 0)
det += matrix[0][j] * GetSubmatrix(this, 0, j).Determinant();
else
det -= matrix[0][j] * GetSubmatrix(this, 0, j).Determinant();
}
return det;
}
/// <summary>
/// Return a new Matrix with:
/// dimension = passed matrix dimension - 1
/// elements = all element of the original matrix, except row and column specified
/// </summary>
/// <param name="m"></param>
/// <param name="rowToExclude"></param>
/// <param name="colToExclude"></param>
/// <returns></returns>
public Matrix GetSubmatrix(Matrix m, int rowToExclude, int colToExclude)
{
double[][] values = new double[m.dimension - 1][];
for (int i = 0; i < m.dimension; i++)
{
// create row array:
if (i < m.dimension - 1)
values[i] = new double[m.dimension - 1];
// copy values:
for (int j = 0; j < m.dimension; j++)
if (i != rowToExclude && j != colToExclude)
values[i < rowToExclude ? i : i - 1][j < colToExclude ? j : j - 1] = m.matrix[i][j];
}
return new Matrix(values);
}
internal class Program
{
private static void Main(string[] args)
{
Matrix mat02 = new Matrix(
new double[][] {
new double[] { 1.0, 2.0},
new double[] { -2.0, -5.0} });
Matrix mat03 = new Matrix(
new double[][] {
new double[] { 1.0, 2.0, -1.0 },
new double[] { -2.0, -5.0, -1.0},
new double[] { 1.0, -1.0, -2.0 } });
Matrix mat04 = new Matrix(
new double[][] {
new double[] {1.0, 2.0, 1.0, 3.0},
new double[] { -2.0, -5.0, -2.0, 1.0 },
new double[] { 1.0, -1.0, -3.0, 2.0 },
new double[] {4.0, -1.0, -3.0, 1.0} });
Matrix mat05 = new Matrix(
new double[][] {
new double[] {1.0, 2.0, 1.0, 2.0, 3.0},
new double[] {2.0, 1.0, 2.0, 2.0, 1.0},
new double[] {3.0, 1.0, 3.0, 1.0, 2.0},
new double[] {1.0, 2.0, 4.0, 3.0, 2.0},
new double[] {2.0, 2.0, 1.0, 2.0, 1.0} });
double determinant = mat02.Determinant();
Console.WriteLine("determinant is: {0}", determinant);
determinant = mat03.Determinant();
Console.WriteLine("determinant is: {0}", determinant);
determinant = mat04.Determinant();
Console.WriteLine("determinant is: {0}", determinant);
determinant = mat05.Determinant();
Console.WriteLine("determinant is: {0}", determinant);
Console.ReadLine();
}
}
}
}
答案 6 :(得分：0)
对于任何可能在寻找矩阵计算行列式算法时遇到此问题的人，请注意以上发布的解决方案，其中包含此代码：
static double DeterminantGaussElimination(double[,] matrix)
{
int n = int.Parse(System.Math.Sqrt(matrix.Length).ToString());
int nm1 = n - 1;
int kp1;
double p;
double det=1;
for (int k = 0; k < nm1; k++)
{
kp1 = k + 1;
for(int i=kp1;i<n;i++)
{
p = matrix[i, k] / matrix[k, k];
for (int j = kp1; j < n; j++)
matrix[i, j] = matrix[i, j] - p * matrix[k, j];
}
}
for (int i = 0; i < n; i++)
det = det * matrix[i, i];
return det;
}
适用于3x3和4x4但不适用于5x5等，
这是一个证据：
using System;
public class Matrix
{
private int row_matrix; //number of rows for matrix
private int column_matrix; //number of colums for matrix
private double[,] matrix; //holds values of matrix itself
//create r*c matrix and fill it with data passed to this constructor
public Matrix(double[,] double_array)
{
matrix = double_array;
row_matrix = matrix.GetLength(0);
column_matrix = matrix.GetLength(1);
Console.WriteLine("Contructor which sets matrix size {0}*{1} and fill it with initial data executed.", row_matrix, column_matrix);
}
//returns total number of rows
public int countRows()
{
return row_matrix;
}
//returns total number of columns
public int countColumns()
{
return column_matrix;
}
//returns value of an element for a given row and column of matrix
public double readElement(int row, int column)
{
return matrix[row, column];
}
//sets value of an element for a given row and column of matrix
public void setElement(double value, int row, int column)
{
matrix[row, column] = value;
}
public double deterMatrix()
{
int n = int.Parse(System.Math.Sqrt(matrix.Length).ToString());
int nm1 = n - 1;
int kp1;
double p;
double det = 1;
for (int k = 0; k < nm1; k++)
{
kp1 = k + 1;
for (int i = kp1; i < n; i++)
{
p = matrix[i, k] / matrix[k, k];
for (int j = kp1; j < n; j++)
matrix[i, j] = matrix[i, j] - p * matrix[k, j];
}
}
for (int i = 0; i < n; i++)
det = det * matrix[i, i];
return det;
}
}
internal class Program
{
private static void Main(string[] args)
{
Matrix mat03 = new Matrix(new[,]
{
{1.0, 2.0, -1.0},
{-2.0, -5.0, -1.0},
{1.0, -1.0, -2.0},
});
Matrix mat04 = new Matrix(new[,]
{
{1.0, 2.0, 1.0, 3.0},
{-2.0, -5.0, -2.0, 1.0},
{1.0, -1.0, -3.0, 2.0},
{4.0, -1.0, -3.0, 1.0},
});
Matrix mat05 = new Matrix(new[,]
{
{1.0, 2.0, 1.0, 2.0, 3.0},
{2.0, 1.0, 2.0, 2.0, 1.0},
{3.0, 1.0, 3.0, 1.0, 2.0},
{1.0, 2.0, 4.0, 3.0, 2.0},
{2.0, 2.0, 1.0, 2.0, 1.0},
});
double determinant = mat03.deterMatrix();
Console.WriteLine("determinant is: {0}", determinant);
determinant = mat04.deterMatrix();
Console.WriteLine("determinant is: {0}", determinant);
determinant = mat05.deterMatrix();
Console.WriteLine("determinant is: {0}", determinant);
}
}
但是，由于4x4的具体问题，我发现算法正确（至少在我测试过的几个案例中）。
如果您运行上面的代码，您将获得：
行列式是：-8 决定因素是：-142 决定因素是：NaN
答案 7 :(得分：0)
我知道这个问题已经解决了，但是对于任何可能需要一种可以计算任意维度矩阵的行列式算法的人，这就是我想出的。
此类使用许多不同的方法使矩阵为三角形，然后计算其行列式。它可以用于500 x 500甚至更大的高尺寸矩阵。此类的优点是，您可以获得 BigDecimal的结果，因此没有无限，并且您将始终获得准确的答案。顺便说一句，使用许多种方法并避免递归会导致更快的方法和更高的答案性能。希望对您有所帮助。
import java.math.BigDecimal;
public class DeterminantCalc {
private double[][] matrix;
private int sign = 1;
DeterminantCalc(double[][] matrix) {
this.matrix = matrix;
}
public int getSign() {
return sign;
}
public BigDecimal determinant() {
BigDecimal deter;
if (isUpperTriangular() || isLowerTriangular())
deter = multiplyDiameter().multiply(BigDecimal.valueOf(sign));
else {
makeTriangular();
deter = multiplyDiameter().multiply(BigDecimal.valueOf(sign));
}
return deter;
}
/* receives a matrix and makes it triangular using allowed operations
on columns and rows
*/
public void makeTriangular() {
for (int j = 0; j < matrix.length; j++) {
sortCol(j);
for (int i = matrix.length - 1; i > j; i--) {
if (matrix[i][j] == 0)
continue;
double x = matrix[i][j];
double y = matrix[i - 1][j];
multiplyRow(i, (-y / x));
addRow(i, i - 1);
multiplyRow(i, (-x / y));
}
}
}
public boolean isUpperTriangular() {
if (matrix.length < 2)
return false;
for (int i = 0; i < matrix.length; i++) {
for (int j = 0; j < i; j++) {
if (matrix[i][j] != 0)
return false;
}
}
return true;
}
public boolean isLowerTriangular() {
if (matrix.length < 2)
return false;
for (int j = 0; j < matrix.length; j++) {
for (int i = 0; j > i; i++) {
if (matrix[i][j] != 0)
return false;
}
}
return true;
}
public BigDecimal multiplyDiameter() {
BigDecimal result = BigDecimal.ONE;
for (int i = 0; i < matrix.length; i++) {
for (int j = 0; j < matrix.length; j++) {
if (i == j)
result = result.multiply(BigDecimal.valueOf(matrix[i][j]));
}
}
return result;
}
// when matrix[i][j] = 0 it makes it's value non-zero
public void makeNonZero(int rowPos, int colPos) {
int len = matrix.length;
outer:
for (int i = 0; i < len; i++) {
for (int j = 0; j < len; j++) {
if (matrix[i][j] != 0) {
if (i == rowPos) { // found "!= 0" in it's own row, so cols must be added
addCol(colPos, j);
break outer;
}
if (j == colPos) { // found "!= 0" in it's own col, so rows must be added
addRow(rowPos, i);
break outer;
}
}
}
}
}
//add row1 to row2 and store in row1
public void addRow(int row1, int row2) {
for (int j = 0; j < matrix.length; j++)
matrix[row1][j] += matrix[row2][j];
}
//add col1 to col2 and store in col1
public void addCol(int col1, int col2) {
for (int i = 0; i < matrix.length; i++)
matrix[i][col1] += matrix[i][col2];
}
//multiply the whole row by num
public void multiplyRow(int row, double num) {
if (num < 0)
sign *= -1;
for (int j = 0; j < matrix.length; j++) {
matrix[row][j] *= num;
}
}
//multiply the whole column by num
public void multiplyCol(int col, double num) {
if (num < 0)
sign *= -1;
for (int i = 0; i < matrix.length; i++)
matrix[i][col] *= num;
}
// sort the cols from the biggest to the lowest value
public void sortCol(int col) {
for (int i = matrix.length - 1; i >= col; i--) {
for (int k = matrix.length - 1; k >= col; k--) {
double tmp1 = matrix[i][col];
double tmp2 = matrix[k][col];
if (Math.abs(tmp1) < Math.abs(tmp2))
replaceRow(i, k);
}
}
}
//replace row1 with row2
public void replaceRow(int row1, int row2) {
if (row1 != row2)
sign *= -1;
double[] tempRow = new double[matrix.length];
for (int j = 0; j < matrix.length; j++) {
tempRow[j] = matrix[row1][j];
matrix[row1][j] = matrix[row2][j];
matrix[row2][j] = tempRow[j];
}
}
//replace col1 with col2
public void replaceCol(int col1, int col2) {
if (col1 != col2)
sign *= -1;
System.out.printf("replace col%d with col%d, sign = %d%n", col1, col2, sign);
double[][] tempCol = new double[matrix.length][1];
for (int i = 0; i < matrix.length; i++) {
tempCol[i][0] = matrix[i][col1];
matrix[i][col1] = matrix[i][col2];
matrix[i][col2] = tempCol[i][0];
}
}
}
然后，此类从用户那里接收n x n的矩阵，或者可以生成nxn的随机矩阵，然后计算它的行列式。它还显示了解和最终的三角矩阵。
import java.math.BigDecimal;
import java.security.SecureRandom;
import java.text.NumberFormat;
import java.util.Scanner;
public class DeterminantTest {
public static void main(String[] args) {
String determinant;
//generating random numbers
int len = 500;
SecureRandom random = new SecureRandom();
double[][] matrix = new double[len][len];
for (int i = 0; i < len; i++) {
for (int j = 0; j < len; j++) {
matrix[i][j] = random.nextInt(500);
System.out.printf("%15.2f", matrix[i][j]);
}
}
System.out.println();
/*double[][] matrix = {
{1, 5, 2, -2, 3, 2, 5, 1, 0, 5},
{4, 6, 0, -2, -2, 0, 1, 1, -2, 1},
{0, 5, 1, 0, 1, -5, -9, 0, 4, 1},
{2, 3, 5, -1, 2, 2, 0, 4, 5, -1},
{1, 0, 3, -1, 5, 1, 0, 2, 0, 2},
{1, 1, 0, -2, 5, 1, 2, 1, 1, 6},
{1, 0, 1, -1, 1, 1, 0, 1, 1, 1},
{1, 5, 5, 0, 3, 5, 5, 0, 0, 6},
{1, -5, 2, -2, 3, 2, 5, 1, 1, 5},
{1, 5, -2, -2, 3, 1, 5, 0, 0, 1}
};
double[][] matrix = menu();*/
DeterminantCalc deter = new DeterminantCalc(matrix);
BigDecimal det = deter.determinant();
determinant = NumberFormat.getInstance().format(det);
for (int i = 0; i < matrix.length; i++) {
for (int j = 0; j < matrix.length; j++) {
System.out.printf("%15.2f", matrix[i][j]);
}
System.out.println();
}
System.out.println();
System.out.printf("%s%s%n", "Determinant: ", determinant);
System.out.printf("%s%d", "sign: ", deter.getSign());
}
public static double[][] menu() {
Scanner scanner = new Scanner(System.in);
System.out.print("Matrix Dimension: ");
int dim = scanner.nextInt();
double[][] inputMatrix = new double[dim][dim];
System.out.println("Set the Matrix: ");
for (int i = 0; i < dim; i++) {
System.out.printf("%5s%d%n", "row", i + 1);
for (int j = 0; j < dim; j++) {
System.out.printf("M[%d][%d] = ", i + 1, j + 1);
inputMatrix[i][j] = scanner.nextDouble();
}
System.out.println();
}
scanner.close();
return inputMatrix;
}
}