给定维数为N的正方形矩阵mat [] [] ，任务是使用枢轴凝聚法找到矩阵的行列式。
例子：
Input: mat[][] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}
Output: 0
Explanation:
Performing R3 = R3 – R2 modifies the matrix mat[][] to {{1, 2, 3}, {4, 5, 6}, {1, 1, 1}}.
Performing R2 = R2 – R1 modifies the matrix mat[][] to {{1, 2, 3}, {1, 1, 1}, {1, 1, 1}}.
Now, the rows R2 and R3 are equal.
Therefore, the determinant will of the matrix becomes equal to zero (using the property of matrix).
Input: mat[][] = {{1, 0, 2, -1}, {3, 0, 0, 5}, {2, 1, 4, -3}, {1, 0, 5, 0}}
Output: 30
方法：想法是使用“ Pivotal Condensation”方法来计算矩阵mat [] []的行列式。下面是对提出的方法的详细说明：
在这种计算维数N×N的行列式的方法中，平方矩阵为：
- 首先，将尺寸为N * N的矩阵A [] []简化为尺寸为(N – 1)*(N – 1)的矩阵B [] [] ，使得：
- 然后，可以使用公式从矩阵B [] []中找出A [] []的行列式值，
- 现在进一步将矩阵简化为(N – 2)*(N – 2)并计算矩阵B [] []的行列式。
- 并重复上述过程，直到矩阵的尺寸变为2 * 2为止。
- 然后，使用公式det(A)= ad-bc计算矩阵的行列式为2×2 说A [] []为{{a，b}，{c，d}} 。
请按照以下步骤解决问题：
- 初始化一个变量D ，以存储矩阵的行列式。
- 在N大于2时进行迭代并检查以下情况：
- 检查mat [0] [0]是否为0 ，然后使用矩阵属性将当前行与下一行交换，以使mat [i] [0]> 0 。
- 否则，如果未找到mat [i] [0]> 0的行，则打印零。
- 现在，将D乘以pow(1 / / mat [0] [0]，N – 2) 。
- 使用公式b [i – 1] [j – 1] = mat [0] [0 * mat [i]计算下一个矩阵，即B [] [] ，其尺寸为(N – 1)x(N – 1) ] [i] – mat [0] [j] * mat [i] [0] 。
- 分配垫= B。
- D乘以尺寸为2×2的矩阵mat [] []的行列式，即mat [0] [0] * mat [1] [1] – mat [0] [j] * mat [i] [0 ]。
- 最后，打印存储在D中的值。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to swap values
void swap(float& i, float& j)
{
float temp = i;
i = j;
j = temp;
}
// Function to find the determinant
// of matrix M[][]
float determinantOfMatrix(
vector > mat, int N)
{
float mul = 1;
// Iterate over N while N > 2
while (N > 2) {
// Store the reduced matrix
// of dimension (N-1)x(N-1)
float M[N - 1][N - 1];
int next_index = 1;
// Check if first element
// of first row is zero
while (mat[0][0] == 0) {
if (mat[next_index][0] > 0) {
// For swapping
for (int k = 0; k < N; k++) {
swap(mat[0][k],
mat[next_index][k]);
}
// Update mul
mul = mul * pow((-1),
(next_index));
}
else if (next_index == (N - 1))
return 0;
next_index++;
}
// Store the first element
// of the matrix
float p = mat[0][0];
// Multiply the mul by
// (1/p) to the power n-2
mul = mul * pow(1 / p, N - 2);
// Calculate the next matrix
// of dimension (N-1) x (N-1)
for (int i = 1; i < N; i++) {
for (int j = 1; j < N; j++) {
// Calculate each element of
// the matrix from previous
// matrix
M[i - 1][j - 1] = mat[0][0]
* mat[i][j]
- mat[i][0]
* mat[0][j];
}
}
// Copy elements of the matrix
// M into mat to use it in
// next iteration
for (int i = 0;
i < (N - 1); i++) {
for (int j = 0;
j < (N - 1); j++) {
mat[i][j] = M[i][j];
}
}
// Decrement N by one
N--;
}
// Calculate the determinant
// of reduced 2x2 matrix and
// multiply it with factor mul
float D = mul * (mat[0][0]
* mat[1][1]
- mat[0][1]
* mat[1][0]);
// Print the determinant
cout << D;
}
// Driver Code
int main()
{
// Given matrix
vector > mat = { { 1, 0, 2, -1 },
{ 3, 0, 0, 5 },
{ 2, 1, 4, -3 },
{ 1, 0, 5, 0 } };
// Size of the matrix
int N = mat.size();
// Function Call
determinantOfMatrix(mat, N);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG
{
// Function to find the determinant
// of matrix M[][]
static void determinantOfMatrix(int[][] mat, int N)
{
int mul = 1;
// Iterate over N while N > 2
while (N > 2)
{
// Store the reduced matrix
// of dimension (N-1)x(N-1)
int [][]M = new int[N - 1][N - 1];
int next_index = 1;
// Check if first element
// of first row is zero
while (mat[0][0] == 0)
{
if (mat[next_index][0] > 0)
{
// For swapping
for (int k = 0; k < N; k++)
{
int temp = mat[0][k];
mat[0][k] = mat[next_index][k];
mat[next_index][k] = temp;
}
// Update mul
mul = (int) (mul * Math.pow((-1),
(next_index)));
}
else if (next_index == (N - 1))
return;
next_index++;
}
// Store the first element
// of the matrix
int p = mat[0][0];
// Multiply the mul by
// (1/p) to the power n-2
mul = (int) (mul * Math.pow(1 / p, N - 2));
// Calculate the next matrix
// of dimension (N-1) x (N-1)
for (int i = 1; i < N; i++)
{
for (int j = 1; j < N; j++)
{
// Calculate each element of
// the matrix from previous
// matrix
M[i - 1][j - 1] = mat[0][0]
* mat[i][j]
- mat[i][0]
* mat[0][j];
}
}
// Copy elements of the matrix
// M into mat to use it in
// next iteration
for (int i = 0;
i < (N - 1); i++)
{
for (int j = 0;
j < (N - 1); j++)
{
mat[i][j] = M[i][j];
}
}
// Decrement N by one
N--;
}
// Calculate the determinant
// of reduced 2x2 matrix and
// multiply it with factor mul
int D = mul * (mat[0][0]
* mat[1][1]
- mat[0][1]
* mat[1][0]);
// Print the determinant
System.out.print(D);
}
// Driver Code
public static void main(String[] args)
{
// Given matrix
int[][] mat = { { 1, 0, 2, -1 },
{ 3, 0, 0, 5 },
{ 2, 1, 4, -3 },
{ 1, 0, 5, 0 } };
// Size of the matrix
int N = mat.length;
// Function Call
determinantOfMatrix(mat, N);
}
}
// This code is contributed by 29AjayKumarPython3
# Pyhton 3 program for the above approach
# Function to find the determinant
# of matrix M[][]
def determinantOfMatrix(mat, N):
mul = 1
# Iterate over N while N > 2
while (N > 2):
# Store the reduced matrix
# of dimension (N-1)x(N-1)
M = [[0 for i in range(N-1)] for j in range(N-1)]
next_index = 1
# Check if first element
# of first row is zero
while (mat[0][0] == 0):
if (mat[next_index][0] > 0):
# For swapping
for k in range(N):
temp = mat[0][k]
mat[0][k] = mat[next_index][k]
mat[next_index][k] = temp
# Update mul
mul = mul * pow((-1),(next_index))
elif (next_index == (N - 1)):
return 0;
next_index += 1
# Store the first element
# of the matrix
p = mat[0][0]
# Multiply the mul by
# (1/p) to the power n-2
mul = mul * pow(1 / p, N - 2)
# Calculate the next matrix
# of dimension (N-1) x (N-1)
for i in range(1,N):
for j in range(1,N,1):
# Calculate each element of
# the matrix from previous
# matrix
M[i - 1][j - 1] = mat[0][0] * mat[i][j] - mat[i][0] * mat[0][j]
# Copy elements of the matrix
# M into mat to use it in
# next iteration
for i in range(N - 1):
for j in range(N - 1):
mat[i][j] = M[i][j]
# Decrement N by one
N -= 1
# Calculate the determinant
# of reduced 2x2 matrix and
# multiply it with factor mul
D = mul * (mat[0][0] * mat[1][1] - mat[0][1] * mat[1][0])
# Print the determinant
print(int(D))
# Driver Code
if __name__ == '__main__':
# Given matrix
mat = [[1, 0, 2, -1],[3, 0, 0, 5], [2, 1, 4, -3], [1, 0, 5, 0]]
# Size of the matrix
N = len(mat)
# Function Call
determinantOfMatrix(mat, N)
# This code is contributed by bgangwar59.C#
// C# program for the above approach
using System;
public class GFG
{
// Function to find the determinant
// of matrix [,]M
static void determinantOfMatrix(int[,] mat, int N)
{
int mul = 1;
// Iterate over N while N > 2
while (N > 2)
{
// Store the reduced matrix
// of dimension (N-1)x(N-1)
int [,]M = new int[N - 1,N - 1];
int next_index = 1;
// Check if first element
// of first row is zero
while (mat[0,0] == 0)
{
if (mat[next_index,0] > 0)
{
// For swapping
for (int k = 0; k < N; k++)
{
int temp = mat[0,k];
mat[0,k] = mat[next_index,k];
mat[next_index,k] = temp;
}
// Update mul
mul = (int) (mul * Math.Pow((-1),
(next_index)));
}
else if (next_index == (N - 1))
return;
next_index++;
}
// Store the first element
// of the matrix
int p = mat[0,0];
// Multiply the mul by
// (1/p) to the power n-2
mul = (int) (mul * Math.Pow(1 / p, N - 2));
// Calculate the next matrix
// of dimension (N-1) x (N-1)
for (int i = 1; i < N; i++)
{
for (int j = 1; j < N; j++)
{
// Calculate each element of
// the matrix from previous
// matrix
M[i - 1,j - 1] = mat[0,0]
* mat[i,j]
- mat[i,0]
* mat[0,j];
}
}
// Copy elements of the matrix
// M into mat to use it in
// next iteration
for (int i = 0;
i < (N - 1); i++)
{
for (int j = 0;
j < (N - 1); j++)
{
mat[i,j] = M[i,j];
}
}
// Decrement N by one
N--;
}
// Calculate the determinant
// of reduced 2x2 matrix and
// multiply it with factor mul
int D = mul * (mat[0,0]
* mat[1,1]
- mat[0,1]
* mat[1,0]);
// Print the determinant
Console.Write(D);
}
// Driver Code
public static void Main(String[] args)
{
// Given matrix
int[,] mat = { { 1, 0, 2, -1 },
{ 3, 0, 0, 5 },
{ 2, 1, 4, -3 },
{ 1, 0, 5, 0 } };
// Size of the matrix
int N = mat.GetLength(0);
// Function Call
determinantOfMatrix(mat, N);
}
}
// This code is contributed by Rajput-Ji输出：
30时间复杂度： O(N 3 )
辅助空间： O(N 2 )