2019年5月29日星期三

Deteminant of a matrix 行列式值

Deteminant of a matrix 行列式值

#include <stdio.h>
#include <math.h>

#define N 3

int determinantOfMatrix(int mat[N][N], int n);
void display(int mat[N][N], int row, int col);

int main(void)
{
int matrix[N][N] =
{
{6, 1, 1 },
{4, -2, 5},
{2, 8, 7 }
};

display(matrix, N, N);

int determinant = determinantOfMatrix(matrix, N);

printf("Determinant of the matrix is: %d\n", determinant);

return 0;
}

int determinantOfMatrix(int mat[N][N], int n)
{
int det=0, p, h, k, i, j, temp[N][N];

if (n == 1)
{
return mat[0][0];
}
else if (n == 2)
{
det = (mat[0][0] * mat[1][1] - mat[0][1] * mat[1][0]);

return det;
}
else {
for (p = 0; p < n; p++) {
h = 0;
k = 0;

for (i = 1; i < n; i++) {
for (j = 0; j < n; j++) {
if (j==p) {
continue;
}

temp[h][k] = mat[i][j];
k++;

if (k == n-1) {
h++;
k = 0;
}
}
}

det = det + mat[0][p] * pow(-1, p) * determinantOfMatrix(temp, n-1);
}

return det;
}
}

void display(int mat[N][N], int row, int col)
{
int i;
int j;

for (i = 0; i < row; i++)
{
for (j = 0; j < col; j++)
{
printf("%d\t", mat[i][j]);
}

printf("\n");
}
}

輸出畫面
6 1 1
4 -2 5
2 8 7
Determinant of the matrix is: -306

Deteminant of a matrix 行列式值

Deteminant of a matrix 行列式值

//C program to find determinant of nxn matrix.
#include <stdio.h>
#include <math.h>

int a[20][20],m;
int determinant(int f[20][20],int a);
int main()
{
int i,j;
printf("\n\nEnter order of matrix : ");
scanf("%d",&m);
printf("\nEnter the elements of matrix\n");
for(i=1;i<=m;i++)
{
for(j=1;j<=m;j++)
{
printf("a[%d][%d] = ",i,j);
scanf("%d",&a[i][j]);
}
}
printf("\n\n---------- Matrix A is --------------\n");
for(i=1;i<=m;i++)
{
printf("\n");
for(j=1;j<=m;j++)
{
printf("\t%d \t",a[i][j]);
}
}
printf("\n \n");
printf("\n Determinant of Matrix A is %d .",determinant(a,m));
return 0;
}

int determinant(int f[20][20],int x)
{
int pr,c[20],d=0,b[20][20],j,p,q,t;
if(x==2)
{
d=0;
d=(f[1][1]*f[2][2])-(f[1][2]*f[2][1]);
return(d);
}
else
{
for(j=1;j<=x;j++)
{
int r=1,s=1;
for(p=1;p<=x;p++)
{
for(q=1;q<=x;q++)
{
if(p!=1&&q!=j)
{
b[r][s]=f[p][q];
s++;
if(s>x-1)
{
r++;
s=1;
}
}
}
}
for(t=1,pr=1;t<=(1+j);t++)
pr=(-1)*pr;
c[j]=pr*determinant(b,x-1);
}
for(j=1,d=0;j<=x;j++)
{
d=d+(f[1][j]*c[j]);
}
return(d);
}
}

輸入資料
4
1 0 2 -1
3 0 0 5
2 1 4 -3
1 0 5 0

輸出畫面
Enter order of matrix :
Enter the elements of matrix
a[1][1] = a[1][2] = a[1][3] = a[1][4] = a[2][1] = a[2][2] = a[2][3] = a[2][4] = a[3][1] = a[3][2] = a[3][3] = a[3][4] = a[4][1] = a[4][2] = a[4][3] = a[4][4] =

---------- Matrix A is --------------

1 0 2 -1
3 0 0 5
2 1 4 -3
1 0 5 0

Determinant of Matrix A is 30 .

Deteminant of a matrix 行列式值

Deteminant of a matrix 行列式值

源自於
https://www.bragitoff.com/2018/02/determinant-matrix-c-program/

/**************************************************
******DETERMINANT FROM GAUSS ELIMINATION***********
**************************************************/
#include<stdio.h>
#include<math.h>
/*******
Function that calculates the determinant of a square matrix using Gauss-Elimination :
Pass the square matrix as a parameter, and calculate and return the dete
Parameters: order(n),matrix[n][n]
********/
double determinant(int n, double a[n][n]){
double det=1;
int i,swapCount;
swapCount=gauEl(n,n,a);
for(i=0;i<n;i++){
det =det*a[i][i];
}
return det*pow(-1,swapCount);
}
/********
Function that perform Gauss Elimination
Pass the square matrix as a parameter, and calculate and store the upperTriangular(Gauss-Eliminated Matrix) in it
Parameters: rows(m),columns(n),matrix[m][n]
********/
int gauEl(int m, int p, double a[m][p]){
int i,j,k;
int swapCount=0;
for(i=0;i<m-1;i++){
//Partial Pivoting
for(k=i+1;k<m;k++){
//If diagonal element(absolute vallue) is smaller than any of the terms below it
if(fabs(a[i][i])<fabs(a[k][i])){
//Swap the rows
swapCount++;
for(j=0;j<p;j++){
double temp;
temp=a[i][j];
a[i][j]=a[k][j];
a[k][j]=temp;
}
}
}
//Begin Gauss Elimination
for(k=i+1;k<m;k++){
double term=a[k][i]/ a[i][i];
for(j=0;j<p;j++){
a[k][j]=a[k][j]-term*a[i][j];
}
}
}
return swapCount;
}

/*******
Function that prints the elements of a matrix row-wise
Parameters: rows(m),columns(n),matrix[m][n]
*******/
void printMatrix(int m, int n, double matrix[m][n]){
int i,j;
for(i=0;i<m;i++){
for(j=0;j<n;j++){
printf("%0.2lf\t",matrix[i][j]);
}
printf("\n");
}
}
/*******
Function that copies the elements of a matrix to another matrix
Parameters: rows(m),columns(n),matrix1[m][n] , matrix2[m][n]
*******/
void copyMatrix(int m, int n, double matrix1[m][n], double matrix2[m][n]){
int i,j;
for(i=0;i<m;i++){
for(j=0;j<n;j++){
matrix2[i][j]=matrix1[i][j];
}
}
}

int main(){
int n,i,j;
n=4;
printf("Enter the order of the matrix:\n(No. of rows/columns (n))\n");
printf("%d",n);
//Declare a matrix to store the user given matrix
/*
double a[3][3]= {{6., 1, 1},
{4., -2, 5},
{2., 8, 7}};
*/
double a[4][4] = { {1, 0, 2, -1},
{3, 0, 0, 5},
{2, 1, 4, -3},
{1, 0, 5, 0}};

/*double a[n][n] = {{1, 2},
{3, 2} }; */

printf("\nEnter the elements of matrix:\n");
printMatrix(n,n,a);
printf("\nThe determinant using Gauss Eliminiation is:\n\n%0.2lf\n",determinant(n,a));

}

輸出畫面
Enter the order of the matrix:
(No. of rows/columns (n))
4
Enter the elements of matrix:
1.00 0.00 2.00 -1.00
3.00 0.00 0.00 5.00
2.00 1.00 4.00 -3.00
1.00 0.00 5.00 0.00

The determinant using Gauss Eliminiation is:

30.00

2019年5月28日星期二

Deteminant of a matrix 行列式值

Deteminant of a matrix 行列式值

源自於https://www.geeksforgeeks.org/determinant-of-a-matrix/

// C program to find Deteminant of a matrix
#include<stdio.h>
#include<math.h>

// Dimension of input square matrix
#define N 4

// Function to get determinant of matrix
int determinantOfMatrix(int mat[N][N], int n)
{
int num1,num2,det = 1,index,total = 1; // Initialize result

// temporary array for storing row
int temp[n + 1];

//loop for traversing the diagonal elements
for(int i = 0; i < n; i++)
{
index = i; // intialize the index

//finding the index which has non zero value
while(mat[index][i] == 0 && index < n) {
index++;

}
if(index == n) // if there is non zero element
{
// the determinat of matrix as zero
continue;

}
if(index != i)
{
//loop for swaping the diagonal element row and index row
for(int j = 0; j < n; j++)
{
//swap(mat[index][j],mat[i][j]);
int tmp=mat[i][j];
mat[i][j]=mat[index][j];
mat[index][j]=tmp;

//determinant sign changes when we shift rows
//go through determinant properties
det = det*pow(-1,index-i);

}
}

//storing the values of diagonal row elements
for(int j = 0; j < n; j++)
{
temp[j] = mat[i][j];

}
//traversing every row below the diagonal element
for(int j = i+1; j < n; j++)
{
num1 = temp[i]; //value of diagonal element
num2 = mat[j][i]; //value of next row element

//traversing every column of row
// and multiplying to every row
for(int k = 0; k < n; k++)
{
//multiplying to make the diagonal
// element and next row element equal
mat[j][k] = (num1 * mat[j][k]) - (num2 * temp[k]);

}
total = total * num1; // Det(kA)=kDet(A);
}

}

//mulitplying the diagonal elements to get determinant
for(int i = 0; i < n; i++)
{
det = det * mat[i][i];

}
return (det/total); //Det(kA)/k=Det(A);
}

// Driver code
int main()
{
/* int mat[N][N] = {{6, 1, 1},
{4, -2, 5},
{2, 8, 7}}; */

int mat[N][N] = {{1, 0, 2, -1},
{3, 0, 0, 5},
{2, 1, 4, -3},
{1, 0, 5, 0}
};

/*int mat[N][N] = {{1, 2},
{3, 2} }; */

printf("Determinant of the matrix is : %d",
determinantOfMatrix(mat, N));
return 0;
}

輸出結果
Determinant of the matrix is : -30

Deteminant of a matrix 行列式值

Deteminant of a matrix

矩陣 A 是方陣(Square Matrix)時才會有行列式值，而其行列式是表為 A 或 det A，其運算之結果是一個純量(Scalar)，而非一個矩陣也不是一個向量。

// C program to find Deteminant of a matrix
#include<stdio.h>
#include<math.h>

// Dimension of input square matrix
#define N 2

// Function to get cofactor of mat[p][q] in temp[][]. n is current
// dimension of mat[][]
void getCofactor(int mat[N][N], int temp[N][N], int p, int q, int n)
{
int i = 0, j = 0;

// Looping for each element of the matrix
for (int row = 0; row < n; row++)
{
for (int col = 0; col < n; col++)
{
// Copying into temporary matrix only those element
// which are not in given row and column
if (row != p && col != q)
{
temp[i][j++] = mat[row][col];

// Row is filled, so increase row index and
// reset col index
if (j == n - 1)
{
j = 0;
i++;
}
}
}
}
}

/* Recursive function for finding determinant of matrix.
n is current dimension of mat[][]. */
int determinantOfMatrix(int mat[N][N], int n)
{
int D = 0; // Initialize result

// Base case : if matrix contains single element
if (n == 1)
return mat[0][0];

int temp[N][N]; // To store cofactors

int sign = 1; // To store sign multiplier

// Iterate for each element of first row
for (int f = 0; f < n; f++)
{
// Getting Cofactor of mat[0][f]
getCofactor(mat, temp, 0, f, n);
D += sign * mat[0][f] * determinantOfMatrix(temp, n - 1);

// terms are to be added with alternate sign
sign = -sign;
}

return D;
}

/* function for displaying the matrix */
void display(int mat[N][N], int row, int col)
{
for (int i = 0; i < row; i++)
{
for (int j = 0; j < col; j++)
printf(" %d", mat[i][j]);
printf("n");
}
}

// Driver program to test above functions
int main()
{
/* int mat[N][N] = {{6, 1, 1},
{4, -2, 5},
{2, 8, 7}}; */

/*int mat[N][N] = {{1, 0, 2, -1},
{3, 0, 0, 5},
{2, 1, 4, -3},
{1, 0, 5, 0}
}; */

int mat[N][N] = {{1, 2},
{3, 2} };

printf("Determinant of the matrix is : %d",
determinantOfMatrix(mat, N));
return 0;
}

輸出畫面
Determinant of the matrix is : -4

C語言例題6-8 LU分解法求線性代數解

Solve the following system of equations using LU Decomposition method:

$\begin{equation*} x_1 + x_2 + x_3 = 1 \end{equation*} \begin{equation*} 4x_1 + 3x_2 - x_3 = 6 \end{equation*} \begin{equation*} 3x_1 + 5x_2 + 3x_3 = 4 \end{equation*}$

Solution: Here, we have

A = $\begin{bmatrix} 1 & 1 & 1 \\ 4 & 3 & -1 \\ 3 & 5 & 3 \end{bmatrix} , X = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$ and $C = \begin{bmatrix} 1 \\ 6 \\ 4 \end{bmatrix}$ such that A X = C.

Mathematics | L U Decomposition of a System of Linear Equations

源自於 https://www.geeksforgeeks.org/l-u-decomposition-system-linear-equations/

Example:
Solve the following system of equations using LU Decomposition method:

$\begin{equation*} x_1 + x_2 + x_3 = 1 \end{equation*} \begin{equation*} 4x_1 + 3x_2 - x_3 = 6 \end{equation*} \begin{equation*} 3x_1 + 5x_2 + 3x_3 = 4 \end{equation*}$

Solution: Here, we have

A = $\begin{bmatrix} 1 & 1 & 1 \\ 4 & 3 & -1 \\ 3 & 5 & 3 \end{bmatrix} , X = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$ and $C = \begin{bmatrix} 1 \\ 6 \\ 4 \end{bmatrix}$ such that A X = C.

Now, we first consider $\begin{bmatrix} 1 & 1 & 1 \\ 4 & 3 & -1 \\ 3 & 5 & 3 \end{bmatrix}$ and convert it to row echelon form using Gauss Elimination Method.

So, by doing

(1) $\begin{equation*} R_2 \to R_2 - 4R_1 \end{equation*}$

(2) $\begin{equation*} R_3 \to R_3 - 3R_1 \end{equation*}$

we get

$\begin{bmatrix} 1 & 1 & 1 \\ 4 & 3 & -1 \\ 3 & 5 & 3 \end{bmatrix} \sim$ $\begin{bmatrix} 1 & 1 & 1 \\ 0 & -1 & -5 \\ 0 & 2 & 0 \end{bmatrix}$

Now, by doing

(3) $\begin{equation*} R_3 \to R_3 - (-2)R_2 \end{equation*}$

we get

$\sim \begin{bmatrix} 1 & 1 & 1 \\ 0 & -1 & -5 \\ 0 & 0 & -10 \end{bmatrix}$

(Remember to always keep ‘ – ‘ sign in between, replace ‘ + ‘ sign by two ‘ – ‘ signs)

Hence, we get L = $\begin{bmatrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 3 & -2 & 1 \end{bmatrix}$ and U = $\begin{bmatrix} 1 & 1 & 1 \\ 0 & -1 & -5 \\ 0 & 0 & -10 \end{bmatrix}$

(notice that in L matrix, $l_{21} = 4$ is from (1), $l_{31} = 3$ is from (2) and $l_{32} = -2$ is from (3))

Now, we assume Z $= \begin{bmatrix} z_1 \\ z_2 \\ z_3 \end{bmatrix}$ and solve L Z = C.

$\begin{bmatrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 3 & -2 & 1 \end{bmatrix} \begin{bmatrix} z_1 \\ z_2 \\ z_3 \end{bmatrix}$ $= \begin{bmatrix} 1 \\ 6 \\ 4 \end{bmatrix}$

So, we have $z_1 = 1 ,$ $4z_1 + z_2 = 6 ,$ $3z_1 - 2z_2 + z_3 = 4 .$

Solving, we get $z_1 = 1$ , $z_2 = 2$ and $z_3 = 5$ .

Now, we solve U X = Z

$\begin{bmatrix} 1 & 1 & 1 \\ 0 & -1 & -5 \\ 0 & 0 & -10 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$ $= \begin{bmatrix} 1 \\ 2 \\ 5 \end{bmatrix}$

Therefore, we get $x_1 + x_2 + x_3 = 1 ,$ $-x_2 - 5x_3 = 2$ , $-10x_3 = 5 .$

Thus, the solution to the given system of linear equations is $x_1 = 1$ , $x_2 = 0.5$ , $x_3 = -0.5$ and hence the matrix X = $\begin{bmatrix} 1 \\ 0.5 \\ -0.5 \end{bmatrix}$

/************** LU Decomposition for solving linear equations ***********/

[A] = [L][U]

where, L is the lower triangular matrix and

U is the upper triangular matrix.

The elements of these three matrices row-wise are:

[A] = { a11, a12, a13,

a21, a22, a23,

a31, a32, a33 }

[L] = { 1, 0, 0,

l21, 1, 0,

l31, l32, 1}

[U] = {u11, u12, u13,

0 , u22, u23,

0 , 0, u33}

-1.00x1 1.00x2 2.00x3 = 2.00

3.00x1 -1.00x2 1.00x3 = 6.00

-1.00x1 3.00x2 4.00x3 = 4.00

x3 = 2.000 x2 = -1.000 x1 = 1.000

#include<stdio.h>

#include<math.h>

int main()

{

int n,i,k,j,p;

float l[10][10]={0},u[10][10]={0},sum,z[10]={0},x[10]={0};

n=3;

printf("The order of square matrix: %2d\n",n);

float a[3][3]= { { 1, 1, 1 },

{ 4, 3, -1 },

{ 3, 5, 3 } };

float b[3]= {1,6,4};

for(i=0;i<n;i++)

{

printf("\nRow%1d\t",i);

for(j=0;j<n;j++)

printf("%0.2lf\t\t",a[i][j]);

}

printf("\n\nelements of b matrix\n");

for(i=0;i<n;i++)

printf("%0.2lf\t\t",b[i]);

//********** LU decomposition *****//

for(k=0;k<n;k++)

{

u[k][k]=1;

for(i=k;i<n;i++)

{

sum=0;

for(p=0;p<k;p++)

sum+=l[i][p]*u[p][k];

l[i][k]=a[i][k]-sum;

}

for(j=k+1;j<n;j++)

{

sum=0;

for(p=0;p<k;p++)

sum+=l[k][p]*u[p][j];

u[k][j]=(a[k][j]-sum)/l[k][k];

}

//******** Displaying LU matrix**********//

printf("\n\nLU matrix is : \n");

//==================L matrix===============

for(i=0;i<n;i++)

{

for(j=0;j<n;j++)

printf("%0.2lf\t",l[i][j]);

printf("\n");

}

printf("\n\n");

//==================U matrix===============

for(i=0;i<n;i++)

{

for(j=0;j<n;j++)

printf("%0.2lf\t",u[i][j]);

printf("\n");

}

//***** FINDING Z; LZ=b*********//

for(i=0;i<n;i++)

{ //forward subtitution method

sum=0;

for(p=0;p<i;p++)

sum+=l[i][p]*z[p];

z[i]=(b[i]-sum)/l[i][i];

}

//********** FINDING X; UX=Z***********//

for(i=n-1;i>=0;i--)

{

sum=0;

for(p=n-1;p>=i;p--)

sum+=u[i][p]*x[p];

x[i]=(z[i]-sum)/u[i][i];

}

//*********** DISPLAYING SOLUTION**************//

printf("\n\nSet of solution is\n");

for(i=0;i<n;i++)

printf("%0.2lf\t",x[i]);

return 0;

}

輸出畫面

The order of square matrix: 3

Row0 1.00 1.00 1.00

Row1 4.00 3.00 -1.00

Row2 3.00 5.00 3.00

elements of b matrix

1.00 6.00 4.00

LU matrix is :

1.00 0.00 0.00

4.00 -1.00 0.00

3.00 2.00 -10.00

1.00 1.00 1.00

0.00 1.00 5.00

0.00 0.00 1.00

Set of solution is

1.00 0.50 -0.50

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alex9ufo 聰明人求知心切

2019年5月29日星期三

Deteminant of a matrix 行列式值

Deteminant of a matrix 行列式值

Deteminant of a matrix 行列式值

2019年5月28日星期二

Deteminant of a matrix 行列式值

Deteminant of a matrix 行列式值

C語言例題6-8 LU分解法求線性代數解

C語言例題6-8 LU分解法求線性代數解

Mathematics | L U Decomposition of a System of Linear Equations

2024_09 作業3 以Node-Red 為主

2019年5月29日 星期三

2019年5月28日 星期二

C語言 例題6-8 LU分解法求線性代數解

Mathematics | L U Decomposition of a System of Linear Equations

2019年5月29日星期三

2019年5月28日星期二

C語言例題6-8 LU分解法求線性代數解