矩阵的秩是多少?
大小为M x N的矩阵A的秩定义为
(a)矩阵中线性独立列向量的最大数目，或
(b)矩阵中线性独立的行向量的最大数目。
我们强烈建议您单击此处并进行实践，然后再继续解决方案。
例子：
Input: mat[][] = {{10, 20, 10},
{20, 40, 20},
{30, 50, 0}}
Output: Rank is 2
Explanation: Ist and IInd rows are linearly dependent.
But Ist and 3rd or IInd and IIIrd are
independent.
Input: mat[][] = {{10, 20, 10},
{-20, -30, 10},
{30, 50, 0}}
Output: Rank is 2
Explanation: Ist and IInd rows are linearly independent.
So rank must be atleast 2. But all three rows
are linearly dependent (the first is equal to
the sum of the second and third) so the rank
must be less than 3. 换句话说，A的秩是A中所有非零未成年人的最大阶，其中次幂的阶次是确定它的平方子矩阵的边长。
因此，如果M
如何找到等级?
这个想法是基于对Row梯形形式的转换。
1) Let the input matrix be mat[][]. Initialize rank equals
to number of columns
// Before we visit row 'row', traversal of previous
// rows make sure that mat[row][0],....mat[row][row-1]
// are 0.
2) Do following for row = 0 to rank-1.
a) If mat[row][row] is not zero, make all elements of
current column as 0 except the element mat[row][row]
by finding appropriate multiplier and adding a the
multiple of row 'row'
b) Else (mat[row][row] is zero). Two cases arise:
(i) If there is a row below it with non-zero entry in
same column, then swap current 'row' and that row.
(ii) If all elements in current column below mat[r][row]
are 0, then remove this column by swapping it with
last column and reducing number of rank by 1.
Reduce row by 1 so that this row is processed again.
3) Number of remaining columns is rank of matrix.例子：
Input: mat[][] = {{10, 20, 10},
{-20, -30, 10},
{30, 50, 0}}
row = 0:
Since mat[0][0] is not 0, we are in case 2.a of above algorithm.
We set all entries of 0'th column as 0 (except entry mat[0][0]).
To do this, we subtract R1*(-2) from R2, i.e., R2 --> R2 - R1*(-2)
mat[][] = {{10, 20, 10},
{ 0, 10, 30},
{30, 50, 0}}
And subtract R1*3 from R3, i.e., R3 --> R3 - R1*3
mat[][] = {{10, 20, 10},
{ 0, 10, 30},
{ 0, -10, -30}}
row = 1:
Since mat[1][1] is not 0, we are in case 2.a of above algorithm.
We set all entries of 1st column as 0 (except entry mat[1][1]).
To do this, we subtract R2*2 from R1, i.e., R1 --> R1 - R2*2
mat[][] = {{10, 0, -50},
{ 0, 10, 30},
{ 0, -10, -30}}
And subtract R2*(-1) from R3, i.e., R3 --> R3 - R2*(-1)
mat[][] = {{10, 0, -50},
{ 0, 10, 30},
{ 0, 0, 0}}
row = 2:
Since Since mat[2][2] is 0, we are in case 2.b of above algorithm.
Since there is no row below it swap. We reduce the rank by 1 and
keep row as 2.
The loop doesn't iterate next time because loop termination condition
row <= rank-1 returns false.以下是上述想法的实现。
C++
// C++ program to find rank of a matrix
#include
using namespace std;
#define R 3
#define C 3
/* function for exchanging two rows of
a matrix */
void swap(int mat[R][C], int row1, int row2,
int col)
{
for (int i = 0; i < col; i++)
{
int temp = mat[row1][i];
mat[row1][i] = mat[row2][i];
mat[row2][i] = temp;
}
}
// Function to display a matrix
void display(int mat[R][C], int row, int col);
/* function for finding rank of matrix */
int rankOfMatrix(int mat[R][C])
{
int rank = C;
for (int row = 0; row < rank; row++)
{
// Before we visit current row 'row', we make
// sure that mat[row][0],....mat[row][row-1]
// are 0.
// Diagonal element is not zero
if (mat[row][row])
{
for (int col = 0; col < R; col++)
{
if (col != row)
{
// This makes all entries of current
// column as 0 except entry 'mat[row][row]'
double mult = (double)mat[col][row] /
mat[row][row];
for (int i = 0; i < rank; i++)
mat[col][i] -= mult * mat[row][i];
}
}
}
// Diagonal element is already zero. Two cases
// arise:
// 1) If there is a row below it with non-zero
// entry, then swap this row with that row
// and process that row
// 2) If all elements in current column below
// mat[r][row] are 0, then remvoe this column
// by swapping it with last column and
// reducing number of columns by 1.
else
{
bool reduce = true;
/* Find the non-zero element in current
column */
for (int i = row + 1; i < R; i++)
{
// Swap the row with non-zero element
// with this row.
if (mat[i][row])
{
swap(mat, row, i, rank);
reduce = false;
break ;
}
}
// If we did not find any row with non-zero
// element in current columnm, then all
// values in this column are 0.
if (reduce)
{
// Reduce number of columns
rank--;
// Copy the last column here
for (int i = 0; i < R; i ++)
mat[i][row] = mat[i][rank];
}
// Process this row again
row--;
}
// Uncomment these lines to see intermediate results
// display(mat, R, C);
// printf("\n");
}
return rank;
}
/* function for displaying the matrix */
void display(int mat[R][C], 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[][3] = {{10, 20, 10},
{-20, -30, 10},
{30, 50, 0}};
printf("Rank of the matrix is : %d",
rankOfMatrix(mat));
return 0;
} Java
// Java program to find rank of a matrix
class GFG {
static final int R = 3;
static final int C = 3;
// function for exchanging two rows
// of a matrix
static void swap(int mat[][],
int row1, int row2, int col)
{
for (int i = 0; i < col; i++)
{
int temp = mat[row1][i];
mat[row1][i] = mat[row2][i];
mat[row2][i] = temp;
}
}
// Function to display a matrix
static void display(int mat[][],
int row, int col)
{
for (int i = 0; i < row; i++)
{
for (int j = 0; j < col; j++)
System.out.print(" "
+ mat[i][j]);
System.out.print("\n");
}
}
// function for finding rank of matrix
static int rankOfMatrix(int mat[][])
{
int rank = C;
for (int row = 0; row < rank; row++)
{
// Before we visit current row
// 'row', we make sure that
// mat[row][0],....mat[row][row-1]
// are 0.
// Diagonal element is not zero
if (mat[row][row] != 0)
{
for (int col = 0; col < R; col++)
{
if (col != row)
{
// This makes all entries
// of current column
// as 0 except entry
// 'mat[row][row]'
double mult =
(double)mat[col][row] /
mat[row][row];
for (int i = 0; i < rank; i++)
mat[col][i] -= mult
* mat[row][i];
}
}
}
// Diagonal element is already zero.
// Two cases arise:
// 1) If there is a row below it
// with non-zero entry, then swap
// this row with that row and process
// that row
// 2) If all elements in current
// column below mat[r][row] are 0,
// then remvoe this column by
// swapping it with last column and
// reducing number of columns by 1.
else
{
boolean reduce = true;
// Find the non-zero element
// in current column
for (int i = row + 1; i < R; i++)
{
// Swap the row with non-zero
// element with this row.
if (mat[i][row] != 0)
{
swap(mat, row, i, rank);
reduce = false;
break ;
}
}
// If we did not find any row with
// non-zero element in current
// columnm, then all values in
// this column are 0.
if (reduce)
{
// Reduce number of columns
rank--;
// Copy the last column here
for (int i = 0; i < R; i ++)
mat[i][row] = mat[i][rank];
}
// Process this row again
row--;
}
// Uncomment these lines to see
// intermediate results display(mat, R, C);
// printf("\n");
}
return rank;
}
// Driver code
public static void main (String[] args)
{
int mat[][] = {{10, 20, 10},
{-20, -30, 10},
{30, 50, 0}};
System.out.print("Rank of the matrix is : "
+ rankOfMatrix(mat));
}
}
// This code is contributed by Anant Agarwal.Python3
# Python 3 program to find rank of a matrix
class rankMatrix(object):
def __init__(self, Matrix):
self.R = len(Matrix)
self.C = len(Matrix[0])
# Function for exchanging two rows of a matrix
def swap(self, Matrix, row1, row2, col):
for i in range(col):
temp = Matrix[row1][i]
Matrix[row1][i] = Matrix[row2][i]
Matrix[row2][i] = temp
# Function to Display a matrix
def Display(self, Matrix, row, col):
for i in range(row):
for j in range(col):
print (" " + str(Matrix[i][j]))
print ('\n')
# Find rank of a matrix
def rankOfMatrix(self, Matrix):
rank = self.C
for row in range(0, rank, 1):
# Before we visit current row
# 'row', we make sure that
# mat[row][0],....mat[row][row-1]
# are 0.
# Diagonal element is not zero
if Matrix[row][row] != 0:
for col in range(0, self.R, 1):
if col != row:
# This makes all entries of current
# column as 0 except entry 'mat[row][row]'
multiplier = (Matrix[col][row] /
Matrix[row][row])
for i in range(rank):
Matrix[col][i] -= (multiplier *
Matrix[row][i])
# Diagonal element is already zero.
# Two cases arise:
# 1) If there is a row below it
# with non-zero entry, then swap
# this row with that row and process
# that row
# 2) If all elements in current
# column below mat[r][row] are 0,
# then remvoe this column by
# swapping it with last column and
# reducing number of columns by 1.
else:
reduce = True
# Find the non-zero element
# in current column
for i in range(row + 1, self.R, 1):
# Swap the row with non-zero
# element with this row.
if Matrix[i][row] != 0:
self.swap(Matrix, row, i, rank)
reduce = False
break
# If we did not find any row with
# non-zero element in current
# columnm, then all values in
# this column are 0.
if reduce:
# Reduce number of columns
rank -= 1
# copy the last column here
for i in range(0, self.R, 1):
Matrix[i][row] = Matrix[i][rank]
# process this row again
row -= 1
# self.Display(Matrix, self.R,self.C)
return (rank)
# Driver Code
if __name__ == '__main__':
Matrix = [[10, 20, 10],
[-20, -30, 10],
[30, 50, 0]]
RankMatrix = rankMatrix(Matrix)
print ("Rank of the Matrix is:",
(RankMatrix.rankOfMatrix(Matrix)))
# This code is contributed by Vikas ChitturiC#
// C# program to find rank of a matrix
using System;
class GFG {
static int R = 3;
static int C = 3;
// function for exchanging two rows
// of a matrix
static void swap(int [,]mat,
int row1, int row2, int col)
{
for (int i = 0; i < col; i++)
{
int temp = mat[row1,i];
mat[row1,i] = mat[row2,i];
mat[row2,i] = temp;
}
}
// Function to display a matrix
static void display(int [,]mat,
int row, int col)
{
for (int i = 0; i < row; i++)
{
for (int j = 0; j < col; j++)
Console.Write(" "
+ mat[i,j]);
Console.Write("\n");
}
}
// function for finding rank of matrix
static int rankOfMatrix(int [,]mat)
{
int rank = C;
for (int row = 0; row < rank; row++)
{
// Before we visit current row
// 'row', we make sure that
// mat[row][0],....mat[row][row-1]
// are 0.
// Diagonal element is not zero
if (mat[row,row] != 0)
{
for (int col = 0; col < R; col++)
{
if (col != row)
{
// This makes all entries
// of current column
// as 0 except entry
// 'mat[row][row]'
double mult =
(double)mat[col,row] /
mat[row,row];
for (int i = 0; i < rank; i++)
mat[col,i] -= (int) mult
* mat[row,i];
}
}
}
// Diagonal element is already zero.
// Two cases arise:
// 1) If there is a row below it
// with non-zero entry, then swap
// this row with that row and process
// that row
// 2) If all elements in current
// column below mat[r][row] are 0,
// then remvoe this column by
// swapping it with last column and
// reducing number of columns by 1.
else
{
bool reduce = true;
// Find the non-zero element
// in current column
for (int i = row + 1; i < R; i++)
{
// Swap the row with non-zero
// element with this row.
if (mat[i,row] != 0)
{
swap(mat, row, i, rank);
reduce = false;
break ;
}
}
// If we did not find any row with
// non-zero element in current
// columnm, then all values in
// this column are 0.
if (reduce)
{
// Reduce number of columns
rank--;
// Copy the last column here
for (int i = 0; i < R; i ++)
mat[i,row] = mat[i,rank];
}
// Process this row again
row--;
}
// Uncomment these lines to see
// intermediate results display(mat, R, C);
// printf("\n");
}
return rank;
}
// Driver code
public static void Main ()
{
int [,]mat = {{10, 20, 10},
{-20, -30, 10},
{30, 50, 0}};
Console.Write("Rank of the matrix is : "
+ rankOfMatrix(mat));
}
}
// This code is contributed by nitin mittalPHP
Javascript
输出：
Rank of the matrix is : 2