In mathematics, a square matrix is said to be diagonally dominant if for every row of the matrix, the magnitude of the diagonal entry in a row is larger than or equal to the sum of the magnitudes of all the other (non-diagonal) entries in that row. More precisely, the matrix A is diagonally dominant if
For example, The matrix
is diagonally dominant because
|a11| ? |a12| + |a13| since |+3| ? |-2| + |+1|
|a22| ? |a21| + |a23| since |-3| ? |+1| + |+2|
|a33| ? |a31| + |a32| since |+4| ? |-1| + |+2|
Given a matrix A of n rows and n columns. The task is to check whether matrix A is diagonally dominant or not.
Examples :
Input : A = { { 3, -2, 1 }, { 1, -3, 2 }, { -1, 2, 4 } }; Output : YES Given matrix is diagonally dominant because absolute value of every diagonal element is more than sum of absolute values of corresponding row. Input : A = { { -2, 2, 1 }, { 1, 3, 2 }, { 1, -2, 0 } }; Output : NO
The idea is to run a loop from i = 0 to n-1 for the number of rows and for each row, run a loop j = 0 to n-1 find the sum of non-diagonal element i.e i != j. And check if diagonal element is greater than or equal to sum. If for any row, it is false, then return false or print “No”. Else print “YES”.
# Python Program to check # whether given matrix is # Diagonally Dominant Matrix. # check the given # matrix is Diagonally # Dominant Matrix or not. def isDDM(m, n) :
# for each row
for i in range ( 0 , n) :
# for each column, finding
# sum of each row.
sum = 0
for j in range ( 0 , n) :
sum = sum + abs (m[i][j])
# removing the
# diagonal element.
sum = sum - abs (m[i][i])
# checking if diagonal
# element is less than
# sum of non-diagonal
# element.
if ( abs (m[i][i]) < sum ) :
return False
return True
# Driver Code n = 3
m = [[ 3 , - 2 , 1 ],
[ 1 , - 3 , 2 ],
[ - 1 , 2 , 4 ]]
if ((isDDM(m, n))) :
print ( "YES" )
else :
print ( "NO" )
# This code is contributed by # Manish Shaw(manishshaw1) |
Output :
YES
Time Complexity: O(N2)
Auxiliary Space: O(1)
Please refer complete article on Diagonally Dominant Matrix for more details!