The number of different n × n symmetric matrices with each element being either 0 or 1 is: (Note: power(2, x) is same as 2x)
(A) power(2, n)
(B) power(2, n2)
(C) power(2, (n2 + n)/2)
(D) power(2, (n2 – n)/2)
Explanation: Given matrix is Symmetric, i.e, A[i][j] = A[j][i] property of Symmetric matrix. That means we can change the elements of upper with diagonal and lower will be fixed. Or we can change the elements of lower with diagonal and upper will be fixed. So, total elements will be
= 1+2+3+ … + (n-1) + n = n(n+1)/2 = (n2+n)/2
Now there are two choices either 0 or 1 for given places ((n2+n)/2).
Therefore, total number of such possible Symmetric matrix will be Power(2, (n2+n)/2) which is answer.
Note that diagonal elements of Skew-Symmetric matrix always are 0, that means fixed and cannot be counted. So, total elements of Skew Symmetric matrix will be (n2 – n)/2
This solution is contributed by Mohit Gupta.
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