The Gateway to Computer Science Excellence
First time here? Checkout the FAQ!
x
+13 votes
1.8k views

The number of different $n \times n $ symmetric matrices with each element being either 0 or 1 is: (Note: $\text{power} \left(2, X\right)$ is same as $2^X$)

  1. $\text{power} \left(2, n\right)$
  2. $\text{power} \left(2, n^2\right)$
  3. $\text{power} \left(2,\frac{ \left(n^2+ n \right) }{2}\right)$
  4. $\text{power} \left(2, \frac{\left(n^2 - n\right)}{2}\right)$
asked in Linear Algebra by Veteran (69k points)
edited by | 1.8k views

This might help ...

3 Answers

+24 votes
Best answer
In symmetric matrix, $A[i][j] = A[j][i]$. So, we have choice only for either the upper triangular elements or the lower triangular elements. Number of such elements will be $n + (n-1) + (n-2) + \cdots + 1 = n\frac{(n+1)}{2} = \frac{(n^2+n)}{2}$. Now, each element being either 0 or 1 means, we have 2 choices for each element and thus for $\frac{(n^2+n)}{2}$ elements we have $2^{\frac{(n^2+n)}{2}}$ possibilities.

Choice C.
answered by Veteran (346k points)
selected by

The problem is similar to finding the number of symmetric relations possible on an n element set.
Now if we consider the n pairs of reflexive relation, those n pairs also satisfy the property of symmetric relation and each of these n pair may or may not be present in a symmetric relation.

Moreover there are (n2  - n)/2 diagonal pairs in an nxn matrix each of which should be present to have relation symmetric.
Means if we have pair (x,y) we need to have pair (y,x)

So, total symmetric relations possible on an n element set :

2n . 2 (n2-n)/2  = 2 n(n+1)/2  

+5 votes
Matrices with 0 or 1 can be viewed as adjacency matrix of an undirected graph. There can be at max $n(n-1)/2$ edges in such graph.

So each edge can either be present or not. That gives total of $2^{(n^{2}-n)/2}$ , but this does not include n combination of self loops. so the total number of combincation is $2^{(n^{2}+n)/2}$.
Answer C
answered by Active (2.2k points)
+3 votes
for diagnol elements we have 2 choices either 0 or 1 So , 2^n  for rest of the elements we can pair which will have 2 choices either 0 or 1  total pairs =(n^2-n)/2

2^n * 2^((n^2-n)/2)  = Option C
answered by Loyal (4.7k points)


Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true

33,593 questions
40,128 answers
114,021 comments
38,389 users