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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. $\text{power} \left(2, \frac{\left(n^2 - n\right)}{2}\right)$
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consider a 5*5 square matrix, where entries in each cell represent the number of ways we can fill it. Since only 0 or 1 can be inserted therefore 2 ways.

$\bigl(\begin{smallmatrix} 2& 1 & 1& 1 &1\\ 2& 2 &1 &1 &1 \\ 2&2 &2 &1 & 1\\ 2&2 & 2 &2 &1 \\ 2&2 &2 & 2 & 2 \end{smallmatrix}\bigr)$

Now diagonal entries can be filled by any of 0 or 1 , so 2 ways for diagonal elements

but for non-diagonal we need to fix the elements in either upper/lower triangle of matrix, Because its a symmetric matrix hence it will be mirror image about diagonal.

for e.g. if A[2,1]= 1, then A[1,2]=1 OR  if A[2,1]= 0, then A[1,2]=0

fixing lower triangle: number of ways to fill   A[2,1] is 2 , so number of ways to fill   A[1,2] is 1

otherwise fixing upper triangle: number of ways to fill A[1,2] is 2, so number of ways to fill A[2,1] is 1

number of n*n symmetric = $2^{n}*2^{n-1}*2^{n-1}$............$2^{1}$

                                         =$2^{\frac{n(n+1)}{2}}$
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In a symmetric matrix, the lower triangle must be the mirror image of upper triangle using the diagonal as mirror. Diagonal elements may be anything. Therefore, when we are counting symmetric matrices we count how many ways are there to fill the upper triangle and diagonal elements. Since the first row has n elements, second (n-1) elements, third row (n-2) elements and so on upto last row, one element

 

Answer:

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