Number of different 2 × 2 symmetric matrices with elements being either 0 or 1 is: 

Concept:- Number of different n × n symmetric matrices with each element being 0 or 1 is equal to \({2^{\frac{{n\left( {n + 1} \right)}}{2}}}\)

Eg. For n = 1 possible matrices [0], [1] only 2 matrices are possible

For n = 2

 \(\left[ {\begin{array}{*{20}{c}} 0&0\\ 0&0 \end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}} 1&0\\ 0&0 \end{array}} \right]\),\(\left[ {\begin{array}{*{20}{c}} 0&1\\ 0&0 \end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}} 0&0\\ 1&0 \end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}} 0&0\\ 0&1 \end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}} 1&1\\ 0&0 \end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}} 1&0\\ 1&0 \end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}} 0&1\\ 0&1 \end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}} 0&0\\ 1&1 \end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}} 0&1\\ 1&0 \end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}} 1&1\\ 1&0 \end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}} 1&0\\ 1&1 \end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}} 0&1\\ 1&1 \end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}} 1&1\\ 0&1 \end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}} 1&1\\ 1&1 \end{array}} \right]\)

Maximum 16 matrices are possible out of which 8 are symmetric

Application:-    ∴ for n = 2     \({2^{\frac{{n\left( {n + 1} \right)}}{2}}}\) = 2³ 

hence 8 different matrices are possible. so option (2) is correct.