An easy IIT-JEE problem

For the matrix $\textbf{A}$ to be symmetric it must be equal to its transpose, i.e., the entries must be such that $a_{ij} = a_{ji}$ .

For a $3$ x $3$ matrix this means that we require the $3$ "pairings" $a_{12} = a_{21}, a_{13} = a_{31}$ and $a_{23} = a_{32}$ , while $a_{11}, a_{22}$ and $a_{33}$ are unchanged by the transposing.

Now, $4$ of the entries can be $0$ (and the remaining entries $1$ ), in the following ways:

• (i) precisely one of the pairings can be composed of $0$ 's and two of the $a_{ii}$ entries can be $0$ , or

• (ii) precisely two of the pairings can be composed of $0$ 's, with all the remaining entires being $1$ 's.

Case (i) gives us $\binom{3}{1} \binom{3}{2} = 9$ possible matrices, and case (ii) gives us just $\binom{3}{2} = 3$ possible matrices.

Thus the total number of possible matrices that meet the given restrictions is $9 + 3 = \boxed{12}$ .