Number of $3 \times 3$ symmetric matrices with entries five ones and four zeros which are Non Singular Find Number of $3 \times 3$ symmetric matrices with entries five ones and four zeros which are Non Singular
My try:  
The only possibilities of symmetric matrices with given entries are as follows:
$1.$ With diagonal entries $1,1,1$ one of which is
$$\begin{bmatrix}
1 & 0 &0 \\ 
 0&1  &1 \\ 
 0&1 & 1
\end{bmatrix}$$ and Non Diagonal entries can be Permuted in $3$ ways
$2.$ With Diagonal entries $0,0,1$ one of which is
\begin{bmatrix}
0 & 1 &1 \\ 
 1&0  &0 \\ 
 1&0 & 1
\end{bmatrix}
Here Number of Matrices possible are $3 \times 3=9$
Any clue of how to check  Non singular matrices among these $12$ matrices?
 A: Let us remember that $3\times 3$ matrix $A=(a_{ij})$ is no-singular if, only if,
\begin{align}
\det (A)=
&+a_{11}\cdot  ( a_{22} \cdot a_{33} - a_{23}\cdot a_{32}) \\
&-a_{12}\cdot  ( a_{21} \cdot a_{33} - a_{31}\cdot a_{23})\\
&+a_{13}\cdot  ( a_{21} \cdot a_{32} - a_{31}\cdot a_{22})\neq 0
\end{align}
Affirmation 1: If $a_{11}=1$ then $a_{22}\cdot a_{33} - a_{23}\cdot a_{32}\neq 0$ is non-zero if, and only if,
$$
-a_{12}\cdot ( a_{21} \cdot a_{33} - a_{31}\cdot a_{23}) +a_{13}\cdot ( a_{21} \cdot a_{32} - a_{31}\cdot a_{22})=0.
$$
Affirmation 2: If $a_{11}=1$ then $a_{22}\cdot a_{33} - a_{23}\cdot a_{32}\neq 0$ if, and only if,
$3$ of the $4$ entries are equal to $1$ and exactly $1$ entry equals $0$. And it is easy to see that there are $4$ ways for this to happen under these conditions.
Doing this process for the remaining $2$ products
$$
-a_{12}\cdot  ( a_{21} \cdot a_{33} - a_{31}\cdot a_{23})
\qquad 
+a_{13}\cdot  ( a_{21} \cdot a_{32} - a_{31}\cdot a_{22})
$$
gives a total of $4+4+4=12$ non-sigular matrices of the required type.
