Let $T$ be the set of all $3 × 3$ symmetric matrices all of whose entries are either $0$ or $1$. Answer following Let $T$ be the set of all $3 × 3$ symmetric matrices all of whose entries are either $0$ or $1$. Five of these entries are $1$ and four of them are $0$.
i) The number of matrices in $T$ is
ii)The number of matrices in $T$ for which the system of linear equations has  unique solution where $A\in T$
$$A\begin{bmatrix}
x\\
y\\
z
\end{bmatrix}=\begin{bmatrix}
1\\
0\\
0
\end{bmatrix}$$
iii) The number of matrices in $T$ for which the system of linear equations is inconsistent where $A\in T$
$$A\begin{bmatrix}
x\\
y\\
z
\end{bmatrix}=\begin{bmatrix}
1\\
0\\
0
\end{bmatrix}$$
My attempt is as follows:-
i) $A=\begin{bmatrix}
a&g&h\\
g&b&i\\
h&i&c
\end{bmatrix}$
It is given that five of the entries are $1$ and four of them are $0$
Case $1$: Among elements $g$,$h$,$i$ only one element is $1$, so automatically we have $2$ elements in A which are $1$ as A is symmetric. Remaining $3$ ones will be along the diagonal. So $\dbinom{3}{1}\dbinom{3}{3}$ matrices will be there for case $1$
Case $2$: Among elements $g$,$h$,$i$ only $2$ elements are $1$, so automatically we have $4$ elements in A which are $1$ as A is symmetric. Remaining $1$ one  will be along the diagonal. So $\dbinom{3}{2}\dbinom{3}{1}$ matrices will be there for case $1$
So in total we would have $\dbinom{3}{1}\dbinom{3}{3}+\dbinom{3}{2}\dbinom{3}{1}=12$ matrices
ii) For given system of linear equations to have a unique solution, we need to count the matrices which have determinant non-zero.
Now here I was not getting any clever way to count the matrices which have determinant non-zero. So I decided to write all $12$ matrices and check their determinant.
I got $6$ matrices which had their determinant non-zero. But I am not satisfied with my way. Any other way?
iii) Now here we first need to know the number of matrices which have determinant zero, so with the help of computation in part ii), we can say there are $6$ matrices with determinant zero.
$$A\begin{bmatrix}
x\\
y\\
z
\end{bmatrix}=\begin{bmatrix}
1\\
0\\
0
\end{bmatrix}$$
Taking $adj(A)$ on both sides
$$adj(A)(A)\begin{bmatrix}
x\\
y\\
z
\end{bmatrix}=adj(A)\begin{bmatrix}
1\\
0\\
0
\end{bmatrix}$$
L.H.S would be zero matrix as $|A|=0$, so R.H.S should be non-zero matrix as we have to count the matrices with inconsistent solutions. For matrix $A$ to have inconsistent solution, at-least one of $C_{11},C_{12},C_{13}$ should be non-zero or in other words we can say first column of $adj(A)$ should have one non-zero element.
So I was getting $4$ matrices out of $6$ which had at-least one of $C_{11},C_{12},C_{13}$ non-zero.
 A: There are so few matrices satisfying the conditions that writing them out explicitly is probably the fastest and most reliable way of solving this. Elegant solutions are great, but if there is any kind of time pressure, it is usually best to go with the first correct solution!
$A_1=\begin{pmatrix}1&1&1\\1&0&0\\1&0&0\end{pmatrix},$
$A_2=\begin{pmatrix}0&1&1\\1&1&0\\1&0&0\end{pmatrix},$
$A_3=\begin{pmatrix}0&1&1\\1&0&0\\1&0&1\end{pmatrix},$
$A_4=\begin{pmatrix}1&0&1\\0&0&1\\1&1&0\end{pmatrix},$
$A_5=\begin{pmatrix}0&0&1\\0&1&1\\1&1&0\end{pmatrix},$
$A_6=\begin{pmatrix}0&0&1\\0&0&1\\1&1&1\end{pmatrix},$
$A_7=\begin{pmatrix}1&1&0\\1&0&1\\0&1&0\end{pmatrix},$
$A_8=\begin{pmatrix}0&1&0\\1&1&1\\0&1&0\end{pmatrix},$
$A_9=\begin{pmatrix}0&1&0\\1&0&1\\0&1&1\end{pmatrix},$
$A_{10}=\begin{pmatrix}1&1&0\\1&1&0\\0&0&1\end{pmatrix},$
$A_{11}=\begin{pmatrix}1&0&1\\0&1&0\\1&0&1\end{pmatrix},$
$A_{12}=\begin{pmatrix}1&0&0\\0&1&1\\0&1&1\end{pmatrix},$ 
$A_1,A_{12}$ have many solutions; $A_6,A_8,A_{10},A_{11}$ have no solutions; $A_2,A_3,A_4,A_5,A_7,A_9$ each have a unique solution.
It is easy to see whether the zero-determinant cases have 0 or many solutions because they have a repeated row. Where the rows 2 and 3 are the same there are many solutions (because we want a value 0 in each case for that component of the vector). Where one of the repeat rows is 1 we have zero solutions. 
