Number of matrices with positive determinant whose entries are {1,-1} 
How many different matrices  $3 \times 3$ matrix, whose every entry is 1 or -1, have positive determinant?

My approach:
I found out the possible values of determinant of all such $3 \times 3$ matrices whose entires are 1 or -1: {-4,-2,0,2,4}.
The number of matrices with positive determinant = number of matrices with negative determinant by symmetry, (as we can just multiply each entry with -1, and sign of determinant would change as it's an odd degree).
So the answer would be $\frac{2^9-n(0)}{2}$ where n(0) is the number of singular matrices(determinant 0).
For n(0), any two rows/columns must be proportional to each other i.e. all 1 or all -1 or one of them 1 and the other -1.(4 possibilities for two such choosen rows/columns).
So I tried to count it as follows:
$({3 \choose 2} \cdot 4 \cdot 2^3)\cdot 2=192$ for rows and columns. However this includes cases with three rows /three column repeated as well, for which we must subtract the over counted cases. 
For all 3 rows/columns to be proportional there are $(3\cdot 2 +2) \cdot 2 = 16$ ways of which each is counted thrice.
So a total of 48 repeated cases.
So n(0)=192-48=144.
This gives number of matrices with positive determinant as $\frac{512-144}{2}=184$, 
However the answer is given as $96$.
Where am I going wrong?
Also is there a more systematic way of doing this question? 
 A: For counting the singular cases, here's a geometric interpretation . . .

Consider the solid cube $C=\big{\{}(x,y,z)\in \mathbb{R}^3\mid x,y,z\in [-1,1]\big\}$.

Let $S$ be the set of $3{\,\times\,}3$ matrices with all entries in $\{\pm 1\}$.

For $A\in S$ we can regard the $3$ rows of $A$ as vertices $V_1,V_2,V_3$ of $C$, hence $\det(A)=0$ if and only if the vectors $\vec{V_1},\vec{V_2},\vec{V_3}$ are coplanar.

To count the number of matrices $A\in S$ such that $\det(A)=0$, consider $3$ cases . . .

Case $(1)$:$\;V_1,V_2,V_3$ are all equal.


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The count for this case is $8$.


Case $(2)$:$\;$Exactly two of $V_1,V_2,V_3$ are all equal.


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The count for this case is $8{\,\cdot\,}7{\,\cdot\,}3=168$, since there $8$ choices for the vertex with multiplicity $2$, $7$ choices for the non-duplicated vertex, and $3$ ways of arranging the rows of $A$.


Case $(3)$:$\;\vec{V_1},\vec{V_2},\vec{V_3}$ are coplanar and $V_1,V_2,V_3$ are distinct.


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Then the plane containing the triangle with vertices $V_1,V_2,V_3$ must pass through the origin. There are $6$ such planes, each of which passes through two diagonally opposite edges of $C$. Each such plane contains $4$ vertices of $C$, so for each such plane there are $4$ choices for $\{V_1,V_2,V_3\}$. Thus there are $6{\,\cdot\,}4=24$ possible choices for $\{V_1,V_2,V_3\}$, and for each such choice there are $3!=6$ was of arranging the rows of $A$. Hence the count for this case is $24{\,\cdot\,}6=144$.


Summing the counts for the $3$ cases, we get a total count of $8+168+144=320$ for the count of the number of singular matrices.

Hence we get ${\Large{\frac{2^9-320}{2}}}=96$ for the count of the number of matrices $A\in S$ with $\det(A) > 0$.
