Number of $\pm 1$ matrices whose determinant is negative or zero

Find the number of $$3 \times 3$$ matrices whose determinant is $$(a)$$ positive $$(b)$$ negative $$(c)$$ zero and whose elements are taken from $$\{-1,1\}$$.

This is what I tried:

The total number of $$3\times 3$$ matrices with entries in $$\{ 1 , -1\}$$ is $$2^9$$. Now counting each rows or column of matrices as a vectors with entries in $$\{-1,1\}$$, there are $$8$$ such vectors and if any two vectors are same then its determinant is $$0$$. There are such $$24+24=48$$ ways.

Matrices with $$\det <0$$ are exactly those resulting from swapping the first two rows of matrices with $$\det >0$$ because this multiplies the determinant by $$-1$$, and so the number with $$\det >0$$ is the same as the number with $$\det <0$$. So you just need to count the matrices with $$\det =0$$.

First we count how many of the $$2^4 =16$$ matrices of the form $$A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & a & b \\ 1 & c & d \end{pmatrix}$$ have $$\det A =0$$.

The determinant of $$A$$ is $$\det A = \left| \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \right| - \left| \begin{smallmatrix} 1 & b \\ 1 & d \end{smallmatrix} \right| + \left| \begin{smallmatrix} 1 & a \\ 1 & c \end{smallmatrix} \right| = (ad -a -d) - (bc -b -c).$$

We have the first term equal to $$3$$ when $$(a,d)=(-1,-1)$$, and equal to $$-1$$ in the other three cases- and the same for the second term. We therefore have one in the case $$a=b=c=d=-1$$ and $$3 \times 3 = 9$$ for the other cases, totaling $$10$$ such matrices $$A$$.

Each matrix with determinant $$0$$ can be uniquely made into the above form by multiplying a choice of the rows and columns by $$-1$$ (which doesn't change the determinant). There are $$10$$ matrices of determinant $$0$$ for each of the $$2^5 = 32$$ ways to choose those entries in the first row and column.

Hence, out of the $$2^9 = 512$$ matrices with entries in $$\{ 1, -1 \}$$, $$320$$ have $$\det = 0$$, $$96$$ have $$\det > 0$$ and $$96$$ have $$\det <0$$.

Since interchanging two rows multiplies the determinant by $$-1$$, the number with positive determinant is the same as the number with negative determinant. So you just need to count the matrices with determinant $$0$$. Moreover, since multiplying a row or column by $$-1$$ multiplies the determinant by $$-1$$, you can assume the first row and column are all $$1$$, and multiply the resulting count by $$2^5$$. There are just $$2^4$$ cases to check, taking all possibilities for the entries not in the first row or column.