<s>[INST] 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.
 [/INST] 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 $\det <0$.
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