<s>[INST] prove that $\{\det(A) : A\in S\} = \{-4,-3,-2,-1,0,1,2,3,4\}$ 
Let $S$ be the set of all $3\times 3$ matrices with entries in $\{0,1,-1\}$. Prove that $\{\det(A) : A\in S\} = \{-4,-3,-2,-1,0,1,2,3,4\}$.

Let $R = [-4,4]\cap \mathbb{Z}$, $T =\{\det(A) : A\in S\}$. To show $R\subseteq T,$ it suffices to show that $R\cap \mathbb{Z}^+\subseteq T,$ since clearly any matrix with an all-zero row or column (or a repeated row or column) has determinant zero and swapping any two distinct rows or columns of a matrix multiplies the determinant by $-1$ (the determinant is an alternating linear map). I know the determinant of an n by n matrix is also $n$-linear, meaning that it's linear with respect to any single row or column. Obviously the identity matrix has determinant 1. Let $A_2 = \begin{pmatrix}1 & 1 & 0\\
-1 & 1 & 0\\
0 & 0 & 1\end{pmatrix}, A_3 = \begin{pmatrix}1 & 1 & 0\\
-1 & 1 & 1\\
1 & 0 & 1\end{pmatrix},A_4 = \begin{pmatrix}1 & 1 & 0\\
-1 & 1 & 1\\
1 & -1 & 1\end{pmatrix}.$ Then using Laplace expansion along the first column, we can easily see that $\det A_i = i$ for $2\leq i\leq 4.$ Thus we just need to show $T\subseteq R$ to solve the question. Let $A = \begin{pmatrix}a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}\end{pmatrix}$. If $a_{i1}=0$ for $1\leq i\leq 3,$ then $\det(A) = 0\in R,$ so suppose otherwise. We just need to show that $|\det(A)|\leq 4$ since it is an integer (being a (multivariate) polynomial in the entries). WLOG, assume $a_{11} = 1$ (we may swap row $i\neq 1$ with row 1 and multiply the first row by $-1$ if necessary, which doesn't change the absolute value of the determinant). Perform the elementary row operations $R_i\mapsto R_i - R_1$ for $i=2,3$. Now note that if $a_{22} = a_{32} = 0,$ rows 2 and 3 are scalar multiples of each other or one of them equals the zero row, so the determinant is zero, which is in S. Hence, by swapping the two and multiplying row 2 by $-1$ if necessary, assume WLOG that $a_{22} > 0.$ Suppose first that $a_{22} = 1.$ Then $a_{12} = 0$ and $|a_{32}|\leq 1.$ Note that $|a_{23}|\leq 2,$ since $a_{13}$ was subtracted from it and has absolute value at most 1 and $a_{23}$ initially had absolute value at most 1. Also, we similarly have $|a_{33}|\leq 2.$ Subtracting row 2 from row 3 if necessary (i.e. if $a_{32}$ is nonzero), we get that $|a_{33}|\leq 4.$ But now the determinant equals $|a_{33}|\leq 4.$ Finally suppose $a_{22}= 2.$ Then $|a_{12}|=1$. Note that $|a_{23}| = 2$ only if $a_{23}' a_{13} = -1,$ where $a_{23}'$ is the original value of $a_{23}$ before subtracting from row 1. Similarly, $a_{22} = 2\Rightarrow a_{12}a_{22}' = -1.$ So $|a_{33}|=2$ and $|a_{23}|=2$ implies that $a_{33}' = a_{23}'$ and so $a_33 = a_{23}$. However, this approach seems very tedious and error-prone, so there's probably a better method.
 [/INST] The main task is to prove that $T\subseteq R$.
$\det A$ is the sum of six elements of $\{0,1,-1\}$ but if some entry of $A$ is $0$, at least two of these six elements are $0$ and you have won.
If all entries of $A$ are $\pm1$ then $\det A$ is even and wlog (up to changing some rows to their opposite), $A = \begin{pmatrix}1 &a&x\\
1&b&y\\
1 &c&z\end{pmatrix}$.
Then for $\det A$ to be $6$, we should have $b=z$, $c=x$, $a=y$, $b=-x$, $c=-y$, $a=-z$, but theses six equations are incompatible. Similarly, the six equations necessary for $\det A$ to be $-6$ are incompatible, which ends the proof.
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