I have adapted an idea I have found in Stackexchange. It is based on the following facts:

Let $J_k$ be the identity matrix of order $n$ with the null elements in the $k$-th row replaced by random integers. Then $J_k$ is unimodular, i.e. $\lvert J_k\rvert=1$.

If $A$ is an unimodular symmetric matrix, so it is $J_k^T A J_k$.

Thus, one can start with $A=I_n$, choose a row $k$ at random, construct $J_k$, compute $J_k^T A J_k$ and replace $A$ by that matrix. This process can be repeated as many times as wanted. The result is an unimodular symmetric matrix. The following code implements this idea:

```
def unimod_sym_matrix(n, niter=1, min_int=-1, max_int=1):
A = identity_matrix(n)
rows = random_vector(niter, x=0, y=n)
for k in rows:
J = identity_matrix(n)
J[k,:] = random_vector(n, x=min_int, y=max_int+1)
J[k,k] = 1
A = J.transpose()*A*J
return A
```

Observe that, to get $J_k$ , random integers are taken from `min_int`

to `max_int`

, both included. Likewise, `niter`

is the number of matrices $J_k$ that are constructed.

For example, the following code

```
set_random_seed(1000)
A = unimod_sym_matrix(6,3); show(A)
A = unimod_sym_matrix(6); show(A)
```

yields

$\begin{pmatrix}
2 & -2 & -1 & 0 & 1 & 0 \\
-2 & 9 & 4 & 1 & -3 & 3 \\
-1 & 4 & 2 & 0 & -1 & 2 \\
0 & 1 & 0 & 2 & -1 & -1 \\
1 & -3 & -1 & -1 & 2 & 1 \\
0 & 3 & 2 & -1 & 1 & 6
\end{pmatrix}$

$\begin{pmatrix}
2 & 0 & 1 & -1 & -1 & -1 \\
0 & 1 & 0 & 0 & 0 & 0 \\
1 & 0 & 2 & -1 & -1 & -1 \\
-1 & 0 & -1 & 1 & 1 & 1 \\
-1 & 0 & -1 & 1 & 2 & 1 \\
-1 & 0 & -1 & 1 & 1 & 2
\end{pmatrix}$

I have included `set_random_seed`

just for the sake of reproducibility of the above output.

Since you want matrix entries between $-2$ and $2$, you had better use the default values for `niter`

, `min_int`

and `max_int`

.

Are you sure that your restrictions are correct? For example, let $$A=\begin{pmatrix} 2 & a & b \\ a & 2 & c \\ b & c & 2 \end{pmatrix},$$ with $a,b,c\in\mathbb{Z}$. Since $$\det(A)= 8+2(abc-a^2-b^2-c^2),$$ $A$ will be unimodular if and only if $$a^2+b^2+c^2-abc=\frac{7}{2},$$ which is impossible because $a^2+b^2+c^2-abc\in\mathbb{Z}$.

Good point. This certainly works if the dimension is 8 e.g. the $E_8$ root lattice gram matrix: E8 lattice wiki. I edited the post. Thank you