The eigenvalues $\lambda_1,\lambda_2,\lambda_3$ of a matrix $M\in SL_3(\mathbb{Z})$ satisfy:
$\lambda_1\lambda_2\lambda_3=\det M=1$.
Let $f(x)\in\mathbb{Z}[x]$ be the characteristic polynomial of $M$, then $\lambda_i$ is a root of $f(x)$.
Let $f(x)=x^3+tx-1$ for some $t\in\mathbb{Z}$, then the discriminant $d_{f}=-4*t^3 - 27$. Now if you pick $t>0$, then $d_f<0$, then $f(x)$ has two complex root and one real root.
To find a choice for $M$, let $K$ be the number field defined by $K(\alpha)=\mathbb{Q}[x]/(f(x))$.
Then the ring of integer $\mathcal{O}_K$ is a free $\mathbb{Z}$-module of rank $3$. Now multiplication by $\alpha$ induces an Endomorphism $\phi(\alpha)\in End_{\mathbb{Z}}(\mathcal{O}_K)$. Since $\mathcal{O}_K$ is free $\mathbb{Z}$ module of rank $3$, we have $\phi(\alpha)\in GL_{3}(\mathbb{Z})$ since $\alpha$ is invertible in $\mathcal{O}_K$. Choose the sign of $\alpha$ such that $\det(\phi(\alpha))=1$, then we have $\phi(\alpha)\in SL_3(\mathbb{Z})$. Note that the characteristic polynomial of $\phi(\alpha)$ has $\alpha$ as a root and hence an irreducible polynomial. Then the conjugate of $\alpha$ also satisfies it. Now you can take $M=\phi(\alpha)$.