Unipotent matrix with all block matrices having non-zero determinant

Question

Let $M\in \mathrm{SL}(4, \mathbb{Z})$ with all eigenvalues equal to $1$ (i.e. $M$ is a unipotent matrix).

Write $M=\begin{bmatrix} A_1&A_2\\ A_3&A_4 \end{bmatrix}, $ where each $A_i$ is a $2$ by $2$ sumbatrix of $M$.

Let $a_i = \mathrm{det}(A_i)$.

Consider the matrix $A = \begin{bmatrix} a_1&a_2\\ a_3&a_4 \end{bmatrix}$,

Question: Is it possible that

1. All $a_i$'s are non-zero?
2. The matrix $A$ is in $\mathrm{GL}(2, \mathbb{Z})$, and has one eigenvalue with an absolute value not equal to $1$?

Question 1 has been answered by Dietrich Burde, any hint with question 2 would be really appreciated.

Answer 1

With the ideas from the previous answer we immediately find examples, e.g., $$ M=\begin{pmatrix} 0 & 4 & 1 & 0 \cr 36 & 0 & 49 & 1 \cr 2 & -3 & 2 & 0 \cr 1 & -1 & 0 & 2 \end{pmatrix}. $$ Here $M$ has characteristic polynomial $(t-1)^4$. The matrix of determinants is given by $$ A=\begin{pmatrix} -144 & 1 \cr 1 & 4 \end{pmatrix}. $$ For the second question, take $$ M=\begin{pmatrix} 0 & 0 & -1 & 0 \cr 1 & 0 & 0 & 1 \cr 1 & 0 & 0 & 2 \cr 0 & 1 & -3 & 4 \end{pmatrix}. $$ Then the matrix $A$ of block determinants is given by $$ A=\begin{pmatrix} 0 & -1 \cr 1 & 6 \end{pmatrix}\in SL_2(\Bbb Z). $$ The eigenvalues do not have absolute value $1$.