Condition on a matrix so that its determinant is $2$ 
Is the following statement true or false? There exists $A \in M_{3,3}(\mathbb{Z})$ with determinant $2$ such that $$A\begin{pmatrix}2\\1\\4\end{pmatrix} = \begin{pmatrix}4\\-8\\16\end{pmatrix}$$

I first thought of eigenvalues but it doesn't look like the second vector is a multiple of the first. What are the conditions on $A$ so that it's "true"? Or is it always false?
 A: False. Let us say $A=(a_{i,j})$ with $1\le i,j\le 3$. Then for all $i$ 
$$
4\,\,\text{ divides }\,\, 2a_{i,1}+a_{i,2}+4a_{i,3},
$$
which happens if and only if $a_{i,2}$ is even and $a_{i,1}$ has the same parity of $a_{i,2}/2$.
Then 
\begin{align}
\mathrm{det}(A)&=
\begin{bmatrix}a_{1,1}&a_{1,2}&a_{1,3}\\
a_{2,1}&a_{2,2}&a_{2,3} \\ 
a_{3,1}&a_{3,2}&a_{3,3}
\end{bmatrix}
=2\mathrm{det}
\begin{bmatrix}a_{1,1}&a_{1,2}/2&a_{1,3}\\
a_{2,1}&a_{2,2}/2&a_{2,3} \\ 
a_{3,1}&a_{3,2}/2&a_{3,3}
\end{bmatrix}\\
&=2\mathrm{det}
\begin{bmatrix}a_{1,1}&a_{1,2}/2-a_{1,1}&a_{1,3}\\
a_{2,1}&a_{2,2}/2-a_{2,1}&a_{2,3} \\ 
a_{3,1}&a_{3,2}/2-a_{3,1}&a_{3,3}
\end{bmatrix}\\
&=4\mathrm{det}
\begin{bmatrix}a_{1,1}&\frac{1}{2}(a_{1,2}/2-a_{1,1})&a_{1,3}\\
a_{2,1}&\frac{1}{2}(a_{2,2}/2-a_{2,1})&a_{2,3} \\ 
a_{3,1}&\frac{1}{2}(a_{3,2}/2-a_{3,1})&a_{3,3}
\end{bmatrix}.\\
\end{align}
A: Such $A$ does not exist. Suppose the contrary that $A$ exists. Then
$$
A\pmatrix{0\\ 1\\ 0}\equiv\pmatrix{0\\ 0\\ 0}\mod2,
$$
meaning that the second column of $A$ is entrywise an integer multiple of $2$. Divide the second column of $A$ by $2$ to obtain a new integer matrix $B$. Then
$$
B\pmatrix{2\\ \color{red}{2}\\ 4}=\pmatrix{4\\ -8\\ 16}
\ \Rightarrow\ B\pmatrix{1\\ 1\\ 2}=\pmatrix{2\\ -4\\ 8}
\ \Rightarrow\ B\pmatrix{1\\ 1\\ 0}\equiv\pmatrix{0\\ 0\\ 0}\mod2,
$$
i.e. $B$ is singular in modulo-2 arithmetic. Yet, this is impossible because $\det B=1$. Therefore $A$ does not exist.
