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?
On
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}
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.