= Solution
Let $v\in\mathbb Q^2$ be nonzero. If the columns $v,Av$ were <linearly dependent> over the <rational numbers>, then $Av=\lambda v$ for some $\lambda\in\mathbb Q$. Applying $A$ once more would give
$$
-v=A^2v=\lambda Av=\lambda^2v,
$$
so $(\lambda^2+1)v=0$. Since $v\ne0$, this forces $\lambda^2=-1$, impossible in $\mathbb Q$. Thus the two columns are <linearly independent>, and the square <matrix>
$$
\boxed{E=\begin{pmatrix}|&|\\v&Av\\|&|\end{pmatrix}}
$$
is an <invertible matrix>.
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