 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 giveso $({\lambda }^2+1)v=0$. Since $v\ne 0$, this forces ${\lambda }^2=-1$, impossible in $\mathbb{Q}$. Thus the two columns are linearly independent, and the square matrixis an invertible matrix.