Let \(T:\mathbb{Q}^2\rightarrow \mathbb{Q}^2\) be given by
\(T\left(\begin{bmatrix} x\\y\end{bmatrix}\right)
=\begin{bmatrix} 2x-y\\x+y\end{bmatrix}.\)
Is \(T\) invertible?
The answer is “Yes”.
Note that \(T\) is given by
\(T\left(\begin{bmatrix} x\\y\end{bmatrix}\right) =
\begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix}
\begin{bmatrix} x\\y\end{bmatrix}\).
The matrix
\(\begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix}\)
is invertible and so \(T\) is invertible.