Archetype Q is the linear transformation \begin{equation*} \ltdefn{T}{\complex{5}}{\complex{5}},\quad \lt{T}{\colvector{x_1\\x_2\\x_3\\x_4\\x_5}}= \colvector{-2 x_1 + 3 x_2 + 3 x_3 - 6 x_4 + 3 x_5\\ -16 x_1 + 9 x_2 + 12 x_3 - 28 x_4 + 28 x_5\\ -19 x_1 + 7 x_2 + 14 x_3 - 32 x_4 + 37 x_5\\ -21 x_1 + 9 x_2 + 15 x_3 - 35 x_4 + 39 x_5\\ -9 x_1 + 5 x_2 + 7 x_3 - 16 x_4 + 16 x_5} \end{equation*}
Notice that for \begin{align*} \vect{x}&=\colvector{1\\3\\-1\\2\\4}& \vect{y}&=\colvector{4\\7\\0\\5\\7} \end{align*} we have \begin{align*} \lt{T}{\colvector{1\\3\\-1\\2\\4}}&=\colvector{4\\55\\72\\77\\31}& \lt{T}{\colvector{4\\7\\0\\5\\7}}&=\colvector{4\\55\\72\\77\\31} \end{align*}
So we have two vectors from the domain, $\vect{x}\neq\vect{y}$, yet $\lt{T}{\vect{x}}=\lt{T}{\vect{y}}$, in violation of Definition ILT. This is another example where you should not concern yourself with how $\vect{x}$ and $\vect{y}$ were selected, as this will be explained shortly. However, do understand why these two vectors provide enough evidence to conclude that $T$ is not injective.