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*}
We will demonstrate that \begin{equation*} \vect{v}=\colvector{-1\\2\\3\\-1\\4} \end{equation*} is an unobtainable element of the codomain. Suppose to the contrary that $\vect{u}$ is an element of the domain such that $\lt{T}{\vect{u}}=\vect{v}$.
Then \begin{align*} \colvector{-1\\2\\3\\-1\\4}&=\vect{v}=\lt{T}{\vect{u}} =\lt{T}{\colvector{u_1\\u_2\\u_3\\u_4\\u_5}}\\ &=\colvector{ -2 u_1 + 3 u_2 + 3 u_3 - 6 u_4 + 3 u_5\\ -16 u_1 + 9 u_2 + 12 u_3 - 28 u_4 + 28 u_5\\ -19 u_1 + 7 u_2 + 14 u_3 - 32 u_4 + 37 u_5\\ -21 u_1 + 9 u_2 + 15 u_3 - 35 u_4 + 39 u_5\\ -9 u_1 + 5 u_2 + 7 u_3 - 16 u_4 + 16 u_5}\\ &= \begin{bmatrix} -2&3&3&-6&3\\ -16&9&12&-28&28\\ -19&7&14&-32&37\\ -21&9&15&-35&39\\ -9&5&7&-16&16 \end{bmatrix} \colvector{u_1\\u_2\\u_3\\u_4\\u_5} \end{align*}
Now we recognize the appropriate input vector $\vect{u}$ as a solution to a linear system of equations. Form the augmented matrix of the system, and row-reduce to \begin{equation*} \begin{bmatrix} \leading{1} & 0 & 0 & 0 & -1 & 0\\ 0 & \leading{1} & 0 & 0 & -\frac{4}{3} & 0\\ 0 & 0 & \leading{1} & 0 & -\frac{1}{3} & 0\\ 0 & 0 & 0 & \leading{1} & -1 & 0\\ 0 & 0 & 0 & 0 & 0 & \leading{1}\\ \end{bmatrix} \end{equation*}
With a leading 1 in the last column, Theorem RCLS tells us the system is inconsistent. From the absence of any solutions we conclude that no such vector $\vect{u}$ exists, and by Definition SLT, $T$ is not surjective.
Again, do not concern yourself with how $\vect{v}$ was selected, as this will be explained shortly. However, do understand why this vector provides enough evidence to conclude that $T$ is not surjective.