Archetype S is the linear transformation defined by \begin{equation*} \ltdefn{T}{\complex{3}}{M_{22}},\quad \lt{T}{\colvector{a\\b\\c}}= \begin{bmatrix} a-b&2a+2b+c\\ 3a+b+c&-2a-6b-2c \end{bmatrix} \end{equation*}
We could compute a pre-image for every element of the codomain $M_{22}$. However, even in a free textbook, we do not have the room to do that, so we will compute just two.
Choose \begin{equation*} \vect{v}= \begin{bmatrix} 2&1\\3&2 \end{bmatrix} \in M_{22} \end{equation*} for no particular reason. What is $\preimage{T}{\vect{v}}$? Suppose $\vect{u}=\colvector{u_1\\u_2\\u_3}\in\preimage{T}{\vect{v}}$. The condition that $\lt{T}{\vect{u}}=\vect{v}$ becomes \begin{equation*} \begin{bmatrix} 2&1\\3&2 \end{bmatrix} =\vect{v} =\lt{T}{\vect{u}} =\lt{T}{\colvector{u_1\\u_2\\u_3}}\\ =\begin{bmatrix} u_1-u_2&2u_1+2u_2+u_3\\ 3u_1+u_2+u_3&-2u_1-6u_2-2u_3 \end{bmatrix} \end{equation*}
Using matrix equality (Definition ME), we arrive at a system of four equations in the three unknowns $u_1,\,u_2,\,u_3$ with an augmented matrix that we can row-reduce in the hunt for solutions, \begin{equation*} \begin{bmatrix} 1 & -1 & 0 & 2\\ 2 & 2 & 1 & 1\\ 3 & 1 & 1 & 3\\ -2 & -6 & -2 & 2 \end{bmatrix} \rref \begin{bmatrix} \leading{1} & 0 & \frac{1}{4} & \frac{5}{4}\\ 0 & \leading{1} & \frac{1}{4} & -\frac{3}{4}\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix} \end{equation*}
We recognize this system as having infinitely many solutions described by the single free variable $u_3$. Eventually obtaining the vector form of the solutions (Theorem VFSLS), we can describe the preimage precisely as, \begin{align*} \preimage{T}{\vect{v}}&=\setparts{\vect{u}\in\complex{3}}{\lt{T}{\vect{u}}=\vect{v}}\\ &=\setparts{\colvector{u_1\\u_2\\u_3}}{u_1=\frac{5}{4}-\frac{1}{4}u_3,\,u_2=-\frac{3}{4}-\frac{1}{4}u_3}\\ &=\setparts{\colvector{\frac{5}{4}-\frac{1}{4}u_3\\-\frac{3}{4}-\frac{1}{4}u_3\\u_3}}{u_3\in\complexes}\\ &=\setparts{\colvector{\frac{5}{4}\\-\frac{3}{4}\\0}+u_3\colvector{-\frac{1}{4}\\-\frac{1}{4}\\1}}{u_3\in\complexes}\\ &=\colvector{\frac{5}{4}\\-\frac{3}{4}\\0}+\spn{\set{\colvector{-\frac{1}{4}\\-\frac{1}{4}\\1}}} \end{align*}
This last line is merely a suggestive way of describing the set on the previous line. You might create three or four vectors in the preimage, and evaluate $T$ with each. Was the result what you expected? For a hint of things to come, you might try evaluating $T$ with just the lone vector in the spanning set above. What was the result? Now take a look back at Theorem PSPHS. Hmmmm.
OK, let us compute another preimage, but with a different outcome this time. Choose \begin{equation*} \vect{v}= \begin{bmatrix} 1&1\\2&4 \end{bmatrix} \in M_{22} \end{equation*}
What is $\preimage{T}{\vect{v}}$? Suppose $\vect{u}=\colvector{u_1\\u_2\\u_3}\in\preimage{T}{\vect{v}}$. That $\lt{T}{\vect{u}}=\vect{v}$ becomes \begin{equation*} \begin{bmatrix} 1&1\\2&4 \end{bmatrix} =\vect{v} =\lt{T}{\vect{u}} =\lt{T}{\colvector{u_1\\u_2\\u_3}}\\ =\begin{bmatrix} u_1-u_2&2u_1+2u_2+u_3\\ 3u_1+u_2+u_3&-2u_1-6u_2-2u_3 \end{bmatrix} \end{equation*}
Using matrix equality (Definition ME), we arrive at a system of four equations in the three unknowns $u_1,\,u_2,\,u_3$ with an augmented matrix that we can row-reduce in the hunt for solutions, \begin{equation*} \begin{bmatrix} 1 & -1 & 0 & 1\\ 2 & 2 & 1 & 1\\ 3 & 1 & 1 & 2\\ -2 & -6 & -2 & 4 \end{bmatrix} \rref \begin{bmatrix} \leading{1} & 0 & \frac{1}{4} & 0\\ 0 & \leading{1} & \frac{1}{4} & 0\\ 0 & 0 & 0 & \leading{1}\\ 0 & 0 & 0 & 0 \end{bmatrix} \end{equation*}
By Theorem RCLS we recognize this system as inconsistent. So no vector $\vect{u}$ is a member of $\preimage{T}{\vect{v}}$ and so \begin{equation*} \preimage{T}{\vect{v}}=\emptyset \end{equation*}