Consider the coefficient matrix from Archetype A, \begin{equation*} A=\begin{bmatrix} 1 & -1 & 2\\ 2 & 1 & 1\\ 1 & 1 & 0 \end{bmatrix} \end{equation*}
Suppose that $A$ is invertible and does have an inverse, say $B$. Choose the vector of constants \begin{equation*} \vect{b}=\colvector{1\\3\\2} \end{equation*} and consider the system of equations $\linearsystem{A}{\vect{b}}$. Just as in Example SABMI, this vector equation would have the unique solution $\vect{x}=B\vect{b}$.
However, the system $\linearsystem{A}{\vect{b}}$ is inconsistent. Form the augmented matrix $\augmented{A}{\vect{b}}$ and row-reduce to \begin{equation*} \begin{bmatrix} \leading{1} & 0 & 1 & 0\\ 0 & \leading{1} & -1 & 0\\ 0 & 0 & 0 & \leading{1} \end{bmatrix} \end{equation*} which allows us to recognize the inconsistency by Theorem RCLS.
So the assumption of $A$'s inverse leads to a logical inconsistency (the system cannot be both consistent and inconsistent), so our assumption is false. $A$ is not invertible.
It is possible this example is less than satisfying. Just where did that particular choice of the vector $\vect{b}$ come from anyway? Stay tuned for an application of the future Theorem CSCS in Example CSAA.