The inverse
The Collapse preset, $\begin{pmatrix}1 & 0.5\\ 2 & 1\end{pmatrix}$, flattens the whole lattice onto a single line. Can some second matrix now move every dot back where it started?
The shear runs out and comes home: every dot returns, and the stats bar names the undo — A⁻¹ reads $1.00$, $-0.90$ over $0.00$, $1.00$, the same shear run backwards. Collapse behaves differently. At blend $t = 0.50$, det A reads $0.75$; at $t = 1$ it reads $0.00$, the lattice lies on one dashed line, the panel stamps INVERSE DOES NOT EXIST, and the run never returns.
For $2\times2$ matrices the undo is a recipe — swap the diagonal entries, negate the other two, divide everything by the determinant:
The division is the whole story. $A$ scales every area by $\det A$, so its undo must scale areas by $1/\det A$. The collapse has $\det A = 1 \cdot 1 - 0.5 \cdot 2 = 0$: its columns $(1, 2)$ and $(0.5, 1)$ point down one line, every area flattens to zero, and $1/0$ has no answer. A matrix like this is called singular. Whole families of inputs now share each output, and no function can send one shared output back to two inputs — the information is not hidden, it is destroyed.
The Shear preset is $A = \begin{pmatrix}1 & 0.9\\ 0 & 1\end{pmatrix}$. Using the recipe, what is $A^{-1}$?
For the Collapse preset $\begin{pmatrix}1 & 0.5\\ 2 & 1\end{pmatrix}$, what does the inverse recipe return?
Transfer: a matrix $A$ scales every area by $\det A = 4$. What must $A^{-1}$ do to areas, and what does that make $\det(A^{-1})$?
Invertibility is an area statement. $A^{-1}$ must scale areas by $1/\det A$, which is why the recipe divides by $\det A$ — and when the determinant is zero the plane has been flattened, the division has no answer, and no matrix of any kind can bring the lattice back.