Determinant zero
The Collapse preset $\begin{pmatrix}1 & 0.5\\ 2 & 1\end{pmatrix}$ has $\det = 1 \cdot 1 - 0.5 \cdot 2 = 0$. What does an area factor of zero do to the unit square?
At $t = 0.50$ the square is a thinning parallelogram and $\det A = 0.75$. By $t = 0.90$ it reads $0.19$. At $t = 1.00$ the two columns $c_1 = (1, 2)$ and $c_2 = (0.5, 1)$ point down the same line, every lattice dot lands on that line, and the stats read $\det A = 0.00$, area $= 0.00$, inverse NONE.
Determinant zero is what collapse looks like as a number. The columns say why: $c_2$ is exactly half of $c_1$, so the parallelogram they span has no width, and a plane built from two parallel directions is only a line. Every region's area gets multiplied by $0$.
A zero area factor is fatal to undoing. An inverse would have to scale areas back up by $1/\det = 1/0$, and no number does that. Distinct points that landed on the same spot cannot be told apart again, so no matrix — no map of any kind — reverses a collapse. That gives the working test: $A^{-1}$ exists exactly when $\det A \neq 0$. Compute $ad - bc$ and you know before you ever try.
APPLICATIONYou saw this collapse before — now you know the number that predicts it.About this model
What it represents: the exact theorem for square matrices — invertible precisely when the determinant is nonzero. For this $2\times2$ family, $\det = 1 - t^2$ really does hit zero only at $t = 1$, the instant the columns become parallel.
What it leaves out: numerical practice. In floating-point arithmetic a tiny computed determinant does not by itself mean nearly singular — rescaling a perfectly invertible matrix can make its determinant arbitrarily small — so software judges invertibility with condition numbers instead of determinants.
Source: Margalit and Rabinoff, Interactive Linear Algebra, The Invertible Matrix Theorem (Georgia Tech, free online) — the full list of conditions equivalent to $\det A \neq 0$.
Verify the collapse by hand: what is $\det\begin{pmatrix}1 & 0.5\\ 2 & 1\end{pmatrix}$?
At $t = 0.99$ the stats read $\det A = 0.02$. Is this matrix invertible?
Transfer: without computing any inverse, which of these matrices has none?
One number answers the reversibility question. $\det A \neq 0$ means every region keeps some area and $A^{-1}$ exists; $\det A = 0$ means the plane has been flattened onto a line and nothing brings it back. Area factor, orientation bit, invertibility test — the determinant carries all three.