Geometric intuition behind inverse matrix restrictions We know that for an invertible matrix $A$, $A^{-1} = \frac{\mathrm{adj}(A)}{\det{(A)}}$. We can see algebraically that this is undefined when $\det(A)=0$, but it is cryptic what this means geometrically. Is the following intuition valid? This is not meant to be rigorous.
Define a matrix $A:\mathbb{R}^n \rightarrow \mathbb{R}^{n}$ (e.g. mapping 3D space onto the X-Y plane). Because this maps the $n$-dimensional parallelepiped such that it does not occupy the $n$-th dimension, it's $n$-dimensional "volume" is $0$. Hence, $\det(A)=0$. Now, define another matrix $T:\mathbb{R}^{n-1}\rightarrow\mathbb{R}^{n-1}$ that maps the $(n-1)$-dimensional parallelepiped such that it coincides with the parallelepiped that $A$ acted on. When viewing in $\mathbb{R}^{n-1}$ space, $A$ and $T$ are indistinguishable, so there is ambiguity as to what the initial state was. Therefore, $A^{-1}$ is undefined. $\blacksquare$
EDIT: 3Blue1Brown articulates this point better than I do. I just found this. I was trying to say that essentially, if you squish space down into a smaller dimension, there is no way to distinguish squishing higher dimensional space and transforming space of the same dimension. There is ambiguity as to what the initial state was, so $A$ is not a bijection.
https://youtu.be/uQhTuRlWMxw?t=403
 A: It's not clear what you mean by "Now, define another matrix $T:\mathbb{R}^{n-1}\rightarrow\mathbb{R}^{n-1}$ that maps the $(n-1)$-dimensional parallelepiped such that it coincides with the parallelepiped that $A$ acted on."
That said, you seem to have already accepted that if $\det(A) = 0$, then the image of $A$ is at most $(n-1)$-dimensional (i.e. $A$ fails to be surjective/onto). From this alone, it is clear that $A$ cannot have an inverse.  After all: for any $b \in \Bbb R^n$, $x = A^{-1}b$ is supposed to satisfy $T(x) = b$. However, if $b$ lies outside the "squished" subspace of outputs from $A$, then we can never have $T(x) = b$.

Another geometric argument: note that $|\det(A)|$ is the "volume stretch factor" of the transformation associated with $A$. It follows that for $A,B$, we have $|\det(AB)| = |\det(A)| \cdot |\det(B)|$.  Now, because the identity mapping has a stretch factor of $1$ and $AA^{-1} = I$, it follows that $|\det(A)|\cdot |\det(A^{-1})| = 1$.
So, if $\det(A) = 0$, then whatever the inverse $A^{-1}$ is would have to satisfy $0 \cdot |\det(A^{-1})| = 1$, but that's impossible.
