Let $A$ be a matrix where each entry is a polynomial.
For example,
$A = \begin{bmatrix} t & t^2 \\ t^3 & t^4 \end{bmatrix}$
Does it make sense to talk about the inverse of $A$, i.e., $A^{-1} = \dfrac{\text{adj}(A)}{\det(A)}$
Wikipedia says that I could do it for any $n \times n$ matrix. https://en.wikipedia.org/wiki/Invertible_matrix#In_relation_to_its_adjugate
Want a second opinion on this claim.
There are two meanings for "inverse" here, since you're introducing a parameter $t$.
The inverse matrix of $$A(t) = \begin{bmatrix} t & t^2 \\ t^3 & t^4 \end{bmatrix}$$will exist if and only if $\det A(t) \neq 0$, and in this case, the usual formula for the inverse of a $2 \times 2$ matrix applied for each fixed $t$ would give you $A(t)^{-1}$. But you're out of luck: $\det A(t) = tt^4 - t^2t^3 = t^5-t^5=0$.
Regard $A$ as a function $A \colon \Bbb R \to {\rm Mat}(2,\Bbb R)$, given by $$A(t) = \begin{bmatrix} t & t^2 \\ t^3 & t^4 \end{bmatrix}.$$This function is injective because of the very first entry, but it is not surjective, as there are several matrices outside its image (e.g., anything with negative $(1,1)$ entry). You will have more than one left inverse for $A$.
Yes, it does for general matrices, but to define an inverse your way you need $\det A \ne 0$. In your case, however, $$\det A = t^5 - t^5 = 0.$$
Generally, if $a,b,c,d$ are functions of $t$ with $a(t)d(t) \ne b(t) c(t)$, you can write $$ \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} ad-bc & 0 \\ 0 & ad-bc \end{bmatrix}, $$ so $$ \begin{bmatrix} a & b \\ c & d \end{bmatrix}^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} $$
