In a proof of “If $\det(A)\ne 0$, then an inverse of $A$ exists” in my textbook for a Linear Algebra course, there is a step that is unclear to me:
It states following known equivalence:
$A^{-1} = \frac{1}{\det(A)} \operatorname{adj}(A) \DeclareMathOperator{\cof}{Cof}$
And then the following equivalence is obtained from this fact:
$$ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \begin{pmatrix} \cof(a) & \cof(d) & \cof(g) \\ \cof(b) & \cof(e) & \cof(h) \\ \cof(c) & \cof(f) & \cof(i) \end{pmatrix} = det(A)*I_3 $$
So the second factor in this last equation is $\operatorname{adj}(A)$, but I'm wondering how the matrix $A$ is equal to $\frac{1}{\det(A)}$
I don't know how the text you mention is written, but it looks like you are not following the flow of the argument.
The point is that the equality $$A\ \text {adj}(A)=\det (A)\,I_n $$ holds for any $n\times n $ matrix $A $. So if $\det (A)\neq0$ you can divide both sides of the equality by it, to get $$A\ \frac1 {\det (A)}\text {adj}(A)=I_n.$$ So now you have that the matrix $B=\frac1 {\det (A)}\text {adj}(A)$ has the property that $AB=I $, so it is an inverse for $A $ (you also need $BA=I$; this follows automatically from $AB=I $, but requires a short proof).
The equality you wrote is the case $n=3$, but it's not really different for other $n $.