What are the properties of this cousin to the characteristic equation: $\det (xA-I)=0$ The characteristic polynomial, defined for a matrix $A$:
$
c(x; A) = \det (A-I x ) = 0
$
has nice properties related to the eigenvalues $\lambda$, of the matrix:
$
c(x; A) = (x-\lambda_1)(x-\lambda_2)(x-\lambda_3) \ldots
$
What, if any, is the connection to the eigenvalues of a matrix to this function:
$
p(x; A) = \det (x A-I )
$
 A: If $x$ is a root of $p(x;A)$, then $Ax-I$ is not invertible, and therefore has a nonzero null space. So there exists a nonzero vector $\vec{v}$ such that $$(Ax-I)\vec{v}=\vec{0}$$
So $Ax\vec{v}=\vec{v}$. Note that $x$ cannot be zero for two reasons - we have specified $\vec{v}\neq\vec{0}$ and $\det(A\cdot0-I)\neq0$. We then have that $A\vec{v}=x^{-1}\vec{v}$, and $x^{-1}$ is an eigenvalue for $A$.
So the roots of $p(x;A)$ are inverses of the nonzero eigenvalues of $A$. (If $A$ has zero as an eigenvalue, then $p(x;A)$ has degree less than $n$.)

A more complete picture added later:
If the Jordan canonical form of $A$ (over $\mathbb{C}$) is given by $A=PJP^{-1}$, with $J$ a composite of Jordan blocks, then 
$$
\begin{align}
\det(Ax-I) & =\det(PJP^{-1}x-I)\\
&=\det(PJP^{-1}x-PIP^{-1})\\
&=\det(P)\det(Jx-I)\det(P^{-1})\\
&=\det(Jx-I)
\end{align}
$$
so let's assume that $A$ is already in its Jordan canonical form. Some of the Jordan blocks of $A$ have eigenvalue $0$ and some do not. Write $$A=\begin{bmatrix}
Z & 0\\
0 & Y
\end{bmatrix}
$$
where $Z$ has the Jordan blocks with eigenvalue $0$, and $Y$ has the other (nonzero eigenvalued) Jordan blocks. It's important to understand that $Z$ has $0$'s everywhere except for some $1$s at selected places along the $+1$-off-diagonal. Then 
$$
\begin{align}
\det(Ax-I)&=\det\left(\begin{bmatrix}Zx & 0\\0 & Yx\end{bmatrix}-I\right)\\
&=\det(Zx-I_{z\times z})\det(Yx-I_{y\times y})\\
&=(-1)^z\det(Yx-I_{y\times y})\\
&=(-1)^zp(x;Y)
\end{align}
$$
where $z$ is the multiplicity of $0$ as an eigenvalue of $A$ and $y$ is the complement: $n-z$.
Since $Y$ is invertible, Alex Becker's answer can be applied. In summary: 
$$\begin{align}p(x;A)&=(-1)^zp(x;Y)\\&=(-1)^z(-1)^y\det(Y)c(x;Y^{-1})\\&=(-1)^n\det(Y)c(x;Y^{-1})\end{align}$$
That is, $p(x;A)$ is a certain multiple of the characteristic polynomial of $Y^{-1}$, where $Y$ is the composite of $A$'s invertible Jordan blocks. Said one more way, $p$ is a polynomial whose roots are the inverses of $A$'s nonzero eigenvalues, and the multiplicities are respected.
A: If $A$ is an $n\times n$ invertible matrix, then 
$$\begin{eqnarray}
p(x;A)&=&\det(Ax-I)\\
&=&\det(A)\det(Ix-A^{-1})\\
&=&(-1)^n\det(A)\det(A^{-1}-Ix)\\
&=&(-1)^n\det(A)c(x;A^{-1})\end{eqnarray}$$
and so the roots are the eigenvalues of $A^{-1}$, which are the inverses of the eigenvalues of $A$.
