Relation between $A^{-1}$ and $\det A$ I know there is a relationship between an $n\times n$ matrix $A^{-1}$ and $(\det A)^{-1}$.
That is,
$A^{-1}$ is equal to $(\det A)^{-1}$ times what? How to use a formula to express the relationship between  $A^{-1}$ and $(\det A)^{-1}$? And why?
Thanks in advance.
 A: The determinant of $A^{-1}$ is $(\det A)^{-1}$. This follows from the general identity $\det(AB) = (\det A)(\det B)$.
You may be looking for Cramer's rule which expresses $A^{-1}$ as $(\det A)^{-1}$ times the adjoint of $A$.
A: I think one of the most elegant and conceptually clean ways to view the relationship 'twixt $A^{-1}$ and $\det A$ is to use the characteristic polynomial $p_A(\lambda)$ of $A$:
$p_A(\lambda) = \det(\lambda I_N - A). \tag{1}$
Here $\deg p_A(\lambda) = N$, and $I_N$ is the $N \times N$ identity matrix.  When this determinant is expanded out, the degree zero term of the ensuing polynomial is
$(-1)^N \det A$.  If we write
$p_A(\lambda) = \sum_0^N c_{N - i}\lambda^{N - i}, \tag{2}$
then we have
$c_0 = (-1)^N \det A. \tag{3}$
Now consider the polynomial $q_A(\lambda)$ such that
$\lambda q_A(\lambda) = p_A(\lambda) - c_0; \tag{4}$
from (2), it is easy to see that
$q_A(\lambda) = \sum_0^{N - 1} c_{N - i}\lambda^{N - i - 1}, \tag{5}$
so $q_A(\lambda)$ is easily had once $p_A(\lambda)$ is known.  If we evaluate (4) on the matrix $A$, we obtain
$Aq_A(A) = p_A(A) - c_0 I_N = - c_0 I_N = (-1)^{N + 1}(\det A) I_N, \tag{6}$
since $ p_A(A) = 0$ by the Hamilton-Cayley theorem.  If $\det A \ne 0$ then from (6)
$A((-1)^{N + 1} (\det A)^{-1}q_A(A)) = I_N, \tag{7}$
showing
$A^{-1} = (-1)^{N + 1} (\det A)^{-1}q_A(A); \tag{8}$
the computation of $q_A(\lambda)$, via $p_A(\lambda)$, provides all the determinantal data obtained by the calculation of the adjoint; indeed, (8) shows that $\text{adj} A = (-1)^{N + 1}q_A(A)$, at least when $\det A \ne 0$.  In the event that $\det A = 0$, we see that $q_A(A)$ annihilates $A$; but as long as $\det A \ne 0$, (8) provide answers to our OP Ian's "times what" etc. questions.  To wit:
1.)  "$A^{−1}$ is equal to $(\det A)^{−1}$ times what?"
(8) is a matrix formula for $A^{-1}$ of the form $(\det) A)^{-1} \times \text{what}$, where "what" is the matrix $(-1)^{N + 1}q_A(A)$; one might even write
$\text{what} \equiv (-1)^{N + 1}q_A(A); \tag{9}$
2.)  How to use a formula to express the relationship between $A^{−1}$ and $(\det A)^{−1}$?
See item (1) above.  (8) gives such a formula.
3.)  If  by "why" one means "why does this formula hold?", then the answer is provided in the above derivation.  If the "why" is of a more metaphysical ilk, you might want to check out Sefer Yetzirah and the Ten Sefirot! ;) ;) ;) !!!
Hope this helps.  Cheerio, and as always,
Fiat Lux!!!
A: I do not entirely understand you question, but I think you seek this relation:
$$A^{-1} = \frac{\operatorname{adj} A}{\det A}$$
if and only $\det A \ne 0$.
