Understanding determinants and matrices My question is, looking at a determinant geometrically as a 'volume', is dividing by the determinant similar to getting unit values for the elements of a matrix?
 A: First, the determinant, geometrically, as a volume is only true for a 3 by 3 determinant.  In that case, we can think of the matrix $\begin{bmatrix}2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4\end{bmatrix}$ as made from the three vectors <2, 0, 0>, <0, 3, 0>, and <0, 0, 4> which form sides of a rectangle with sides of length 2, 3, and 4, lying along the three axes so of volume $abc= 2(3)(4)= 24$.  Dividing the matrix by 24, we get $\begin{bmatrix}\frac{1}{12} & 0 & 0 \\ 0 & \frac{1}{8} & 0 \\ 0 & 0 & \frac{1}{6}\end{bmatrix}$ which has determinant $\frac{1}{56}$ NOT 1.
A: First off, let's go over what a determinant is.  Given a linear transformation $f: \Bbb R^n \to \Bbb R^n$ (determinants are defined on more general spaces, but I can only visualize it geometrically for $\Bbb R^n$) defined by $f(x) = Ax$ for $A \in M_{n\times n}(\Bbb R)$, we know that volumes (technically generalized volumes) in the domain of the $f$ are transformed in a multilinear fashion into volumes in the image of $f$.
For example, let's say we have a linear transformation $f: \Bbb R^3 \to \Bbb R^3$.  Then, if we know that the volume of the parallelotope with sides $v_1$, $v_2$, and $v_3$ (where $v_1, v_2, v_3 \in \Bbb R^3$) is 12, then we can be sure that the volume of the parallelotope with sides $f(v_1)$, $f(v_2)$, and $f(v_3)$ is $\det(f)\cdot 12$, where $\det(f) := \det(A)$.

Dividing a matrix by its determinant is tantamount to defining a new function $f^*$ given by $f^*(x) = \left(\frac 1{\det(A)}A\right)x$.  But remember that $$\det(kA) = k^n\det(A)$$ where $k$ is a constant and $A$ is $n\times n$.  Thus this new transformation $f^*$ doesn't keep volumes the same (unless $\det(A)=1$) -- it transforms volumes by $$\det(f^*) = \det\left(\frac 1{\det(A)}A\right) = \frac{1}{\det(A)^n}\det(A) = \frac{1}{\det(A)^{n-1}}$$
So I think what you really wanted to do is ask about dividing a matrix by $\det(A)^{1/n}$.  Confirm for yourself that the determinant of the matrix $\frac{1}{\det(A)^{1/n}}A$ (assuming $\det(A) \ne 0$) is $1$.

Okay.  What does that mean geometrically?  It means that the transformation $g^*$ given by $g^*(x) = \frac{1}{\det(B)^{1/n}}Bx$ doesn't change volumes of parallelotopes at all.
And what does that mean for the columns of this matrix $\frac{1}{\det(B)^{1/n}}B$?  Well remember that the columns are just $g^*(e_1)$, $g^*(e_2)$, etc. (where $\{e_1, \dots, e_n\}$ are the standard basis for $\Bbb R^n$). What it doesn't mean is that each one of those vectors is unit-length.  Remember if we have the vectors $(1,1)$ and $(2,3)$, then the area of the parallelogram of those two vectors is $1$.  But of course, neither $(1,1)$ nor $(2,3)$ have unit lengths.
So all scaling the matrix does is scale the lengths of the image vectors proportionally so that they preserve volumes from the domain of the transformation.
