Is the determinant a scale factor for any shape? My linear algebra text states that the determinant of a matrix A is always the scaling factor that you will get by left multiplying a coordinate matrix B by A. But the only examples they give are triangles and squares. Is this true regardless of the shape described by B? Even if it's not regular, not convex, intersects with itself, etc.?
 A: The answer depends on your definition of volume.
In an introductory analysis class, we typically define the volume of a set $A$ lying in an $n$-dimensional Euclidean space $E$ by the integral $\int_A|dV|$, where $|dV|$ is the Euclidean density and $\int$ is the Riemann integral. (This definition of volume is called the "content" or the "Jordan measure" of the set $A$.) If you perform a linear transformation $L$ on $E$, then the volume of the image of $A$, i.e. the volume of $L(A)$, is $\int_{L(A)}|dV|$. (It turns out that if $A$ has volume according to this definition, then $L(A)$ must have volume, too.)
But by the key step in the proof of the multivariable change of variables theorem, $\int_{L(A)}|dV|=\int_A|\det L|\,|dV|$. 
Therefore $\text{Vol}(L(A))=|\det L|\cdot\text{Vol}(A)$, assuming that $A$ has volume in the first place.
Your book focuses on the case of triangles and squares because linear algebra isn't analysis. But to answer your question, you must do some analysis in order to get precise about what you mean by the volume of an arbitrarily shaped subset of a Euclidean space.
A: Yes. One way to see this is to divide your shape into triangles. Since $A$ scales each triangle by the same factor (namely, $\det A$) it will scale the entire shape described by $B$ by this factor too.
