This was one of the more delightful aspects of linear algebra, when I first learned it properly. It could be because my initial introduction to linear algebra was purely computational, with no discussion of the geometric meaning of the procedures we learned. Sure, we learned how to calculate determinants, eigenvectors, eigenvalues, reduced row echelon forms, and such, but it was a mere drilling of procedures till we could do them blindfolded.

So here’s an ‘intuitive’ explanation of why the determinant of a matrix represents the ‘oriented volume’ of the parallelpiped spanned by the column vectors that constitute the matrix. The rigorous proofs can be found in any good linear algebra text — I am more interested in offering an *explanation* of why we might want to relate determinants to oriented volume.

The standard definition of the determinant, taught to most students of subjects where mathematics is used as a tool and not understood for its own sake, is as follows:

Here, is the entry of the matrix in the ith and 1st column, and is the matrix obtained by striking out the ith row and jth column of .

We can start with the simplest example: the identity matrix . The determinant of this matrix is 1, and the column vectors constituting it are , , and . The volume that these three vectors span is the cube with vertices at (0, 0, 0), (1, 1, 0), (1, 0, 1), (0, 1, 1), (1, 0, 0), (0, 1, 0), (0, 0, 1), and (1, 1, 1). It is, in short, the cube with a corner at the origin and unit lengths stretching along the x, y and z axes. This cube has side of length 1 each, so it has a volume of 1, the same value as the determinant.

Now consider a similar unit cube, but spanned instead by the vectors , , and . This cube has vertices (0, 0, 0), (1, 1, 0), (1, 0, -1), (0, 1, -1), (1, 0, 0), (0, 1, 0), (0, 0, -1), and (1, 1, -1). If we consider volume to have the standard property of being a positive number, then clearly the volume of this cube is 1, just like the previous cube. *Oriented* volume, though, as its name suggests, is a kind of volume that is dependent on the ‘directions’ possessed by the object with the volume in question. In this case, since we have fliipped one vector to its ‘negative’, the oriented volume is also ‘negated’, so although the cube has conventional volume, its oriented volume is -1. -1 also ‘happens’ to be the determinant of the matrix composed of , , and .

We can say the same for the matrices composed of , , and of , , . Both are something like the unit cube of , , and “flipped” (“negated”) once; both have an oriented volume of -1, and both have determinants of -1.

We can abstract this property of the determinant as such: when one of the column vectors of the matrix is negated, the entire determinant is negated. The absolute value of the determinant is unaltered. This can be interpreted as showing that when you ‘flip’ a vector about the origin to point in the ‘opposite’ direction, it together with its two other companions then ‘span’ a parallelpiped ‘pointing’ in the opposite direction from the original but having the same *absolute* volume. The change in the ‘direction’ of the parallelpiped is manifested as a change in the sign of its oriented volume, just as the change in the sign of one of the matrix’s column vectors is manifested as a change in the sign of its determinant.

To summarise, we have seen that negating one of the column vectors of a matrix also negates our intuitive notion of ‘oriented volume’, and negates its determinant. This is just a special case of a property shared by both oriented volumes and determinants: multiplying a vector constituent of either by a scalar results in a multiplication of either (oriented volume or determinant) by a factor of .

Another common feature shared by both the oriented volume and the determinant is that both are invariant under the addition of a scalar multiple of a constituent vector to any of the other vectors. For example, if we transform the matrix to by adding to the original middle vector , the determinant remains the same: the extra “1” in the x-direction is irrelevant somehow.

This irrelevance makes sense in the ‘oriented volume’ interpretation. There, the volume represented by the second matrix is simply the first volume, the unit cube along all the positive axes, with four faces ‘skewed’ from squares to parallelograms so that it’s a parallelpiped with a 45^{o} skew in one direction but with two square faces in the ‘unskewed’ directions. However, since the y-value of the middle vector is unchanged at 1, the ‘component’ that matters for calculating parallelpiped volume is unchanged — this ‘skewed’ parallelpiped still has volume 1, in the same way that a right angled triangle skewed to a scalene triangle of the same base length and height retains its original volume. Parallelograms are really just triangles with twice the volume, so for them as well, it doesn’t matter how we angle their sides as long as they retain the same ‘base’ and ‘height’. And parallelpipeds are just higher dimension analogues of parallelograms. So it should be intuitively plausible that we should expect their volume to remain the same under certain ‘skew’ transformations in which the ‘components’ added to one of their ‘edges’ only result in them becoming ‘more skewed’ but essentially the same ‘size’.