Determinants and volume of parallelotopes The absolute value of a $2 \times 2$ matrix determinant is the area of a corresponding parallelogram with the $2$ row vectors as sides.
The absolute value of a $3 \times 3$ matrix determinant is the volume of a corresponding parallelepiped with the $3$ row vectors as sides.
Can it be generalized to $n-D$? The absolute value of an $n \times n$ matrix determinant is the volume of a corresponding $n-$parallelotope?

Yes it can. In fact, as Jamie Banks noted, a determinant is an intuitive way of thinking about volumes. To summarise the argument, if we consider the vectors as a matrix, switching two rows, multiplying one by a constant or adding a linear combination will have the same effect on the volume as on the determinate. We can use these operations to transform any n-parallelotope to cube and note that the determinate matches the signed volume here, so it will match it everywhere as well.
