# Interpretations of Higher Order Determinants

For determinants of $2 \times 2$ and $3 \times 3$ matrices, there is a simple geometric interpretation of the determinant. It is, respectively, the area of parallelogram or parallelepiped formed by the transformed basis vectors of a vector space following a linear transformation.

For determinants of $n \times n$ matrices with $n \geq 4$, there is obviously no simple geometric interpretation. One possible interpretation for the determinants of these higher order matrices involves linear systems of equations, where, for a system of $n$ linear equations, if the coefficient matrix $A$ has a determinant det$(A) = 0$, the system has no solution.

Are there any other interesting interpretations or applications of higher order determinants?

• Actually - it's still the area of the relevant $n$-dimensional parallelogram if I recall correctly. – Artimis Fowl May 29 '17 at 4:37
• In all dimensions, it's the signed volume of the (possibly degenerate) n-dimensional parallelopiped determined by the columns, and it's also the signed volume of the n-dimensional parallelopiped determined by the rows. See this thread: math.stackexchange.com/questions/427528 – quasi May 29 '17 at 4:42
• @quasi I do not have any experience with geometry past $\mathbb{R}^3$. What exactly is the understanding of volume and area in $\mathbb{R}^n$ with $n \geq 4$? – nguzman May 29 '17 at 4:48
• It's inductive. If $P$ is a subset of an $(n-1)$-dimensional hyperplane of $\mathbb{R}^n$, and if $P$ has $(n-1)$-dimensional volume $a$, then if $P$ is extended by a vector $v$ of length $h$, where $v$ is perpendicular to the hyperplane, then the volume of the extended region is $ah$. In particular, the $n$-dimensional volume of the region contained in the standard unit $n$-cube is $1$. – quasi May 29 '17 at 4:59
• A box $[0,1]^n$ has measure $1$. Scaling it scales the measure in the usual ways. Nice subsets of $\Bbb R^n$ are disjoint unions of boxes. – arctic tern May 29 '17 at 5:00