Let $\Lambda$ be a lattice in $\mathbb R^n$ generated by the columns of a square matrix $B$ of order $n$.
If $\Lambda$ is full rank, then the quantity $d(\Lambda)=|\det B|$ has a simple geometrical meaning: it is the volume of a fundamental parallelepiped of $\Lambda$. Thus, for instance, the condition $|\det B|\le 1$ means that the volume of the fundamental parallelepiped of $\Lambda$ is at most $1$.
Suppose however that the rank of $\Lambda$ is $r<n$; therefore, $B$ is degenerate. What does it mean that every square submatrix of $B$ of order $r$ has determinant not exceeding $1$ in absolute value?
A rank-$r$ matrix will span a dimension-$r$ subspace of $\Bbb R^n$; since $r < n$, the volume of the fundamental parallelipiped is $0$.
As an example, the vectors $\pmatrix{1\\2}, \pmatrix{2\\ 4}$ generate the lattice $$ \{ \pmatrix{k\\2k} \mid k \in \Bbb Z \}, $$ and the fundamental parallelipiped for that lattice consists of just the line segment from the origin to $\pmatrix{1\\2}$, whose two-dimensional volume (i.e., area) is zero.