Let $A$ be a matrix in the special orthogonal group, $A \in \mathrm{SO}_n$. This means that $A$ is real, $n \times n$, $A^t A = I$ and $\det(A)=1$, that is, the column vectors of $A$ make a positively-oriented orthonormal basis for $\mathbb R^n$.

Decompose $A$ as a block matrix

$$A = \begin{pmatrix} B & C \\ D & E\end{pmatrix}$$

where $B$ is $k \times k$ and $E$ is $(n-k)\times (n-k)$.

I'm looking for a basic linear-algebra argument that $\det(B) = \det(E)$, ideally something that could be presented in a 2nd or 3rd year undergraduate course. So I do not want people to invoke anything like tensor products or differential forms.