We denote $E_{n} = \mathcal{M}_{n,1}(\mathbb{R})$ equipped with the inner product $(X \mid Y) = X^{\top}Y$. We assume $n \geqslant 3$. Let $F$ be a vector subspace of $E_{n}$ of dimension $n-2$. We consider $(N_{1}, N_{2})$ a basis of $F^{\perp}$ and we set $N = \begin{pmatrix} N_{1} & N_{2} \end{pmatrix} \in \mathcal{M}_{n,2}(\mathbb{R})$. $A$ is an invertible matrix in $\mathcal{M}_{n}(\mathbb{R})$, and $A_{N} = \begin{pmatrix} A & N \\ N^{\top} & 0_{2} \end{pmatrix}$. Deduce that $\operatorname{det}(A_{N}) = \operatorname{det}(N^{\top}A^{-1}N)\operatorname{det}(A)$.
We denote $E_{n} = \mathcal{M}_{n,1}(\mathbb{R})$ equipped with the inner product $(X \mid Y) = X^{\top}Y$. We assume $n \geqslant 3$. Let $F$ be a vector subspace of $E_{n}$ of dimension $n-2$. We consider $(N_{1}, N_{2})$ a basis of $F^{\perp}$ and we set $N = \begin{pmatrix} N_{1} & N_{2} \end{pmatrix} \in \mathcal{M}_{n,2}(\mathbb{R})$. $A$ is an invertible matrix in $\mathcal{M}_{n}(\mathbb{R})$, and $A_{N} = \begin{pmatrix} A & N \\ N^{\top} & 0_{2} \end{pmatrix}$.
Deduce that $\operatorname{det}(A_{N}) = \operatorname{det}(N^{\top}A^{-1}N)\operatorname{det}(A)$.