Proof of a Determinant Identity

Solution 1:

It seems the formula you want is the Desnanot-Jacobi identity.

Solution 2:

Use the formula given previously where this $$\pmatrix{\mathbf A & \mathbf b\\ \mathbf c^T & d}^{-1} = \pmatrix{\mathbf E & \mathbf f\\ \mathbf g^T & h}$$ gives this $$\mathbf{A}^{-1} = E - \frac{\mathbf{f}\mathbf{g}^T}{h}$$

As hinted, using your matrix $$M=\left(\matrix{a & \vec{m}^\top & b \\ \vec{g} & N & \vec{h} \\ c & \vec{k}^\top & d}\right)$$ look at the inverse times the determinant, so that we have the matrix of cofactors $$ \Delta M^{-1} = \left(\matrix{a_c & \vec{m}_c^\top & b_c \\ \vec{g}_c & N_c & \vec{h}_c \\ c_c & \vec{k}_c^\top & d_c}\right)$$ and thus that $$ M^{-1} = \frac{1}{\Delta}\left(\matrix{a_c & \vec{m}_c^\top & b_c \\ \vec{g}_c & N_c & \vec{h}_c \\ c_c & \vec{k}_c^\top & d_c}\right)$$ Use the formula to find the matrix of cofactors of the sub-matrix. Remember that the matrix of cofactors is obtained from the inverse times the determinant, so we are looking for an expression for $$d_c\left(\matrix{a & \vec{m}^\top \\ \vec{g} & N}\right)^{-1}$$ The top left element of this will give you the determinant for $N$.

Here we go: $$\left(\matrix{a & \vec{m}^\top \\ \vec{g} & N}\right)^{-1} =\frac{1}{\Delta}\left(\matrix{a_c - \frac{b_cc_c}{d_c}& \star \\ \star & \star \\ }\right)= \frac{1}{\Delta d_c}\left(\matrix{d_c a_c - b_cc_c& \star \\ \star & \star \\ }\right)$$ $$\Rightarrow d_c\left(\matrix{a & \vec{m}^\top \\ \vec{g} & N}\right)^{-1} =\frac{1}{\Delta}\left(\matrix{d_c a_c - b_cc_c& \star \\ \star & \star \\ }\right) $$

And you have your formula $$\operatorname{det}(N)=\frac{d_c a_c - b_cc_c}{\Delta}$$