Proof of a Determinant Identity

Question

I have a matrix identity that I wish to prove, which relates the determinant of a matrix to determinants of sub-matrices (essentially, cofactors of the larger matrix). In general terms, consider the matrix:

$$M=\left(\matrix{a & \vec{m}^\top & b \\ \vec{g} & N & \vec{h} \\ c & \vec{k}^\top & d}\right),$$

and let us write $\Delta=\det(M)$. We can also write the determinants of various sub-matrices of $M$ as cofactors, such as

$$a_c = \left|\matrix{N & \vec{h} \\ \vec{k}^\top & d}\right|,$$

for example. Under this formulation, the identity that I wish to prove is

$$\det(N) = \frac{a_cd_c - b_cc_c}{\Delta}.$$

I understand that this can be done by making use of a slight modification of the method employed in this answer to another matrix question, where the procedure is performed not on the inverse but rather on the adjoint (the inverse multiplied by the determinant).

However, after a fruitless attempt to carry this out myself I feel that it would be a better use of my time to allow someone with a deeper understanding of the matter to fill in the appropriate steps that I'm clearly missing.

Answer 1

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

Answer 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 $$\operatorname{det}(N)=\frac{d_c a_c - b_cc_c}{\Delta}$$