Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

Thank you in advance.

share|improve this question
You have seen this, I presume? –  J. M. Apr 3 '13 at 10:40
Thanks, I have. Unfortunately something so simple hasn't worked for me so far. I'd be interested if you knew of a straightforward use of block matrix determinants prove the result in question. –  Daniel Apr 3 '13 at 11:20

2 Answers 2

up vote 2 down vote accepted

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

share|improve this answer
Wow, so it is! I thought I'd looked everywhere and couldn't locate what I had as a named identity. This will certainly open up a whole new avenue of investigation! –  Daniel Apr 3 '13 at 13:18

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

share|improve this answer
Thank you for all your assistance Adam, that has certainly made clear for me what's going on with the cofactors now. –  Daniel Apr 3 '13 at 23:45

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.