# Determinant of Matrix Computed by Expanding Down the Diagonal?

Going through this paper:

http://dx.doi.org/10.1016/S0893-9659(00)00169-5

at the bottom of page 407, the authors seem to compute the determinant of a matrix by expanding down the diagonal.

The authors discuss a matrix

$A = \left[ \begin{array}{ccc} a_{11} a_{12} a_{13} \\ a_{21} a_{22} a_{23} \\ a_{31} a_{32} a_{33} \end{array}\right]$

and call $M_{ij}$ the cofactor of $a_{ij}$.

Then at the bottom of page 407 they write

$\det(A) = a_{11} M_{11} + a_{22} M_{22} + a_{33} M_{33}.$

Is expansion down the diagonal possible? Is it possible in some special cases?

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This type of expansion is certainly not possible for all matrices. Consider $2\times 2$ matrices for instance... –  Abel Jun 11 '13 at 14:11
The equation on page 407 has some $d$'s in it, which I don't understand. They might represent differentials rather than factors, but the equation doesn't look correct under that interpretation either. –  Andreas Blass Jun 11 '13 at 14:53
@AndreasBlass I thought the d's were diffusion coefficients but you may be correct -- however, as you say, this would indicate that det(A) is somehow a function of the diagonal entries of A and that doesn't seem correct. –  Name Jun 11 '13 at 15:13

You are not quoting the formula correctly. It says $$\color{red}{d}[\det(A)] = M_{11}\color{red}{d}a_{11} + M_{22}\color{red}{d}a_{22} + M_{33}\color{red}{d}a_{33},\tag{1}$$ where the $\color{red}{d}$ means derivative. The authors are being sloppy here. They actually consider a matrix function $A(t)=A(0)-tD$, where $A(0)$ is a constant and stable matrix, $D$ is a constant positive diagonal matrix and $t=k^2\ge0$. The $A$ in $(1)$ is actually $A(t)$ and the $d$ means derivative with respect to $t$. Formula $(1)$ is a direct consequence of Jacobi's formula.