Determinants of minors occurring 'within' determinant of full matrix

Question

$A= (a_{ij})$ is an $n\times n$ symmetric positive matrix. It induces a quadratic form $f(x):= x^tAx$ on $\mathbb{R}^n$. $D_m$ denotes the determinant of the top left $m\times m$ submatrix of $A$ (or rather of $f$). What does the following highlighted sentence mean?

This is from page 781 of Hancock's book on 'Minkowski's geometry of numbers'. It was written in the 1930's and I've been having a hard time with the language.

E.g. When $A$ is $3\times 3$ and $m=2$, we have \begin{equation} D_3 = \det\left[ {\begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{12} & a_{22} & a_{23} \\ a_{13} & a_{23} & a_{33} \end{array} } \right],\ D_m = D_2 = \det\left[ {\begin{array}{cc} a_{11} & a_{12} \\ a_{12} & a_{22} \end{array} } \right],\ D_{n-m}=D_1 = a_{11}. \end{equation}

Answer 1

If the $n\times n$ matrix $M$ is decomposed into submatrices, $$M=\begin{pmatrix}A&B\\ C&D\end{pmatrix},$$ where $A$ has dimension $m\times m$, then the determinant of $M$ can be decomposed as $$\det M=\det A\det D+X.$$ The multinomial $X$ in the matrix elements of $M$ contains $n!-m!(n-m)!$ terms, for a general matrix $M$. If the matrix is symmetric, the number of distinct terms is less.

In the $n=3$, $m=2$ example given in the OP, this gives for $X$ the four terms $$X=a_{13} a_{22} a_{31} + a_{12} a_{23} a_{31} + a_{13} a_{21} a_{32} - a_{11} a_{23} a_{32}.$$ Notice that the indices of $X$ follow Hancock's description.

So I would paraphrase the sentence highlighted in yellow as "Write down the determinant $D_n$ of $f$ and within that expression single out the product of the principal minors $D_m$ and $D_{n-m}$."

Answer 2

Hancock seems to do not such a good job with the translation of Minkowski's work. I've posted the original here. See for example the double occurrence of $a_{mm}$ in the last inequality above versus what's written in the English version.