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From the question here and solution posted, we see that $$\begin{matrix} \big((x-c)&(y-r)&1\big)\\\\\\\\\\\end{matrix}\;\; \left(\begin{matrix} p^2 &-ap &\;\;\scriptsize\frac q2\left|\begin{matrix} a &b\\p&q\end{matrix}\right|\\ -ap &a^2 &\scriptsize-\frac b2\left|\begin{matrix}a&b\\p&q\end{matrix}\right|\\ \scriptsize\frac q2\left|\begin{matrix}a&b\\p&q\end{matrix}\right| &\scriptsize-\frac b2 \left|\begin{matrix} a&b\\p&q\end{matrix}\right| &\;\;\;0 \end{matrix}\right)\;\; \left(\begin{matrix}x-c\\\\y-r\\\\1\end{matrix}\right)$$

is equivalent to

$$\hspace{2cm}\left|\;\;\begin{matrix} \left|\;\;\;\begin{matrix} -a&&-p\\ -b&&-q \end{matrix}\;\;\;\right| &&\left|\begin{matrix} -a&-p\\ (x-c)&(y-r)\end{matrix}\right| \\\\ \left|\begin{matrix} -a&-p\\ (x-c)&(y-r)\end{matrix}\right| &&\left|\begin{matrix} -b&-q\\ (x-c)&(y-r)\end{matrix}\right| \end{matrix}\;\;\right|$$

Question:
Is it always possible to express the quadratic form $\mathbf {x^T Q x}$, where $\mathbf Q$ is symmetric, in the form of a determinant of determinants, i.e.
$$\left|\begin{matrix}|\mathbf A|&|\mathbf B|\\|\mathbf B|&|\mathbf C|\end{matrix}\right|$$ ? If so, express $\mathbf {A, B, C}$ in terms of $\mathbf {x, Q}$.

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