A deteminant related to the value of a symmetric bilinear map 
If $Q$ is a symmetric $q\times q$ matrix, then for $\vec v,\vec w\in \mathbb R^q,$ the value of the $(q+1)\times(q+1)$ determinant $$\det\pmatrix{Q&\vec w\\\vec v^t&0}$$
  is $(-1)$ times the value of the symmetric bilinear map associated to the "classical adjoint" or "cofactor" matrix of $Q$ applied to $\vec v$ and $\vec w$.

Why?
 A: Since $Q$ is symmetric we can find an orthogonal matrix $P$ such that $P^tDP=Q$, where $D$ is diagonal. Therefore $$\det\pmatrix{Q&w\\\ v^t&0}=\det\pmatrix{P^t&0\\\ 0&1}\pmatrix{Q&w\\\ v^t&0}\pmatrix{P&0\\\ 0&1}=\det\pmatrix{D&P^tw\\\ v^t P&0},$$
so we just have to deal with the case $Q$ diagonal. We can write $D=\operatorname{diag}(\alpha_q,\ldots,\alpha_1)$ and let $D':=\operatorname{diag}(\alpha_{q-1},\ldots,\alpha_1)$, . We expand the last determinant with respect to the first line: 
$$\det\pmatrix{Q&w\\\ v^t&0}=\alpha_q\det\pmatrix{D'&(P^tw)'\\\ (v^tP)'&0}+(P^tw)_1(-1)^{q+1}\cdot (-1)^{q-1+1}(v^tP)_1\alpha_{q-1}\ldots \alpha_1,$$
so the result follows by induction on $q$, after having checked it for $q=1$.
A: Apply the Laplace expansion to the added row, and in each of the determinants occurring in the expansion apply it to the added column. (You can also do it the other way around; the order doesn't matter.) Then each of the terms you get will be one element of $v$ times one element of $w$ times the determinant of the corresponding minor, which is the corresponding element of the classical adjoint. The sign factors for the column index in the row expansion and for the row index in the column expansion are the sign factors in the classical adjoint, and since the index of the added column gets shifted before the second expansion is applied (because a row before it gets removed), the sign factors for the row index in the row expansion and for the column index in the column expansion are opposite, leaving a minus sign in all.
