Let $a=[1,2]^t$ ( $^t$ means transpose) and let $A=\begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}$. One can verify that $a^t a = 5$, $a^t A a = 14$, and $a^t a \cdot \det(A)= 15$. So obviously, $a^t A a \neq a^t a \cdot \det(A)$.
However, I wish to balance out the insert of $A$. Namely, I wish to figure out if it is possible to get the value of $a^t A a$ based on $ \left(\text{something} \times a^t a \right)$. The elements in $a$ are all positive and $A$ is symmetric and positive definite.
Is there a way to do that? Thanks.
You have $a^t a = \|a\|^2$, where $\|a\|$ is the vector norm of $a$. Now define $b = A a$ and consider
$\|a +b \|^2 - \|a-b\|^2 = \|a\|^2 + \|b\|^2 + 2 a^t b - (\|a\|^2 + \|b\|^2 - 2 a^t b)$.
Thus
$a^t b = a^t A a = \frac{1}{4}( \|a +b \|^2 - \|a-b\|^2) = \frac{1}{2} (\|a+b\|^2 - \|a\|^2 - \|b\|^2)$.
Not of the form that you asked for (I don't think that's possible), but maybe this helps.