Assume I have the pdf of a random vector $Y_{3\times 1}$ as $$ \frac{1}{(2\pi)^{3/2} \sqrt{|\Sigma|}} \text{exp}\left\{ -\frac{1}{2} \left( 2y^2_1 + y^2_2 + y_3^2 + 2y_1y_2 - 8y_1 - 4y_2 + 8 \right)\right\} $$ and want to find $\Sigma^{-1}$ and $\mu.$
I know one possible approach is to write out the exponent part into the quadratic form $$q = a_1y_1^2 + a_2 y_2^2 + a_3y_3^2 + a_4y_1y_2 + a_5y_1y_2 +a_6y_2y_3$$ equivalently, $$Y'AY = \begin{bmatrix} y_{1} & y_{2} & y_{3} \end{bmatrix} A \begin{bmatrix} y_{1} \\ y_{2} \\ y_{3} \end{bmatrix}.$$ Where $$A = \begin{bmatrix} a_{1} & \frac{1}{2}a_4 & \frac{1}{2} a_{5} \\ \frac{1}{2}a_4 & a_{2} & \frac{1}{2} a_{6}\\ \frac{1}{2}a_5 & \frac{1}{2} a_{6} & a_{3} \end{bmatrix}$$ So, we let say $A = \Sigma^{-1} = 2y^2_1 + y^2_2 + y_3^2 + 2y_1y_2 - 8y_1 - 4y_2 + 8, $ then we have
$$A = \Sigma^{-1} = \begin{bmatrix} 2 & 1 & 0 \\ 1 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}$$ But $$ Y'AY = 2^1y_1 + y^2_2 + y^3_3 + 2y_1y_2 \neq 2y^2_1 + y^2_2 + y_3^2 + 2y_1y_2 - 8y_1 - 4y_2 + 8$$
So what am I missing in order to find $\Sigma^{-1}$ and $\mu?$
$\textbf{Update - credit due to @Hume2 comments}$
Clearly $$ -2\mu^TA Y = -2 \left[ (2\mu_1 + \mu_2)y_1 + (\mu_1 + \mu_2)y_2 + y_3 \mu_3 \right]$$ So, $$-8y_1 - 4y_2 + 8 = -2 \left[ (2\mu_1 + \mu_2)y_1 + (\mu_1 + \mu_2)y_2 + y_3 \mu_3\right]$$ Equating $y_i$ terms and solving for $\mu_i,$ we get $\begin{bmatrix} \mu_1\\ \mu_2\\ \mu_3\end{bmatrix} = \begin{bmatrix} 2\\ 0 \\ 0 \end{bmatrix}$
In the exxponent, the quadratic function can be shifted by a vector $\mu$, which is the mean of the distribution. For zero mean, it is, as you wrote $Y^TAY$. For mean $\mu$, it is $(Y-\mu)^TA(Y-\mu)$. You can figure out that this equals $Y^TAY-2\mu^TAY+\mu^TA\mu$. To figure out what is the matrix $A$, you do exactly the same what you wrote, you just ignore all linear and constant terms. To figure out the $\mu$, you take the linear terms and put them equal to $-2\mu^TAY$. You get: $-8y_1-4y_1=-2\mu^TAY$. You know $AY$, so this is a linear equation. And you are done.