Prove GDA decision boundary is linear My attempt: 
(a) I solved that $a=\ln{\frac{P(X|C_0)P(C_0)}{P(X|C_1)P(C_1)}}$  
(b) Here is where I'm running into trouble. I'm plugging the distributions into $\ln{\frac{P(X|C_0)P(C_0)}{P(X|C_1)P(C_1)}}$ and I get $a=\ln{\frac{P(C_0)}{P(C_1)}}+\frac{1}{2}(x-\mu_1)^T\Sigma^{-1}(x-\mu_1)-\frac{1}{2}(x-\mu_0)^T\Sigma^{-1}(x-\mu_0)$. 
I can see that $b=\ln{\frac{P(C_0)}{P(C_1)}}$ and $w^Tx=\frac{1}{2}(x-\mu_1)^T\Sigma^{-1}(x-\mu_1)-\frac{1}{2}(x-\mu_0)^T\Sigma^{-1}(x-\mu_0)$. 
I'm not sure how to simplify $w^Tx$ so that I can solve for $w$. Or is there something that I did wrong?
If you expand the terms, you can see that the quadratic terms cancel out.

\begin{align}
a &amp;= \ln \frac{P(C_0)}{P(C_1)} + \frac12(x - \mu_1)^T\Sigma^{-1}(x - \mu_1) - \frac12(x-\mu_0)^T\Sigma^{-1}(x-\mu_0)\\
&amp;=\ln \frac{P(C_0)}{P(C_1)} + \frac12\left[x^T\Sigma^{-1}x-2x^T\Sigma^{-1}\mu_1+\mu_1^T\Sigma^{-1}\mu_1\right]\\&amp; - \frac12\left[x^T\Sigma^{-1}x-2x^T\Sigma^{-1}\mu_0+\mu_0^T\Sigma^{-1}\mu_0\right]\\
&amp;= (\mu_0-\mu_1)^T\Sigma^{-1}x+\ln \frac{P(C_0)}{P(C_1)} +\frac12\left[\mu_1^T\Sigma^{-1}\mu_1-\mu_0^T\Sigma^{-1}\mu_0\right] \\
\end{align}