How does one apply the Lagrange method to vectors?
The problem I have (it's financial) is $$\max_w w^T r - w^T\Sigma ^{-1} w $$ under the condition that $w_1 + w_2 = 1$. $r$ is a known vector with 2 elements, and $\Sigma$ is a 2 by 2 matrix. When I get the two first-order conditions, I don't know how to isolate $\lambda$ since I am unsure if I can just "divide" away similar matrix/vector-expressions? Surely not?
You shouldn't be solving this with a Lagrange multiplier. You only have two variables and one linear condition. Substitute $w_2=1-w_1$ and optimise with respect to $w_1$ without constraints.
If you really want to do it with a Lagrange multiplier, write $w^\top e=1$, where $e$ is the vector with entries $1$; then differentiating
$$ w^\top r-w^\top\Sigma^{-1}w-\lambda w^\top e $$
with respect to $w$ yields
$$ r-2\Sigma^{-1}w-\lambda e=0\;, $$
and solving for $w$ yields
$$ w=\frac12\Sigma\left(r-\lambda e\right)\;. $$
Substituting into the constraint yields
$$e^\top\left(\frac12\Sigma\left(r-\lambda e\right)\right)=1\;,$$
solving for $\lambda$ yields
$$ \lambda=\frac{e^\top\Sigma r-2}{e^\top\Sigma e} $$
and thus
$$ w=\frac12\Sigma\left(r-\frac{e^\top\Sigma r-2}{e^\top\Sigma e}e\right)\;. $$