Matrix of Killing form a Lie algebra Let $L$ be the Lie algebra with basis $B = \{u,v,w\}$, with $[u,v] = w, [v,w] = u, [w,u] = v$.
Question : Find the matrix of the Killing form $\kappa$ of $L$ with respect to $B$.
I have come across this question in a past exam paper for my Lie algebras course. It is not worth many marks so I am guessing this is an easy question, however I can't seem to convince myself of an answer. 
My guess is that the matrix of $\kappa$ is given by
\begin{pmatrix} 0 & 1 & 0 \\
0 & 0 & 1 \\
1 & 0 & 0\end{pmatrix}
Can anyone confirm this is correct, or tell me the correct way to do it?
Edit: After thinking about this again, I am pretty sure my guess is incorrect. Any guidance is appreciated.

Following the definition/notation here, we need to compute the matrices for
$\DeclareMathOperator{\ad}{ad}$
$\DeclareMathOperator{\tr}{trace}$ 
$\ad(u),\ad(v),\ad(w)$ with respect to $B$.  In particular, we have
$$
\ad(u) = 
\pmatrix{
0&0&0\\
0&0&-1\\
0&1&0}\\
\ad(v) = 
\pmatrix{
0&0&1\\
0&0&0\\
-1&0&0}\\
\ad(w) = 
\pmatrix{
0&-1&0\\
1&0&0\\
0&0&0}
$$
The entries of the Killing form can then be found as $\tr(\ad(e_i)\ad(e_j))$.  In particular, we have
$$
L = \pmatrix{
-2&0&0\\
0&-2&0\\
0&0&-2
}
$$
for this computation, it helps to note that $\tr(A^TB)$ is the "dot-product" of the matrices $A$ and $B$.  Note that the Killing form should always be symmetric.
