Finding the matrix of a linear transformation with respect to given (non-standard) ordered bases I am working on the following question from a study guide for my exam on Friday. 

Let $L: ℝ^3 → ℝ^3$ be a linear transformation, represented in standard basis by: $$ L(e_1) = \begin{pmatrix}-3 \\ -1 \\ -2  \end{pmatrix}, L(e_2) = \begin{pmatrix}1 \\ 4 \\ 2  \end{pmatrix}, L(e_3) = \begin{pmatrix}-5 \\ -6 \\ -5  \end{pmatrix}$$ 
  Represent the same transformation from $B$ to $B'$ basis where:
  $$B = \begin{bmatrix}\begin{pmatrix}0 \\ -1 \\ 2  \end{pmatrix} \begin{pmatrix} 1 \\ -2 \\ 3  \end{pmatrix} \begin{pmatrix}3 \\ -4 \\ 6  \end{pmatrix}\end{bmatrix}$$
  $$ B' = \begin{bmatrix}\begin{pmatrix}1 \\ 1 \\ 0  \end{pmatrix} \begin{pmatrix} -2 \\ -1 \\ 2  \end{pmatrix} \begin{pmatrix}8 \\ 3 \\ -9  \end{pmatrix}\end{bmatrix}$$

The first thing I thought was: $$L(e_1) = L\begin{pmatrix}1 \\ 0 \\ 0  \end{pmatrix}, L(e_2) = L\begin{pmatrix}0 \\ 1 \\ 0  \end{pmatrix}, L(e_3) = L\begin{pmatrix}0 \\ 0 \\ 1  \end{pmatrix}$$So then I was thinking, that means that I can just apply it to $B$, as $L(B)$, for each vector in $B$. But then I got lost. I am now thinking that this is not what I should do, because apparently there needs to be a linear transformation from $B → B'$. But I don't know what to do now. How should I go about solving this?

You should apply $L$ to each column of $B$, then write them in linear combination of $B'$. The coefficients are the columns of the transformation. For example, 
$$LB_1=\begin{pmatrix}-3&1&-5\\-1&4&-6\\-2&2&-5\end{pmatrix}\begin{pmatrix}0\\-1\\2\end{pmatrix}=\begin{pmatrix}-11\\-16\\-12\end{pmatrix}=-25\begin{pmatrix}1\\1\\0\end{pmatrix}-15\begin{pmatrix}-2\\-1\\2\end{pmatrix}-2\begin{pmatrix}8\\3\\-9\end{pmatrix}$$
So $\begin{pmatrix}-25\\-15\\-2\end{pmatrix}$ is the first column of your transformation. I believe you can continue from here.
