= Solution
Let $P_B,P_{B'},P_C,P_{C'}$ have the corresponding basis vectors as columns in standard coordinates. From the PDF,
$$
P_B=\begin{pmatrix}0&-2\\2&0\end{pmatrix},
\quad
P_{B'}=\begin{pmatrix}1&1\\1&-1\end{pmatrix},
\quad
P_C=\begin{pmatrix}1&0&0\\1&1&0\\0&0&1\end{pmatrix},
\quad
P_{C'}=I_3.
$$
The <matrix representation of a linear map> in standard coordinates is $P_CAP_B^{-1}$. The <change of basis> formula therefore gives
$$
A'=P_{C'}^{-1}P_CAP_B^{-1}P_{B'}.
$$
Multiplication yields
$$
\boxed{
A'=
\begin{pmatrix}
3/2&-1/2\\
5/2&-3/2\\
-1&0
\end{pmatrix}}.
$$
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