Let $B = \{(1,2),(2,3)\}$ and $C = \{(-1,1),(2,1)\}$ be bases for $\Bbb{R}^2$. If $[v]_{C} = (1,2)$ what is $[v]_{B}$?
The problem doesn't specify what $v$ is, so I'm assuming it is some abstract vector. To find $[v]_B$, do I just compute $[I_{\Bbb{R}^2}]_{C}^{B}[v]_{C}$, where $I_{\Bbb{R}^2}(x) = I_2x = x$? According to my calculations
$$[I_{\Bbb{R}^2}]_{C}^{B} = \begin{pmatrix} 5 & -4 \\ -3 & 3 \\ \end{pmatrix}, $$ so $[v]_B = (-3,3)$. Does this seem right?
On
Well, $v$ is explicitly determined, since $[v]_C=(1, 2)$ means that $$v={\bf 1}\cdot(-1,1)+{\bf2}\cdot(2,1)$$ which is $(3,3)$ in the standard coordinates.
Now, $v=(3, 3)$ and we want to coordinate it w.r.t. $B$.
Your solution is thus correct, as we indeed have
$$v=(3,3)=(-3)\cdot(1,2)+3\cdot(2,3)$$
$v$ is $1\cdot (-1,1)+2\cdot (2,1)=(3,3)$. Then $[v]_B$ is $(a,b)$ where $(3,3)=a\cdot (1,2)+b\cdot (2,3)$. So, $\begin{cases} a+2b=3\\2a+3b=3\end{cases}$. Solving, $b=3$ and $a=-3$. Thus you are indeed correct.
Alternatively, you could take $[I]_B^S=\begin{pmatrix}-1&2\\1&1\end{pmatrix}$ and $[I]_C^S=\begin{pmatrix}1&2\\2&3\end{pmatrix}$, which are the two change of basis matrices, and work with them. Your change of basis matrix should be the product of the first and the inverse of the second. Indeed it is: $$\begin{pmatrix}1&2\\2&3\end{pmatrix}^{-1}\cdot \begin{pmatrix}-1&2\\1&1\end{pmatrix}=\begin{pmatrix}5&-4\\-3&3\end{pmatrix}$$.