$P$ is a change of basis matrix from basis $B$ to basis $C$. Given $P$ and $C$ find $B$ 
$$
P=\begin{pmatrix}1&1&0\\0&3&2\\0&1&1\end{pmatrix}
$$ is a change of basis matrix from  basis $B$ to basis $C$.
  $$
C=\begin{pmatrix}1&1&1\\1&1&0\\1&0&0\end{pmatrix}
$$
  Find $B$.

The way I understand this is that $C=PBP^{-1}$, then $B=P^{-1}CP$. So I need to find $P^{-1}$ which is:
$$
P^{-1}=\begin{pmatrix}1&-1&2\\0&1&-2\\0&-1&3\end{pmatrix}
$$
However when I compute $B=P^{-1}CP$ my solution the result is wrong when compared with the correct solution.
The correct solution also finds the same matrix $P^{-1}$ but after that it is as follows:
let $B=(b_1,b_2,b_3), C=(c_1,c_2,c_3)$ then:
$$
b_1=c_1=\begin{pmatrix}1\\1\\1\end{pmatrix}\\
b_2=-c_1+c_2-v_3=\begin{pmatrix}-1\\0\\-1\end{pmatrix}\\
b_3=2c_1-2c_2+3c_3=\begin{pmatrix}3\\0\\2\end{pmatrix}
$$
What's wrong with my solution?
 A: To put it shortly, $C=BP$ because of how the change of basis matrix is defined as well as how matrices $B$ and $C$ represent their respective bases.
I'll include the long version below, however.

For bases $B=\{\vec b_1,...,\vec b_n\}$ and $C=\{\vec c_1,...,\vec c_n\}$, the change of basis matrix, $P$, is defined such that
$$(\forall j)\,\,\vec c_j=\sum_{i=1}^nP_{ij}\vec b_i$$
Let $[\vec v]_A$ represent the components of some vector $\vec v$ in some basis $A=\{\vec a_1,...,\vec a_n\}$, and ${([\vec v]_A)}_i$ represent the $i$th component. Then
$$\vec v=\sum_{i=1}^n{([\vec v]_A)}_i\vec a_i$$
$\implies$
$$(\forall j)\,\,\sum_{k=1}^n{([\vec c_j]_E)}_k\vec e_k=\sum_{i=1}^nP_{ij}\sum_{k=1}^n{([\vec b_i]_E)}_k\vec e_k$$
Note that (what I assume to be) your book lets matrices $B=([\vec b_1]_E\,|\,...\,|\,[\vec b_n]_E)$ and $C=([\vec c_1]_E|...|[\vec c_n]_E)$ (where '$|$' concatenates vectors and $E=\{\vec e_1,...,\vec e_n\}$ is the standard basis).
Using that, we get
$$(\forall j)\,\,\sum_{k=1}^nC_{kj}\vec e_k=\sum_{i=1}^nP_{ij}\sum_{k=1}^nB_{ki}\vec e_k$$
$$=\sum_{k=1}^n\left(\sum_{i=1}^nB_{ki}P_{ij}\right)\vec e_k$$
$\implies$ (using ${(AB)}_{ij}=\sum_kA_{ik}B_{kj}$ from matrix multiplication)
$$(\forall j)\,\,\sum_{k=1}^nC_{kj}\vec e_k=\sum_{k=1}^n{(BP)}_{kj}\vec e_k$$
As $E$ is a basis, and so the vectors $\vec e_1,...,\vec e_n$ are linearly independent, we can conclude
$$(\forall j,k)\,\,C_{kj}={(BP)}_{kj}$$
$\implies$
$$C=BP$$
