In $P_2$, find the change-of-coordinates matrix In $P_2$, find the change-of-coordinates matrix from the basis $B=\{1-2t+t^2, 3-5t+4t^2, 2t+3t^2\}$ to the standard basis $C=\{1, t, t^2\}$. Then find the B-coordinate vector for $-1+2t$
I know how to do the first part. $P$ from $B$ to $C: \begin{bmatrix}1& 3& 0\\ -2 &-5& 2\\ 1& 4& 3\end{bmatrix}$. I do not know what the process is for finding the B coordinate vector though. Can someone give me a place to start for doing that?
 A: So, you have $P_{C \leftarrow B} = \begin{bmatrix}1&3&0\\-2&-5&2\\1&4&3\\ \end{bmatrix}$. So to get the second part, you need to row reduce the augmented matrix $ \begin{bmatrix} P_{C \leftarrow B} & [x]_{C} \\ \end{bmatrix}$. 
So, what is $[x]_{C}$? Take $-1+2t$ and plug it in:
$$[x]_{C} = \begin{bmatrix} ?\\ ?\\ ?\\ \end{bmatrix} = \begin{bmatrix} -1 \\ 2 \\ 0\\ \end{bmatrix}$$
If you don't see where the $0$ came from, think about the coefficient for $t^{2}$.
So now, we have
$$\begin{bmatrix} P_{C \leftarrow B}& [x]_{C}\\ \end{bmatrix} = \begin{bmatrix} 1&3&0&-1\\-2&-5&2&2\\1&4&3&0\\ \end{bmatrix}$$
Now row reduce, and you'll have your answer.
A: The basis vectors of B expressed with respect to the standard basis are:
$$v_1=(1,-2,1), v_2=(3,-5,4), v_3=(0,2,3)$$
Thus, the following matrix M, which colums are the component of the basis vectors B with respect to the standard basis:
$$M=\begin{bmatrix}1& 3& 0\\ -2 &-5& 2\\ 1& 4& 3\end{bmatrix}$$
represent the change of coordinates from the basis B to the standard basis C, that is:
$$v_C=Mv_B\implies v_B=M^{-1}v_C$$
where
$$M^{-1}=\begin{bmatrix}-23& -9& 6\\ 8 &3& -2\\ -3& -1& 1\end{bmatrix}$$
represent the change of coordinates from the standard basis C to the basis B. 
Thus for the vector $-1+2t=v_C=(-1,2,0)$ in the standard basis C, we obtain:
$$v_B=v_CM^{-1}=\begin{bmatrix}-23& -9& 6\\ 8 &3& -2\\ -3& -1& 1\end{bmatrix}\begin{bmatrix}-1\\ 2 \\ 0\end{bmatrix}=\begin{bmatrix}5\\ -2 \\ 1\end{bmatrix}$$
You can check indeed that:
$$v_C=5v_1-2v_2+v_3$$
