Change of basis in a polynomial space

Transition matrix. In basis $\mathcal{B}$:
$$1=(1,0,0), \quad X-1=(-1,1,0), \quad (X-1)^2=(1,-2,1).$$
So
$$P_{\mathcal{B}\to\mathcal{B}'} = \begin{pmatrix} 1 & -1 & 1 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{pmatrix}.$$

Inverse. Since $P$ is triangular with diagonal $1$, its inverse is
$$P^{-1} = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{pmatrix}.$$

Coordinates in $\mathcal{B}'$. $[P]_{\mathcal{B}}=(1,-3,2)$. Then
$$[P]_{\mathcal{B}'} = P^{-1}\begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix} = \begin{pmatrix} 1-3+2 \\ -3+4 \\ 2 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix}.$$

Verification. $P=0\cdot 1+1\cdot(X-1)+2\cdot(X-1)^2 = X-1+2(X^2-2X+1) = 2X^2-3X+1$. Correct.