Let $\phi: \mathbb{R}^3 \to \mathbb{R}^3$ and $E$ the standard basis with
$$M^E_E(\phi)=\begin{pmatrix}1 & 2 & 3\\ 3 & 2 & 1\\ 1 & 1 & 1\end{pmatrix}$$ and $$A:=((1,1,0),(0,1,1),(1,1,1)) \text{ and } B:=((-1,1,1),(1,1,1),(2,2,0))$$ are further bases of $\mathbb{R}^3$. Calculate $M^A_B(\phi)$.
Notation: $\color{red}{I^A_B}$ means we change from $\color{red}A$ to $\color{red}B$. $$\begin{align} M^A_B(\phi)&=T^E_B\cdot M^E_E(\phi)\cdot T^A_E\\ &=\begin{pmatrix}-1/2 & 1/2 & 0\\ -1/2 & -1/2 & 1\\ 1/2 & 1/2 & -1/2 \end{pmatrix}\cdot \begin{pmatrix}1 & 2 & 3\\ 3 & 2 & 1\\ 1 & 1 & 1\end{pmatrix} \cdot \begin{pmatrix}1 & 0 & 1\\ 1 & 1 & 1\\ 0 & 1 & 1\end{pmatrix}\\ &=\begin{pmatrix}1 & -1 & 0\\ 1 & 3 & 3\\ 3/2 & 1/2 & 3/2\end{pmatrix} \end{align}$$
or is it $$\begin{align} &=\begin{pmatrix}1 & 0 & 1\\ 1 & 1 & 1\\ 0 & 1 & 1\end{pmatrix}\cdot \begin{pmatrix}1 & 2 & 3\\ 3 & 2 & 1\\ 1 & 1 & 1\end{pmatrix}\cdot \begin{pmatrix}-1/2 & 1/2 & 0\\ -1/2 & -1/2 & 1\\ 1/2 & 1/2 & -1/2 \end{pmatrix}\\ &=\begin{pmatrix}-1/2 & 3/2 & 1\\ -5/2 &5/2 &5/2\\ -5/2& 3/2 & 2\end{pmatrix} \end{align}$$ ? I'm not really sure, which one is correct.
The former $M^A_B(\phi)=T^E_B\cdot M^E_E(\phi)\cdot T^A_E$ you recive vector $v_A$ representation of $v$ in basis $A$ $T^E_B\cdot M^E_E(\phi)\cdot T^A_E v_A = (T^E_B\cdot M^E_E(\phi)) v_E = T^E_B (\phi ( v) )_E = (\phi( v))_B $ as desired