Let $T=\begin{pmatrix} 3x+3y+3z \\ 4x+7y+4z\\ 6x+8y+6z \end{pmatrix}$ And basis $B=\left\{\begin{pmatrix} 3 \\ 1\\ 2 \end{pmatrix},\begin{pmatrix} 4 \\ 2\\ 3 \end{pmatrix},\begin{pmatrix} 2 \\ 1\\ 2 \end{pmatrix}\right\}$ Find $[T]_B$
Now to find it directly I applied T on the elements of $B$ and wrote the coordinates vectors of the images with respect to basis $B$ and got
$$\begin{pmatrix} -3 & -5 & -3 \\ 10 & 15 & 8\\ -4 & -6 & -3 \end{pmatrix}$$
But how can I get the answer using basis change matrices?
I first need to find $[T]_B^{E}$ the transformation form the standard basis to $B$ which is
$$\begin{pmatrix} -5 & -1 & -1 \\ 2 & 2 & 1\\ 5 & -1 & 0 \end{pmatrix}$$
And multiply it by the matrix of basis change $[I]_B^E$ and $[I]_E^B$? Meaning $[I]_E^B\cdot[T]^E_B\cdot [I]_B^E$?
Note that
$$[T]_E= \begin{pmatrix} 3 & 3 & 3 \\ 4 & 7 & 4\\ 6 & 8& 6 \end{pmatrix}$$
and
$$[M]_{EB}= \begin{pmatrix} 3 & 4 & 2 \\ 1 & 2 & 1\\ 2 & 3& 2 \end{pmatrix}$$
$$v_E=[M]_{EB} \cdot v_B\implies v_B=[M]_{EB}^{-1}\cdot v_E \implies v_B= [M]_{BE}\cdot v_E$$
thus for $w=T(v)$ we have
$$w_E=[T]_{E}\cdot v_E\implies [M]_{EB}\cdot w_B=[T]_{E}\cdot [M]_{EB}\cdot v_B \implies w_B=[M]_{BE}\cdot [T]_{E}\cdot[M]_{EB}\cdot v_B $$