Let be $B=(\vec e_1,\vec e_2,\vec e_3)$, $B'=(\vec e_1+2\vec e_2+\vec e_3, \vec e_1 +2\vec e_2+2\vec e_3,\vec e_2+2\vec e_3)$,$\phi $ an endomorphism of $\mathbb R^3$ such that $M_B(\phi)$= \begin{pmatrix} 5 & -4 & 2\\ 14 & -10 & 4\\ 16 & -10 & 3 \end{pmatrix}
Find $M_B'(\phi)$.
I found the inverted matrix $Q^{-1}$:
$$M_B(\phi)= \begin{pmatrix} -2 & 2 & -1\\ 3 & -2 & 1\\ -2 & 1 & 0 \end{pmatrix}$$
Yet I wasn't able to find the right
\begin{align} M_B'(\phi)&= \begin{pmatrix} -2 & 2 & -1\\ 3 & -2 & 1\\ -2 & 1 & 0 \end{pmatrix} \begin{pmatrix} 5 & -4 & 2\\ 14 & -10 & 4\\ 16 & -10 & 3 \end{pmatrix} \begin{pmatrix} 1 & 1 & 0\\ 2 & 2 & 1\\ 1 & 2 & 2 \end{pmatrix}\\ M_B'(\phi)&= \begin{pmatrix} -2 & 2 & -1\\ 3 & -2 & 1\\ -2 & 1 & 0 \end{pmatrix} \begin{pmatrix} -1 & 1 & 0\\ 8 & 14 & -2\\ 7 & 10 & -4 \end{pmatrix}\\ M_B'(\phi)&= \begin{pmatrix} -9 & & \\ 12 & &\\ & & \end{pmatrix} \end{align}
I stopped here because the given answer is: $$M_B'(\phi)=\begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0\\ 0 & 0 & -2 \end{pmatrix}$$
What did I missed? I know I'm not very good at matrix multiplication. do you have advises? You can tell me off, I'm okay with that, if it can help me never forget rules of matrices multiplication.
Furthermore I 've been told that one could have did it this way: we could have calculated $\phi (\vec I)=-\vec I, \phi (\vec J)=\vec J$ and $\phi (\vec K)=-2K$
I don't understand what happened here, I tried:
\begin{align} M_B'(\phi)&= \begin{pmatrix} 5 & -4 & 2\\ 14 & -10 & 4\\ 16 & -10 & 3 \end{pmatrix}\begin{pmatrix} 1 & 0 & 0\\ 2 & 0 & 0\\ 1 & 0 & 0 \end{pmatrix}\\ &=\begin{pmatrix} -3 \\-2\\ -1\end{pmatrix} \end{align}
Which is not what we should find...
In general, if you have a linear map $\phi:\mathbb{R}^n \to \mathbb{R}^n$ and a base $B = (b_1,...,b_n)$, it's always helpful to remind yourself of what an entry in $(a_{ij})_{ij} := A := M_{B}(\phi)$ stands for. $A$ satisfies
$$\phi(b_j) = \sum_{i=1}^{n}{a_{ij} b_i}$$
So what does this tell you? It tells you that the $j$-th column of $A$ is the image of $b_j$ expressed in coordinates of $b_1, \dots, b_n$.
Now in order to find $M_{B'}(\phi)$ you will need to see what $\phi(b_i')$ is expressed in the basis $B'$.
Let's do this exemplary for the first basis vector of $B'$:
\begin{equation*} \phi(b_1') = \phi(e_1 + 2e_2 + e_3) = \phi(e_1) + 2\phi(e_2) + \phi(e_3) \underset{\phi(e_1) = 5e_1+14e_2+16e_3}{=} \\ = 5e_1+14e_2+16e_3 + 2\phi(e_2) + \phi(e_3) \underset{\phi(e_2) = -4e_1-10e_2-10e_3}{=} \\ = 5e_1+14e_2+16e_3 - 8e_1-20e_2-20e_3 + \phi(e_3) \underset{\phi(e_3) = 2e_1+4e_2+3e_3}{=} \\ = 5e_1+14e_2+16e_3 - 8e_1-20e_2-20e_3 + 2e_1+4e_2+3e_3 = \\ = -e_1-2e_2-e_3 = \\ = -b_1' \end{equation*}
So $\phi(b_1') = -1 \cdot b_1' + 0 \cdot b_2' + 0 \cdot b_3'$, which will give you the first column vector of $M_{B'}(\phi)$.