How to obtain a certain transition matrix from a non-standard basis to another non-standard basis? The transition matrix, P, from the basis B=$\{(0, 1, 2)^t ,(1, 2, 0)^t ,(2, 0, 1)^t\}$ to the basis B'=$\{(9, 0, 0)^t ,(0, 9, 0)^t ,(0, 0, 9)^t \}$ of $\mathbb R^3$ ?
The solution I got was 
 $$P=\begin{pmatrix} 0 & 1/9 & 2/9 \\ 1/9 & 2/9 & 0 \\ 2/9 & 0 & 1/9\end{pmatrix}$$
I got the above solution using the general augmented matrices method
[B'|B] -> [I | P] (via Gaussian elimination) where P is the transition matrix.
However this is none of the answers in the problem sheet I am doing.
The matrices in the options are
$\begin{pmatrix} 0 & 3 & 3 \\ 3 & 3 & 0 \\ 3 & 0 & 3\end{pmatrix}$
$\begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 0\end{pmatrix}$
$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}$
$\begin{pmatrix} -2 & 1 & 4 \\ 1 & 4 & -2 \\ 4 & -2 & 1\end{pmatrix}$
 A: When a vector $v$ is written in the first basis like this 
$$v=a_1e_1+a_2e_2+a_3e_3, \ \ \text{which can be written} \ \ v=\underbrace{[e_1|e_2|e_3]}_E\begin{pmatrix}a_1\\b_1\\c_1\end{pmatrix}$$
and in the second basis like this
$$v=a'_1e'_1+a'_2e'_2+a'_3e'_3, \ \ \text{which can be written} \ \ v=\underbrace{[e'_1|e'_2|e'_3]}_{E'}\begin{pmatrix}a'_1\\b'_1\\c'_1\end{pmatrix}$$
Equating the two expressions for $v$, we get:
$$E\begin{pmatrix}a_1\\b_1\\c_1\end{pmatrix}=E'\begin{pmatrix}a'_1\\b'_1\\c'_1\end{pmatrix}$$
which is equivalent to :
$$\begin{pmatrix}a'_1\\b'_1\\c'_1\end{pmatrix}=\underbrace{(E')^{-1}E}_P\begin{pmatrix}a_1\\b_1\\c_1\end{pmatrix}$$
And in this way, I find back your transition matrix !
A: Let me explain this way.
You have, for the basis $B$
$$b_1=e_2+2e_3,$$
$$b_2=e_1+2e_2,$$
$$b_3=2e_1+e_3,$$
and for the basis $B'$
$$c_1=9e_1,$$
$$c_2=9e_2,$$
$$c_3=9e_3.$$
So solving for $e_i$ you have
$$e_1=1/9c_1,$$
$$e_2=1/9c_2,$$
$$e_3=1/9c_3,$$
which subbed into the $b_i$, you get
$$b_1=1/9c_2+2/9c_3,$$
$$b_2=1/9c_1+2/9c_2,$$
$$b_3=2/9c_1+1/9c_3,$$
with matrix specification $P$ as you calculate
but
$$P^{-1}=
\left[\begin{array}{ccc}
-2&1&4\\1&4&-2\\4&-2&1
\end{array}\right],$$
which means that
$$c_1=-2b_1+b_2+4b_3,$$
$$c_2=b_1+4b_2-2b_3,$$
$$c_3=4b_1-2b_2+b_3.$$
