Change of basis for dual bases The formulation of the problem is the following:

Let $u_1=(1,1,1)$, $u_2=(2,0,0)$, and $u_3=(0,3,0)$ be a basis of $\mathbb{R}^3$ and let $\{e_1, e_2, e_3\}$ be the canonical basis of $\mathbb{R}^3$. Calculate the coordinates of $w=2e^{*2}+3e^{*3}$ in dual basis $\{u^{*1}, u^{*2}, u^{*3}\}$.

So I knew that if you let $P$ be the change of basis matrix from $\{u_i\}$ to $\{e_i\}$, then, the change of basis matrix from $\{u^{*i}\}$ to $\{e^{*i}\}$, would be $(P^{-1})^{\top}$. If the problem asks me the opposite way, from $\{e^{*i}\}$ to $\{u^{*i}\}$, the matrix would be $((P^{-1})^{\top})^{-1}=P^{\top}$. Thus, the solution should be
$$(0,2,3)\begin{pmatrix}
1 & 2 & 0\\
1 & 0 & 3\\
1 & 0 & 0
\end{pmatrix}^{\top}=(4,9,0)$$.
The issue is that the solution says it's
$$(0,2,3)\begin{pmatrix}
1 & 2 & 0\\
1 & 0 & 3\\
1 & 0 & 0
\end{pmatrix}=(5,0,6)$$.
Where am I mistaken?
 A: Let us write
\begin{eqnarray*}
u_1&=&e_1+e_2+e_3\\
u_2&=&2e_1\\
u_3&=&3e_2
\end{eqnarray*}
so we can associate the matrix
$$B=\left(
\begin{array}{ccc}
1&2&0\\
1&0&3\\
1&0&0\\
\end{array}
\right).$$
That $B$ has determinant different from zero is due that those $u_i$ are linearly independent and they form a new basis.
Then $B^{-1}$ exists, and in fact obeys $B^{-1}B=1\!\!1$, that is
$$
\left(
\begin{array}{ccc}
0&0&1\\
\frac{1}{2}&0&-\frac{1}{2}\\
0&\frac{1}{3}&-\frac{1}{3}\\
\end{array}
\right)
\left(
\begin{array}{ccc}
1&2&0\\
1&0&3\\
1&0&0\\
\end{array}
\right)=
\left(
\begin{array}{ccc}
1&0&0\\
0&1&0\\
0&0&1\\
\end{array}
\right).$$
Here, you can see how the rows of $B^{-1}$ behave as the duals $u^{*s}$ i.e.
$$u^{*i}(u_j)=\delta^i_j.$$
as well as the canonical basis $e^{*i}(e_j)=\delta^i_j$.
Then
\begin{eqnarray*}
u^{*1}&=&e^{*3}\\
u^{*2}&=&\frac{1}{2}e^{*1}-\frac{1}{2}e^{*3}\\
u^{*3}&=&\frac{1}{3}e^{*2}-\frac{1}{3}e^{*3}
\end{eqnarray*}
that immediately gives
\begin{eqnarray*}
e^{*1}&=&u^{*1}+2u^{*2}\\
e^{*2}&=&u^{*1}+3u^{*3}\\
e^{*3}&=&u^{*1}.
\end{eqnarray*}
Finally, subbing into you covector $w=2e^{*2}+3e^{*3}$, you'll get
\begin{eqnarray*}
w&=&2(u^{*1}+3u^{*3})+3u^{*1},\\
&=&5u^{*1}+6u^{*3}.
\end{eqnarray*}
