$e^i.e_j=\delta_{ij}$ not true in a two dimensional coordinate system with dual. Suppose vector space $\mathbb{R}^2$ with standard basis $E=\{e_1,e_2\}$, vector space $V$ with basis $E^\prime=\{e^\prime_1,e^\prime_2\}$, $V^*$ (dual of vector space $V$) with basis $E^*=\{e^1,e^2\}$  and a linear transformation $A^\prime_X:\mathbb{R}^2\rightarrow V$.
suppose we have vector $u$ with components $(x,y)$, $(x^\prime,y^\prime)$ and $(x^1,y^1)$ for $\mathbb{R}^2$, $V$ and $V^*$ respectively, and :
$$A^\prime_X:\left[\begin{array}{} x^\prime \\ y^\prime\end{array} \right]=
\frac{1}{\sin\theta}\left[ \begin{array}{} \sin\gamma & -\cos\gamma \\ -\sin\beta & \cos\beta \\ \end{array} \right]\left[\begin{array}{} x\\ y\end{array} \right]$$
In this figure, the red lines and blue lines represent coordinate system for $V$ and $V^*$ respectively.
I found :
$$\left[\begin{array}{} x^\prime \\ y^\prime\end{array} \right]=
\frac{1}{\sin\theta}\left[ \begin{array}{} \sin\gamma & -\cos\gamma \\ -\sin\beta & \cos\beta \\ \end{array} \right]\left[\begin{array}{} x\\ y\end{array} \right]$$
$$\left[\begin{array}{} x^1 \\ y^1 \end{array} \right]=
\frac{1}{\sin\theta}\left[ \begin{array}{} \cos\beta & \sin\beta\\ -\cos\gamma &-\sin\gamma\end{array} \right]\left[\begin{array}{} x\\ y\end{array} \right]$$
$$\left[\begin{array}{} x^1 \\ y^1 \end{array} \right]=
\frac{1}{\sin\theta}\left[ \begin{array}{} 1 & \cos\theta\\ -\cos\theta &-1\end{array} \right]\left[\begin{array}{} x^\prime\\ y^\prime\end{array} \right]$$
$$\left[\begin{array}{} e^\prime_1 \\ e^\prime_2\end{array} \right]=
\frac{1}{\sin\theta}\left[ \begin{array}{} \sin\gamma & -\sin\beta\\ -\cos\gamma & \cos\beta \\ \end{array} \right]\left[\begin{array}{} e_1\\ e_2\end{array} \right]$$
$$\left[\begin{array}{} e^1 \\ e^2\end{array} \right]=
\frac{1}{\sin\theta}\left[ \begin{array}{} \cos\beta & -\cos\gamma \\ \sin\beta & -\sin\gamma \\ \end{array} \right]\left[\begin{array}{} e_1\\ e_2\end{array} \right] .$$
My problem: I know $e^i(e^\prime_j)=\delta_{ij}$ but it is not true in this example, I do not know why. where and what are my mistakes?
Figure source


I get
$$
\left[\begin{matrix} e^\prime_1 \\ e^\prime_2\end{matrix} \right]=
\left[ \begin{matrix} 
\cos\beta & \sin\beta \\ 
\cos\gamma & \sin\gamma
\end{matrix} \right]
\left[\begin{matrix} e_1\\ e_2\end{matrix} \right]
$$
At least that makes
$$
\left[\begin{matrix} e^\prime_1 & e^\prime_2\end{matrix} \right]
\left[\begin{matrix} x' \\ y' \end{matrix} \right]
=
\left[\begin{matrix} e_1 & e_2\end{matrix} \right]
\left[ \begin{matrix} 
\cos\beta & \cos\gamma \\ 
\sin\beta & \sin\gamma
\end{matrix} \right]
\frac{1}{\sin\theta}
\left[ \begin{matrix} 
\sin\gamma & -\cos\gamma \\
-\sin\beta & \cos\beta
\end{matrix} \right]
\left[\begin{matrix} x \\ y \end{matrix} \right]
\\
=
\frac{1}{\sin\theta}
\left[\begin{matrix} e_1 & e_2\end{matrix} \right]
\left[ \begin{matrix} 
\cos\beta\sin\gamma - \cos\gamma \sin\beta & -\cos\beta \cos\gamma + \cos\beta \cos\gamma \\ 
\sin\beta \sin\gamma - \sin\beta \sin\gamma & -\sin\beta \cos\gamma + \sin\gamma \cos\beta
\end{matrix} \right]
\left[\begin{matrix} x \\ y \end{matrix} \right]
\\
=
\frac{1}{\sin\theta}
\left[\begin{matrix} e_1 & e_2\end{matrix} \right]
\left[ \begin{matrix} 
\sin(\gamma-\beta) & 0 \\ 
0 & \sin(\gamma-\beta)
\end{matrix} \right]
\left[\begin{matrix} x \\ y \end{matrix} \right]
\\
=
\left[\begin{matrix} e_1 & e_2\end{matrix} \right]
\left[\begin{matrix} x \\ y \end{matrix} \right]
$$
since $\gamma-\beta = \theta$.

How did I find the equation at the top of my post?
We wanted $\left[\begin{matrix} e^\prime_1 & e^\prime_2\end{matrix} \right]
\left[\begin{matrix} x' \\ y' \end{matrix} \right]
= \left[\begin{matrix} e_1 & e_2\end{matrix} \right]
\left[\begin{matrix} x \\ y \end{matrix} \right]$. 
Inserting $\left[\begin{array}{} x^\prime \\ y^\prime\end{array} \right]=
\frac{1}{\sin\theta}\left[ \begin{array}{} \sin\gamma & -\cos\gamma \\ -\sin\beta & \cos\beta \\ \end{array} \right]\left[\begin{array}{} x\\ y\end{array} \right]$ gives
$$\left[\begin{matrix} e^\prime_1 & e^\prime_2\end{matrix} \right]
\left[\begin{matrix} x' \\ y' \end{matrix} \right]
=
\left[\begin{matrix} e^\prime_1 & e^\prime_2\end{matrix} \right]
\frac{1}{\sin\theta}
\left[ \begin{matrix} 
\sin\gamma & -\cos\gamma \\
-\sin\beta & \cos\beta
\end{matrix} \right]
\left[\begin{matrix} x \\ y \end{matrix} \right]
$$
so we must have
$$\left[\begin{matrix} e_1 & e_2\end{matrix} \right] = 
\left[\begin{matrix} e^\prime_1 & e^\prime_2\end{matrix} \right]
\frac{1}{\sin\theta}
\left[ \begin{matrix} 
\sin\gamma & -\cos\gamma \\
-\sin\beta & \cos\beta
\end{matrix} \right]
$$
Solving for $\left[\begin{matrix} e^\prime_1 & e^\prime_2\end{matrix} \right]$ gives
$$
\left[\begin{matrix} e^\prime_1 & e^\prime_2\end{matrix} \right]
=
\left[\begin{matrix} e_1 & e_2\end{matrix} \right]
\frac{\sin\theta}{\sin\gamma \cos\beta - \sin\beta \cos\gamma}
\left[\begin{matrix} \cos\beta & \cos\gamma \\ \sin\beta & \sin\gamma \end{matrix}\right]
=
\left[\begin{matrix} e_1 & e_2\end{matrix} \right]
\left[\begin{matrix} \cos\beta & \cos\gamma \\ \sin\beta & \sin\gamma \end{matrix}\right]
$$
Finally, taking the transpose results in
$$
\left[\begin{matrix} e^\prime_1 \\ e^\prime_2\end{matrix} \right]
=
\left[\begin{matrix} \cos\beta & \sin\beta \\ \cos\gamma & \sin\gamma \end{matrix}\right]
\left[\begin{matrix} e_1 \\ e_2\end{matrix} \right]
$$
One can also see from the picture that we must have
$$
\left[\begin{matrix} e^\prime_1 \\ e^\prime_2\end{matrix} \right]
=
\left[\begin{matrix} a \cos\beta & a \sin\beta \\ b \cos\gamma & b \sin\gamma \end{matrix}\right]
\left[\begin{matrix} e_1 \\ e_2\end{matrix} \right]
$$
for some constants $a$, $b$.
