Finding $\mathbf{E}$ and $\mathbf{e}$ for a Deformation Field

[Dustinsfl]

For the deformation field given by
$$
x_1 = X_1 + \alpha X_2,\quad x_2 = X_2 - \alpha X_1,\quad x_3 = X_3
$$
where ##\alpha## is a constant, determine the matrix form of the tensors ##\mathbf{E}## and ##\mathbf{e}##, and show that the circle of particles ##X_1^2+ X_2^2 = 1## deforms into the circle ##x_1^2 + x_2^2 = 1 + \alpha^2##.

How do I find ##\mathbf{e}## and the solution is
$$
\frac{1}{2(1+\alpha^2)}\begin{bmatrix}
-\alpha^2 & 0 & 0\\
0 & -\alpha^2 & 0\\
0 & 0 & 0
\end{bmatrix}
$$
First, we will find ##\mathbf{E}##.
$$
\mathbf{F} =
\begin{bmatrix}
1 & \alpha & 0\\
-\alpha & 1 & 0\\
0 & 0 & 1
\end{bmatrix}
$$
Then ##\mathbf{C} = \mathbf{F}^T\mathbf{F}##. So we have that
$$
\mathbf{C} =
\begin{bmatrix}
1 + \alpha^2 & 0 & 0\\
0 & 1 + \alpha^2 & 0\\
0 & 0 & 1
\end{bmatrix}
$$
which leads to
$$
\mathbf{E} = \frac{1}{2}
\begin{bmatrix}
\alpha^2 & 0 & 0\\
0 & \alpha^2 & 0\\
0 & 0 & 0
\end{bmatrix}.
$$

 

[Dustinsfl]

I found ##\mathbf{e} = \mathbf{F}^{-T}\mathbf{E}\mathbf{F}^{-1}## but this returns the negative of the book answer for ##\mathbf{e}##.
Assuming the book has a typo, how do I show ##X_1^2+X_2^2 = 1## deforms into ##x_1^2 + x_2^2 = 1 +\alpha^2##?