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 $x_1^2 + x_2^2 = 1 +\alpha^2$?