Electromagnetic Tensor: Calculating $\det{F^{\mu}}_\nu$

[Aleolomorfo]

Homework Statement

Given an electromagnetic tensor $F^{\mu\nu}$, showing that:
$$\det{F^{\mu}}_\nu=-(\vec{B}\cdot\vec{E})^2$$

Homework Equations

The Attempt at a Solution

I had only the (stupid) idea of writing explictly the matrix associated with the electromagnetic tensor and calculating the determinant. But I think this is not the way to do this exercise. Even because in the exercise there is the "post-scriptum": given a tensor $F^{\mu\nu}$, firstly you have to do a Lorentz-transformation in order to simply the calculus. Can someone give me a hint, please?

 

[mark726]

First, let's recall the definition of the electromagnetic tensor:
$$F^{\mu\nu}=\begin{pmatrix}
0 & -E_x & -E_y & -E_z\\
E_x & 0 & -B_z & B_y\\
E_y & B_z & 0 & -B_x\\
E_z & -B_y & B_x & 0
\end{pmatrix}$$
Now, let's consider a Lorentz transformation with velocity $v$ along the x-axis:
$$\Lambda=\begin{pmatrix}
\gamma & -\gamma v & 0 & 0\\
-\gamma v & \gamma & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{pmatrix}$$
where $\gamma=\frac{1}{\sqrt{1-v^2}}$. Applying this transformation to the electromagnetic tensor, we get:
$$F'^{\mu\nu}=\Lambda^\mu_\rho \Lambda^\nu_\sigma F^{\rho\sigma}$$
$$=\begin{pmatrix}
0 & -E_x & -E_y & -E_z\\
E_x & 0 & -B_z & B_y\\
E_y & B_z & 0 & -B_x\\
E_z & -B_y & B_x & 0
\end{pmatrix}$$
$$=\begin{pmatrix}
0 & -\gamma(E_x-vB_y) & -\gamma(E_y+vB_x) & -E_z\\
\gamma(E_x-vB_y) & 0 & -\gamma(B_z-vE_y) & B_y\\
\gamma(E_y+vB_x) & \gamma(B_z-vE_y) & 0 & -B_x\\
E_z & -B_y & B_x & 0
\end{pmatrix}$$
Now, let's calculate the determinant of this transformed tensor:
$$\det{F'^{\mu\nu}}=\gamma^2(E_x^2-v^2B_y^2)(E_y^2+v^2B_x^2)-E_z^2(B_x^2+B_y^2)$$
$$=\gamma^2(E_x^2-v^2B_y^2)(