How to calculate a determinant of a 2x2 symmetry block matrix? I'd like to calculate the determinant of the matrix:
$$
        \begin{pmatrix}
        -A & B^\star  \\
        -B & A^\star  \\
        \end{pmatrix}
$$
$A$, $B$ are $L\times L$ complex matrix.
I know that if $A$ and $B$ are real matrix, the determinant can be easily calculated:
$$
        \begin{pmatrix}
        -A & B  \\
        -B & A  \\
        \end{pmatrix}
=
        \begin{pmatrix}
        A & B  \\
        B & A  \\
        \end{pmatrix} 
.
        \begin{pmatrix}
        -1 & 0  \\
        0 & 1  \\
        \end{pmatrix}
$$
and 
$$
\det
        \begin{pmatrix}
        A & B  \\
        B & A  \\
        \end{pmatrix}
=\det(A-B)\det(A+B)
$$
Is there a similar formula when $A$ and $B$ are complex matrix? Thank you for your help.

Hint.
If $\mathbf {A^*}$ is invertible you can use the general result:
$$
\det
\begin{bmatrix}
\mathbf A&\mathbf B\\
\mathbf C&\mathbf D
\end{bmatrix}= \det (\mathbf D) \det(\mathbf A-\mathbf B\mathbf D^{-1}\mathbf C)
$$
that is a consequence of the identity:
$$
\begin{bmatrix}
\mathbf A&\mathbf B\\
\mathbf C&\mathbf D
\end{bmatrix}
\begin{bmatrix}
\mathbf I&\mathbf 0\\
\mathbf{-D^{-1}}\mathbf C&\mathbf D
\end{bmatrix}=
\begin{bmatrix}
\mathbf A -\mathbf B\mathbf {D^{-1}}\mathbf C&\mathbf B\\
\mathbf 0&\mathbf D
\end{bmatrix}=
\begin{bmatrix}
\mathbf I&\mathbf B\\
\mathbf 0&\mathbf D
\end{bmatrix}
\begin{bmatrix}
\mathbf A -\mathbf B\mathbf {D^{-1}}\mathbf C&\mathbf 0\\
\mathbf 0&\mathbf D
\end{bmatrix}
$$
