Determinant of a specific block matrix Let $A$ be the following block matrix
$$A = \begin{pmatrix}
    0 & A_{12} & A_{13} \\
    A_{21} & A_{22} & 0 \\
    A_{31} & 0 & A_{33}\\
    \end{pmatrix}$$
I am finding difficulty to show that 
$$\det A = \det \begin{pmatrix}
    0 & A_{12} \\
    A_{21} & A_{22} \\
    \end{pmatrix} \times \det(A_{33}) + \det \begin{pmatrix}
    0 & A_{13} \\
    A_{31} & A_{33} \\
    \end{pmatrix} \times \det(A_{22})$$
I tried using the Schur complement but got nowhere. Any hints/suggestions?
Diagonal blocks are square and $A_{21} = A_{12}^t$, $A_{13} = A_{31}^t$, where $A_{21},A_{13}$ are column vectors.

This is not true. Counterexample:
$$
A=\left[\begin{array}{cc|cc|cc}
0&0&1&0&0&0\\
0&0&0&0&0&1\\
\hline
1&0&0&0&0&0\\
0&0&0&1&0&0\\
\hline
0&0&0&0&1&0\\
0&1&0&0&0&0
\end{array}\right].
$$
We have $\det(A)\ne0$ because $A$ is a permutation matrix, but as both $A_{22}$ and $A_{33}$ are singular,
$$
\det\pmatrix{0&A_{12}\\ A_{21}&A_{22}}\det(A_{33})
+\det\pmatrix{0&A_{13}\\ A_{31}&A_{33}}\det(A_{22})=0.
$$
