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, $$ \det\pmatrix{0&A_{12}\\ A_{21}&A_{22}}\det(A_{33}) +\det\pmatrix{0&A_{13}\\ A_{31}&A_{33}}\det(A_{22})=0. $$