# How to prove that $\det \left(\begin{array}{cc} 0 & A\\ -B & I \end{array}\right) = \det(AB)$ [closed]

Show that $$\det \left(\begin{array}{cc} 0 & A\\ -B & I \end{array}\right) = \det(AB)$$ where A, B are compatible matrices, 0 and I are zero and identity matrices of the appropriate size.

I don't know where to start for this proof. Any help would be appreciated.

## closed as off-topic by user296602, Saad, B. Mehta, Xander Henderson, rtybaseMay 18 '18 at 7:59

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HINT: Try multiplying by a block matrix (whose determinant will turn out to be $1$) to turn this into block diagonal form $$\begin{pmatrix} AB & 0 \\ -B & I\end{pmatrix}.$$
HINT: $$\det \begin{bmatrix}A & B \\ C & D \end{bmatrix} = \det\left[D\right] \det\left[A - BD^{-1} C \right]$$
$$\det[I]=1.$$