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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.

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closed as off-topic by user296602, Saad, B. Mehta, Xander Henderson, rtybase May 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 $$\begin{pmatrix} AB & 0 \\ -B & I\end{pmatrix}.$$

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  • $\begingroup$ Thanks for the answer! It really helps! $\endgroup$ – M. W May 18 '18 at 0:01
  • $\begingroup$ You're welcome. If you have follow-up questions, let me know. Otherwise, when you get it figured out, you can accept the answer :P $\endgroup$ – Ted Shifrin May 18 '18 at 0:14
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HINT: $$\det \begin{bmatrix}A & B \\ C & D \end{bmatrix} = \det\left[D\right] \det\left[A - BD^{-1} C \right]$$

Also,

$$\det[I]=1.$$

Now simply do the pattern matching.

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