Invertible = nonzero det?

Question

Let $A$, $B$ be $n\times n$ complex matrices and $I$ be the $n\times n$ identity matrix.

Is $\left(\begin{array}{cc}A&I\\I&B\end{array}\right)$ being invertible the same as $\det(AB-I)\ne 0$?

If not, what is the right condition for this $2n\times 2n$ matrix to be invertible? Thank you.

Answer 1

This follows from $ \begin{pmatrix} I & 0 \\ - A & I \end{pmatrix} \begin{pmatrix} 0 & I \\ I & 0 \end{pmatrix} \begin{pmatrix} A & I \\ I & B \end{pmatrix} = \begin{pmatrix} I & B \\ 0 & I - AB \end{pmatrix}. $