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Let $A$ be a square matrix such that $AA^T=I$ and $\lvert A\rvert < 0$, find $\lvert A \rvert$ and $\lvert I + A \rvert$

My progress was:

$\lvert A \rvert \lvert A^T \rvert = \lvert I \rvert \to \lvert A \rvert ^ 2 = 1 \to \lvert A \rvert = -1$

Not sure how to solve for $\lvert I + A \rvert$, are there properties of matrix with determinant -1 i can make use of?


SOLVED

$\lvert I + A \rvert \lvert A^T \rvert = \lvert (A+I)^T \rvert \to \lvert I + A \rvert \lvert A \rvert = \lvert I + A \rvert \to \lvert I + A \rvert (-1) = \lvert I + A \rvert \to \lvert I + A \rvert = 0$

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Hint: $(I+A)A^T=A^T+AA^T=A^T+I=(A+I)^T$.

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